mit2 852s10 probability

7/27/2019 MIT2 852S10 Probability

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MIT 2.852

Manufacturing Systems AnalysisLectures 25: Probability

Basic probability, Markov processes, M/M/1 queues, and more

Stanley B. Gershwin

http://web.mit.edu/manuf-sysMassachusetts InstituteofTechnology

Spring, 2010

2.852ManufacturingSystemsAnalysis 1/128 Copyright c2010StanleyB.Gershwin.
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Probability and StatisticsTrick Question

I flip a coin 100 times, and it shows heads every time.

Question: What is the probability that it will show heads on the next flip?

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Probability and Statistics

Probability= Statistics

Probability: mathematical theory that describes uncertainty.

Statistics: set of techniques for extracting useful information from data.

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Interpretations of probabilityFrequency

The probability that the outcome of an experiment is A is prob (A)

if the experiment is performed a large number of times and the fraction oftimes that the observed outcome is A is prob (A).

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Interpretations of probabilityParallel universes


if the experiment is performed in each parallel universe and the fraction ofuniverses in which the observed outcome is A is prob (A).



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Interpretations of probabilityBetting Odds

The probability that the outcome of an experiment is A isprob(A) = P(A)

ifbefore the experiment is performed a risk-neutral observer would bewilling to bet $1 against more than $ 1P(A)

P(A) .

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Interpretations of probabilityState of belief


if that is the opinion (ie, belief or state of mind) of an observer before theexperiment is performed.

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Interpretations of probabilityAbstract measure


if prob () satisfies a certain set of axioms.

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Interpretations of probabilityAbstract measure

Axioms of probability

Let Ube a set ofsamples. Let E1, E

2, ... be subsets ofU. Let be the

null set (the set that has no elements).

0 prob (Ei) 1 prob (U) = 1

prob () = 0 IfEi Ej = , then prob (Ei Ej) = prob (Ei) + prob (Ej)

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Probability Basics

Subsets ofUare called events.

prob (E) is the probability ofE.

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Probability Basics

If

iEi = U, and Ei Ej = for

all iandj,

then i prob (Ei) = 1

Ei

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Probability BasicsSet Theory

Venn diagrams

A

A

U

prob (A) = 1 prob (A)



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Probability BasicsSet Theory

Venn diagrams

U

AUB

U

A B

A B

prob (A B) = prob (A) + prob (B) prob (A B)

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Probability BasicsIndependence

A and Bare independent if

prob (A B) = prob (A) prob (B).

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Probability BasicsConditional Probability

prob (A|B) =prob (A B)

prob (B)

U

AUB

U

A B

A B

prob (A B) = prob (A|B) prob (B).

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ExampleThrow a die.

A is the event of getting an odd number (1, 3, 5). B is the event of getting a number less than or equal to 3 (1, 2, 3).

Then prob (A) = prob (B) = 1/2 and

prob (A B) = prob (1, 3) = 1/3.Also, prob (A|B) = prob (A B)/ prob (B) = 2/3.

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Note: prob (A|B) being large does not mean that Bcauses A. It onlymeans that ifBoccurs it is probable that A also occurs. This could bedue to A and Bhaving similar causes.

Similarly prob (A|B) being small does not mean that Bprevents A.

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Probability BasicsLaw of Total Probability

U

C

U U

B

A D

A C A D

Let B= C Dand assume C D= . We haveprob (A|C) =

prob (A C)

prob (C)and prob (A|D) =

prob (A D)

prob (D).

Alsoprob (C|B) =

prob (C B)

prob (B)=

prob (C)

prob (B)because C B= C.

Similarly, prob (D|B) =prob (D)

prob (B)2.852ManufacturingSystemsAnalysis 18/128 Copyright c2010StanleyB.Gershwin.
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U

UA B

A B= A (C D) =

A C+ A D A (C D) =

A C+ A D

Therefore,

prob (A B) = prob (A C) + prob (A D)2.852ManufacturingSystemsAnalysis 19/128 Copyright c2010StanleyB.Gershwin.
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An important case is when C D= B= U, so that A B= A. Then

prob (A)

= prob (A C) + prob (A D)= prob (A|C) prob (C) + prob (A|D) prob (D).

D = C

U

U

U

CA

A D

A C

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More generally, ifA and E1, . . . Ek areevents and

Ei and Ej = , for all i=jand

j

Ej = the universal set

(ie, the set ofEj sets is mutually exclu-sive and collectively exhaustive) then...

