1 birth and death process n(t) depends on how fast arrivals or departures occur objective n(t) = #...
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Birth and death process
N(t) Depends on how fast arrivals or departures occur
Objective
N(t) = # of customersat time t.
λ
arrivals(births)
departures(deaths)
μ
))(Pr( itN
2
Behavior of the system
λ>μ
λ<μ
Possible evolution of N(t)
Time 1 2 3 4 5 6 7 8 9 10 11
123
busy idleN(t)
3
General arrival and departure rates
λn
Depends on the number of customers (n) in the system
Example
μn
Depends on the number of customers in the system
Example
Mn
Mnn ,0
,
n
Or
n
n
n
4
Changing the scale of a unit time
Number of arrivals/unit time Follows the Poisson distribution with rate λn
Inter-arrival time of successive arrivals is exponentially distributed
Average inter-arrival time = 1/ λn
What is the avg. # of customers arriving in dt?
Time
min/2/1min/16.030/30:
hourExample
dtAverage
n
n
5
Probability of one arrival in dt
dt so small Number of arrivals in dt, X is a r.v.
X=1 with probability p
X=0 with probability 1-p
Average number of arrivals in dt
Prob (having one arrival in dt) = λn dt
dtXEbut
pppXE
n][,
)1.(0.1][
dt
6
Probability of having 2 events in dt
Departure rate in dt μn dt
Arrival rate in dt λn dt
What is the probability Of having an (arrival+departure), (2 arrivals or departures)
222
222
2
)()()2Pr(
)()()2Pr(
)(.)11Pr(
dtdtdepartures
dtdtarrivals
dtdtdtdeparturearrival
nn
nn
nnnn
7
Probability distribution of N(t)
Pn (t) The probability of getting n customers by time t
The distribution of the # of customers in systemt+dtt
? nn-1: arrivaln+1: departuren: none of the above
)1)((
)()()( 1111
dtdttP
dttPdttPdttP
nnn
nnnnn
8
Differential equation monitoring evolution of # customers
?)(
))(()()()(
))(()()()()(
)1)(()()()(
1111'
1111
1111
tP
tPtPtPtP
tPtPtPdt
tPdttP
dtdttPdttPdttPdttP
n
nnnnnnnn
nnnnnnnnn
nnnnnnnn
These are solved Numerically using MATLAB
We will explore the cases Of pure death
And pure birth
9
Pure birth process
In this case μn =0, n >= 0
λn = λ, n >= 0
!
)()(;)(
0,)()()(
0),(.)(
0,)()()(
0
1'
0'0
1'
n
ettPetP
ntPtPtP
ntPtP
ntPtPtP
tn
nt
nnn
nnn
Hence,
First order differential equation
10
11
Pure death process
In this case λn =0, n >= 0
μn = μ
)!(
)()(;)(
),(.)(
0,)()()(
0,)()()(
'
1'
1'
nM
ettPetP
MntPtP
ntPtPtP
ntPtPtP
tnM
nt
M
MM
nnn
nnn
12
Queuing system
Transient phase
Steady state Behavior is independent of t
Pn (t)
λ μ
nnt
PtP
)(lim
Pn (t)
t
transient Steady state
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Differential equation: steady state analysis
Limiting case
01
01
1111
1100
1111
'
0,)(
0,)(0
0)(
)(
lim
lim
PP
nPPP
PP
nPPP
tP
PtP
nnnnnnn
nnnnnnn
nt
nnt
14
Solving the equations
n=1
n=22200111 )( PPP
3311222 )( PPP
(1)
(1) =>
021
102
12
122211
22111111
PP
PPPP
PPPP
15
Pn
What about P0
1,...
...0
21
110
0321
2102
23
233322
33222222
nPP
PP
PPPP
PPPP
n
nn
16
Normalization equation
1
1
1
0
0
021
100
1
00
10
1
1
1....
1.......
ni
n
i
i
n
i
n
P
PPP
ionNormilizat
PPP
17
Conditional probability and conditional expectation: d.r.v.
