1Mao W07

TCP

EECS 489 Computer Networkshttpwwweecsumicheducourseseecs489w07

Z Morley MaoWednesday Jan 31 2007

Acknowledgement Some slides taken from KuroseampRoss and KatzampStoica

2Mao W07

TCP Overview RFCs 793 1122 1323 2018 2581

full duplex data- bi-directional data flow in

same connection- MSS maximum segment size

connection-oriented- handshaking (exchange of

control msgs) initrsquos sender receiver state before data exchange

flow controlled- sender will not overwhelm

receiver

point-to-point- one sender one receiver

reliable in-order byte steam- no ldquomessage boundariesrdquo

pipelined- TCP congestion and flow

control set window sizesend amp receive buffers

socketdoor

TCPsend buffer

TCPreceive buffer

socketdoor

segment

applicationwrites data

applicationreads data

3Mao W07

TCP segment structure

source port dest port

32 bits

applicationdata

(variable length)

sequence numberacknowledgement number

Receive windowUrg data pnterchecksum

FSRPAUheadlen

notused

Options (variable length)

URG urgent data (generally not used)

ACK ACK valid

PSH push data now(generally not used)

RST SYN FINconnection estab(setup teardown

commands)

bytes rcvr willingto accept

countingby bytes of data(not segments)

Internetchecksum

(as in UDP)

4Mao W07

TCP seq rsquos and ACKs

Seq rsquos- byte stream

ldquonumberrdquo of first byte in segmentrsquos data

ACKs- seq of next byte

expected from other side

- cumulative ACKQ how receiver handles

out-of-order segments- A TCP spec doesnrsquot

say - up to implementor

Host A Host B

Seq=42 ACK=79 data = lsquoCrsquo

Seq=79 ACK=43 data = lsquoCrsquo

Seq=43 ACK=80

Usertypes

lsquoCrsquo

host ACKsreceipt

of echoedlsquoCrsquo

host ACKsreceipt oflsquoCrsquo echoes

back lsquoCrsquo

timesimple telnet scenario

5Mao W07

TCP Round Trip Time and TimeoutQ how to set TCP timeout

valuelonger than RTT

- but RTT variestoo short premature timeout

- unnecessary retransmissions

too long slow reaction to segment loss

Q how to estimate RTTSampleRTT measured time from segment transmission until ACK receipt

- ignore retransmissionsSampleRTT will vary want estimated RTT ldquosmootherrdquo

- average several recent measurements not just current SampleRTT

6Mao W07

TCP Round Trip Time and TimeoutEstimatedRTT = (1- α)EstimatedRTT + αSampleRTT

Exponential weighted moving averageinfluence of past sample decreases exponentially fasttypical value α = 0125

7Mao W07

Example RTT estimation

RTT gaiacsumassedu to fantasiaeurecomfr

100

150

200

250

300

350

1 8 15 22 29 36 43 50 57 64 71 78 85 92 99 106

time (seconnds)

RTT

(mill

iseco

nds)

SampleRTT Estimated RTT

8Mao W07

TCP Round Trip Time and TimeoutSetting the timeout

EstimtedRTT plus ldquosafety marginrdquo- large variation in EstimatedRTT -gt larger safety margin

first estimate of how much SampleRTT deviates from EstimatedRTT

TimeoutInterval = EstimatedRTT + 4DevRTT

DevRTT = (1-β)DevRTT +β|SampleRTT-EstimatedRTT|

(typically β = 025)

Then set timeout interval

9Mao W07

TCP reliable data transfer

TCP creates rdt service on top of IPrsquos unreliable servicePipelined segmentsCumulative acksTCP uses single retransmission timer

Retransmissions are triggered by

- timeout events- duplicate acks

Initially consider simplified TCP sender

- ignore duplicate acks- ignore flow control

congestion control

10Mao W07

TCP sender eventsdata rcvd from app

Create segment with seq seq is byte-stream number of first data byte in segmentstart timer if not already running (think of timer as for oldest unacked segment)expiration interval TimeOutInterval

timeoutretransmit segment that caused timeoutrestart timer

Ack rcvdIf acknowledges previously unacked segments

- update what is known to be acked

- start timer if there are outstanding segments

11Mao W07

TCP sender(simplified)

NextSeqNum = InitialSeqNumSendBase = InitialSeqNum

loop (forever) switch(event)

event data received from application above create TCP segment with sequence number NextSeqNumif (timer currently not running)

start timerpass segment to IP NextSeqNum = NextSeqNum + length(data)

event timer timeoutretransmit not-yet-acknowledged segment with

smallest sequence numberstart timer

event ACK received with ACK field value of y if (y gt SendBase)

SendBase = yif (there are currently not-yet-acknowledged segments)

start timer

end of loop forever

Commentbull SendBase-1 last cumulatively ackrsquoed byteExamplebull SendBase-1 = 71y= 73 so the rcvrwants 73+ y gt SendBase sothat new data is acked

12Mao W07

TCP retransmission scenariosHost A

Seq=100 20 bytes data

ACK=100

timepremature timeout

Host B

Seq=92 8 bytes data

ACK=120

Seq=92 8 bytes data

Seq=

92 t

imeo

ut

ACK=120

Host A

Seq=92 8 bytes data

ACK=100

loss

tim

eout

lost ACK scenario

Host B

X

Seq=92 8 bytes data

ACK=100

time

Seq=

92 t

imeo

utSendBase

= 100

SendBase= 120

SendBase= 120

Sendbase= 100

13Mao W07

TCP retransmission scenarios (more)Host A

Seq=92 8 bytes data

ACK=100

loss

tim

eout

Cumulative ACK scenario

Host B

X

Seq=100 20 bytes data

ACK=120

time

SendBase= 120

14Mao W07

TCP ACK generation [RFC 1122 RFC 2581]

Event at Receiver

Arrival of in-order segment withexpected seq All data up toexpected seq already ACKed

Arrival of in-order segment withexpected seq One other segment has ACK pending

Arrival of out-of-order segmenthigher-than-expect seq Gap detected

Arrival of segment that partially or completely fills gap

TCP Receiver action

Delayed ACK Wait up to 500msfor next segment If no next segmentsend ACK

Immediately send single cumulative ACK ACKing both in-order segments

Immediately send duplicate ACK indicating seq of next expected byte

Immediate send ACK provided thatsegment startsat lower end of gap

15Mao W07

Fast Retransmit

Time-out period often relatively long

- long delay before resending lost packet

Detect lost segments via duplicate ACKs

- Sender often sends many segments back-to-back

- If segment is lost there will likely be many duplicate ACKs

If sender receives 3 ACKs for the same data it supposes that segment after ACKed data was lost

- fast retransmit resend segment before timer expires

16Mao W07

event ACK received with ACK field value of y if (y gt SendBase)

SendBase = yif (there are currently not-yet-acknowledged segments)

start timer

else increment count of dup ACKs received for yif (count of dup ACKs received for y = 3)

resend segment with sequence number y

Fast retransmit algorithm

a duplicate ACK for already ACKed segment

fast retransmit

17Mao W07

TCP Flow Control

receive side of TCP connection has a receive buffer

speed-matching service matching the send rate to the receiving apprsquos drain rate

app process may be slow at reading from buffer

sender wonrsquot overflowreceiverrsquos buffer by

transmitting too muchtoo fast

flow control

18Mao W07

TCP Flow control how it works

(Suppose TCP receiver discards out-of-order segments)spare room in buffer

= RcvWindow= RcvBuffer-[LastByteRcvd -

LastByteRead]

Rcvr advertises spare room by including value of RcvWindow in segmentsSender limits unACKed data to RcvWindow

- guarantees receive buffer doesnrsquot overflow

19Mao W07

TCP Connection Management

Recall TCP sender receiver establish ldquoconnectionrdquo before exchanging data segmentsinitialize TCP variables

- seq s- buffers flow control info

(eg RcvWindow)client connection initiatorSocket clientSocket = new Socket(hostnameport

number)

server contacted by clientSocket connectionSocket = welcomeSocketaccept()

Three way handshakeStep 1 client host sends TCP

SYN segment to server- specifies initial seq - no data

Step 2 server host receives SYN replies with SYNACK segment

- server allocates buffers- specifies server initial seq

Step 3 client receives SYNACK replies with ACK segment which may contain data

20Mao W07

TCP Connection Management (cont)

Closing a connection

client closes socketclientSocketclose()

Step 1 client end system sends TCP FIN control segment to server

Step 2 server receives FIN replies with ACK Closes connection sends FIN

client

FIN

server

ACK

ACK

FIN

close

close

closed

tim

ed w

ait

21Mao W07

TCP Connection Management (cont)

Step 3 client receives FIN replies with ACK

- Enters ldquotimed waitrdquo - will respond with ACK to received FINs

Step 4 server receives ACK Connection closed

Note with small modification can handle simultaneous FINs

client

FIN

server

ACK

ACK

FIN

closing

closing

closed

tim

ed w

ait

closed

22Mao W07

TCP Connection Management (cont)

TCP clientlifecycle

TCP serverlifecycle

23Mao W07

Principles of Congestion Control

Congestioninformally ldquotoo many sources sending too much data too fast for network to handlerdquodifferent from flow controlmanifestations

- lost packets (buffer overflow at routers)- long delays (queueing in router buffers)

a top-10 problem

24Mao W07

Causescosts of congestion scenario 1

two senders two receiversone router infinite buffers no retransmission

large delays when congestedmaximum achievable throughput

unlimited shared output link buffers

Host Aλin original data

Host B

λout

25Mao W07

Causescosts of congestion scenario 2

one router finite buffers sender retransmission of lost packet

finite shared output link buffers

Host A λin original data

Host B

λout

λin original data plus retransmitted data

26Mao W07

Causescosts of congestion scenario 2

always (goodput)

ldquoperfectrdquo retransmission only when loss

retransmission of delayed (not lost) packet makes larger (than perfect case) for same

λin

λout=

λin

λoutgtλ

inλout

ldquocostsrdquo of congestionmore work (retrans) for given ldquogoodputrdquounneeded retransmissions link carries multiple copies of pkt

R2

R2λin

λ out

b

R2

R2λin

λ out

a

R2

R2λin

λ out

c

R4

R3

27Mao W07

Causescosts of congestion scenario 3

four sendersmultihop pathstimeoutretransmit

λin

Q what happens as and increase λ

in

finite shared output link buffers

Host Aλin original data

Host B

λout

λin original data plus retransmitted data

28Mao W07

Causescosts of congestion scenario 3

Another ldquocostrdquo of congestionwhen packet dropped any ldquoupstream transmission capacity used for that packet was wasted

Host A

Host B

λou

t

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

2Mao W07

TCP Overview RFCs 793 1122 1323 2018 2581

full duplex data- bi-directional data flow in

same connection- MSS maximum segment size

connection-oriented- handshaking (exchange of

control msgs) initrsquos sender receiver state before data exchange

flow controlled- sender will not overwhelm

receiver

point-to-point- one sender one receiver

reliable in-order byte steam- no ldquomessage boundariesrdquo

pipelined- TCP congestion and flow

control set window sizesend amp receive buffers

socketdoor

TCPsend buffer

TCPreceive buffer

socketdoor

segment

applicationwrites data

applicationreads data

3Mao W07

TCP segment structure

source port dest port

32 bits

applicationdata

(variable length)

sequence numberacknowledgement number

Receive windowUrg data pnterchecksum

FSRPAUheadlen

notused

Options (variable length)

URG urgent data (generally not used)

ACK ACK valid

PSH push data now(generally not used)

RST SYN FINconnection estab(setup teardown

commands)

bytes rcvr willingto accept

countingby bytes of data(not segments)

Internetchecksum

(as in UDP)

4Mao W07

TCP seq rsquos and ACKs

Seq rsquos- byte stream

ldquonumberrdquo of first byte in segmentrsquos data

ACKs- seq of next byte

expected from other side

- cumulative ACKQ how receiver handles

out-of-order segments- A TCP spec doesnrsquot

say - up to implementor

Host A Host B

Seq=42 ACK=79 data = lsquoCrsquo

Seq=79 ACK=43 data = lsquoCrsquo

Seq=43 ACK=80

Usertypes

lsquoCrsquo

host ACKsreceipt

of echoedlsquoCrsquo

host ACKsreceipt oflsquoCrsquo echoes

back lsquoCrsquo

timesimple telnet scenario

5Mao W07

TCP Round Trip Time and TimeoutQ how to set TCP timeout

valuelonger than RTT

- but RTT variestoo short premature timeout

- unnecessary retransmissions

too long slow reaction to segment loss

Q how to estimate RTTSampleRTT measured time from segment transmission until ACK receipt

- ignore retransmissionsSampleRTT will vary want estimated RTT ldquosmootherrdquo

- average several recent measurements not just current SampleRTT

6Mao W07

TCP Round Trip Time and TimeoutEstimatedRTT = (1- α)EstimatedRTT + αSampleRTT

Exponential weighted moving averageinfluence of past sample decreases exponentially fasttypical value α = 0125

7Mao W07

Example RTT estimation

RTT gaiacsumassedu to fantasiaeurecomfr

100

150

200

250

300

350

1 8 15 22 29 36 43 50 57 64 71 78 85 92 99 106

time (seconnds)

RTT

(mill

iseco

nds)

SampleRTT Estimated RTT

8Mao W07

TCP Round Trip Time and TimeoutSetting the timeout

EstimtedRTT plus ldquosafety marginrdquo- large variation in EstimatedRTT -gt larger safety margin

first estimate of how much SampleRTT deviates from EstimatedRTT

TimeoutInterval = EstimatedRTT + 4DevRTT

DevRTT = (1-β)DevRTT +β|SampleRTT-EstimatedRTT|

(typically β = 025)

Then set timeout interval

9Mao W07

TCP reliable data transfer

TCP creates rdt service on top of IPrsquos unreliable servicePipelined segmentsCumulative acksTCP uses single retransmission timer

Retransmissions are triggered by

- timeout events- duplicate acks

Initially consider simplified TCP sender

- ignore duplicate acks- ignore flow control

congestion control

10Mao W07

TCP sender eventsdata rcvd from app

Create segment with seq seq is byte-stream number of first data byte in segmentstart timer if not already running (think of timer as for oldest unacked segment)expiration interval TimeOutInterval

timeoutretransmit segment that caused timeoutrestart timer

Ack rcvdIf acknowledges previously unacked segments

- update what is known to be acked

- start timer if there are outstanding segments

11Mao W07

TCP sender(simplified)

NextSeqNum = InitialSeqNumSendBase = InitialSeqNum

loop (forever) switch(event)

event data received from application above create TCP segment with sequence number NextSeqNumif (timer currently not running)

start timerpass segment to IP NextSeqNum = NextSeqNum + length(data)

event timer timeoutretransmit not-yet-acknowledged segment with

smallest sequence numberstart timer

event ACK received with ACK field value of y if (y gt SendBase)

SendBase = yif (there are currently not-yet-acknowledged segments)

start timer

end of loop forever

Commentbull SendBase-1 last cumulatively ackrsquoed byteExamplebull SendBase-1 = 71y= 73 so the rcvrwants 73+ y gt SendBase sothat new data is acked

12Mao W07

TCP retransmission scenariosHost A

Seq=100 20 bytes data

ACK=100

timepremature timeout

Host B

Seq=92 8 bytes data

ACK=120

Seq=92 8 bytes data

Seq=

92 t

imeo

ut

ACK=120

Host A

Seq=92 8 bytes data

ACK=100

loss

tim

eout

lost ACK scenario

Host B

X

Seq=92 8 bytes data

ACK=100

time

Seq=

92 t

imeo

utSendBase

= 100

SendBase= 120

SendBase= 120

Sendbase= 100

13Mao W07

TCP retransmission scenarios (more)Host A

Seq=92 8 bytes data

ACK=100

loss

tim

eout

Cumulative ACK scenario

Host B

X

Seq=100 20 bytes data

ACK=120

time

SendBase= 120

14Mao W07

TCP ACK generation [RFC 1122 RFC 2581]

Event at Receiver

Arrival of in-order segment withexpected seq All data up toexpected seq already ACKed

Arrival of in-order segment withexpected seq One other segment has ACK pending

Arrival of out-of-order segmenthigher-than-expect seq Gap detected

Arrival of segment that partially or completely fills gap

TCP Receiver action

Delayed ACK Wait up to 500msfor next segment If no next segmentsend ACK

Immediately send single cumulative ACK ACKing both in-order segments

Immediately send duplicate ACK indicating seq of next expected byte

Immediate send ACK provided thatsegment startsat lower end of gap

15Mao W07

Fast Retransmit

Time-out period often relatively long

- long delay before resending lost packet

Detect lost segments via duplicate ACKs

- Sender often sends many segments back-to-back

- If segment is lost there will likely be many duplicate ACKs

If sender receives 3 ACKs for the same data it supposes that segment after ACKed data was lost

- fast retransmit resend segment before timer expires

16Mao W07

event ACK received with ACK field value of y if (y gt SendBase)

SendBase = yif (there are currently not-yet-acknowledged segments)

start timer

else increment count of dup ACKs received for yif (count of dup ACKs received for y = 3)

resend segment with sequence number y

Fast retransmit algorithm

a duplicate ACK for already ACKed segment

fast retransmit

17Mao W07

TCP Flow Control

receive side of TCP connection has a receive buffer

speed-matching service matching the send rate to the receiving apprsquos drain rate

app process may be slow at reading from buffer

sender wonrsquot overflowreceiverrsquos buffer by

transmitting too muchtoo fast

flow control

18Mao W07

TCP Flow control how it works

(Suppose TCP receiver discards out-of-order segments)spare room in buffer

= RcvWindow= RcvBuffer-[LastByteRcvd -

LastByteRead]

Rcvr advertises spare room by including value of RcvWindow in segmentsSender limits unACKed data to RcvWindow

- guarantees receive buffer doesnrsquot overflow

19Mao W07

TCP Connection Management

Recall TCP sender receiver establish ldquoconnectionrdquo before exchanging data segmentsinitialize TCP variables

- seq s- buffers flow control info

(eg RcvWindow)client connection initiatorSocket clientSocket = new Socket(hostnameport

number)

server contacted by clientSocket connectionSocket = welcomeSocketaccept()

Three way handshakeStep 1 client host sends TCP

SYN segment to server- specifies initial seq - no data

Step 2 server host receives SYN replies with SYNACK segment

- server allocates buffers- specifies server initial seq

Step 3 client receives SYNACK replies with ACK segment which may contain data

20Mao W07

TCP Connection Management (cont)

Closing a connection

client closes socketclientSocketclose()

