Post on 22-Feb-2016
description
transcript
1
Birth and death process
N(t) Depends on how fast arrivals or departures occur
Objective
N(t) = # of customersat time t.
λarrivals(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
MnMn
n ,0,
n
Orn
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:
hourExampledtAverage
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
dtXEbutpppXE
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
dtdttPdttPdttPdttP
nnn
nnnnn
8
Differential equation monitoring evolution of # customers
?)())(()()()(
))(()()()()()1)(()()()(
1111'
1111
1111
tPtPtPtPtP
tPtPtPdt
tPdttPdtdttPdttPdttPdttP
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'
nettPetP
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'
nMettPetP
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
13
Differential equation: steady state analysis Limiting case
01
01
1111
1100
1111
'
0,)(
0,)(0
0)(
)(
limlim
PP
nPPPPP
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,......
021
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
ionNormilizatPPP
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
)(),(
)(),(
)|()|(|
ypyxp
yYPyYxXP
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
)(),()|(| yfyxfyxf
YYX
dxyxfxyYXE YX )|(.]|[ |
19
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
yYPyYPyYxXPx
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?
22
Solution
27)
21(2)
21(5
]2[].2|[]1[].1|[][_,2__,1
_
YPYXEYPYXEXEchosenyprobabilitchosenbookhistory
Y
mistakesnumberX
23
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?
24
Solution
10][][5][32(31][
][5]3|[];[3]2|[;2)1|[
])3|[]2|[]1|[(31
]3[]3|[]2[]2|[]1[]1|[][
____
XEXEXEXE
XEYXEXEYXEYXE
YXEYXEYXE
YPYXEYPYXEYPYXEXE
choseninitiallydoorYsafetyuntiltimeX
25
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
27
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
mpe
npe
mnepp
nmn
mNnNP
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
mpemNP
andnpemNnNPnNP
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?
31
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
nnMEPPMEP
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
1111)|(
nnn
nnc
Pn
PnnP
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(....
!41
!31
!21
!41
!31
!21;
!31
!21
43
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
10110).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)
31
301
301
)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
)(),(
)(),(
)|()|(|
ypyxp
yYPyYxXP
yYxXPyxp
Y
YX
53
)1()1,1()1|1(,
52
)1()1,0()1|0(,
5.0)1,1()1,0()1,()1(
|
|
YYX
YYX
xY
PPPand
PPPhence
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,!)())()(( nntensNstNPn
t
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