A

Ei

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j prob (Ej) = 1

and

prob (A) =j

prob (A|Ej) prob (Ej).

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Some useful generalizations:

prob (A|B) =j prob (A|Band Ej) prob (Ej|B),

prob (A and B) =

j prob (A|Band Ej) prob (Ej and B).

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Probability BasicsRandom Variables

Let V be a vector space. Then a random variable X is a mapping (a

function) from Uto V.

If Uand x= X() V, then X is a random variable.

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Flip of One Coin

Let U=H,T. Let = H if we flip a coin and get heads; = T if we flip acoin and get tails.

Let X() be the number of times we get heads. Then X() = 0 or 1.

prob ( = T ) = prob (X= 0) = 1/2

prob ( = H ) = prob (X= 1) = 1/2


P b b l B
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Flip of Three Coins

Let U=HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.Let = HHH if we flip 3 coins and get 3 heads; = HHT if we flip 3 coins and

get 2 heads and then tails, etc. The order matters!

prob () = 1/8 for all .

Let Xbe the number of heads. The order does not matter! Then X= 0, 1, 2,or 3.

prob (X= 0)=1/8; prob (X= 1)=3/8; prob (X= 2)=3/8;prob (X= 3)=1/8.


P b bili B i
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Probability Distributions Let X() be a random variable. Thenprob (X() = x) is the probability distribution ofX (usually writtenP(x)). For three coin flips:

0 1 2

1/8

1/4

3/8

P(x)

3 x


D i S
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Dynamic Systems

tis the time index, a scalar. It can be discrete or continuous. X(t) is the state.

The state can be scalar or vector. The state can be discrete or continuous or mixed. The state can be deterministic or random.

X is a stochastic process ifX(t) is a random variable for every t.

The value ofX is sometimes written explicitly as X(t, ) or X(t).


Di t R d V i bl
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Discrete Random VariablesBernoulli

Flip a biased coin. IfXB is Bernoulli, then there is a psuch thatprob(XB = 0) = p.prob(XB = 1) = 1 p.


Di t R d V i bl
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Discrete Random VariablesBinomial

The sum ofn independent Bernoulli random variables XBi with the sameparameter pis a binomial random variable Xb.

Xb =n

i=0XBi

prob (Xb

= x) =

n!

x!(n x)!px(1 p)

(nx)


Discrete Random Variables
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Discrete Random VariablesGeometric

The number of independent Bernoulli random variables XBi tested untilthe first 0 appears is a geometricrandom variable Xg.

Xg = mini {X

Bi = 0}

To calculate prob (Xg = t): For t= 1, we know prob (XB = 0) = p.

Therefore prob (Xg > 1) = 1 p.


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For t> 1,

prob (Xg > t)= prob (Xg > t|Xg > t 1) prob (Xg > t 1)

= (1 p) prob (Xg > t 1),so

prob (Xg

> t) = (1 p)t

andprob (Xg = t) = (1 p)t1p


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Alternative view

1 0

p

1p 1

Consider a two-state system. The system can go from 1 to 0, but not from0 to 1.

Let pbe the conditional probability that the system is in state 0 at timet+ 1, given that it is in state 1 at time t. That is,

p= prob

(t+ 1) = 0

(t) = 1

.


Discrete Random Variablesp
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1 0

p

Let p(, t) be the probability of the system being in state at time t.

Then, since

p(0, t+ 1) = prob

(t+ 1) = 0

(t) = 1

prob [(t) = 1]

+ prob(t+ 1) = 0

(t) = 0

prob [(t) = 0],

(Why?)we have

p(0, t+ 1) = pp(1, t) + p(0, t),p(1, t+ 1) = (1 p)p(1, t),

and the normalization equation

p(1, t) + p(0, t) = 1.


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1 0

p

Assume that p(1, 0) = 1. Then the solution is

p(0, t) = 1 (1 p)t,

p(1, t) = (1 p)t.


Discrete Random Variables1p 1
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1 0

p

Geometric Distribution

t

probability

0 10 20 300

0.2

0.4

0.6

0.8

1

p(0,t)p(1,t)


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1 0

p

Recall that once the system makes the transition from 1 to 0 it can nevergo back. The probability that the transition takes place at time t is

prob [(t) = 0 and (t 1) = 1] = (1 p)t1p.