Step 1 client end system sends TCP FIN control segment to server

Step 2 server receives FIN replies with ACK Closes connection sends FIN

client

FIN

server

ACK

ACK

FIN

close

close

closed

tim

ed w

ait

21Mao W07

TCP Connection Management (cont)

Step 3 client receives FIN replies with ACK

- Enters ldquotimed waitrdquo - will respond with ACK to received FINs

Step 4 server receives ACK Connection closed

Note with small modification can handle simultaneous FINs

client

FIN

server

ACK

ACK

FIN

closing

closing

closed

tim

ed w

ait

closed

22Mao W07

TCP Connection Management (cont)

TCP clientlifecycle

TCP serverlifecycle

23Mao W07

Principles of Congestion Control

Congestioninformally ldquotoo many sources sending too much data too fast for network to handlerdquodifferent from flow controlmanifestations

- lost packets (buffer overflow at routers)- long delays (queueing in router buffers)

a top-10 problem

24Mao W07

Causescosts of congestion scenario 1

two senders two receiversone router infinite buffers no retransmission

large delays when congestedmaximum achievable throughput

unlimited shared output link buffers

Host Aλin original data

Host B

λout

25Mao W07

Causescosts of congestion scenario 2

one router finite buffers sender retransmission of lost packet

finite shared output link buffers

Host A λin original data

Host B

λout

λin original data plus retransmitted data

26Mao W07

Causescosts of congestion scenario 2

always (goodput)

ldquoperfectrdquo retransmission only when loss

retransmission of delayed (not lost) packet makes larger (than perfect case) for same

λin

λout=

λin

λoutgtλ

inλout

ldquocostsrdquo of congestionmore work (retrans) for given ldquogoodputrdquounneeded retransmissions link carries multiple copies of pkt

R2

R2λin

λ out

b

R2

R2λin

λ out

a

R2

R2λin

λ out

c

R4

R3

27Mao W07

Causescosts of congestion scenario 3

four sendersmultihop pathstimeoutretransmit

λin

Q what happens as and increase λ

in

finite shared output link buffers

Host Aλin original data

Host B

λout

λin original data plus retransmitted data

28Mao W07

Causescosts of congestion scenario 3

Another ldquocostrdquo of congestionwhen packet dropped any ldquoupstream transmission capacity used for that packet was wasted

Host A

Host B

λou

t

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

3Mao W07

TCP segment structure

source port dest port

32 bits

applicationdata

(variable length)

sequence numberacknowledgement number

Receive windowUrg data pnterchecksum

FSRPAUheadlen

notused

Options (variable length)

URG urgent data (generally not used)

ACK ACK valid

PSH push data now(generally not used)

RST SYN FINconnection estab(setup teardown

commands)

bytes rcvr willingto accept

countingby bytes of data(not segments)

Internetchecksum

(as in UDP)

4Mao W07

TCP seq rsquos and ACKs

Seq rsquos- byte stream

ldquonumberrdquo of first byte in segmentrsquos data

ACKs- seq of next byte

expected from other side

- cumulative ACKQ how receiver handles

out-of-order segments- A TCP spec doesnrsquot

say - up to implementor

Host A Host B

Seq=42 ACK=79 data = lsquoCrsquo

Seq=79 ACK=43 data = lsquoCrsquo

Seq=43 ACK=80

Usertypes

lsquoCrsquo

host ACKsreceipt

of echoedlsquoCrsquo

host ACKsreceipt oflsquoCrsquo echoes

back lsquoCrsquo

timesimple telnet scenario

5Mao W07

TCP Round Trip Time and TimeoutQ how to set TCP timeout

valuelonger than RTT

- but RTT variestoo short premature timeout

- unnecessary retransmissions

too long slow reaction to segment loss

Q how to estimate RTTSampleRTT measured time from segment transmission until ACK receipt

- ignore retransmissionsSampleRTT will vary want estimated RTT ldquosmootherrdquo

- average several recent measurements not just current SampleRTT

6Mao W07

TCP Round Trip Time and TimeoutEstimatedRTT = (1- α)EstimatedRTT + αSampleRTT

Exponential weighted moving averageinfluence of past sample decreases exponentially fasttypical value α = 0125

7Mao W07

Example RTT estimation

RTT gaiacsumassedu to fantasiaeurecomfr

100

150

200

250

300

350

1 8 15 22 29 36 43 50 57 64 71 78 85 92 99 106

time (seconnds)

RTT

(mill

iseco

nds)

SampleRTT Estimated RTT

8Mao W07

TCP Round Trip Time and TimeoutSetting the timeout

EstimtedRTT plus ldquosafety marginrdquo- large variation in EstimatedRTT -gt larger safety margin

first estimate of how much SampleRTT deviates from EstimatedRTT

TimeoutInterval = EstimatedRTT + 4DevRTT

DevRTT = (1-β)DevRTT +β|SampleRTT-EstimatedRTT|

(typically β = 025)

Then set timeout interval

9Mao W07

TCP reliable data transfer

TCP creates rdt service on top of IPrsquos unreliable servicePipelined segmentsCumulative acksTCP uses single retransmission timer

Retransmissions are triggered by

- timeout events- duplicate acks

Initially consider simplified TCP sender

- ignore duplicate acks- ignore flow control

congestion control

10Mao W07

TCP sender eventsdata rcvd from app

Create segment with seq seq is byte-stream number of first data byte in segmentstart timer if not already running (think of timer as for oldest unacked segment)expiration interval TimeOutInterval

timeoutretransmit segment that caused timeoutrestart timer

Ack rcvdIf acknowledges previously unacked segments

- update what is known to be acked

- start timer if there are outstanding segments

11Mao W07

TCP sender(simplified)

NextSeqNum = InitialSeqNumSendBase = InitialSeqNum

loop (forever) switch(event)

event data received from application above create TCP segment with sequence number NextSeqNumif (timer currently not running)

start timerpass segment to IP NextSeqNum = NextSeqNum + length(data)

event timer timeoutretransmit not-yet-acknowledged segment with

smallest sequence numberstart timer

event ACK received with ACK field value of y if (y gt SendBase)

SendBase = yif (there are currently not-yet-acknowledged segments)

start timer

end of loop forever

Commentbull SendBase-1 last cumulatively ackrsquoed byteExamplebull SendBase-1 = 71y= 73 so the rcvrwants 73+ y gt SendBase sothat new data is acked

12Mao W07

TCP retransmission scenariosHost A

Seq=100 20 bytes data

ACK=100

timepremature timeout

Host B

Seq=92 8 bytes data

ACK=120

Seq=92 8 bytes data

Seq=

92 t

imeo

ut

ACK=120

Host A

Seq=92 8 bytes data

ACK=100

loss

tim

eout

lost ACK scenario

Host B

X

Seq=92 8 bytes data

ACK=100

time

Seq=

92 t

imeo

utSendBase

= 100

SendBase= 120

SendBase= 120

Sendbase= 100

13Mao W07

TCP retransmission scenarios (more)Host A

Seq=92 8 bytes data

ACK=100

loss

tim

eout

Cumulative ACK scenario

Host B

X

Seq=100 20 bytes data

ACK=120

time

SendBase= 120

14Mao W07

TCP ACK generation [RFC 1122 RFC 2581]

Event at Receiver

Arrival of in-order segment withexpected seq All data up toexpected seq already ACKed

Arrival of in-order segment withexpected seq One other segment has ACK pending

Arrival of out-of-order segmenthigher-than-expect seq Gap detected

Arrival of segment that partially or completely fills gap

TCP Receiver action

Delayed ACK Wait up to 500msfor next segment If no next segmentsend ACK

Immediately send single cumulative ACK ACKing both in-order segments

Immediately send duplicate ACK indicating seq of next expected byte

Immediate send ACK provided thatsegment startsat lower end of gap

15Mao W07

Fast Retransmit

Time-out period often relatively long

- long delay before resending lost packet

Detect lost segments via duplicate ACKs

- Sender often sends many segments back-to-back

- If segment is lost there will likely be many duplicate ACKs

If sender receives 3 ACKs for the same data it supposes that segment after ACKed data was lost

- fast retransmit resend segment before timer expires

16Mao W07

event ACK received with ACK field value of y if (y gt SendBase)

SendBase = yif (there are currently not-yet-acknowledged segments)

start timer

else increment count of dup ACKs received for yif (count of dup ACKs received for y = 3)

resend segment with sequence number y

Fast retransmit algorithm

a duplicate ACK for already ACKed segment

fast retransmit

17Mao W07

TCP Flow Control

receive side of TCP connection has a receive buffer

speed-matching service matching the send rate to the receiving apprsquos drain rate

app process may be slow at reading from buffer

sender wonrsquot overflowreceiverrsquos buffer by

transmitting too muchtoo fast

flow control

18Mao W07

TCP Flow control how it works

(Suppose TCP receiver discards out-of-order segments)spare room in buffer

= RcvWindow= RcvBuffer-[LastByteRcvd -

LastByteRead]

Rcvr advertises spare room by including value of RcvWindow in segmentsSender limits unACKed data to RcvWindow

- guarantees receive buffer doesnrsquot overflow

19Mao W07

TCP Connection Management

Recall TCP sender receiver establish ldquoconnectionrdquo before exchanging data segmentsinitialize TCP variables

- seq s- buffers flow control info

(eg RcvWindow)client connection initiatorSocket clientSocket = new Socket(hostnameport

number)

server contacted by clientSocket connectionSocket = welcomeSocketaccept()

Three way handshakeStep 1 client host sends TCP

SYN segment to server- specifies initial seq - no data

Step 2 server host receives SYN replies with SYNACK segment

- server allocates buffers- specifies server initial seq

Step 3 client receives SYNACK replies with ACK segment which may contain data

20Mao W07

TCP Connection Management (cont)

Closing a connection

client closes socketclientSocketclose()

Step 1 client end system sends TCP FIN control segment to server

Step 2 server receives FIN replies with ACK Closes connection sends FIN

client

FIN

server

ACK

ACK

FIN

close

close

closed

tim

ed w

ait

21Mao W07

TCP Connection Management (cont)

Step 3 client receives FIN replies with ACK

- Enters ldquotimed waitrdquo - will respond with ACK to received FINs

Step 4 server receives ACK Connection closed

Note with small modification can handle simultaneous FINs

client

FIN

server

ACK

ACK

FIN

closing

closing

closed

tim

ed w

ait

closed

22Mao W07

TCP Connection Management (cont)

TCP clientlifecycle

TCP serverlifecycle

23Mao W07

Principles of Congestion Control

Congestioninformally ldquotoo many sources sending too much data too fast for network to handlerdquodifferent from flow controlmanifestations

- lost packets (buffer overflow at routers)- long delays (queueing in router buffers)

a top-10 problem

24Mao W07

Causescosts of congestion scenario 1

two senders two receiversone router infinite buffers no retransmission

large delays when congestedmaximum achievable throughput

unlimited shared output link buffers

Host Aλin original data

Host B

λout

25Mao W07

Causescosts of congestion scenario 2

one router finite buffers sender retransmission of lost packet

finite shared output link buffers

Host A λin original data

Host B

λout

λin original data plus retransmitted data

26Mao W07

Causescosts of congestion scenario 2

always (goodput)

ldquoperfectrdquo retransmission only when loss

retransmission of delayed (not lost) packet makes larger (than perfect case) for same

λin

λout=

λin

λoutgtλ

inλout

ldquocostsrdquo of congestionmore work (retrans) for given ldquogoodputrdquounneeded retransmissions link carries multiple copies of pkt

R2

R2λin

λ out

b

R2

R2λin

λ out

a

R2

R2λin

λ out

c

R4

R3

27Mao W07

Causescosts of congestion scenario 3

four sendersmultihop pathstimeoutretransmit

λin

Q what happens as and increase λ

in

finite shared output link buffers

Host Aλin original data

Host B

λout

λin original data plus retransmitted data

28Mao W07

Causescosts of congestion scenario 3

Another ldquocostrdquo of congestionwhen packet dropped any ldquoupstream transmission capacity used for that packet was wasted

Host A

Host B

λou

t

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

4Mao W07

TCP seq rsquos and ACKs

Seq rsquos- byte stream

ldquonumberrdquo of first byte in segmentrsquos data

ACKs- seq of next byte

expected from other side

- cumulative ACKQ how receiver handles

out-of-order segments- A TCP spec doesnrsquot

say - up to implementor

Host A Host B

Seq=42 ACK=79 data = lsquoCrsquo

Seq=79 ACK=43 data = lsquoCrsquo

Seq=43 ACK=80

Usertypes

lsquoCrsquo

host ACKsreceipt

of echoedlsquoCrsquo

host ACKsreceipt oflsquoCrsquo echoes

back lsquoCrsquo

timesimple telnet scenario

5Mao W07

TCP Round Trip Time and TimeoutQ how to set TCP timeout

valuelonger than RTT

- but RTT variestoo short premature timeout

- unnecessary retransmissions

too long slow reaction to segment loss

Q how to estimate RTTSampleRTT measured time from segment transmission until ACK receipt

- ignore retransmissionsSampleRTT will vary want estimated RTT ldquosmootherrdquo

- average several recent measurements not just current SampleRTT

6Mao W07

TCP Round Trip Time and TimeoutEstimatedRTT = (1- α)EstimatedRTT + αSampleRTT

Exponential weighted moving averageinfluence of past sample decreases exponentially fasttypical value α = 0125

7Mao W07

Example RTT estimation

RTT gaiacsumassedu to fantasiaeurecomfr

100

150

200

250

300

350

1 8 15 22 29 36 43 50 57 64 71 78 85 92 99 106

time (seconnds)

RTT

(mill

iseco

nds)

SampleRTT Estimated RTT

8Mao W07

TCP Round Trip Time and TimeoutSetting the timeout

EstimtedRTT plus ldquosafety marginrdquo- large variation in EstimatedRTT -gt larger safety margin

first estimate of how much SampleRTT deviates from EstimatedRTT

TimeoutInterval = EstimatedRTT + 4DevRTT

DevRTT = (1-β)DevRTT +β|SampleRTT-EstimatedRTT|

(typically β = 025)

Then set timeout interval

9Mao W07

TCP reliable data transfer

TCP creates rdt service on top of IPrsquos unreliable servicePipelined segmentsCumulative acksTCP uses single retransmission timer

Retransmissions are triggered by

- timeout events- duplicate acks

Initially consider simplified TCP sender

- ignore duplicate acks- ignore flow control

congestion control

10Mao W07

TCP sender eventsdata rcvd from app

Create segment with seq seq is byte-stream number of first data byte in segmentstart timer if not already running (think of timer as for oldest unacked segment)expiration interval TimeOutInterval

timeoutretransmit segment that caused timeoutrestart timer

Ack rcvdIf acknowledges previously unacked segments

- update what is known to be acked

- start timer if there are outstanding segments

11Mao W07

TCP sender(simplified)

NextSeqNum = InitialSeqNumSendBase = InitialSeqNum

loop (forever) switch(event)

event data received from application above create TCP segment with sequence number NextSeqNumif (timer currently not running)

start timerpass segment to IP NextSeqNum = NextSeqNum + length(data)

event timer timeoutretransmit not-yet-acknowledged segment with

smallest sequence numberstart timer

event ACK received with ACK field value of y if (y gt SendBase)

SendBase = yif (there are currently not-yet-acknowledged segments)

start timer

end of loop forever

Commentbull SendBase-1 last cumulatively ackrsquoed byteExamplebull SendBase-1 = 71y= 73 so the rcvrwants 73+ y gt SendBase sothat new data is acked

12Mao W07

TCP retransmission scenariosHost A

Seq=100 20 bytes data

ACK=100

timepremature timeout

Host B

Seq=92 8 bytes data

ACK=120

Seq=92 8 bytes data

Seq=

92 t

imeo

ut

ACK=120

Host A

Seq=92 8 bytes data

ACK=100

loss

tim

eout

lost ACK scenario

Host B

X

Seq=92 8 bytes data

ACK=100

time

Seq=

92 t

imeo

utSendBase

= 100

SendBase= 120

SendBase= 120

Sendbase= 100

13Mao W07

TCP retransmission scenarios (more)Host A

Seq=92 8 bytes data

ACK=100

loss

tim

eout

Cumulative ACK scenario

Host B

X

Seq=100 20 bytes data

ACK=120

time

SendBase= 120

14Mao W07

TCP ACK generation [RFC 1122 RFC 2581]

Event at Receiver

Arrival of in-order segment withexpected seq All data up toexpected seq already ACKed

Arrival of in-order segment withexpected seq One other segment has ACK pending

Arrival of out-of-order segmenthigher-than-expect seq Gap detected

Arrival of segment that partially or completely fills gap

TCP Receiver action

Delayed ACK Wait up to 500msfor next segment If no next segmentsend ACK

Immediately send single cumulative ACK ACKing both in-order segments

Immediately send duplicate ACK indicating seq of next expected byte

Immediate send ACK provided thatsegment startsat lower end of gap

15Mao W07

Fast Retransmit

Time-out period often relatively long

- long delay before resending lost packet

Detect lost segments via duplicate ACKs

- Sender often sends many segments back-to-back

- If segment is lost there will likely be many duplicate ACKs

If sender receives 3 ACKs for the same data it supposes that segment after ACKed data was lost

- fast retransmit resend segment before timer expires

16Mao W07

event ACK received with ACK field value of y if (y gt SendBase)

SendBase = yif (there are currently not-yet-acknowledged segments)

start timer

else increment count of dup ACKs received for yif (count of dup ACKs received for y = 3)

resend segment with sequence number y

Fast retransmit algorithm

a duplicate ACK for already ACKed segment

fast retransmit

17Mao W07

TCP Flow Control

receive side of TCP connection has a receive buffer

speed-matching service matching the send rate to the receiving apprsquos drain rate

app process may be slow at reading from buffer

sender wonrsquot overflowreceiverrsquos buffer by

transmitting too muchtoo fast

flow control

18Mao W07

TCP Flow control how it works

(Suppose TCP receiver discards out-of-order segments)spare room in buffer

= RcvWindow= RcvBuffer-[LastByteRcvd -

LastByteRead]

Rcvr advertises spare room by including value of RcvWindow in segmentsSender limits unACKed data to RcvWindow

- guarantees receive buffer doesnrsquot overflow

19Mao W07

TCP Connection Management

Recall TCP sender receiver establish ldquoconnectionrdquo before exchanging data segmentsinitialize TCP variables

- seq s- buffers flow control info

(eg RcvWindow)client connection initiatorSocket clientSocket = new Socket(hostnameport

number)

server contacted by clientSocket connectionSocket = welcomeSocketaccept()

