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1 Birth and death process

N(t). Depends on how fast arrivals or departures occur . Objective . N(t) = # of customers. at time t.. λ. arrivals. (births). departures. (deaths). μ. 2. Behavior of the system. λ. >. μ. λ. <.

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1 Birth and death process






Presentation on theme: "1 Birth and death process"— Presentation transcript:

Slide1

1

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)

μSlide2

2

Behavior of the system

λ>μλ<μ

Possible evolution of N(t)

Time

1 2 3 4 5 6 7 8 9 10 11

1

2

3

busy

idle

N(t)Slide3

3

General arrival and departure rates

λnDepends on the number of customers (n) in the systemExample

μn Depends on the number of customers in the systemExampleSlide4

4

Changing the scale of a unit time

Number of arrivals/unit timeFollows the Poisson distribution with rate λn

Inter-arrival time of successive arrivalsis exponentially distributed Average inter-arrival time = 1/ λnWhat is the avg. # of customers arriving in dt?

TimeSlide5

5

Probability of one arrival in dt

dt so smallNumber of arrivals in dt, X is a r.v.X=1 with probability pX=0 with probability 1-p

Average number of arrivals in dtProb (having one arrival in dt) = λn dt

dtSlide6

6

Probability of having 2 events in dt

Departure rate in dtμn dt Arrival rate in dt

λn dtWhat is the probabilityOf having an (arrival+departure), (2 arrivals or departures)Slide7

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Probability distribution of N(t)

Pn (t)The probability of getting n customers by time tThe distribution of the # of customers in system

t+dt

t

?

n

n-1: arrival

n+1: departure

n: none of the aboveSlide8

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Differential equation monitoring evolution of # customers

These are solved

Numerically using MATLAB

We will explore the cases

Of pure death And pure birthSlide9

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Pure birth process

In this caseμn =0, n >= 0λn

= λ, n >= 0

Hence,Slide10

First order differential equation

 

10Slide11

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Pure death process

In this caseλn =0, n >= 0μn

= μSlide12

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Queuing system

Transient phaseSteady stateBehavior is independent of t

Pn (t)

λ

μ

P

n

(t)

t

transient

Steady stateSlide13

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Differential equation: steady state analysis

Limiting caseSlide14

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Solving the equations

n=1n=2

(1)

(1) => Slide15

15

P

nWhat about P0Slide16

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Normalization equation Slide17

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Conditional probability and conditional expectation: d.r.v.

X and Y are discrete r.v.Conditional probability mass functionOf X given that Y=y

Conditional expectation of X given that Y=ySlide18

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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=yThe conditional expectation Of X given that Y=y Slide19

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Computing expectations by conditioning

Denote E[X|Y]: function of the r.v. YWhose value at Y=y is E[X|Y=y]

E[X|Y]: is itself a random variableProperty of conditional expectationif Y is a discrete r.v.if Y is continuous with density fY (y) =>

(1)

(2)

(3)Slide20

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Proof of equation when X and Y are discreteSlide21

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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 chapterOf his probability bookis Poisson distributed with mean 2Of his history bookis Poisson distributed with mean 5Assuming Sam equally likely to choose either book

What is the expected number of misprints he comes across?Slide22

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SolutionSlide23

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Problem 2

A miner is trapped in a mine containing three doorsFirst doorleads to a tunnel that takes him to safety

After 2 hours of travelSecond door leads to a tunnel that returns him to the mineAfter 3 hours of travel

Third doorLeads to a tunnel that returns him to the mineAfter 5 hours Assuming he is equally likely to choose any doorWhat is the expected length of time until he reaches safety?Slide24

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SolutionSlide25

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Computing probabilities by conditioning

Let E denote an arbitrary eventX is a random variable defined by

It follows from the definition of X Slide26

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Problem 3

Suppose that the number of peopleWho visit a yoga studio each day is a Poisson random variable with mean λ

Suppose further that each person who visitis, independently, female with probability pOr male with probability 1-pFind the joint probability That n women and m men visit the academy todaySlide27

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Solution

Let N1 denote the number of women, N2 the number of men

Who visit the academy todayN= N1 +N2 : total number of people who visit Conditioning on N givesBecause P(N1=n,N2=m|N=i)=0 when i != n+mSlide28

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Solution (cont’d)

Each of the n+m visit is independently a woman with probability pThe conditional probabilityThat n of them are women is

The binomial probability of n successes in n+m trialsSlide29

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Solution: analysis

When each of a Poisson number of eventsis independently classifiedAs 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 variablesSlide30

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Problem 4

At a party N men take off their hatsThe hats are then mixed up and Each man randomly selects one

A match occurs if a man selects his own hatWhat is the probability of no matches?Slide31

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Solution

E = event that no matches occurP(E) = Pn : explicit dependence on n

Start by conditioningWhether or not the first man selects his own hatM: 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 menSlide32

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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 matchesExtra man does select the extra hat=> (1/n-1)xPn-2Slide33

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Solution (cont’d)

Pn is the probability of no matchesWhen n men select among their own hats=> P1 =0 and P

2 = ½=> Slide34

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Problem 5:

continuous random variablesThe probability density function of a non-negative random variable X is given by Compute the constant λ?Slide35

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Problem 6:

continuous random variablesBuses arrives at a specified stop at 15 min intervalsStarting at 7:00 AMThey arrive at 7:00, 7:15, 7:30, 7:45If the passenger arrives at the stop at a time

Uniformly distributed between 7:00 and 7:30Find the probability that he waits less than 5 min?SolutionLet X denote the number of minutes past 7 That the passenger arrives at the stop =>X is uniformly distributed over (0, 30)Slide36

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Problem 7:

conditional probabilitySuppose that p(x,y) the joint probability mass function of X and Y is given byP(0,0) = .4, P(0,1) = .2, P(1,0) = .1, P(1,1) = .3Calculate the conditional probability mass function of X given Y = 1 Slide37

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counting process

A stochastic process {N(t), t>=0}is said to be a counting process ifN(t) represents the total number of events that occur by time t

N(t) must satisfyN(t) >= 0N(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 independentSlide38

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Poisson process

The counting process {N(t), t>=0} is Said to be a Poisson process having rate λ, ifN(0) = 0

The process has independent incrementsThe # of events in any interval of length t is Poisson distributed with mean λt, that isSlide39

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Properties of the Poisson process

Superposition propertyIf k independent Poisson processesA1, A2, …, AnAre combined into a single process A

=> A is still Poisson with rate Equal to the sum of individual λi of AiSlide40

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Properties of the Poisson process (cont’d)

Decomposition propertyJust the reverse process “A” is a Poisson process split into n processesUsing probability Pi

The other processes are Poisson With rate Pi.λ