The time of the transition from 1 to 0 is said to be geometricallydistributed with parameter p. The expected transition time is 1/p. (Proveit!)

Note: If the transition represents a machine failure, then 1/pis the MeanTime to Fail (MTTF). The Mean Time to Repair (MTTR) is similarlycalculated.


Discrete Random Variables p
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Discrete Random VariablesGeometric 1 0

p

p

Memorylessness: ifT is the transition time,

prob (T> t+ x|T> x) = prob (T> t).


Digression: Difference Equations
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Digression: Difference EquationsDefinition

A difference equation is an equation of the form

x(t+ 1) = f(x(t), t)

where t is an integer and x(t) is a real or complex vector.

To determine x(t), we must also specify additional information, for exampleinitial conditions:

x(0) = c

Difference equations are similar to differential equations. They are easier to solve

numerically because we can iterate the equation to determine x(1), x(2), .... In fact,

numerical solutions of differential equations are often obtained by approximating them

as difference equations.


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Digression: Difference EquationsSpecial Case

A linear difference equation with constant coefficients is one of the form

x(t+ 1) = Ax(t)

where A is a square matrix of appropriate dimension.

Solution:x(t) = Atc

However, this form of the solution is not always convenient.


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g qSpecial Case

We can also write

x(t) = b1t1+ b2t2+ ... + bktk

where k is the dimensionality ofx, 1, 2,...,k are scalars andb1, b2, ...,bk are vectors. The bj satisfy

c= b1+ b2+ ... + bk1, 2, ..., k are the eigenvalues ofAand b1, b2, ..., bk are its eigenvectors, but we dontalways have to use that explicitly to determine them. This is very similar to the solution

of linear differential equations with constant coefficients.


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g qSpecial Case

The typical solution technique is to guess a solution of the form

x(t) = btand plug it into the difference equation. We find that must satisfy a kthorder polynomial, which gives us the ks. We also find that bmustsatisfy a set of linear equations which depends on .

Examples and variations will follow.


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p

A Markov process is a stochastic process in which the probability offinding Xat some value at time t+ tdepends only on the value ofXat time t.

Or, let x(s), s t, be the history of the values ofXbefore time tandlet A be a set of possible values ofX(t+ t). Then

prob {X(t+ t) A|X(s) = x(s), s t} =

prob {X(t+ t) A|X(t) = x(t)}

In words: if we know what Xwas at time t, we dont gain any moreuseful information about X(t+ t) by also knowing what Xwas atany time earlier than t.


Markov processes
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pStates and transitions

Discrete state, discretetime

States can be numbered 0, 1, 2, 3, ... (or with multiple indices if thatis more convenient).

Time can be numbered 0, 1, 2, 3, ... (or 0, , 2, 3, ... if moreconvenient).

The probability of a transition fromjto i in one time unit is oftenwritten Pij, wherePij = prob{X(t+ 1) = i|X(t) =j}


Markov processes
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States and transitions

Discrete state, discretetimeTransition graph

14P14

P24

P64

P45

14P14

P24

P6

1

1

2

3

4

5

6

7

Pij is a probability. Note that Pii = 1 m,m=iPmi.


Markov processes
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Define pi(t) = prob{X(t) = i}. {pi(t)for all i} is the probability distribution at time t. Transition equations: pi(t+ 1) = jPijpj(t). Initial condition: pi(0) specified. For example, if we observe that the system

is in statejat time 0, then pj(0) = 1 and pi(0) = 0 for all i=j. Let the current time be 0. The probability distribution at time t> 0

describes our state of knowledge at time 0 about what state the system willbe in at time t.

Normalization equation: ipi(t) = 1.


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Steady state: pi = limt pi(t), if it exists. Steady-state transition equations: pi = jPijpj. Steady state probability distribution:

Very important concept, but different from the usual concept of steadystate.

The system does not stop changing or approach a limit. The probability distribution stops changing and approaches a limit.


Markov processes
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Steady state probability distribution: Consider a typical (?) Markov process.Look at a system at time 0.

Pick a state. Any state.

The probability of the system being in that state at time 1 is very different fromthe probability of it being in that state at time 2, which is very different from itbeing in that state at time 3.

The probability of the system being in that state at time 1000 is very

close

to

theprobability of it being in that state at time 1001, which is very close to the

probability of it being in that state at time 2000.