Three way handshakeStep 1 client host sends TCP

SYN segment to server- specifies initial seq - no data

Step 2 server host receives SYN replies with SYNACK segment

- server allocates buffers- specifies server initial seq

Step 3 client receives SYNACK replies with ACK segment which may contain data

20Mao W07

TCP Connection Management (cont)

Closing a connection

client closes socketclientSocketclose()

Step 1 client end system sends TCP FIN control segment to server

Step 2 server receives FIN replies with ACK Closes connection sends FIN

client

FIN

server

ACK

ACK

FIN

close

close

closed

tim

ed w

ait

21Mao W07

TCP Connection Management (cont)

Step 3 client receives FIN replies with ACK

- Enters ldquotimed waitrdquo - will respond with ACK to received FINs

Step 4 server receives ACK Connection closed

Note with small modification can handle simultaneous FINs

client

FIN

server

ACK

ACK

FIN

closing

closing

closed

tim

ed w

ait

closed

22Mao W07

TCP Connection Management (cont)

TCP clientlifecycle

TCP serverlifecycle

23Mao W07

Principles of Congestion Control

Congestioninformally ldquotoo many sources sending too much data too fast for network to handlerdquodifferent from flow controlmanifestations

- lost packets (buffer overflow at routers)- long delays (queueing in router buffers)

a top-10 problem

24Mao W07

Causescosts of congestion scenario 1

two senders two receiversone router infinite buffers no retransmission

large delays when congestedmaximum achievable throughput

unlimited shared output link buffers

Host Aλin original data

Host B

λout

25Mao W07

Causescosts of congestion scenario 2

one router finite buffers sender retransmission of lost packet

finite shared output link buffers

Host A λin original data

Host B

λout

λin original data plus retransmitted data

26Mao W07

Causescosts of congestion scenario 2

always (goodput)

ldquoperfectrdquo retransmission only when loss

retransmission of delayed (not lost) packet makes larger (than perfect case) for same

λin

λout=

λin

λoutgtλ

inλout

ldquocostsrdquo of congestionmore work (retrans) for given ldquogoodputrdquounneeded retransmissions link carries multiple copies of pkt

R2

R2λin

λ out

b

R2

R2λin

λ out

a

R2

R2λin

λ out

c

R4

R3

27Mao W07

Causescosts of congestion scenario 3

four sendersmultihop pathstimeoutretransmit

λin

Q what happens as and increase λ

in

finite shared output link buffers

Host Aλin original data

Host B

λout

λin original data plus retransmitted data

28Mao W07

Causescosts of congestion scenario 3

Another ldquocostrdquo of congestionwhen packet dropped any ldquoupstream transmission capacity used for that packet was wasted

Host A

Host B

λou

t

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

5Mao W07

TCP Round Trip Time and TimeoutQ how to set TCP timeout

valuelonger than RTT

- but RTT variestoo short premature timeout

- unnecessary retransmissions

too long slow reaction to segment loss

Q how to estimate RTTSampleRTT measured time from segment transmission until ACK receipt

- ignore retransmissionsSampleRTT will vary want estimated RTT ldquosmootherrdquo

- average several recent measurements not just current SampleRTT

6Mao W07

TCP Round Trip Time and TimeoutEstimatedRTT = (1- α)EstimatedRTT + αSampleRTT

Exponential weighted moving averageinfluence of past sample decreases exponentially fasttypical value α = 0125

7Mao W07

Example RTT estimation

RTT gaiacsumassedu to fantasiaeurecomfr

100

150

200

250

300

350

1 8 15 22 29 36 43 50 57 64 71 78 85 92 99 106

time (seconnds)

RTT

(mill

iseco

nds)

SampleRTT Estimated RTT

8Mao W07

TCP Round Trip Time and TimeoutSetting the timeout

EstimtedRTT plus ldquosafety marginrdquo- large variation in EstimatedRTT -gt larger safety margin

first estimate of how much SampleRTT deviates from EstimatedRTT

TimeoutInterval = EstimatedRTT + 4DevRTT

DevRTT = (1-β)DevRTT +β|SampleRTT-EstimatedRTT|

(typically β = 025)

Then set timeout interval

9Mao W07

TCP reliable data transfer

TCP creates rdt service on top of IPrsquos unreliable servicePipelined segmentsCumulative acksTCP uses single retransmission timer

Retransmissions are triggered by

- timeout events- duplicate acks

Initially consider simplified TCP sender

- ignore duplicate acks- ignore flow control

congestion control

10Mao W07

TCP sender eventsdata rcvd from app

Create segment with seq seq is byte-stream number of first data byte in segmentstart timer if not already running (think of timer as for oldest unacked segment)expiration interval TimeOutInterval

timeoutretransmit segment that caused timeoutrestart timer

Ack rcvdIf acknowledges previously unacked segments

- update what is known to be acked

- start timer if there are outstanding segments

11Mao W07

TCP sender(simplified)

NextSeqNum = InitialSeqNumSendBase = InitialSeqNum

loop (forever) switch(event)

event data received from application above create TCP segment with sequence number NextSeqNumif (timer currently not running)

start timerpass segment to IP NextSeqNum = NextSeqNum + length(data)

event timer timeoutretransmit not-yet-acknowledged segment with

smallest sequence numberstart timer

event ACK received with ACK field value of y if (y gt SendBase)

SendBase = yif (there are currently not-yet-acknowledged segments)

start timer

end of loop forever

Commentbull SendBase-1 last cumulatively ackrsquoed byteExamplebull SendBase-1 = 71y= 73 so the rcvrwants 73+ y gt SendBase sothat new data is acked

12Mao W07

TCP retransmission scenariosHost A

Seq=100 20 bytes data

ACK=100

timepremature timeout

Host B

Seq=92 8 bytes data

ACK=120

Seq=92 8 bytes data

Seq=

92 t

imeo

ut

ACK=120

Host A

Seq=92 8 bytes data

ACK=100

loss

tim

eout

lost ACK scenario

Host B

X

Seq=92 8 bytes data

ACK=100

time

Seq=

92 t

imeo

utSendBase

= 100

SendBase= 120

SendBase= 120

Sendbase= 100

13Mao W07

TCP retransmission scenarios (more)Host A

Seq=92 8 bytes data

ACK=100

loss

tim

eout

Cumulative ACK scenario

Host B

X

Seq=100 20 bytes data

ACK=120

time

SendBase= 120

14Mao W07

TCP ACK generation [RFC 1122 RFC 2581]

Event at Receiver

Arrival of in-order segment withexpected seq All data up toexpected seq already ACKed

Arrival of in-order segment withexpected seq One other segment has ACK pending

Arrival of out-of-order segmenthigher-than-expect seq Gap detected

Arrival of segment that partially or completely fills gap

TCP Receiver action

Delayed ACK Wait up to 500msfor next segment If no next segmentsend ACK

Immediately send single cumulative ACK ACKing both in-order segments

Immediately send duplicate ACK indicating seq of next expected byte

Immediate send ACK provided thatsegment startsat lower end of gap

15Mao W07

Fast Retransmit

Time-out period often relatively long

- long delay before resending lost packet

Detect lost segments via duplicate ACKs

- Sender often sends many segments back-to-back

- If segment is lost there will likely be many duplicate ACKs

If sender receives 3 ACKs for the same data it supposes that segment after ACKed data was lost

- fast retransmit resend segment before timer expires

16Mao W07

event ACK received with ACK field value of y if (y gt SendBase)

SendBase = yif (there are currently not-yet-acknowledged segments)

start timer

else increment count of dup ACKs received for yif (count of dup ACKs received for y = 3)

resend segment with sequence number y

Fast retransmit algorithm

a duplicate ACK for already ACKed segment

fast retransmit

17Mao W07

TCP Flow Control

receive side of TCP connection has a receive buffer

speed-matching service matching the send rate to the receiving apprsquos drain rate

app process may be slow at reading from buffer

sender wonrsquot overflowreceiverrsquos buffer by

transmitting too muchtoo fast

flow control

18Mao W07

TCP Flow control how it works

(Suppose TCP receiver discards out-of-order segments)spare room in buffer

= RcvWindow= RcvBuffer-[LastByteRcvd -

LastByteRead]

Rcvr advertises spare room by including value of RcvWindow in segmentsSender limits unACKed data to RcvWindow

- guarantees receive buffer doesnrsquot overflow

19Mao W07

TCP Connection Management

Recall TCP sender receiver establish ldquoconnectionrdquo before exchanging data segmentsinitialize TCP variables

- seq s- buffers flow control info

(eg RcvWindow)client connection initiatorSocket clientSocket = new Socket(hostnameport

number)

server contacted by clientSocket connectionSocket = welcomeSocketaccept()

Three way handshakeStep 1 client host sends TCP

SYN segment to server- specifies initial seq - no data

Step 2 server host receives SYN replies with SYNACK segment

- server allocates buffers- specifies server initial seq

Step 3 client receives SYNACK replies with ACK segment which may contain data

20Mao W07

TCP Connection Management (cont)

Closing a connection

client closes socketclientSocketclose()

Step 1 client end system sends TCP FIN control segment to server

Step 2 server receives FIN replies with ACK Closes connection sends FIN

client

FIN

server

ACK

ACK

FIN

close

close

closed

tim

ed w

ait

21Mao W07

TCP Connection Management (cont)

Step 3 client receives FIN replies with ACK

- Enters ldquotimed waitrdquo - will respond with ACK to received FINs

Step 4 server receives ACK Connection closed

Note with small modification can handle simultaneous FINs

client

FIN

server

ACK

ACK

FIN

closing

closing

closed

tim

ed w

ait

closed

22Mao W07

TCP Connection Management (cont)

TCP clientlifecycle

TCP serverlifecycle

23Mao W07

Principles of Congestion Control

Congestioninformally ldquotoo many sources sending too much data too fast for network to handlerdquodifferent from flow controlmanifestations

- lost packets (buffer overflow at routers)- long delays (queueing in router buffers)

a top-10 problem

24Mao W07

Causescosts of congestion scenario 1

two senders two receiversone router infinite buffers no retransmission

large delays when congestedmaximum achievable throughput

unlimited shared output link buffers

Host Aλin original data

Host B

λout

25Mao W07

Causescosts of congestion scenario 2

one router finite buffers sender retransmission of lost packet

finite shared output link buffers

Host A λin original data

Host B

λout

λin original data plus retransmitted data

26Mao W07

Causescosts of congestion scenario 2

always (goodput)

ldquoperfectrdquo retransmission only when loss

retransmission of delayed (not lost) packet makes larger (than perfect case) for same

λin

λout=

λin

λoutgtλ

inλout

ldquocostsrdquo of congestionmore work (retrans) for given ldquogoodputrdquounneeded retransmissions link carries multiple copies of pkt

R2

R2λin

λ out

b

R2

R2λin

λ out

a

R2

R2λin

λ out

c

R4

R3

27Mao W07

Causescosts of congestion scenario 3

four sendersmultihop pathstimeoutretransmit

λin

Q what happens as and increase λ

in

finite shared output link buffers

Host Aλin original data

Host B

λout

λin original data plus retransmitted data

28Mao W07

Causescosts of congestion scenario 3

Another ldquocostrdquo of congestionwhen packet dropped any ldquoupstream transmission capacity used for that packet was wasted

Host A

Host B

λou

t

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

6Mao W07

TCP Round Trip Time and TimeoutEstimatedRTT = (1- α)EstimatedRTT + αSampleRTT

Exponential weighted moving averageinfluence of past sample decreases exponentially fasttypical value α = 0125

7Mao W07

Example RTT estimation

RTT gaiacsumassedu to fantasiaeurecomfr

100

150

200

250

300

350

1 8 15 22 29 36 43 50 57 64 71 78 85 92 99 106

time (seconnds)

RTT

(mill

iseco

nds)

SampleRTT Estimated RTT

8Mao W07

TCP Round Trip Time and TimeoutSetting the timeout

EstimtedRTT plus ldquosafety marginrdquo- large variation in EstimatedRTT -gt larger safety margin

first estimate of how much SampleRTT deviates from EstimatedRTT

TimeoutInterval = EstimatedRTT + 4DevRTT

DevRTT = (1-β)DevRTT +β|SampleRTT-EstimatedRTT|

(typically β = 025)

Then set timeout interval

9Mao W07

TCP reliable data transfer

TCP creates rdt service on top of IPrsquos unreliable servicePipelined segmentsCumulative acksTCP uses single retransmission timer

Retransmissions are triggered by

- timeout events- duplicate acks

Initially consider simplified TCP sender

- ignore duplicate acks- ignore flow control

congestion control

10Mao W07

TCP sender eventsdata rcvd from app

Create segment with seq seq is byte-stream number of first data byte in segmentstart timer if not already running (think of timer as for oldest unacked segment)expiration interval TimeOutInterval

timeoutretransmit segment that caused timeoutrestart timer

Ack rcvdIf acknowledges previously unacked segments

- update what is known to be acked

- start timer if there are outstanding segments

11Mao W07

TCP sender(simplified)

NextSeqNum = InitialSeqNumSendBase = InitialSeqNum

loop (forever) switch(event)

event data received from application above create TCP segment with sequence number NextSeqNumif (timer currently not running)

start timerpass segment to IP NextSeqNum = NextSeqNum + length(data)

event timer timeoutretransmit not-yet-acknowledged segment with

smallest sequence numberstart timer

event ACK received with ACK field value of y if (y gt SendBase)

SendBase = yif (there are currently not-yet-acknowledged segments)

start timer

end of loop forever

Commentbull SendBase-1 last cumulatively ackrsquoed byteExamplebull SendBase-1 = 71y= 73 so the rcvrwants 73+ y gt SendBase sothat new data is acked

12Mao W07

TCP retransmission scenariosHost A

Seq=100 20 bytes data

ACK=100

timepremature timeout

Host B

Seq=92 8 bytes data

ACK=120

Seq=92 8 bytes data

Seq=

92 t

imeo

ut

ACK=120

Host A

Seq=92 8 bytes data

ACK=100

loss

tim

eout

lost ACK scenario

Host B

X

Seq=92 8 bytes data

ACK=100

time

Seq=

92 t

imeo

utSendBase

= 100

SendBase= 120

SendBase= 120

Sendbase= 100

13Mao W07

TCP retransmission scenarios (more)Host A

Seq=92 8 bytes data

ACK=100

loss

tim

eout

Cumulative ACK scenario

Host B

X

Seq=100 20 bytes data

ACK=120

time

SendBase= 120

14Mao W07

TCP ACK generation [RFC 1122 RFC 2581]

Event at Receiver

Arrival of in-order segment withexpected seq All data up toexpected seq already ACKed

Arrival of in-order segment withexpected seq One other segment has ACK pending

Arrival of out-of-order segmenthigher-than-expect seq Gap detected

Arrival of segment that partially or completely fills gap

TCP Receiver action

Delayed ACK Wait up to 500msfor next segment If no next segmentsend ACK

Immediately send single cumulative ACK ACKing both in-order segments

Immediately send duplicate ACK indicating seq of next expected byte

Immediate send ACK provided thatsegment startsat lower end of gap

15Mao W07

Fast Retransmit

Time-out period often relatively long

- long delay before resending lost packet

Detect lost segments via duplicate ACKs

- Sender often sends many segments back-to-back

- If segment is lost there will likely be many duplicate ACKs

If sender receives 3 ACKs for the same data it supposes that segment after ACKed data was lost

- fast retransmit resend segment before timer expires

16Mao W07

event ACK received with ACK field value of y if (y gt SendBase)

SendBase = yif (there are currently not-yet-acknowledged segments)

start timer

else increment count of dup ACKs received for yif (count of dup ACKs received for y = 3)

resend segment with sequence number y

Fast retransmit algorithm

a duplicate ACK for already ACKed segment

fast retransmit

17Mao W07

TCP Flow Control

receive side of TCP connection has a receive buffer

speed-matching service matching the send rate to the receiving apprsquos drain rate

app process may be slow at reading from buffer

sender wonrsquot overflowreceiverrsquos buffer by

transmitting too muchtoo fast

flow control

18Mao W07

TCP Flow control how it works

(Suppose TCP receiver discards out-of-order segments)spare room in buffer

= RcvWindow= RcvBuffer-[LastByteRcvd -

LastByteRead]

Rcvr advertises spare room by including value of RcvWindow in segmentsSender limits unACKed data to RcvWindow

- guarantees receive buffer doesnrsquot overflow

19Mao W07

TCP Connection Management

Recall TCP sender receiver establish ldquoconnectionrdquo before exchanging data segmentsinitialize TCP variables

- seq s- buffers flow control info

(eg RcvWindow)client connection initiatorSocket clientSocket = new Socket(hostnameport

number)

server contacted by clientSocket connectionSocket = welcomeSocketaccept()

Three way handshakeStep 1 client host sends TCP

SYN segment to server- specifies initial seq - no data

Step 2 server host receives SYN replies with SYNACK segment

- server allocates buffers- specifies server initial seq

Step 3 client receives SYNACK replies with ACK segment which may contain data

20Mao W07

TCP Connection Management (cont)

Closing a connection

client closes socketclientSocketclose()

Step 1 client end system sends TCP FIN control segment to server

Step 2 server receives FIN replies with ACK Closes connection sends FIN

client

FIN

server

ACK

ACK

FIN

close

close

closed

tim

ed w

ait

21Mao W07

TCP Connection Management (cont)

Step 3 client receives FIN replies with ACK

- Enters ldquotimed waitrdquo - will respond with ACK to received FINs

Step 4 server receives ACK Connection closed

Note with small modification can handle simultaneous FINs

client

FIN

server

ACK

ACK

FIN

closing

closing

closed

tim

ed w

ait

closed

22Mao W07

TCP Connection Management (cont)

TCP clientlifecycle

TCP serverlifecycle

23Mao W07

Principles of Congestion Control

Congestioninformally ldquotoo many sources sending too much data too fast for network to handlerdquodifferent from flow controlmanifestations

- lost packets (buffer overflow at routers)- long delays (queueing in router buffers)

a top-10 problem

24Mao W07

Causescosts of congestion scenario 1

two senders two receiversone router infinite buffers no retransmission

large delays when congestedmaximum achievable throughput

unlimited shared output link buffers

Host Aλin original data

Host B

λout

25Mao W07

Causescosts of congestion scenario 2

one router finite buffers sender retransmission of lost packet

finite shared output link buffers

Host A λin original data

Host B

λout

λin original data plus retransmitted data

26Mao W07

Causescosts of congestion scenario 2

always (goodput)

ldquoperfectrdquo retransmission only when loss

retransmission of delayed (not lost) packet makes larger (than perfect case) for same

