5/13/2018 hw-4_ans - slidepdf.com

http://slidepdf.com/reader/full/hw-4ans 1/3

ECE3076 Homework 4 AnswersDue: Wednesday Oct. 5 11:00

Please show calculations under these questions and answers (integers)

in the {___}'s .

1. TCP with no Congestion A 10 Mbps network connection has a 50ms round trip time (RTT). A server begins sending maximum-sizepackets (MSS bytes, after the initial SYN-ACK). The client ACK's everypacket, and has a 7200-byte receiver-window which holds 6 MSS(maximum-size =1400 byte) packets. Show how many TCP segments(packets) with data are sent in each RTT period below:

  Time Period No. Segments sent

Start (ms)0 {0} SYN-ACK has no data50 {1}

100 {2}150 {4}200 {6}250 {6}300 {6}

Note: The "Time Period" that starts at 200 includes data send during

200 to 249.9999 ms. None of this data is sent during the first 200 ms.

If the MSS is 1400 bytes, what is the average transmission rate duringthe first 200 ms (include bytes sent at periods starting 50-150)?

{ 392,000 } (bps)

  (1 + 2 + 4) * 1400 Bytes * 8 bits/Byte / 0.200 s = 392,000 bps

What is the maximum transfer rate in bytes/second? { 168,000 }(bytes/sec)

  6 * 1400 Bytes / 0.050 s = 168,000 Bytes/sec

At what time does the connection become "Transmission-RateLImited"? { never } (ms) 

It is always "Window Limited" since the maximum rate set by the

5/13/2018 hw-4_ans - slidepdf.com

http://slidepdf.com/reader/full/hw-4ans 2/3

receiver window is less than 10 Mbps.

What is the minimum size of the transmission window W in order forthe connection to be "Transmission-Rate LImited"?

{ 62,500 } (bytes)

  W/RTT = 10 Mbps W = 10,000,000 bps * 0.050 s / 8 (bits/byte)= 62,500 bytesQuestion 4 – TCP Flow and Congestion Control

The sender buffer SendBase is 1000. Six 500-byte TCP segmentshave been sent with sequence numbers 1000, 1500, 2000, 2500, 3000,3500.

A segment is received with TCP acknowledgement number 1500 andwindow 7000. The CongWin is now 12,000. MSS is 1400. Untilanother ACK is received:

a. What is the new value of SendBase?

{1500} (bytes)It's the highest (last) ACK number received.

b. What is the last byte (number) that can be sent with certainty that

the receiver's buffer will not overflow?

{8500}  It's the highest (last) ACK number plus the Window value in that 

 ACK.

c. What is the last byte (number) that can be sent without violatingCongestion Control rules?

{13500}

  SendBase + CongWin

d. Three more TCP segments are received with TCP acknowledgementnumber 1500. What is the next byte (sequence number) that will besent?

{1500}   After 3 duplicate ACKs, the missing segment is resent.

5/13/2018 hw-4_ans - slidepdf.com

http://slidepdf.com/reader/full/hw-4ans 3/3

e. What is the next byte (sequence number) that will be sent afterthat?

{4000}  Then the sender continues sending new data.

f. What is the mode for CongWin increase (Multiplicative, Additive, orSlow-Start)?

{Additive}  CongWin adds 1 MSS for every RTT without a problem (AdditiveIncrease).

h. What is the new value of CongWin?{6000}

CongWin is multiplied by 1/2 (Multiplicative Decrease after 3

duplicate ACKs).

i. There is now a RTO expiration (time-out). What is the new value of CongWin? {1400}   After RTO time-out, CongWin becomes 1 MSS.

 j. Now what is the mode for CongWin increase (Multiplicative, Additive,or Slow-Start)?

{Slow-Start}CongWin doubles every RTT until "Threshold" is reached (1/2

half of old CongWin), then it increases by Additive Increase.