Then, the system has reached steady state at time 1000.


Markov processes
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Discrete state, discretetimeTransition equations are valid for steady-state and non-steady-stateconditions.

(Self-loops suppressed for clarity.)


Markov processes
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Discrete state, discretetimeBalance equations steady-state only. Probability of leaving node i=probability of entering node i.

pi

m,m=iPmi =

j,j=i

Pijpj

(Prove it!)


Markov processes
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Unreliable machine

1=up; 0=down.

p

1 0

p rr


Markov processesU l bl h
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Unreliable machine

The probability distribution satisfies

p(0, t+ 1) = p(0, t)(1 r) + p(1, t)p,

p(1, t+ 1) = p(0, t)r+ p(1, t)(1 p).


Markov processesU li bl hi
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Unreliable machine

Solution

Guess

p(0, t) = a(0)Xtp(1, t) = a(1)Xt

Then

a(0)Xt+1 = a(0)Xt(1 r) + a(1)Xtp,a(1)Xt+1 = a(0)Xtr+ a(1)Xt(1 p).


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Unreliable machine

SolutionOr,

a(0)X = a(0)(1r) + a(1)p,a(1)X = a(0)r+ a(1)(1p).

or,

X = 1r+ a(1)a(0)

p,

X = a(0)a(1)

r+ 1p.so

X = 1r+ rpX1 + p

or,(X1 + r)(X1 + p) = rp.


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Unreliable machine

SolutionTwo solutions:

X= 1 and X= 1 r p.

IfX= 1, a(1)a(0) = rp. IfX= 1 r p, a(1)a(0) = 1. Therefore

p(0, t) = a1(0)Xt1 + a2(0)Xt2 = a1(0) + a2(0)(1 r p)t

p(1, t) = a1(1)Xt1 + a2(1)Xt2 = a1(0) r

p a2(0)(1 r p)

t


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Unreliable machine

SolutionTo determine a1(0) and a2(0), note that

p(0, 0) = a1(0) + a2(0)

p(1, 0) = a1(0)

r

p a

2(0)

Therefore

p(0, 0) + p(1, 0) = 1 = a1(0) + a1(0)r

p= a1(0)

r+ p

pSo

a1(0) =p

r+ pand a2(0) = p(0, 0)

p

r+ p


Markov processesUnreliable machine
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Unreliable machine

SolutionAfter more simplification and some beautification,

p(0, t) = p(0, 0)(1 p r)t

+p

r+ p

1 (1 p r)t ,

p(1, t) = p(1, 0)(1 p r)t

+r

r+ p1 (1 p r)t

.


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Unreliable machine

SolutionDiscrete Time Unreliable Machine

t

probability

0 20 40 60 80 1000

0.2

0.4

0.6

0.8

1

p(0,t)p(1,t)


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Unreliable machine

Steady-state solutionAs t ,

p(0) p

r+ p,

p(1) rr+ p

which is the solution of

p(0) = p(0)(1 r) + p(1)p,

p(1) = p(0)r+ p(1)(1 p).


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Unreliable machine

Steady-state solutionIf the machine makes one part per time unit when it is operational, the

average production rate is

p(1) =r

r+ p=

1

1 +p

r.


Markov processesStates and Transitions
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States and Transitions

Classification of statesA chain is irreducibleif and only if each state can be reached from each

other state.

Let fij be the probability that, if the system is in statej, it will at somelater time be in state i. State i is transientiffij < 1. If a steady statedistribution exists, and i is a transient state, its steady state probability is

0.


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Classification of statesThe states can be uniquely divided into sets T, C1, . . . Cn such that T is the setof all transient states and fij = 1 for iandj in the same set Cm and fij = 0 for iin some set Cm andjnot in that set. If there is only one set C, the chain isirreducible. The sets Cm are called final classesor absorbing classesand T is thetransient class.

Transient states cannot be reached from any other states except possibly other

transient states. If state i is in T, there is no statej in any set Cm such thatthere is a sequence of possible transitions (transitions with nonzero probability)fromjto i.


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Classification of states

C2 C3

C4

C5C6C7

C1T


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Discrete state, continuoustime

States can be numbered 0, 1, 2, 3, ... (or with multiple indices if thatis more convenient).