λin

λout=

λin

λoutgtλ

inλout

ldquocostsrdquo of congestionmore work (retrans) for given ldquogoodputrdquounneeded retransmissions link carries multiple copies of pkt

R2

R2λin

λ out

b

R2

R2λin

λ out

a

R2

R2λin

λ out

c

R4

R3

27Mao W07

Causescosts of congestion scenario 3

four sendersmultihop pathstimeoutretransmit

λin

Q what happens as and increase λ

in

finite shared output link buffers

Host Aλin original data

Host B

λout

λin original data plus retransmitted data

28Mao W07

Causescosts of congestion scenario 3

Another ldquocostrdquo of congestionwhen packet dropped any ldquoupstream transmission capacity used for that packet was wasted

Host A

Host B

λou

t

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

7Mao W07

Example RTT estimation

RTT gaiacsumassedu to fantasiaeurecomfr

100

150

200

250

300

350

1 8 15 22 29 36 43 50 57 64 71 78 85 92 99 106

time (seconnds)

RTT

(mill

iseco

nds)

SampleRTT Estimated RTT

8Mao W07

TCP Round Trip Time and TimeoutSetting the timeout

EstimtedRTT plus ldquosafety marginrdquo- large variation in EstimatedRTT -gt larger safety margin

first estimate of how much SampleRTT deviates from EstimatedRTT

TimeoutInterval = EstimatedRTT + 4DevRTT

DevRTT = (1-β)DevRTT +β|SampleRTT-EstimatedRTT|

(typically β = 025)

Then set timeout interval

9Mao W07

TCP reliable data transfer

TCP creates rdt service on top of IPrsquos unreliable servicePipelined segmentsCumulative acksTCP uses single retransmission timer

Retransmissions are triggered by

- timeout events- duplicate acks

Initially consider simplified TCP sender

- ignore duplicate acks- ignore flow control

congestion control

10Mao W07

TCP sender eventsdata rcvd from app

Create segment with seq seq is byte-stream number of first data byte in segmentstart timer if not already running (think of timer as for oldest unacked segment)expiration interval TimeOutInterval

timeoutretransmit segment that caused timeoutrestart timer

Ack rcvdIf acknowledges previously unacked segments

- update what is known to be acked

- start timer if there are outstanding segments

11Mao W07

TCP sender(simplified)

NextSeqNum = InitialSeqNumSendBase = InitialSeqNum

loop (forever) switch(event)

event data received from application above create TCP segment with sequence number NextSeqNumif (timer currently not running)

start timerpass segment to IP NextSeqNum = NextSeqNum + length(data)

event timer timeoutretransmit not-yet-acknowledged segment with

smallest sequence numberstart timer

event ACK received with ACK field value of y if (y gt SendBase)

SendBase = yif (there are currently not-yet-acknowledged segments)

start timer

end of loop forever

Commentbull SendBase-1 last cumulatively ackrsquoed byteExamplebull SendBase-1 = 71y= 73 so the rcvrwants 73+ y gt SendBase sothat new data is acked

12Mao W07

TCP retransmission scenariosHost A

Seq=100 20 bytes data

ACK=100

timepremature timeout

Host B

Seq=92 8 bytes data

ACK=120

Seq=92 8 bytes data

Seq=

92 t

imeo

ut

ACK=120

Host A

Seq=92 8 bytes data

ACK=100

loss

tim

eout

lost ACK scenario

Host B

X

Seq=92 8 bytes data

ACK=100

time

Seq=

92 t

imeo

utSendBase

= 100

SendBase= 120

SendBase= 120

Sendbase= 100

13Mao W07

TCP retransmission scenarios (more)Host A

Seq=92 8 bytes data

ACK=100

loss

tim

eout

Cumulative ACK scenario

Host B

X

Seq=100 20 bytes data

ACK=120

time

SendBase= 120

14Mao W07

TCP ACK generation [RFC 1122 RFC 2581]

Event at Receiver

Arrival of in-order segment withexpected seq All data up toexpected seq already ACKed

Arrival of in-order segment withexpected seq One other segment has ACK pending

Arrival of out-of-order segmenthigher-than-expect seq Gap detected

Arrival of segment that partially or completely fills gap

TCP Receiver action

Delayed ACK Wait up to 500msfor next segment If no next segmentsend ACK

Immediately send single cumulative ACK ACKing both in-order segments

Immediately send duplicate ACK indicating seq of next expected byte

Immediate send ACK provided thatsegment startsat lower end of gap

15Mao W07

Fast Retransmit

Time-out period often relatively long

- long delay before resending lost packet

Detect lost segments via duplicate ACKs

- Sender often sends many segments back-to-back

- If segment is lost there will likely be many duplicate ACKs

If sender receives 3 ACKs for the same data it supposes that segment after ACKed data was lost

- fast retransmit resend segment before timer expires

16Mao W07

event ACK received with ACK field value of y if (y gt SendBase)

SendBase = yif (there are currently not-yet-acknowledged segments)

start timer

else increment count of dup ACKs received for yif (count of dup ACKs received for y = 3)

resend segment with sequence number y

Fast retransmit algorithm

a duplicate ACK for already ACKed segment

fast retransmit

17Mao W07

TCP Flow Control

receive side of TCP connection has a receive buffer

speed-matching service matching the send rate to the receiving apprsquos drain rate

app process may be slow at reading from buffer

sender wonrsquot overflowreceiverrsquos buffer by

transmitting too muchtoo fast

flow control

18Mao W07

TCP Flow control how it works

(Suppose TCP receiver discards out-of-order segments)spare room in buffer

= RcvWindow= RcvBuffer-[LastByteRcvd -

LastByteRead]

Rcvr advertises spare room by including value of RcvWindow in segmentsSender limits unACKed data to RcvWindow

- guarantees receive buffer doesnrsquot overflow

19Mao W07

TCP Connection Management

Recall TCP sender receiver establish ldquoconnectionrdquo before exchanging data segmentsinitialize TCP variables

- seq s- buffers flow control info

(eg RcvWindow)client connection initiatorSocket clientSocket = new Socket(hostnameport

number)

server contacted by clientSocket connectionSocket = welcomeSocketaccept()

Three way handshakeStep 1 client host sends TCP

SYN segment to server- specifies initial seq - no data

Step 2 server host receives SYN replies with SYNACK segment

- server allocates buffers- specifies server initial seq

Step 3 client receives SYNACK replies with ACK segment which may contain data

20Mao W07

TCP Connection Management (cont)

Closing a connection

client closes socketclientSocketclose()

Step 1 client end system sends TCP FIN control segment to server

Step 2 server receives FIN replies with ACK Closes connection sends FIN

client

FIN

server

ACK

ACK

FIN

close

close

closed

tim

ed w

ait

21Mao W07

TCP Connection Management (cont)

Step 3 client receives FIN replies with ACK

- Enters ldquotimed waitrdquo - will respond with ACK to received FINs

Step 4 server receives ACK Connection closed

Note with small modification can handle simultaneous FINs

client

FIN

server

ACK

ACK

FIN

closing

closing

closed

tim

ed w

ait

closed

22Mao W07

TCP Connection Management (cont)

TCP clientlifecycle

TCP serverlifecycle

23Mao W07

Principles of Congestion Control

Congestioninformally ldquotoo many sources sending too much data too fast for network to handlerdquodifferent from flow controlmanifestations

- lost packets (buffer overflow at routers)- long delays (queueing in router buffers)

a top-10 problem

24Mao W07

Causescosts of congestion scenario 1

two senders two receiversone router infinite buffers no retransmission

large delays when congestedmaximum achievable throughput

unlimited shared output link buffers

Host Aλin original data

Host B

λout

25Mao W07

Causescosts of congestion scenario 2

one router finite buffers sender retransmission of lost packet

finite shared output link buffers

Host A λin original data

Host B

λout

λin original data plus retransmitted data

26Mao W07

Causescosts of congestion scenario 2

always (goodput)

ldquoperfectrdquo retransmission only when loss

retransmission of delayed (not lost) packet makes larger (than perfect case) for same

λin

λout=

λin

λoutgtλ

inλout

ldquocostsrdquo of congestionmore work (retrans) for given ldquogoodputrdquounneeded retransmissions link carries multiple copies of pkt

R2

R2λin

λ out

b

R2

R2λin

λ out

a

R2

R2λin

λ out

c

R4

R3

27Mao W07

Causescosts of congestion scenario 3

four sendersmultihop pathstimeoutretransmit

λin

Q what happens as and increase λ

in

finite shared output link buffers

Host Aλin original data

Host B

λout

λin original data plus retransmitted data

28Mao W07

Causescosts of congestion scenario 3

Another ldquocostrdquo of congestionwhen packet dropped any ldquoupstream transmission capacity used for that packet was wasted

Host A

Host B

λou

t

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

8Mao W07

TCP Round Trip Time and TimeoutSetting the timeout

EstimtedRTT plus ldquosafety marginrdquo- large variation in EstimatedRTT -gt larger safety margin

first estimate of how much SampleRTT deviates from EstimatedRTT

TimeoutInterval = EstimatedRTT + 4DevRTT

DevRTT = (1-β)DevRTT +β|SampleRTT-EstimatedRTT|

(typically β = 025)

Then set timeout interval

9Mao W07

TCP reliable data transfer

TCP creates rdt service on top of IPrsquos unreliable servicePipelined segmentsCumulative acksTCP uses single retransmission timer

Retransmissions are triggered by

- timeout events- duplicate acks

Initially consider simplified TCP sender

- ignore duplicate acks- ignore flow control

congestion control

10Mao W07

TCP sender eventsdata rcvd from app

Create segment with seq seq is byte-stream number of first data byte in segmentstart timer if not already running (think of timer as for oldest unacked segment)expiration interval TimeOutInterval

timeoutretransmit segment that caused timeoutrestart timer

Ack rcvdIf acknowledges previously unacked segments

- update what is known to be acked

- start timer if there are outstanding segments

11Mao W07

TCP sender(simplified)

NextSeqNum = InitialSeqNumSendBase = InitialSeqNum

loop (forever) switch(event)

event data received from application above create TCP segment with sequence number NextSeqNumif (timer currently not running)

start timerpass segment to IP NextSeqNum = NextSeqNum + length(data)

event timer timeoutretransmit not-yet-acknowledged segment with

smallest sequence numberstart timer

event ACK received with ACK field value of y if (y gt SendBase)

SendBase = yif (there are currently not-yet-acknowledged segments)

start timer

end of loop forever

Commentbull SendBase-1 last cumulatively ackrsquoed byteExamplebull SendBase-1 = 71y= 73 so the rcvrwants 73+ y gt SendBase sothat new data is acked

12Mao W07

TCP retransmission scenariosHost A

Seq=100 20 bytes data

ACK=100

timepremature timeout

Host B

Seq=92 8 bytes data

ACK=120

Seq=92 8 bytes data

Seq=

92 t

imeo

ut

ACK=120

Host A

Seq=92 8 bytes data

ACK=100

loss

tim

eout

lost ACK scenario

Host B

X

Seq=92 8 bytes data

ACK=100

time

Seq=

92 t

imeo

utSendBase

= 100

SendBase= 120

SendBase= 120

Sendbase= 100

13Mao W07

TCP retransmission scenarios (more)Host A

Seq=92 8 bytes data

ACK=100

loss

tim

eout

Cumulative ACK scenario

Host B

X

Seq=100 20 bytes data

ACK=120

time

SendBase= 120

14Mao W07

TCP ACK generation [RFC 1122 RFC 2581]

Event at Receiver

Arrival of in-order segment withexpected seq All data up toexpected seq already ACKed

Arrival of in-order segment withexpected seq One other segment has ACK pending

Arrival of out-of-order segmenthigher-than-expect seq Gap detected

Arrival of segment that partially or completely fills gap

TCP Receiver action

Delayed ACK Wait up to 500msfor next segment If no next segmentsend ACK

Immediately send single cumulative ACK ACKing both in-order segments

Immediately send duplicate ACK indicating seq of next expected byte

Immediate send ACK provided thatsegment startsat lower end of gap

15Mao W07

Fast Retransmit

Time-out period often relatively long

- long delay before resending lost packet

Detect lost segments via duplicate ACKs

- Sender often sends many segments back-to-back

- If segment is lost there will likely be many duplicate ACKs

If sender receives 3 ACKs for the same data it supposes that segment after ACKed data was lost

- fast retransmit resend segment before timer expires

16Mao W07

event ACK received with ACK field value of y if (y gt SendBase)

SendBase = yif (there are currently not-yet-acknowledged segments)

start timer

else increment count of dup ACKs received for yif (count of dup ACKs received for y = 3)

resend segment with sequence number y

Fast retransmit algorithm

a duplicate ACK for already ACKed segment

fast retransmit

17Mao W07

TCP Flow Control

receive side of TCP connection has a receive buffer

speed-matching service matching the send rate to the receiving apprsquos drain rate

app process may be slow at reading from buffer

sender wonrsquot overflowreceiverrsquos buffer by

transmitting too muchtoo fast

flow control

18Mao W07

TCP Flow control how it works

(Suppose TCP receiver discards out-of-order segments)spare room in buffer

= RcvWindow= RcvBuffer-[LastByteRcvd -

LastByteRead]

Rcvr advertises spare room by including value of RcvWindow in segmentsSender limits unACKed data to RcvWindow

- guarantees receive buffer doesnrsquot overflow

19Mao W07

TCP Connection Management

Recall TCP sender receiver establish ldquoconnectionrdquo before exchanging data segmentsinitialize TCP variables

- seq s- buffers flow control info

(eg RcvWindow)client connection initiatorSocket clientSocket = new Socket(hostnameport

number)

server contacted by clientSocket connectionSocket = welcomeSocketaccept()

Three way handshakeStep 1 client host sends TCP

SYN segment to server- specifies initial seq - no data

Step 2 server host receives SYN replies with SYNACK segment

- server allocates buffers- specifies server initial seq

Step 3 client receives SYNACK replies with ACK segment which may contain data

20Mao W07

TCP Connection Management (cont)

Closing a connection

client closes socketclientSocketclose()

Step 1 client end system sends TCP FIN control segment to server

Step 2 server receives FIN replies with ACK Closes connection sends FIN

client

FIN

server

ACK

ACK

FIN

close

close

closed

tim

ed w

ait

21Mao W07

TCP Connection Management (cont)

Step 3 client receives FIN replies with ACK

- Enters ldquotimed waitrdquo - will respond with ACK to received FINs

Step 4 server receives ACK Connection closed

Note with small modification can handle simultaneous FINs

client

FIN

server

ACK

ACK

FIN

closing

closing

closed

tim

ed w

ait

closed

22Mao W07

TCP Connection Management (cont)

TCP clientlifecycle

TCP serverlifecycle

23Mao W07

Principles of Congestion Control

Congestioninformally ldquotoo many sources sending too much data too fast for network to handlerdquodifferent from flow controlmanifestations

- lost packets (buffer overflow at routers)- long delays (queueing in router buffers)

a top-10 problem

24Mao W07

Causescosts of congestion scenario 1

two senders two receiversone router infinite buffers no retransmission

large delays when congestedmaximum achievable throughput

unlimited shared output link buffers

Host Aλin original data

Host B

λout

25Mao W07

Causescosts of congestion scenario 2

one router finite buffers sender retransmission of lost packet

finite shared output link buffers

Host A λin original data

Host B

λout

λin original data plus retransmitted data

26Mao W07

Causescosts of congestion scenario 2

always (goodput)

ldquoperfectrdquo retransmission only when loss

retransmission of delayed (not lost) packet makes larger (than perfect case) for same

λin

λout=

λin

λoutgtλ

inλout

ldquocostsrdquo of congestionmore work (retrans) for given ldquogoodputrdquounneeded retransmissions link carries multiple copies of pkt

R2

R2λin

λ out

b

R2

R2λin

λ out

a

R2

R2λin

λ out

c

R4

R3

27Mao W07

Causescosts of congestion scenario 3

four sendersmultihop pathstimeoutretransmit

λin

Q what happens as and increase λ

in

finite shared output link buffers

Host Aλin original data

Host B

λout

λin original data plus retransmitted data

28Mao W07

Causescosts of congestion scenario 3

Another ldquocostrdquo of congestionwhen packet dropped any ldquoupstream transmission capacity used for that packet was wasted

Host A

Host B

λou

t

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

9Mao W07

TCP reliable data transfer

TCP creates rdt service on top of IPrsquos unreliable servicePipelined segmentsCumulative acksTCP uses single retransmission timer

Retransmissions are triggered by

- timeout events- duplicate acks

Initially consider simplified TCP sender

- ignore duplicate acks- ignore flow control

congestion control

10Mao W07

TCP sender eventsdata rcvd from app

Create segment with seq seq is byte-stream number of first data byte in segmentstart timer if not already running (think of timer as for oldest unacked segment)expiration interval TimeOutInterval

timeoutretransmit segment that caused timeoutrestart timer

Ack rcvdIf acknowledges previously unacked segments

- update what is known to be acked

- start timer if there are outstanding segments

11Mao W07

TCP sender(simplified)

NextSeqNum = InitialSeqNumSendBase = InitialSeqNum

loop (forever) switch(event)

event data received from application above create TCP segment with sequence number NextSeqNumif (timer currently not running)

start timerpass segment to IP NextSeqNum = NextSeqNum + length(data)

event timer timeoutretransmit not-yet-acknowledged segment with

smallest sequence numberstart timer

event ACK received with ACK field value of y if (y gt SendBase)

SendBase = yif (there are currently not-yet-acknowledged segments)

start timer

end of loop forever

Commentbull SendBase-1 last cumulatively ackrsquoed byteExamplebull SendBase-1 = 71y= 73 so the rcvrwants 73+ y gt SendBase sothat new data is acked

12Mao W07

TCP retransmission scenariosHost A

Seq=100 20 bytes data

ACK=100

timepremature timeout

Host B

Seq=92 8 bytes data

ACK=120

Seq=92 8 bytes data

Seq=

92 t

imeo

ut

ACK=120

Host A

Seq=92 8 bytes data

ACK=100

loss

tim

eout

lost ACK scenario

Host B

X

Seq=92 8 bytes data

ACK=100

time

Seq=

92 t

imeo

utSendBase

= 100

SendBase= 120

SendBase= 120

Sendbase= 100

13Mao W07

TCP retransmission scenarios (more)Host A

Seq=92 8 bytes data

ACK=100

loss

tim

eout

Cumulative ACK scenario

Host B

X

Seq=100 20 bytes data

ACK=120

time

SendBase= 120

14Mao W07

TCP ACK generation [RFC 1122 RFC 2581]