X and Y are discrete r.v. Conditional probability mass function
Of X given that Y=y
Conditional expectation of X given that Y=y
)(
),(
)(
),(
)|()|(|
yp
yxp
yYP
yYxXP
yYxXPyxp
Y
YX
x
yYxXPxyYXE )|(.]|[
18
Conditional probability and expectation: continuous r.v.
If X and Y have a joint pdf fX,Y(x,y) Then, the conditional probability density function
Of X given that Y=y
The conditional expectation Of X given that Y=y
)(
),()|(| yf
yxfyxf
YYX
dxyxfxyYXE YX )|(.]|[ |
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Computing expectations by conditioning Denote
E[X|Y]: function of the r.v. Y Whose value at Y=y is E[X|Y=y]
E[X|Y]: is itself a random variable Property of conditional expectation
if Y is a discrete r.v.
if Y is continuous with density fY (y) =>
]]|[[][ YXEEXE
y
yYPyYXEYXEEXE ][].|[]]|[[][
dyyfyYXEXE Y )(]|[][
(1)
(2)
(3)
20
Proof of equation when X and Y are discrete
][][
],[],[
][][
],[
][].|[
][].|[]]|[[
XExXxP
yYxXPxyYxXxP
yYPyYP
yYxXPx
yYPyYxXxP
yYPyYXEYXEE
x
x yy x
y x
y x
y
21
Problem 1 Sam will read
Either one chapter of his probability book or
One chapter of his history book
If the number of misprints in a chapter Of his probability book
is Poisson distributed with mean 2
Of his history book is Poisson distributed with mean 5
Assuming Sam equally likely to choose either book What is the expected number of misprints he comes across?
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Solution
2
7)2
1(2)
2
1(5
]2[].2|[]1[].1|[][
_,2
__,1
_
YPYXEYPYXEXE
chosenyprobabilit
chosenbookhistoryY
mistakesnumberX
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Problem 2 A miner is trapped in a mine containing three doors
First door leads to a tunnel that takes him to safety
After 2 hours of travel
Second door leads to a tunnel that returns him to the mine
After 3 hours of travel
Third door Leads to a tunnel that returns him to the mine
After 5 hours Assuming he is equally likely to choose any door
What is the expected length of time until he reaches safety?
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Solution
10][][5][32(3
1][
][5]3|[
];[3]2|[;2)1|[
])3|[]2|[]1|[(3
1
]3[]3|[
]2[]2|[]1[]1|[][
__
__
XEXEXEXE
XEYXE
XEYXEYXE
YXEYXEYXE
YPYXE
YPYXEYPYXEXE
choseninitiallydoorY
safetyuntiltimeX
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Computing probabilities by conditioning Let E denote an arbitrary event
X is a random variable defined by
It follows from the definition of X
tdoesnEif
occursEifX
'__,0
__,1
continuousYdyyfyYEPEP
discreteYyYPyYEPEP
yYEPyYXEEPXE
Y
y
_,)()|()(
_,)().|()(
)3(&)2(
)|(]|[)(][
26
Problem 3
Suppose that the number of people Who visit a yoga studio each day
is a Poisson random variable with mean λ
Suppose further that each person who visit is, independently, female with probability p
Or male with probability 1-p
Find the joint probability That n women and m men visit the academy today
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Solution
Let N1 denote the number of women, N2 the number of men
Who visit the academy today
N= N1 +N2 : total number of people who visit Conditioning on N gives
Because P(N1=n,N2=m|N=i)=0 when i != n+m
0
2121 )().|,(),( iNPiNmNnNPmNnNP
)!().|,(
)().|,(),(
21
2121
mnemnNmNnNP
mnNPmnNmNnNPmNnNPmn
28
Solution (cont’d)
Each of the n+m visit is independently a woman with probability p
The conditional probability That n of them are women is
The binomial probability of n successes in n+m trials
!
))1((
!