Event at Receiver

Arrival of in-order segment withexpected seq All data up toexpected seq already ACKed

Arrival of in-order segment withexpected seq One other segment has ACK pending

Arrival of out-of-order segmenthigher-than-expect seq Gap detected

Arrival of segment that partially or completely fills gap

TCP Receiver action

Delayed ACK Wait up to 500msfor next segment If no next segmentsend ACK

Immediately send single cumulative ACK ACKing both in-order segments

Immediately send duplicate ACK indicating seq of next expected byte

Immediate send ACK provided thatsegment startsat lower end of gap

15Mao W07

Fast Retransmit

Time-out period often relatively long

- long delay before resending lost packet

Detect lost segments via duplicate ACKs

- Sender often sends many segments back-to-back

- If segment is lost there will likely be many duplicate ACKs

If sender receives 3 ACKs for the same data it supposes that segment after ACKed data was lost

- fast retransmit resend segment before timer expires

16Mao W07

event ACK received with ACK field value of y if (y gt SendBase)

SendBase = yif (there are currently not-yet-acknowledged segments)

start timer

else increment count of dup ACKs received for yif (count of dup ACKs received for y = 3)

resend segment with sequence number y

Fast retransmit algorithm

a duplicate ACK for already ACKed segment

fast retransmit

17Mao W07

TCP Flow Control

receive side of TCP connection has a receive buffer

speed-matching service matching the send rate to the receiving apprsquos drain rate

app process may be slow at reading from buffer

sender wonrsquot overflowreceiverrsquos buffer by

transmitting too muchtoo fast

flow control

18Mao W07

TCP Flow control how it works

(Suppose TCP receiver discards out-of-order segments)spare room in buffer

= RcvWindow= RcvBuffer-[LastByteRcvd -

LastByteRead]

Rcvr advertises spare room by including value of RcvWindow in segmentsSender limits unACKed data to RcvWindow

- guarantees receive buffer doesnrsquot overflow

19Mao W07

TCP Connection Management

Recall TCP sender receiver establish ldquoconnectionrdquo before exchanging data segmentsinitialize TCP variables

- seq s- buffers flow control info

(eg RcvWindow)client connection initiatorSocket clientSocket = new Socket(hostnameport

number)

server contacted by clientSocket connectionSocket = welcomeSocketaccept()

Three way handshakeStep 1 client host sends TCP

SYN segment to server- specifies initial seq - no data

Step 2 server host receives SYN replies with SYNACK segment

- server allocates buffers- specifies server initial seq

Step 3 client receives SYNACK replies with ACK segment which may contain data

20Mao W07

TCP Connection Management (cont)

Closing a connection

client closes socketclientSocketclose()

Step 1 client end system sends TCP FIN control segment to server

Step 2 server receives FIN replies with ACK Closes connection sends FIN

client

FIN

server

ACK

ACK

FIN

close

close

closed

tim

ed w

ait

21Mao W07

TCP Connection Management (cont)

Step 3 client receives FIN replies with ACK

- Enters ldquotimed waitrdquo - will respond with ACK to received FINs

Step 4 server receives ACK Connection closed

Note with small modification can handle simultaneous FINs

client

FIN

server

ACK

ACK

FIN

closing

closing

closed

tim

ed w

ait

closed

22Mao W07

TCP Connection Management (cont)

TCP clientlifecycle

TCP serverlifecycle

23Mao W07

Principles of Congestion Control

Congestioninformally ldquotoo many sources sending too much data too fast for network to handlerdquodifferent from flow controlmanifestations

- lost packets (buffer overflow at routers)- long delays (queueing in router buffers)

a top-10 problem

24Mao W07

Causescosts of congestion scenario 1

two senders two receiversone router infinite buffers no retransmission

large delays when congestedmaximum achievable throughput

unlimited shared output link buffers

Host Aλin original data

Host B

λout

25Mao W07

Causescosts of congestion scenario 2

one router finite buffers sender retransmission of lost packet

finite shared output link buffers

Host A λin original data

Host B

λout

λin original data plus retransmitted data

26Mao W07

Causescosts of congestion scenario 2

always (goodput)

ldquoperfectrdquo retransmission only when loss

retransmission of delayed (not lost) packet makes larger (than perfect case) for same

λin

λout=

λin

λoutgtλ

inλout

ldquocostsrdquo of congestionmore work (retrans) for given ldquogoodputrdquounneeded retransmissions link carries multiple copies of pkt

R2

R2λin

λ out

b

R2

R2λin

λ out

a

R2

R2λin

λ out

c

R4

R3

27Mao W07

Causescosts of congestion scenario 3

four sendersmultihop pathstimeoutretransmit

λin

Q what happens as and increase λ

in

finite shared output link buffers

Host Aλin original data

Host B

λout

λin original data plus retransmitted data

28Mao W07

Causescosts of congestion scenario 3

Another ldquocostrdquo of congestionwhen packet dropped any ldquoupstream transmission capacity used for that packet was wasted

Host A

Host B

λou

t

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

10Mao W07

TCP sender eventsdata rcvd from app

Create segment with seq seq is byte-stream number of first data byte in segmentstart timer if not already running (think of timer as for oldest unacked segment)expiration interval TimeOutInterval

timeoutretransmit segment that caused timeoutrestart timer

Ack rcvdIf acknowledges previously unacked segments

- update what is known to be acked

- start timer if there are outstanding segments

11Mao W07

TCP sender(simplified)

NextSeqNum = InitialSeqNumSendBase = InitialSeqNum

loop (forever) switch(event)

event data received from application above create TCP segment with sequence number NextSeqNumif (timer currently not running)

start timerpass segment to IP NextSeqNum = NextSeqNum + length(data)

event timer timeoutretransmit not-yet-acknowledged segment with

smallest sequence numberstart timer

event ACK received with ACK field value of y if (y gt SendBase)

SendBase = yif (there are currently not-yet-acknowledged segments)

start timer

end of loop forever

Commentbull SendBase-1 last cumulatively ackrsquoed byteExamplebull SendBase-1 = 71y= 73 so the rcvrwants 73+ y gt SendBase sothat new data is acked

12Mao W07

TCP retransmission scenariosHost A

Seq=100 20 bytes data

ACK=100

timepremature timeout

Host B

Seq=92 8 bytes data

ACK=120

Seq=92 8 bytes data

Seq=

92 t

imeo

ut

ACK=120

Host A

Seq=92 8 bytes data

ACK=100

loss

tim

eout

lost ACK scenario

Host B

X

Seq=92 8 bytes data

ACK=100

time

Seq=

92 t

imeo

utSendBase

= 100

SendBase= 120

SendBase= 120

Sendbase= 100

13Mao W07

TCP retransmission scenarios (more)Host A

Seq=92 8 bytes data

ACK=100

loss

tim

eout

Cumulative ACK scenario

Host B

X

Seq=100 20 bytes data

ACK=120

time

SendBase= 120

14Mao W07

TCP ACK generation [RFC 1122 RFC 2581]

Event at Receiver

Arrival of in-order segment withexpected seq All data up toexpected seq already ACKed

Arrival of in-order segment withexpected seq One other segment has ACK pending

Arrival of out-of-order segmenthigher-than-expect seq Gap detected

Arrival of segment that partially or completely fills gap

TCP Receiver action

Delayed ACK Wait up to 500msfor next segment If no next segmentsend ACK

Immediately send single cumulative ACK ACKing both in-order segments

Immediately send duplicate ACK indicating seq of next expected byte

Immediate send ACK provided thatsegment startsat lower end of gap

15Mao W07

Fast Retransmit

Time-out period often relatively long

- long delay before resending lost packet

Detect lost segments via duplicate ACKs

- Sender often sends many segments back-to-back

- If segment is lost there will likely be many duplicate ACKs

If sender receives 3 ACKs for the same data it supposes that segment after ACKed data was lost

- fast retransmit resend segment before timer expires

16Mao W07

event ACK received with ACK field value of y if (y gt SendBase)

SendBase = yif (there are currently not-yet-acknowledged segments)

start timer

else increment count of dup ACKs received for yif (count of dup ACKs received for y = 3)

resend segment with sequence number y

Fast retransmit algorithm

a duplicate ACK for already ACKed segment

fast retransmit

17Mao W07

TCP Flow Control

receive side of TCP connection has a receive buffer

speed-matching service matching the send rate to the receiving apprsquos drain rate

app process may be slow at reading from buffer

sender wonrsquot overflowreceiverrsquos buffer by

transmitting too muchtoo fast

flow control

18Mao W07

TCP Flow control how it works

(Suppose TCP receiver discards out-of-order segments)spare room in buffer

= RcvWindow= RcvBuffer-[LastByteRcvd -

LastByteRead]

Rcvr advertises spare room by including value of RcvWindow in segmentsSender limits unACKed data to RcvWindow

- guarantees receive buffer doesnrsquot overflow

19Mao W07

TCP Connection Management

Recall TCP sender receiver establish ldquoconnectionrdquo before exchanging data segmentsinitialize TCP variables

- seq s- buffers flow control info

(eg RcvWindow)client connection initiatorSocket clientSocket = new Socket(hostnameport

number)

server contacted by clientSocket connectionSocket = welcomeSocketaccept()

Three way handshakeStep 1 client host sends TCP

SYN segment to server- specifies initial seq - no data

Step 2 server host receives SYN replies with SYNACK segment

- server allocates buffers- specifies server initial seq

Step 3 client receives SYNACK replies with ACK segment which may contain data

20Mao W07

TCP Connection Management (cont)

Closing a connection

client closes socketclientSocketclose()

Step 1 client end system sends TCP FIN control segment to server

Step 2 server receives FIN replies with ACK Closes connection sends FIN

client

FIN

server

ACK

ACK

FIN

close

close

closed

tim

ed w

ait

21Mao W07

TCP Connection Management (cont)

Step 3 client receives FIN replies with ACK

- Enters ldquotimed waitrdquo - will respond with ACK to received FINs

Step 4 server receives ACK Connection closed

Note with small modification can handle simultaneous FINs

client

FIN

server

ACK

ACK

FIN

closing

closing

closed

tim

ed w

ait

closed

22Mao W07

TCP Connection Management (cont)

TCP clientlifecycle

TCP serverlifecycle

23Mao W07

Principles of Congestion Control

Congestioninformally ldquotoo many sources sending too much data too fast for network to handlerdquodifferent from flow controlmanifestations

- lost packets (buffer overflow at routers)- long delays (queueing in router buffers)

a top-10 problem

24Mao W07

Causescosts of congestion scenario 1

two senders two receiversone router infinite buffers no retransmission

large delays when congestedmaximum achievable throughput

unlimited shared output link buffers

Host Aλin original data

Host B

λout

25Mao W07

Causescosts of congestion scenario 2

one router finite buffers sender retransmission of lost packet

finite shared output link buffers

Host A λin original data

Host B

λout

λin original data plus retransmitted data

26Mao W07

Causescosts of congestion scenario 2

always (goodput)

ldquoperfectrdquo retransmission only when loss

retransmission of delayed (not lost) packet makes larger (than perfect case) for same

λin

λout=

λin

λoutgtλ

inλout

ldquocostsrdquo of congestionmore work (retrans) for given ldquogoodputrdquounneeded retransmissions link carries multiple copies of pkt

R2

R2λin

λ out

b

R2

R2λin

λ out

a

R2

R2λin

λ out

c

R4

R3

27Mao W07

Causescosts of congestion scenario 3

four sendersmultihop pathstimeoutretransmit

λin

Q what happens as and increase λ

in

finite shared output link buffers

Host Aλin original data

Host B

λout

λin original data plus retransmitted data

28Mao W07

Causescosts of congestion scenario 3

Another ldquocostrdquo of congestionwhen packet dropped any ldquoupstream transmission capacity used for that packet was wasted

Host A

Host B

λou

t

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

11Mao W07

TCP sender(simplified)

NextSeqNum = InitialSeqNumSendBase = InitialSeqNum

loop (forever) switch(event)

event data received from application above create TCP segment with sequence number NextSeqNumif (timer currently not running)

start timerpass segment to IP NextSeqNum = NextSeqNum + length(data)

event timer timeoutretransmit not-yet-acknowledged segment with

smallest sequence numberstart timer

event ACK received with ACK field value of y if (y gt SendBase)

SendBase = yif (there are currently not-yet-acknowledged segments)

start timer

end of loop forever

Commentbull SendBase-1 last cumulatively ackrsquoed byteExamplebull SendBase-1 = 71y= 73 so the rcvrwants 73+ y gt SendBase sothat new data is acked

12Mao W07

TCP retransmission scenariosHost A

Seq=100 20 bytes data

ACK=100

timepremature timeout

Host B

Seq=92 8 bytes data

ACK=120

Seq=92 8 bytes data

Seq=

92 t

imeo

ut

ACK=120

Host A

Seq=92 8 bytes data

ACK=100

loss

tim

eout

lost ACK scenario

Host B

X

Seq=92 8 bytes data

ACK=100

time

Seq=

92 t

imeo

utSendBase

= 100

SendBase= 120

SendBase= 120

Sendbase= 100

13Mao W07

TCP retransmission scenarios (more)Host A

Seq=92 8 bytes data

ACK=100

loss

tim

eout

Cumulative ACK scenario

Host B

X

Seq=100 20 bytes data

ACK=120

time

SendBase= 120

14Mao W07

TCP ACK generation [RFC 1122 RFC 2581]

Event at Receiver

Arrival of in-order segment withexpected seq All data up toexpected seq already ACKed

Arrival of in-order segment withexpected seq One other segment has ACK pending

Arrival of out-of-order segmenthigher-than-expect seq Gap detected

Arrival of segment that partially or completely fills gap

TCP Receiver action

Delayed ACK Wait up to 500msfor next segment If no next segmentsend ACK

Immediately send single cumulative ACK ACKing both in-order segments

Immediately send duplicate ACK indicating seq of next expected byte

Immediate send ACK provided thatsegment startsat lower end of gap

15Mao W07

Fast Retransmit

Time-out period often relatively long

- long delay before resending lost packet

Detect lost segments via duplicate ACKs

- Sender often sends many segments back-to-back

- If segment is lost there will likely be many duplicate ACKs

If sender receives 3 ACKs for the same data it supposes that segment after ACKed data was lost

- fast retransmit resend segment before timer expires

16Mao W07

event ACK received with ACK field value of y if (y gt SendBase)

SendBase = yif (there are currently not-yet-acknowledged segments)

start timer

else increment count of dup ACKs received for yif (count of dup ACKs received for y = 3)

resend segment with sequence number y

Fast retransmit algorithm

a duplicate ACK for already ACKed segment

fast retransmit

17Mao W07

TCP Flow Control

receive side of TCP connection has a receive buffer

speed-matching service matching the send rate to the receiving apprsquos drain rate

app process may be slow at reading from buffer

sender wonrsquot overflowreceiverrsquos buffer by

transmitting too muchtoo fast

flow control

18Mao W07

TCP Flow control how it works

(Suppose TCP receiver discards out-of-order segments)spare room in buffer

= RcvWindow= RcvBuffer-[LastByteRcvd -

LastByteRead]

Rcvr advertises spare room by including value of RcvWindow in segmentsSender limits unACKed data to RcvWindow

- guarantees receive buffer doesnrsquot overflow

19Mao W07

TCP Connection Management

Recall TCP sender receiver establish ldquoconnectionrdquo before exchanging data segmentsinitialize TCP variables

- seq s- buffers flow control info

(eg RcvWindow)client connection initiatorSocket clientSocket = new Socket(hostnameport

number)

server contacted by clientSocket connectionSocket = welcomeSocketaccept()

Three way handshakeStep 1 client host sends TCP

SYN segment to server- specifies initial seq - no data

Step 2 server host receives SYN replies with SYNACK segment

- server allocates buffers- specifies server initial seq

Step 3 client receives SYNACK replies with ACK segment which may contain data

20Mao W07

TCP Connection Management (cont)

Closing a connection

client closes socketclientSocketclose()

Step 1 client end system sends TCP FIN control segment to server

Step 2 server receives FIN replies with ACK Closes connection sends FIN

client

FIN

server

ACK

ACK

FIN

close

close

closed

tim

ed w

ait

21Mao W07

TCP Connection Management (cont)

Step 3 client receives FIN replies with ACK

- Enters ldquotimed waitrdquo - will respond with ACK to received FINs

Step 4 server receives ACK Connection closed

Note with small modification can handle simultaneous FINs

client

FIN

server

ACK

ACK

FIN

closing

closing

closed

tim

ed w

ait

closed

22Mao W07

TCP Connection Management (cont)

TCP clientlifecycle

TCP serverlifecycle

23Mao W07

Principles of Congestion Control

Congestioninformally ldquotoo many sources sending too much data too fast for network to handlerdquodifferent from flow controlmanifestations

- lost packets (buffer overflow at routers)- long delays (queueing in router buffers)

a top-10 problem

24Mao W07

Causescosts of congestion scenario 1

two senders two receiversone router infinite buffers no retransmission

large delays when congestedmaximum achievable throughput

unlimited shared output link buffers

Host Aλin original data

Host B

λout

25Mao W07

Causescosts of congestion scenario 2

one router finite buffers sender retransmission of lost packet

finite shared output link buffers

Host A λin original data

Host B

λout

λin original data plus retransmitted data

26Mao W07

Causescosts of congestion scenario 2

always (goodput)

ldquoperfectrdquo retransmission only when loss

retransmission of delayed (not lost) packet makes larger (than perfect case) for same

λin

λout=

λin

λoutgtλ

inλout

ldquocostsrdquo of congestionmore work (retrans) for given ldquogoodputrdquounneeded retransmissions link carries multiple copies of pkt

R2

R2λin

λ out

b

R2

R2λin

λ out

a

R2

R2λin

λ out

c

R4

R3

27Mao W07

Causescosts of congestion scenario 3

four sendersmultihop pathstimeoutretransmit

λin

Q what happens as and increase λ

in

finite shared output link buffers

Host Aλin original data

Host B

λout

λin original data plus retransmitted data

28Mao W07

Causescosts of congestion scenario 3

Another ldquocostrdquo of congestionwhen packet dropped any ldquoupstream transmission capacity used for that packet was wasted

Host A

Host B

λou

t

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

12Mao W07

TCP retransmission scenariosHost A

Seq=100 20 bytes data

ACK=100

timepremature timeout

Host B

Seq=92 8 bytes data

ACK=120

Seq=92 8 bytes data

Seq=

92 t

imeo

ut

ACK=120

Host A

Seq=92 8 bytes data

ACK=100

loss

tim

eout

lost ACK scenario

Host B

X

Seq=92 8 bytes data

ACK=100

time

Seq=

92 t

imeo

utSendBase

= 100

SendBase= 120

SendBase= 120

Sendbase= 100

13Mao W07

TCP retransmission scenarios (more)Host A

Seq=92 8 bytes data

ACK=100

loss

tim

eout

Cumulative ACK scenario

Host B

X

Seq=100 20 bytes data

ACK=120

time

SendBase= 120

14Mao W07

TCP ACK generation [RFC 1122 RFC 2581]

Event at Receiver

Arrival of in-order segment withexpected seq All data up toexpected seq already ACKed

Arrival of in-order segment withexpected seq One other segment has ACK pending

Arrival of out-of-order segmenthigher-than-expect seq Gap detected

Arrival of segment that partially or completely fills gap

TCP Receiver action

Delayed ACK Wait up to 500msfor next segment If no next segmentsend ACK

Immediately send single cumulative ACK ACKing both in-order segments

Immediately send duplicate ACK indicating seq of next expected byte

Immediate send ACK provided thatsegment startsat lower end of gap

15Mao W07

Fast Retransmit

Time-out period often relatively long

- long delay before resending lost packet

Detect lost segments via duplicate ACKs

- Sender often sends many segments back-to-back

- If segment is lost there will likely be many duplicate ACKs

If sender receives 3 ACKs for the same data it supposes that segment after ACKed data was lost

- fast retransmit resend segment before timer expires

16Mao W07

event ACK received with ACK field value of y if (y gt SendBase)

SendBase = yif (there are currently not-yet-acknowledged segments)

start timer

else increment count of dup ACKs received for yif (count of dup ACKs received for y = 3)

resend segment with sequence number y

Fast retransmit algorithm

a duplicate ACK for already ACKed segment

fast retransmit

17Mao W07

TCP Flow Control

receive side of TCP connection has a receive buffer

speed-matching service matching the send rate to the receiving apprsquos drain rate

app process may be slow at reading from buffer

sender wonrsquot overflowreceiverrsquos buffer by

transmitting too muchtoo fast

flow control

18Mao W07

TCP Flow control how it works

(Suppose TCP receiver discards out-of-order segments)spare room in buffer

= RcvWindow= RcvBuffer-[LastByteRcvd -

LastByteRead]

Rcvr advertises spare room by including value of RcvWindow in segmentsSender limits unACKed data to RcvWindow

- guarantees receive buffer doesnrsquot overflow

19Mao W07

TCP Connection Management

Recall TCP sender receiver establish ldquoconnectionrdquo before exchanging data segmentsinitialize TCP variables

- seq s- buffers flow control info

(eg RcvWindow)client connection initiatorSocket clientSocket = new Socket(hostnameport

number)

server contacted by clientSocket connectionSocket = welcomeSocketaccept()

Three way handshakeStep 1 client host sends TCP

SYN segment to server- specifies initial seq - no data

Step 2 server host receives SYN replies with SYNACK segment

- server allocates buffers- specifies server initial seq

Step 3 client receives SYNACK replies with ACK segment which may contain data

20Mao W07

TCP Connection Management (cont)

Closing a connection

client closes socketclientSocketclose()

Step 1 client end system sends TCP FIN control segment to server

Step 2 server receives FIN replies with ACK Closes connection sends FIN

client

FIN

server

ACK

ACK

FIN

close

close

closed

tim

ed w

ait

21Mao W07

TCP Connection Management (cont)

Step 3 client receives FIN replies with ACK

- Enters ldquotimed waitrdquo - will respond with ACK to received FINs

Step 4 server receives ACK Connection closed

Note with small modification can handle simultaneous FINs

client

FIN

server

ACK

ACK

FIN

closing

closing

closed

tim

ed w

ait

closed

22Mao W07

TCP Connection Management (cont)

TCP clientlifecycle

TCP serverlifecycle

23Mao W07

Principles of Congestion Control

Congestioninformally ldquotoo many sources sending too much data too fast for network to handlerdquodifferent from flow controlmanifestations

- lost packets (buffer overflow at routers)- long delays (queueing in router buffers)

a top-10 problem

24Mao W07

Causescosts of congestion scenario 1

two senders two receiversone router infinite buffers no retransmission

large delays when congestedmaximum achievable throughput

unlimited shared output link buffers

Host Aλin original data

Host B

λout

25Mao W07

Causescosts of congestion scenario 2

one router finite buffers sender retransmission of lost packet

finite shared output link buffers

Host A λin original data

Host B

λout

λin original data plus retransmitted data

26Mao W07

Causescosts of congestion scenario 2

always (goodput)

ldquoperfectrdquo retransmission only when loss

retransmission of delayed (not lost) packet makes larger (than perfect case) for same

λin

λout=

λin

λoutgtλ

inλout

ldquocostsrdquo of congestionmore work (retrans) for given ldquogoodputrdquounneeded retransmissions link carries multiple copies of pkt

R2

R2λin

λ out

b

R2

R2λin

λ out

a

R2

R2λin

λ out

c

R4

R3

27Mao W07

Causescosts of congestion scenario 3

four sendersmultihop pathstimeoutretransmit

λin

Q what happens as and increase λ

in

finite shared output link buffers

Host Aλin original data

Host B

λout

λin original data plus retransmitted data

28Mao W07

Causescosts of congestion scenario 3

Another ldquocostrdquo of congestionwhen packet dropped any ldquoupstream transmission capacity used for that packet was wasted

Host A

Host B

λou

t

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

13Mao W07

TCP retransmission scenarios (more)Host A

Seq=92 8 bytes data

ACK=100

loss

tim

eout

Cumulative ACK scenario

Host B

X

Seq=100 20 bytes data

ACK=120

time

SendBase= 120

14Mao W07

TCP ACK generation [RFC 1122 RFC 2581]

Event at Receiver

Arrival of in-order segment withexpected seq All data up toexpected seq already ACKed

Arrival of in-order segment withexpected seq One other segment has ACK pending

Arrival of out-of-order segmenthigher-than-expect seq Gap detected

Arrival of segment that partially or completely fills gap

TCP Receiver action

Delayed ACK Wait up to 500msfor next segment If no next segmentsend ACK

Immediately send single cumulative ACK ACKing both in-order segments

Immediately send duplicate ACK indicating seq of next expected byte

Immediate send ACK provided thatsegment startsat lower end of gap

15Mao W07

Fast Retransmit

Time-out period often relatively long

- long delay before resending lost packet

Detect lost segments via duplicate ACKs

- Sender often sends many segments back-to-back

- If segment is lost there will likely be many duplicate ACKs

If sender receives 3 ACKs for the same data it supposes that segment after ACKed data was lost

- fast retransmit resend segment before timer expires

16Mao W07

event ACK received with ACK field value of y if (y gt SendBase)

SendBase = yif (there are currently not-yet-acknowledged segments)

start timer

else increment count of dup ACKs received for yif (count of dup ACKs received for y = 3)

resend segment with sequence number y

Fast retransmit algorithm

a duplicate ACK for already ACKed segment

fast retransmit

17Mao W07

TCP Flow Control

receive side of TCP connection has a receive buffer

speed-matching service matching the send rate to the receiving apprsquos drain rate

app process may be slow at reading from buffer

sender wonrsquot overflowreceiverrsquos buffer by

transmitting too muchtoo fast

flow control

18Mao W07

TCP Flow control how it works

(Suppose TCP receiver discards out-of-order segments)spare room in buffer

= RcvWindow= RcvBuffer-[LastByteRcvd -

LastByteRead]

Rcvr advertises spare room by including value of RcvWindow in segmentsSender limits unACKed data to RcvWindow

- guarantees receive buffer doesnrsquot overflow

19Mao W07

TCP Connection Management

Recall TCP sender receiver establish ldquoconnectionrdquo before exchanging data segmentsinitialize TCP variables

- seq s- buffers flow control info

(eg RcvWindow)client connection initiatorSocket clientSocket = new Socket(hostnameport

number)

server contacted by clientSocket connectionSocket = welcomeSocketaccept()

Three way handshakeStep 1 client host sends TCP

SYN segment to server- specifies initial seq - no data

Step 2 server host receives SYN replies with SYNACK segment

- server allocates buffers- specifies server initial seq

Step 3 client receives SYNACK replies with ACK segment which may contain data

20Mao W07

TCP Connection Management (cont)

Closing a connection

client closes socketclientSocketclose()

Step 1 client end system sends TCP FIN control segment to server

Step 2 server receives FIN replies with ACK Closes connection sends FIN

client

FIN

server

ACK

ACK

FIN

close

close

closed

tim

ed w

ait

21Mao W07

TCP Connection Management (cont)

Step 3 client receives FIN replies with ACK

- Enters ldquotimed waitrdquo - will respond with ACK to received FINs

Step 4 server receives ACK Connection closed

Note with small modification can handle simultaneous FINs

client

FIN

server

ACK

ACK

FIN

closing

closing

closed

tim

ed w

ait

closed

22Mao W07

TCP Connection Management (cont)

TCP clientlifecycle

TCP serverlifecycle

23Mao W07

Principles of Congestion Control

Congestioninformally ldquotoo many sources sending too much data too fast for network to handlerdquodifferent from flow controlmanifestations

- lost packets (buffer overflow at routers)- long delays (queueing in router buffers)

a top-10 problem

24Mao W07

Causescosts of congestion scenario 1

two senders two receiversone router infinite buffers no retransmission

large delays when congestedmaximum achievable throughput

unlimited shared output link buffers

Host Aλin original data

Host B

λout

25Mao W07

Causescosts of congestion scenario 2

one router finite buffers sender retransmission of lost packet

finite shared output link buffers

Host A λin original data

Host B

λout

λin original data plus retransmitted data

26Mao W07

Causescosts of congestion scenario 2

always (goodput)

ldquoperfectrdquo retransmission only when loss

retransmission of delayed (not lost) packet makes larger (than perfect case) for same

λin

λout=

λin

λoutgtλ

inλout

ldquocostsrdquo of congestionmore work (retrans) for given ldquogoodputrdquounneeded retransmissions link carries multiple copies of pkt

R2

R2λin

λ out

b

R2

R2λin

λ out

a

R2

R2λin

λ out

c

R4

R3

27Mao W07

Causescosts of congestion scenario 3

four sendersmultihop pathstimeoutretransmit

λin

Q what happens as and increase λ

in

finite shared output link buffers

Host Aλin original data

Host B

λout

λin original data plus retransmitted data

28Mao W07

Causescosts of congestion scenario 3

Another ldquocostrdquo of congestionwhen packet dropped any ldquoupstream transmission capacity used for that packet was wasted

Host A

Host B

λou

t

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

14Mao W07

TCP ACK generation [RFC 1122 RFC 2581]

Event at Receiver

Arrival of in-order segment withexpected seq All data up toexpected seq already ACKed

Arrival of in-order segment withexpected seq One other segment has ACK pending

Arrival of out-of-order segmenthigher-than-expect seq Gap detected

Arrival of segment that partially or completely fills gap

TCP Receiver action

Delayed ACK Wait up to 500msfor next segment If no next segmentsend ACK

Immediately send single cumulative ACK ACKing both in-order segments

Immediately send duplicate ACK indicating seq of next expected byte

Immediate send ACK provided thatsegment startsat lower end of gap

15Mao W07

Fast Retransmit

Time-out period often relatively long

- long delay before resending lost packet

Detect lost segments via duplicate ACKs

- Sender often sends many segments back-to-back

- If segment is lost there will likely be many duplicate ACKs

If sender receives 3 ACKs for the same data it supposes that segment after ACKed data was lost

- fast retransmit resend segment before timer expires

16Mao W07

event ACK received with ACK field value of y if (y gt SendBase)

SendBase = yif (there are currently not-yet-acknowledged segments)

start timer

else increment count of dup ACKs received for yif (count of dup ACKs received for y = 3)

resend segment with sequence number y

Fast retransmit algorithm

a duplicate ACK for already ACKed segment

fast retransmit

17Mao W07

TCP Flow Control

receive side of TCP connection has a receive buffer

speed-matching service matching the send rate to the receiving apprsquos drain rate

app process may be slow at reading from buffer

sender wonrsquot overflowreceiverrsquos buffer by

transmitting too muchtoo fast

flow control

18Mao W07

TCP Flow control how it works

(Suppose TCP receiver discards out-of-order segments)spare room in buffer

= RcvWindow= RcvBuffer-[LastByteRcvd -

LastByteRead]

Rcvr advertises spare room by including value of RcvWindow in segmentsSender limits unACKed data to RcvWindow

- guarantees receive buffer doesnrsquot overflow

19Mao W07

TCP Connection Management

Recall TCP sender receiver establish ldquoconnectionrdquo before exchanging data segmentsinitialize TCP variables

- seq s- buffers flow control info

(eg RcvWindow)client connection initiatorSocket clientSocket = new Socket(hostnameport

number)

server contacted by clientSocket connectionSocket = welcomeSocketaccept()

Three way handshakeStep 1 client host sends TCP

SYN segment to server- specifies initial seq - no data

Step 2 server host receives SYN replies with SYNACK segment

- server allocates buffers- specifies server initial seq

Step 3 client receives SYNACK replies with ACK segment which may contain data

20Mao W07

TCP Connection Management (cont)

Closing a connection

client closes socketclientSocketclose()

Step 1 client end system sends TCP FIN control segment to server

Step 2 server receives FIN replies with ACK Closes connection sends FIN

client

FIN

server

ACK

ACK

FIN

close

close

closed

tim

ed w

ait

21Mao W07

TCP Connection Management (cont)

Step 3 client receives FIN replies with ACK

- Enters ldquotimed waitrdquo - will respond with ACK to received FINs

Step 4 server receives ACK Connection closed

Note with small modification can handle simultaneous FINs

client

FIN

server

ACK

ACK

FIN

closing

closing

closed

tim

ed w

ait

closed

22Mao W07

TCP Connection Management (cont)

TCP clientlifecycle

TCP serverlifecycle

23Mao W07

Principles of Congestion Control

Congestioninformally ldquotoo many sources sending too much data too fast for network to handlerdquodifferent from flow controlmanifestations

- lost packets (buffer overflow at routers)- long delays (queueing in router buffers)

a top-10 problem

24Mao W07

Causescosts of congestion scenario 1

two senders two receiversone router infinite buffers no retransmission

large delays when congestedmaximum achievable throughput

unlimited shared output link buffers

Host Aλin original data

Host B

λout

25Mao W07

Causescosts of congestion scenario 2

one router finite buffers sender retransmission of lost packet

finite shared output link buffers

Host A λin original data

Host B

λout

λin original data plus retransmitted data

26Mao W07

Causescosts of congestion scenario 2

always (goodput)

ldquoperfectrdquo retransmission only when loss

retransmission of delayed (not lost) packet makes larger (than perfect case) for same

λin

λout=

λin

λoutgtλ

inλout

ldquocostsrdquo of congestionmore work (retrans) for given ldquogoodputrdquounneeded retransmissions link carries multiple copies of pkt

R2

R2λin

λ out

b

R2

R2λin

λ out

a

R2

R2λin

λ out

c

R4

R3

27Mao W07

Causescosts of congestion scenario 3

four sendersmultihop pathstimeoutretransmit

λin

Q what happens as and increase λ

in

finite shared output link buffers

Host Aλin original data

Host B

λout

λin original data plus retransmitted data

28Mao W07

Causescosts of congestion scenario 3

Another ldquocostrdquo of congestionwhen packet dropped any ldquoupstream transmission capacity used for that packet was wasted

Host A

Host B

λou

t

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

15Mao W07

Fast Retransmit

Time-out period often relatively long

- long delay before resending lost packet

Detect lost segments via duplicate ACKs

- Sender often sends many segments back-to-back

- If segment is lost there will likely be many duplicate ACKs

If sender receives 3 ACKs for the same data it supposes that segment after ACKed data was lost

- fast retransmit resend segment before timer expires

16Mao W07

event ACK received with ACK field value of y if (y gt SendBase)

SendBase = yif (there are currently not-yet-acknowledged segments)

start timer

else increment count of dup ACKs received for yif (count of dup ACKs received for y = 3)

resend segment with sequence number y

Fast retransmit algorithm

a duplicate ACK for already ACKed segment

fast retransmit

17Mao W07

TCP Flow Control

receive side of TCP connection has a receive buffer

speed-matching service matching the send rate to the receiving apprsquos drain rate

app process may be slow at reading from buffer

sender wonrsquot overflowreceiverrsquos buffer by

transmitting too muchtoo fast

flow control

18Mao W07

TCP Flow control how it works

(Suppose TCP receiver discards out-of-order segments)spare room in buffer

= RcvWindow= RcvBuffer-[LastByteRcvd -

LastByteRead]

Rcvr advertises spare room by including value of RcvWindow in segmentsSender limits unACKed data to RcvWindow

- guarantees receive buffer doesnrsquot overflow

19Mao W07

TCP Connection Management

Recall TCP sender receiver establish ldquoconnectionrdquo before exchanging data segmentsinitialize TCP variables

- seq s- buffers flow control info

(eg RcvWindow)client connection initiatorSocket clientSocket = new Socket(hostnameport

number)

server contacted by clientSocket connectionSocket = welcomeSocketaccept()

Three way handshakeStep 1 client host sends TCP

SYN segment to server- specifies initial seq - no data

Step 2 server host receives SYN replies with SYNACK segment

- server allocates buffers- specifies server initial seq

Step 3 client receives SYNACK replies with ACK segment which may contain data

20Mao W07

TCP Connection Management (cont)

Closing a connection

client closes socketclientSocketclose()

Step 1 client end system sends TCP FIN control segment to server

Step 2 server receives FIN replies with ACK Closes connection sends FIN

client

FIN

server

ACK

ACK

FIN

close

close

closed

tim

ed w

ait

21Mao W07

TCP Connection Management (cont)

Step 3 client receives FIN replies with ACK

- Enters ldquotimed waitrdquo - will respond with ACK to received FINs

Step 4 server receives ACK Connection closed

Note with small modification can handle simultaneous FINs

client

FIN

server

ACK

ACK

FIN

closing

closing

closed

tim

ed w

ait

closed

22Mao W07

TCP Connection Management (cont)

TCP clientlifecycle

TCP serverlifecycle

23Mao W07

Principles of Congestion Control

Congestioninformally ldquotoo many sources sending too much data too fast for network to handlerdquodifferent from flow controlmanifestations