)(
)!()1(),(
)1(
21
m
pe
n
pe
mnepp
n
mnmNnNP
mp
np
mnmn
29
Solution: analysis
When each of a Poisson number of events is independently classified
As either being type 1 with probability p
Or type 2 with probability (1-p)
=> the numbers of type 1 and 2 events Are independent Poisson random variables
!
))1(()(
!
)(),()(
)1(2
0211
m
pemNP
and
n
pemNnNPnNP
mp
m
np
30
Problem 4
At a party N men take off their hats
The hats are then mixed up and Each man randomly selects one
A match occurs if a man selects his own hat
What is the probability of no matches?
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Solution E = event that no matches occur
P(E) = Pn : explicit dependence on n
Start by conditioning Whether or not the first man selects his own hat
M: if he did, Mc : if he didn’t
P(E|Mc) Probability no matches when n-1 men select of n-1
That does not contain the hat of one of these men
n
nMEPPMEP
MPMEPMPMEPEPP
cn
ccn
1)|(0)|(
);()|()()|()(
32
Solution (cont’d)
P(E|Mc) Either there are no matches and
Extra man does not select the extra hat
=> Pn-1 (as if the extra hat belongs to this man)
Or there are no matches Extra man does select the extra hat
=> (1/n-1)xPn-2
21
21
111
1)|(
nnn
nnc
Pn
Pn
nP
Pn
PMEP
33
Solution (cont’d)
Pn is the probability of no matches When n men select among their own hats
=> P1 =0 and P2 = ½
=>
!
)1(....
!4
1
!3
1
!2
1
!4
1
!3
1
!2
1;!3
1
!2
143
nP
PP
n
n
34
Problem 5: continuous random variables
The probability density function of a non-negative random variable X is given by
Compute the constant λ?
10.)(x
X exf
10
110).10(1
)(1
0
10/
0
10/
x
xX
e
dxedxxf
35
Problem 6: continuous random variables Buses arrives at a specified stop at 15 min intervals
Starting at 7:00 AM They arrive at 7:00, 7:15, 7:30, 7:45
If the passenger arrives at the stop at a time Uniformly distributed between 7:00 and 7:30
Find the probability that he waits less than 5 min? Solution
Let X denote the number of minutes past 7 That the passenger arrives at the stop =>X is uniformly distributed over (0, 30)
3
1
30
1
30
1
)3025Pr()1510Pr(30
25
15
10
dxdx
XX
36
Problem 7: conditional probability
Suppose that p(x,y) the joint probability mass function of X and Y is given by P(0,0) = .4, P(0,1) = .2, P(1,0) = .1, P(1,1) = .3
Calculate the conditional probability mass function of X given Y = 1
)(
),(
)(
),(
)|()|(|
yp
yxp
yYP
yYxXP
yYxXPyxp
Y
YX
5
3
)1(
)1,1()1|1(,
5
2
)1(
)1,0()1|0(,
5.0)1,1()1,0()1,()1(
|
|
YYX
YYX
xY
P
PPand
P
PPhence
PPxPP
37
counting process A stochastic process {N(t), t>=0}
is said to be a counting process if N(t) represents the total number of events that occur by time t
N(t) must satisfy N(t) >= 0
N(t) is integer valued
If s < t, then N(s) <= N(t)
For s < t, N(s) – N(t) = # events in the interval (s,t]
Independent increments # of events in disjoint time intervals are independent
38
Poisson process
The counting process {N(t), t>=0} is Said to be a Poisson process having rate λ, if
N(0) = 0
The process has independent increments
The # of events in any interval of length t is Poisson distributed with mean λt, that is
,...1,0,!
)())()(( n
n
tensNstNP
nt
39
Properties of the Poisson process
Superposition property If k independent Poisson processes
A1, A2, …, An
Are combined into a single process A
=> A is still Poisson with rate Equal to the sum of individual λi of Ai
40
Properties of the Poisson process (cont’d)
Decomposition property Just the reverse process
“A” is a Poisson process split into n processes Using probability Pi
The other processes are Poisson With rate Pi.λ