- lost packets (buffer overflow at routers)- long delays (queueing in router buffers)

a top-10 problem

24Mao W07

Causescosts of congestion scenario 1

two senders two receiversone router infinite buffers no retransmission

large delays when congestedmaximum achievable throughput

unlimited shared output link buffers

Host Aλin original data

Host B

λout

25Mao W07

Causescosts of congestion scenario 2

one router finite buffers sender retransmission of lost packet

finite shared output link buffers

Host A λin original data

Host B

λout

λin original data plus retransmitted data

26Mao W07

Causescosts of congestion scenario 2

always (goodput)

ldquoperfectrdquo retransmission only when loss

retransmission of delayed (not lost) packet makes larger (than perfect case) for same

λin

λout=

λin

λoutgtλ

inλout

ldquocostsrdquo of congestionmore work (retrans) for given ldquogoodputrdquounneeded retransmissions link carries multiple copies of pkt

R2

R2λin

λ out

b

R2

R2λin

λ out

a

R2

R2λin

λ out

c

R4

R3

27Mao W07

Causescosts of congestion scenario 3

four sendersmultihop pathstimeoutretransmit

λin

Q what happens as and increase λ

in

finite shared output link buffers

Host Aλin original data

Host B

λout

λin original data plus retransmitted data

28Mao W07

Causescosts of congestion scenario 3

Another ldquocostrdquo of congestionwhen packet dropped any ldquoupstream transmission capacity used for that packet was wasted

Host A

Host B

λou

t

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

16Mao W07

event ACK received with ACK field value of y if (y gt SendBase)

SendBase = yif (there are currently not-yet-acknowledged segments)

start timer

else increment count of dup ACKs received for yif (count of dup ACKs received for y = 3)

resend segment with sequence number y

Fast retransmit algorithm

a duplicate ACK for already ACKed segment

fast retransmit

17Mao W07

TCP Flow Control

receive side of TCP connection has a receive buffer

speed-matching service matching the send rate to the receiving apprsquos drain rate

app process may be slow at reading from buffer

sender wonrsquot overflowreceiverrsquos buffer by

transmitting too muchtoo fast

flow control

18Mao W07

TCP Flow control how it works

(Suppose TCP receiver discards out-of-order segments)spare room in buffer

= RcvWindow= RcvBuffer-[LastByteRcvd -

LastByteRead]

Rcvr advertises spare room by including value of RcvWindow in segmentsSender limits unACKed data to RcvWindow

- guarantees receive buffer doesnrsquot overflow

19Mao W07

TCP Connection Management

Recall TCP sender receiver establish ldquoconnectionrdquo before exchanging data segmentsinitialize TCP variables

- seq s- buffers flow control info

(eg RcvWindow)client connection initiatorSocket clientSocket = new Socket(hostnameport

number)

server contacted by clientSocket connectionSocket = welcomeSocketaccept()

Three way handshakeStep 1 client host sends TCP

SYN segment to server- specifies initial seq - no data

Step 2 server host receives SYN replies with SYNACK segment

- server allocates buffers- specifies server initial seq

Step 3 client receives SYNACK replies with ACK segment which may contain data

20Mao W07

TCP Connection Management (cont)

Closing a connection

client closes socketclientSocketclose()

Step 1 client end system sends TCP FIN control segment to server

Step 2 server receives FIN replies with ACK Closes connection sends FIN

client

FIN

server

ACK

ACK

FIN

close

close

closed

tim

ed w

ait

21Mao W07

TCP Connection Management (cont)

Step 3 client receives FIN replies with ACK

- Enters ldquotimed waitrdquo - will respond with ACK to received FINs

Step 4 server receives ACK Connection closed

Note with small modification can handle simultaneous FINs

client

FIN

server

ACK

ACK

FIN

closing

closing

closed

tim

ed w

ait

closed

22Mao W07

TCP Connection Management (cont)

TCP clientlifecycle

TCP serverlifecycle

23Mao W07

Principles of Congestion Control

Congestioninformally ldquotoo many sources sending too much data too fast for network to handlerdquodifferent from flow controlmanifestations

- lost packets (buffer overflow at routers)- long delays (queueing in router buffers)

a top-10 problem

24Mao W07

Causescosts of congestion scenario 1

two senders two receiversone router infinite buffers no retransmission

large delays when congestedmaximum achievable throughput

unlimited shared output link buffers

Host Aλin original data

Host B

λout

25Mao W07

Causescosts of congestion scenario 2

one router finite buffers sender retransmission of lost packet

finite shared output link buffers

Host A λin original data

Host B

λout

λin original data plus retransmitted data

26Mao W07

Causescosts of congestion scenario 2

always (goodput)

ldquoperfectrdquo retransmission only when loss

retransmission of delayed (not lost) packet makes larger (than perfect case) for same

λin

λout=

λin

λoutgtλ

inλout

ldquocostsrdquo of congestionmore work (retrans) for given ldquogoodputrdquounneeded retransmissions link carries multiple copies of pkt

R2

R2λin

λ out

b

R2

R2λin

λ out

a

R2

R2λin

λ out

c

R4

R3

27Mao W07

Causescosts of congestion scenario 3

four sendersmultihop pathstimeoutretransmit

λin

Q what happens as and increase λ

in

finite shared output link buffers

Host Aλin original data

Host B

λout

λin original data plus retransmitted data

28Mao W07

Causescosts of congestion scenario 3

Another ldquocostrdquo of congestionwhen packet dropped any ldquoupstream transmission capacity used for that packet was wasted

Host A

Host B

λou

t

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

17Mao W07

TCP Flow Control

receive side of TCP connection has a receive buffer

speed-matching service matching the send rate to the receiving apprsquos drain rate

app process may be slow at reading from buffer

sender wonrsquot overflowreceiverrsquos buffer by

transmitting too muchtoo fast

flow control

18Mao W07

TCP Flow control how it works

(Suppose TCP receiver discards out-of-order segments)spare room in buffer

= RcvWindow= RcvBuffer-[LastByteRcvd -

LastByteRead]

Rcvr advertises spare room by including value of RcvWindow in segmentsSender limits unACKed data to RcvWindow

- guarantees receive buffer doesnrsquot overflow

19Mao W07

TCP Connection Management

Recall TCP sender receiver establish ldquoconnectionrdquo before exchanging data segmentsinitialize TCP variables

- seq s- buffers flow control info

(eg RcvWindow)client connection initiatorSocket clientSocket = new Socket(hostnameport

number)

server contacted by clientSocket connectionSocket = welcomeSocketaccept()

Three way handshakeStep 1 client host sends TCP

SYN segment to server- specifies initial seq - no data

Step 2 server host receives SYN replies with SYNACK segment

- server allocates buffers- specifies server initial seq

Step 3 client receives SYNACK replies with ACK segment which may contain data

20Mao W07

TCP Connection Management (cont)

Closing a connection

client closes socketclientSocketclose()

Step 1 client end system sends TCP FIN control segment to server

Step 2 server receives FIN replies with ACK Closes connection sends FIN

client

FIN

server

ACK

ACK

FIN

close

close

closed

tim

ed w

ait

21Mao W07

TCP Connection Management (cont)

Step 3 client receives FIN replies with ACK

- Enters ldquotimed waitrdquo - will respond with ACK to received FINs

Step 4 server receives ACK Connection closed

Note with small modification can handle simultaneous FINs

client

FIN

server

ACK

ACK

FIN

closing

closing

closed

tim

ed w

ait

closed

22Mao W07

TCP Connection Management (cont)

TCP clientlifecycle

TCP serverlifecycle

23Mao W07

Principles of Congestion Control

Congestioninformally ldquotoo many sources sending too much data too fast for network to handlerdquodifferent from flow controlmanifestations

- lost packets (buffer overflow at routers)- long delays (queueing in router buffers)

a top-10 problem

24Mao W07

Causescosts of congestion scenario 1

two senders two receiversone router infinite buffers no retransmission

large delays when congestedmaximum achievable throughput

unlimited shared output link buffers

Host Aλin original data

Host B

λout

25Mao W07

Causescosts of congestion scenario 2

one router finite buffers sender retransmission of lost packet

finite shared output link buffers

Host A λin original data

Host B

λout

λin original data plus retransmitted data

26Mao W07

Causescosts of congestion scenario 2

always (goodput)

ldquoperfectrdquo retransmission only when loss

retransmission of delayed (not lost) packet makes larger (than perfect case) for same

λin

λout=

λin

λoutgtλ

inλout

ldquocostsrdquo of congestionmore work (retrans) for given ldquogoodputrdquounneeded retransmissions link carries multiple copies of pkt

R2

R2λin

λ out

b

R2

R2λin

λ out

a

R2

R2λin

λ out

c

R4

R3

27Mao W07

Causescosts of congestion scenario 3

four sendersmultihop pathstimeoutretransmit

λin

Q what happens as and increase λ

in

finite shared output link buffers

Host Aλin original data

Host B

λout

λin original data plus retransmitted data

28Mao W07

Causescosts of congestion scenario 3

Another ldquocostrdquo of congestionwhen packet dropped any ldquoupstream transmission capacity used for that packet was wasted

Host A

Host B

λou

t

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

18Mao W07

TCP Flow control how it works

(Suppose TCP receiver discards out-of-order segments)spare room in buffer

= RcvWindow= RcvBuffer-[LastByteRcvd -

LastByteRead]

Rcvr advertises spare room by including value of RcvWindow in segmentsSender limits unACKed data to RcvWindow

- guarantees receive buffer doesnrsquot overflow

19Mao W07

TCP Connection Management

Recall TCP sender receiver establish ldquoconnectionrdquo before exchanging data segmentsinitialize TCP variables

- seq s- buffers flow control info

(eg RcvWindow)client connection initiatorSocket clientSocket = new Socket(hostnameport

number)

server contacted by clientSocket connectionSocket = welcomeSocketaccept()

Three way handshakeStep 1 client host sends TCP

SYN segment to server- specifies initial seq - no data

Step 2 server host receives SYN replies with SYNACK segment

- server allocates buffers- specifies server initial seq

Step 3 client receives SYNACK replies with ACK segment which may contain data

20Mao W07

TCP Connection Management (cont)

Closing a connection

client closes socketclientSocketclose()

Step 1 client end system sends TCP FIN control segment to server

Step 2 server receives FIN replies with ACK Closes connection sends FIN

client

FIN

server

ACK

ACK

FIN

close

close

closed

tim

ed w

ait

21Mao W07

TCP Connection Management (cont)

Step 3 client receives FIN replies with ACK

- Enters ldquotimed waitrdquo - will respond with ACK to received FINs

Step 4 server receives ACK Connection closed

Note with small modification can handle simultaneous FINs

client

FIN

server

ACK

ACK

FIN

closing

closing

closed

tim

ed w

ait

closed

22Mao W07

TCP Connection Management (cont)

TCP clientlifecycle

TCP serverlifecycle

23Mao W07

Principles of Congestion Control

Congestioninformally ldquotoo many sources sending too much data too fast for network to handlerdquodifferent from flow controlmanifestations

- lost packets (buffer overflow at routers)- long delays (queueing in router buffers)

a top-10 problem

24Mao W07

Causescosts of congestion scenario 1

two senders two receiversone router infinite buffers no retransmission

large delays when congestedmaximum achievable throughput

unlimited shared output link buffers

Host Aλin original data

Host B

λout

25Mao W07

Causescosts of congestion scenario 2

one router finite buffers sender retransmission of lost packet

finite shared output link buffers

Host A λin original data

Host B

λout

λin original data plus retransmitted data

26Mao W07

Causescosts of congestion scenario 2

always (goodput)

ldquoperfectrdquo retransmission only when loss

retransmission of delayed (not lost) packet makes larger (than perfect case) for same

λin

λout=

λin

λoutgtλ

inλout

ldquocostsrdquo of congestionmore work (retrans) for given ldquogoodputrdquounneeded retransmissions link carries multiple copies of pkt

R2

R2λin

λ out

b

R2

R2λin

λ out

a

R2

R2λin

λ out

c

R4

R3

27Mao W07

Causescosts of congestion scenario 3

four sendersmultihop pathstimeoutretransmit

λin

Q what happens as and increase λ

in

finite shared output link buffers

Host Aλin original data

Host B

λout

λin original data plus retransmitted data

28Mao W07

Causescosts of congestion scenario 3

Another ldquocostrdquo of congestionwhen packet dropped any ldquoupstream transmission capacity used for that packet was wasted

Host A

Host B

λou

t

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

19Mao W07

TCP Connection Management

Recall TCP sender receiver establish ldquoconnectionrdquo before exchanging data segmentsinitialize TCP variables

- seq s- buffers flow control info

(eg RcvWindow)client connection initiatorSocket clientSocket = new Socket(hostnameport

number)

server contacted by clientSocket connectionSocket = welcomeSocketaccept()

Three way handshakeStep 1 client host sends TCP

SYN segment to server- specifies initial seq - no data

Step 2 server host receives SYN replies with SYNACK segment

- server allocates buffers- specifies server initial seq

Step 3 client receives SYNACK replies with ACK segment which may contain data

20Mao W07

TCP Connection Management (cont)

Closing a connection

client closes socketclientSocketclose()

Step 1 client end system sends TCP FIN control segment to server

Step 2 server receives FIN replies with ACK Closes connection sends FIN

client

FIN

server

ACK

ACK

FIN

close

close

closed

tim

ed w

ait

21Mao W07

TCP Connection Management (cont)

Step 3 client receives FIN replies with ACK

- Enters ldquotimed waitrdquo - will respond with ACK to received FINs

Step 4 server receives ACK Connection closed

Note with small modification can handle simultaneous FINs

client

FIN

server

ACK

ACK

FIN

closing

closing

closed

tim

ed w

ait

closed

22Mao W07

TCP Connection Management (cont)

TCP clientlifecycle

TCP serverlifecycle

23Mao W07

Principles of Congestion Control

Congestioninformally ldquotoo many sources sending too much data too fast for network to handlerdquodifferent from flow controlmanifestations

- lost packets (buffer overflow at routers)- long delays (queueing in router buffers)

a top-10 problem

24Mao W07

Causescosts of congestion scenario 1

two senders two receiversone router infinite buffers no retransmission

large delays when congestedmaximum achievable throughput

unlimited shared output link buffers

Host Aλin original data

Host B

λout

25Mao W07

Causescosts of congestion scenario 2

one router finite buffers sender retransmission of lost packet

finite shared output link buffers

Host A λin original data

Host B

λout

λin original data plus retransmitted data

26Mao W07

Causescosts of congestion scenario 2

always (goodput)

ldquoperfectrdquo retransmission only when loss

retransmission of delayed (not lost) packet makes larger (than perfect case) for same

λin

λout=

λin

λoutgtλ

inλout

ldquocostsrdquo of congestionmore work (retrans) for given ldquogoodputrdquounneeded retransmissions link carries multiple copies of pkt

R2

R2λin

λ out

b

R2

R2λin

λ out

a

R2

R2λin

λ out

c

R4

R3

27Mao W07

Causescosts of congestion scenario 3

four sendersmultihop pathstimeoutretransmit

λin

Q what happens as and increase λ

in

finite shared output link buffers

Host Aλin original data

Host B

λout

λin original data plus retransmitted data

28Mao W07

Causescosts of congestion scenario 3

Another ldquocostrdquo of congestionwhen packet dropped any ldquoupstream transmission capacity used for that packet was wasted

Host A

Host B

λou

t

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

20Mao W07

TCP Connection Management (cont)

Closing a connection

client closes socketclientSocketclose()

Step 1 client end system sends TCP FIN control segment to server

Step 2 server receives FIN replies with ACK Closes connection sends FIN

client

FIN

server

ACK

ACK

FIN

close

close

closed

tim

ed w

ait

21Mao W07

TCP Connection Management (cont)

Step 3 client receives FIN replies with ACK

- Enters ldquotimed waitrdquo - will respond with ACK to received FINs

Step 4 server receives ACK Connection closed

Note with small modification can handle simultaneous FINs

client

FIN

server

ACK

ACK

FIN

closing

closing

closed

tim

ed w

ait

closed

22Mao W07

TCP Connection Management (cont)

TCP clientlifecycle

TCP serverlifecycle

23Mao W07

Principles of Congestion Control

Congestioninformally ldquotoo many sources sending too much data too fast for network to handlerdquodifferent from flow controlmanifestations

- lost packets (buffer overflow at routers)- long delays (queueing in router buffers)

a top-10 problem

24Mao W07

Causescosts of congestion scenario 1

two senders two receiversone router infinite buffers no retransmission

large delays when congestedmaximum achievable throughput

unlimited shared output link buffers

Host Aλin original data

Host B

λout

25Mao W07

Causescosts of congestion scenario 2

one router finite buffers sender retransmission of lost packet

finite shared output link buffers

Host A λin original data

Host B

λout

λin original data plus retransmitted data

26Mao W07

Causescosts of congestion scenario 2

always (goodput)

ldquoperfectrdquo retransmission only when loss

retransmission of delayed (not lost) packet makes larger (than perfect case) for same

λin

λout=

λin

λoutgtλ

inλout

ldquocostsrdquo of congestionmore work (retrans) for given ldquogoodputrdquounneeded retransmissions link carries multiple copies of pkt

R2

R2λin

λ out

b

R2

R2λin

λ out

a

R2

R2λin

λ out

c

R4

R3

27Mao W07

Causescosts of congestion scenario 3

four sendersmultihop pathstimeoutretransmit

λin

Q what happens as and increase λ

in

finite shared output link buffers

Host Aλin original data

Host B

λout

λin original data plus retransmitted data

28Mao W07

Causescosts of congestion scenario 3

Another ldquocostrdquo of congestionwhen packet dropped any ldquoupstream transmission capacity used for that packet was wasted

Host A

Host B

λou

t

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

21Mao W07

TCP Connection Management (cont)

Step 3 client receives FIN replies with ACK

- Enters ldquotimed waitrdquo - will respond with ACK to received FINs

Step 4 server receives ACK Connection closed

Note with small modification can handle simultaneous FINs

client

FIN

server

ACK

ACK

FIN

closing

closing

closed

tim

ed w

ait

closed

22Mao W07

TCP Connection Management (cont)

TCP clientlifecycle

TCP serverlifecycle

23Mao W07

Principles of Congestion Control

Congestioninformally ldquotoo many sources sending too much data too fast for network to handlerdquodifferent from flow controlmanifestations

- lost packets (buffer overflow at routers)- long delays (queueing in router buffers)

a top-10 problem

24Mao W07

Causescosts of congestion scenario 1

two senders two receiversone router infinite buffers no retransmission

large delays when congestedmaximum achievable throughput

unlimited shared output link buffers

Host Aλin original data

Host B

λout

25Mao W07

Causescosts of congestion scenario 2

one router finite buffers sender retransmission of lost packet

finite shared output link buffers

Host A λin original data

Host B

λout

λin original data plus retransmitted data

26Mao W07

Causescosts of congestion scenario 2

always (goodput)

ldquoperfectrdquo retransmission only when loss

retransmission of delayed (not lost) packet makes larger (than perfect case) for same

λin

λout=

λin

λoutgtλ

inλout

ldquocostsrdquo of congestionmore work (retrans) for given ldquogoodputrdquounneeded retransmissions link carries multiple copies of pkt

R2

R2λin

λ out

b

R2

R2λin

λ out

a

R2

R2λin

λ out

c

R4

R3

27Mao W07

Causescosts of congestion scenario 3

four sendersmultihop pathstimeoutretransmit

λin

Q what happens as and increase λ

in

finite shared output link buffers

Host Aλin original data

Host B

λout

λin original data plus retransmitted data

28Mao W07

Causescosts of congestion scenario 3

Another ldquocostrdquo of congestionwhen packet dropped any ldquoupstream transmission capacity used for that packet was wasted

Host A

Host B

λou

t

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

22Mao W07

TCP Connection Management (cont)

TCP clientlifecycle

TCP serverlifecycle

23Mao W07

Principles of Congestion Control

Congestioninformally ldquotoo many sources sending too much data too fast for network to handlerdquodifferent from flow controlmanifestations

- lost packets (buffer overflow at routers)- long delays (queueing in router buffers)

a top-10 problem

24Mao W07

Causescosts of congestion scenario 1

two senders two receiversone router infinite buffers no retransmission

large delays when congestedmaximum achievable throughput

unlimited shared output link buffers

Host Aλin original data

Host B

λout

25Mao W07

Causescosts of congestion scenario 2

one router finite buffers sender retransmission of lost packet

finite shared output link buffers

Host A λin original data

Host B

λout

λin original data plus retransmitted data

26Mao W07

Causescosts of congestion scenario 2

always (goodput)

ldquoperfectrdquo retransmission only when loss

retransmission of delayed (not lost) packet makes larger (than perfect case) for same

λin

λout=

λin

λoutgtλ

inλout

ldquocostsrdquo of congestionmore work (retrans) for given ldquogoodputrdquounneeded retransmissions link carries multiple copies of pkt

R2

R2λin

λ out

b

R2

R2λin

λ out

a

R2

R2λin

λ out

c

R4

R3

27Mao W07

Causescosts of congestion scenario 3

four sendersmultihop pathstimeoutretransmit

λin

Q what happens as and increase λ

in

finite shared output link buffers

Host Aλin original data

Host B

λout

λin original data plus retransmitted data

28Mao W07

Causescosts of congestion scenario 3

Another ldquocostrdquo of congestionwhen packet dropped any ldquoupstream transmission capacity used for that packet was wasted

Host A

Host B

λou

t

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

23Mao W07

Principles of Congestion Control

Congestioninformally ldquotoo many sources sending too much data too fast for network to handlerdquodifferent from flow controlmanifestations

- lost packets (buffer overflow at routers)- long delays (queueing in router buffers)

a top-10 problem

24Mao W07

Causescosts of congestion scenario 1

two senders two receiversone router infinite buffers no retransmission

large delays when congestedmaximum achievable throughput

unlimited shared output link buffers

Host Aλin original data

Host B

λout

25Mao W07

Causescosts of congestion scenario 2

one router finite buffers sender retransmission of lost packet

finite shared output link buffers

Host A λin original data

Host B

λout

λin original data plus retransmitted data

26Mao W07

Causescosts of congestion scenario 2

always (goodput)

ldquoperfectrdquo retransmission only when loss

retransmission of delayed (not lost) packet makes larger (than perfect case) for same

λin

λout=

λin

λoutgtλ

inλout

ldquocostsrdquo of congestionmore work (retrans) for given ldquogoodputrdquounneeded retransmissions link carries multiple copies of pkt

R2

R2λin

λ out

b

R2

R2λin

λ out

a

R2

R2λin

λ out

c

R4

R3

27Mao W07

Causescosts of congestion scenario 3

four sendersmultihop pathstimeoutretransmit

λin

Q what happens as and increase λ

in

finite shared output link buffers

Host Aλin original data

Host B

λout

λin original data plus retransmitted data

28Mao W07

Causescosts of congestion scenario 3

Another ldquocostrdquo of congestionwhen packet dropped any ldquoupstream transmission capacity used for that packet was wasted

Host A

Host B

λou

t

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

24Mao W07

Causescosts of congestion scenario 1

two senders two receiversone router infinite buffers no retransmission

large delays when congestedmaximum achievable throughput

unlimited shared output link buffers

Host Aλin original data

Host B

λout

25Mao W07

Causescosts of congestion scenario 2

one router finite buffers sender retransmission of lost packet

finite shared output link buffers

Host A λin original data

Host B

λout

λin original data plus retransmitted data

26Mao W07

Causescosts of congestion scenario 2

always (goodput)

ldquoperfectrdquo retransmission only when loss

retransmission of delayed (not lost) packet makes larger (than perfect case) for same

λin

λout=

λin

λoutgtλ

inλout

ldquocostsrdquo of congestionmore work (retrans) for given ldquogoodputrdquounneeded retransmissions link carries multiple copies of pkt

R2

R2λin

λ out

b

R2

R2λin

λ out

a

R2

R2λin

λ out

c

R4

R3

27Mao W07

Causescosts of congestion scenario 3

four sendersmultihop pathstimeoutretransmit

λin

Q what happens as and increase λ

in

finite shared output link buffers

Host Aλin original data

Host B

λout

λin original data plus retransmitted data

28Mao W07

Causescosts of congestion scenario 3

Another ldquocostrdquo of congestionwhen packet dropped any ldquoupstream transmission capacity used for that packet was wasted

Host A

Host B

λou

t

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

25Mao W07

Causescosts of congestion scenario 2

one router finite buffers sender retransmission of lost packet

finite shared output link buffers

Host A λin original data

Host B

λout

λin original data plus retransmitted data

26Mao W07

Causescosts of congestion scenario 2

always (goodput)

ldquoperfectrdquo retransmission only when loss

retransmission of delayed (not lost) packet makes larger (than perfect case) for same

λin

λout=

λin

λoutgtλ

inλout

ldquocostsrdquo of congestionmore work (retrans) for given ldquogoodputrdquounneeded retransmissions link carries multiple copies of pkt

R2

R2λin

λ out

b

R2

R2λin

λ out

a

R2

R2λin

λ out

c

R4

R3

27Mao W07

Causescosts of congestion scenario 3

four sendersmultihop pathstimeoutretransmit

λin

Q what happens as and increase λ

in

finite shared output link buffers

Host Aλin original data

Host B

λout

λin original data plus retransmitted data

28Mao W07

Causescosts of congestion scenario 3

Another ldquocostrdquo of congestionwhen packet dropped any ldquoupstream transmission capacity used for that packet was wasted

Host A

Host B

λou

t

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

26Mao W07

Causescosts of congestion scenario 2

always (goodput)

ldquoperfectrdquo retransmission only when loss

retransmission of delayed (not lost) packet makes larger (than perfect case) for same

λin

λout=

λin

λoutgtλ

inλout

ldquocostsrdquo of congestionmore work (retrans) for given ldquogoodputrdquounneeded retransmissions link carries multiple copies of pkt

R2

R2λin

λ out

b

R2

R2λin

λ out

a

R2

R2λin

λ out

c

R4

R3

27Mao W07

Causescosts of congestion scenario 3

four sendersmultihop pathstimeoutretransmit

λin

Q what happens as and increase λ

in

finite shared output link buffers

Host Aλin original data

Host B

λout

λin original data plus retransmitted data

28Mao W07

Causescosts of congestion scenario 3

Another ldquocostrdquo of congestionwhen packet dropped any ldquoupstream transmission capacity used for that packet was wasted

Host A

Host B

λou

t

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

27Mao W07

Causescosts of congestion scenario 3

four sendersmultihop pathstimeoutretransmit

λin

Q what happens as and increase λ

in

finite shared output link buffers

Host Aλin original data

Host B

λout

λin original data plus retransmitted data

28Mao W07

Causescosts of congestion scenario 3

Another ldquocostrdquo of congestionwhen packet dropped any ldquoupstream transmission capacity used for that packet was wasted

Host A

Host B

λou

t

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

28Mao W07

Causescosts of congestion scenario 3

Another ldquocostrdquo of congestionwhen packet dropped any ldquoupstream transmission capacity used for that packet was wasted

Host A

Host B

λou

t

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

29Mao W07

Approaches towards congestion control

End-end congestion controlno explicit feedback from networkcongestion inferred from end-system observed loss delayapproach taken by TCP

Network-assisted congestion controlrouters provide feedback to end systems

- single bit indicating congestion (SNA DECbit TCPIP ECN ATM)

- explicit rate sender should send at

Two broad approaches towards congestion control

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

30Mao W07

Case study ATM ABR congestion control

ABR available bit rateldquoelastic servicerdquoif senderrsquos path ldquounderloadedrdquo

- sender should use available bandwidth

if senderrsquos path congested - sender throttled to

minimum guaranteed rate

RM (resource management) cellssent by sender interspersed with data cellsbits in RM cell set by switches (ldquonetwork-assistedrdquo)

- NI bit no increase in rate (mild congestion)

- CI bit congestion indicationRM cells returned to sender by receiver with bits intact

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

31Mao W07

Case study ATM ABR congestion control

two-byte ER (explicit rate) field in RM cell- congested switch may lower ER value in cell- senderrsquo send rate thus minimum supportable rate on path

EFCI bit in data cells set to 1 in congested switch- if data cell preceding RM cell has EFCI set sender sets CI bit in

returned RM cell

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

32Mao W07

TCP Congestion Control

end-end control (no network assistance)sender limits transmissionLastByteSent-LastByteAcked

le CongWin

Roughly

CongWin is dynamic function of perceived network congestion

How does sender perceive congestionloss event = timeout or 3 duplicate acksTCP sender reduces rate (CongWin) after loss event

three mechanisms- AIMD- slow start- conservative after timeout

events

rate = CongWinRTT Bytessec

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

33Mao W07

TCP AIMD

8 Kbytes

16 Kbytes

24 Kbytes

time

congestionwindow

multiplicative decreasecut CongWin in half after loss event

additive increase increase CongWin by 1 MSS every RTT in the absence of loss events probing

Long-lived TCP connection

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

34Mao W07

TCP Slow Start

When connection begins CongWin = 1 MSS

- Example MSS = 500 bytes amp RTT = 200 msec

- initial rate = 20 kbpsavailable bandwidth may be gtgt MSSRTT

- desirable to quickly ramp up to respectable rate

When connection begins increase rate exponentially fast until first loss event

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

35Mao W07

TCP Slow Start (more)

When connection begins increase rate exponentially until first loss event

- double CongWin every RTT

- done by incrementing CongWin for every ACK received

Summary initial rate is slow but ramps up exponentially fast

Host A

one segment

RTT

Host B

time

two segments

four segments

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

36Mao W07

Refinement

After 3 dup ACKs- CongWin is cut in half- window then grows linearly

But after timeout event- CongWin instead set to 1 MSS - window then grows exponentially- to a threshold then grows linearly

bull 3 dup ACKs indicates network capable of delivering some segmentsbull timeout before 3 dup ACKs is ldquomore alarmingrdquo

Philosophy

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

37Mao W07

Refinement (more)

Q When should the exponential increase switch to linear

A When CongWingets to 12 of its value before timeout

ImplementationVariable Threshold At loss event Threshold is set to 12 of CongWin just before loss event

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

38Mao W07

Summary TCP Congestion Control

When CongWin is below Threshold sender in slow-start phase window grows exponentially

When CongWin is above Threshold sender is in congestion-avoidance phase window grows linearly

When a triple duplicate ACK occurs Threshold set to CongWin2 and CongWin set to Threshold

When timeout occurs Threshold set to CongWin2and CongWin is set to 1 MSS

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

39Mao W07

TCP sender congestion control

CongWin and Threshold not changed

Increment duplicate ACK count for segment being acked

SS or CADuplicate ACK

Enter slow startThreshold = CongWin2 CongWin = 1 MSSSet state to ldquoSlow Startrdquo

SS or CATimeout

Fast recovery implementing multiplicative decrease CongWin will not drop below 1 MSS

Threshold = CongWin2 CongWin = ThresholdSet state to ldquoCongestion Avoidancerdquo

SS or CALoss event detected by triple duplicate ACK

Additive increase resulting in increase of CongWin by 1 MSS every RTT

CongWin = CongWin+MSS (MSSCongWin)

CongestionAvoidance (CA)

ACK receipt for previously unackeddata

Resulting in a doubling of CongWin every RTT

CongWin = CongWin + MSS If (CongWin gt Threshold)

set state to ldquoCongestion Avoidancerdquo

Slow Start (SS)

ACK receipt for previously unackeddata

CommentaryTCP Sender Action StateEvent

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

40Mao W07

TCP throughput

Whatrsquos the average throughout ot TCP as a function of window size and RTT

- Ignore slow start

Let W be the window size when loss occursWhen window is W throughput is WRTTJust after loss window drops to W2 throughput to W2RTT Average throughout 75 WRTT

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

41Mao W07

TCP Futures

Example 1500 byte segments 100ms RTT want 10 Gbps throughputRequires window size W = 83333 in-flight segmentsThroughput in terms of loss rate

L = 210-10 WowNew versions of TCP for high-speed needed

LRTTMSSsdot221

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

42Mao W07

Fairness goal if K TCP sessions share same bottleneck link of bandwidth R each should have average rate of RK

TCP connection 1

bottleneckrouter

capacity R

TCP connection 2

TCP Fairness

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

43Mao W07

Why is TCP fairTwo competing sessions

Additive increase gives slope of 1 as throughout increasesmultiplicative decrease decreases throughput proportionally

R

R

equal bandwidth share

Connection 1 throughput

Conn

ecti

on 2

thr

ough

p ut

congestion avoidance additive increaseloss decrease window by factor of 2

congestion avoidance additive increaseloss decrease window by factor of 2

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

44Mao W07

Fairness (more)

Fairness and UDPMultimedia apps often do not use TCP

- do not want rate throttled by congestion control

Instead use UDP- pump audiovideo at

constant rate tolerate packet loss

Research area TCP friendly

Fairness and parallel TCP connectionsnothing prevents app from opening parallel cnctions between 2 hostsWeb browsers do this Example link of rate R supporting 9 cnctions

- new app asks for 1 TCP gets rate R10

- new app asks for 11 TCPs gets R2

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

45Mao W07

Delay modeling

Q How long does it take to receive an object from a Web server after sending a request

Ignoring congestion delay is influenced byTCP connection establishmentdata transmission delayslow start

Notation assumptionsAssume one link between client and server of rate RS MSS (bits)O object size (bits)no retransmissions (no loss no corruption)

Window sizeFirst assume fixed congestion window W segmentsThen dynamic window modeling slow start

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

46Mao W07

TCP Delay Modeling Slow Start (1)

Now suppose window grows according to slow start

Will show that the delay for one object is

RS

RSRTTP

RORTTLatency P )12(2 minusminus⎥⎦

⎤⎢⎣⎡ +++=

where P is the number of times TCP idles at server

1min minus= KQP

- where Q is the number of times the server idlesif the object were of infinite size

- and K is the number of windows that cover the object

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

47Mao W07

TCP Delay Modeling Slow Start (2)

RTT

initiate TCPconnection

requestobject

first window= SR

second wind= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

Examplebull OS = 15 segmentsbull K = 4 windowsbull Q = 2bull P = minK-1Q = 2

Server idles P=2 times

Delay componentsbull 2 RTT for connection estab and requestbull OR to transmit objectbull time server idles due to slow start

Server idles P = minK-1Q times

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

48Mao W07

TCP Delay Modeling (3)

RS

RSRTTPRTT

RO

RSRTT

RSRTT

RO

idleTimeRTTRO

P

kP

k

P

pp

)12(][2

]2[2

2delay

1

1

1

minusminus+++=

minus+++=

++=

minus

=

=

sum

sum

th window after the timeidle 2 1 kRSRTT

RS k =⎥⎦

⎤⎢⎣⎡ minus+

+minus

ementacknowledg receivesserver until

segment send tostartsserver whenfrom time=+ RTTRS

window kth the transmit totime2 1 =minus

RSk

RTT

initiate TCPconnection

requestobject

first window= SR

second window= 2SR

third window= 4SR

fourth window= 8SR

completetransmissionobject

delivered

time atclient

time atserver

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

49Mao W07

TCP Delay Modeling (4)

⎥⎥⎤

⎢⎢⎡ +=

+ge=

geminus=

ge+++=

ge+++=minus

minus

)1(log

)1(logmin

12min

222min222min

2

2

110

110

SO

SOkk

SOk

SOkOSSSkK

k

k

k

L

L

Calculation of Q number of idles for infinite-size objectis similar (see HW)

Recall K = number of windows that cover object

How do we calculate K

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

50Mao W07

HTTP ModelingAssume Web page consists of

- 1 base HTML page (of size O bits)- M images (each of size O bits)

Non-persistent HTTP - M+1 TCP connections in series- Response time = (M+1)OR + (M+1)2RTT + sum of idle times

Persistent HTTP- 2 RTT to request and receive base HTML file- 1 RTT to request and receive M images- Response time = (M+1)OR + 3RTT + sum of idle times

Non-persistent HTTP with X parallel connections- Suppose MX integer- 1 TCP connection for base file- MX sets of parallel connections for images- Response time = (M+1)OR + (MX + 1)2RTT + sum of idle times

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

51Mao W07

02468

101214161820

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT = 100 msec O = 5 Kbytes M=10 and X=5

For low bandwidth connection amp response time dominated by transmission timePersistent connections only give minor improvement over parallelconnections

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks

52Mao W07

0

10

20

30

40

50

60

70

28Kbps

100Kbps

1 Mbps 10Mbps

non-persistent

persistent

parallel non-persistent

HTTP Response time (in seconds)RTT =1 sec O = 5 Kbytes M=10 and X=5

For larger RTT response time dominated by TCP establishment amp slow start delays Persistent connections now give important improvement particularly in high delaybullbandwidth networks