Time is a real number, defined on (, ) or a smaller interval. The probability of a transition fromjto iduring [t, t+ t] is

approximately ijt, where t is small, andijt= prob{X(t+ t) = i|X(t) =j} + o(t) forj= i.


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Transition graph no self loops!!!!

1

2

3

4

5

6

7

1414

24

45

64

ij is a probability rate. ijt is a probability.2.852ManufacturingSystemsAnalysis 66/128 Copyright c2010StanleyB.Gershwin.

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Define pi(t) = prob{X(t) = i} It is convenient to define ii = j=iji Transition equations: dpi(t)

dt =

jijpj(t). Normalization equation: ipi(t) = 1.


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Steady state: pi = lim

tp

i(t), if it exists.

Steady-state transition equations: 0 = jijpj. Steady-state balance equations: pim,m=imi = j,j=iijpj

Normalization equation:

ipi = 1.


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Sources of confusion in continuous time models:

Never Draw self-loops in continuous time Markov process graphs. Never write 1 14 24 64. Write

1 (14+ 24+ 64)t, or (14+ 24+ 64)

ii = j=iji is NOT a probability rate and NOT a probability. Itis ONLY a convenient notation.


Markov processesExponential
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p

Exponential random variable: the time to move from state 1 to state 0.

1 0

p

pt= prob

(t+ t) = 0

(t) = 1

+ o(t).


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p

1 0

p

p(0, t+ t) =

prob(t+ t) = 0

(t) = 1 prob [(t) = 1]+prob

(t+ t) = 0

(t) = 0

prob[(t) = 0].

or

p(0, t+ t) = ptp(1, t) + p(0, t) + o(t)

ordp(0, t)

dt= pp(1, t).

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1 0

p

Since p(0, t) + p(1, t) = 1,

dp(1, t)

dt= pp(1, t).

Ifp(1, 0) = 1, then

p(1, t) = e

ptand

p(0, t) = 1 ept


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Density function

The probability that the transition takes place in [t, t+ t] is

prob [(t+ t) = 0 and (t) = 1] = eptpt.The exponential density function is pept.The time of the transition from 1 to 0 is said to be exponentially

distributed with rate p. The expected transition time is 1/p. (Prove it!)


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Density function

f(t) = pept for t 0; f(t) = 0 otherwise;F(t) = 1 ept for t 0; F(t) = 0 otherwise.

ET= 1/p,VT = 1/p2. Therefore, cv=1.

t1 p

F(t)

t1 p

f(t)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.5 2 2.5 3 3.5 4 4.51 1.50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.5 2 2.5 3 3.5 4 4.51 1.5


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Density function

Memorylessness: prob (T> t+ x|T> x) = prob (T> t) prob (t T t+ t) tfor small t. IfT1, ...,Tn are exponentially distributed random variables with

parameters 1...,n and T= min(T1, ...,Tn), then T is anexponentially distribution random variable with parameter = 1+ ... + n.


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Density function

Exponential densityfunction and a small

number of actualsamples.


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Continuous time

p

1 0

r


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Continuous timeThe probability distribution satisfies

p(0, t+ t) = p(0, t)(1 rt) + p(1, t)pt+ o(t)p(1, t+ t) = p(0, t)rt+ p(1, t)(1 pt) + o(t)

or

dp(0, t)

dt= p(0, t)r+ p(1, t)p

dp(1, t)

dt= p(0, t)r p(1, t)p.


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Solution

p(0, t) =p

r+ p+

p(0, 0)

p

r+ p

e(r+p)t

p(1, t) = 1 p(0, t).

As t ,

p(0) p

r+ p

,

p(1) r

r+ p


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Steady-state solutionIf the machine makes parts per time unit on the average when it is

operational, the overall average production rate is

p(1) =r

r+ p=

1

1 +p

r.


Markov processesThe M/M/1 Queue
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Simplest model is the M/M/1 queue: Exponentially distributed inter-arrival times mean is 1/; is arrival

rate (customers/time). (Poisson arrival process.) Exponentially distributed service times mean is 1/; is service rate

(customers/time).

1 server. Infinite waiting area.

2.852 Manufacturing Systems Analysis 81/128 Copyright c2010 Stanley B. Gershwin.

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Exponential arrivals: If a part arrives at time s, the probability that the next part arrives

during the interval [s+ t, s+ t+ t] is e

tt+ o(t) t. isthe arrival rate.

Exponential service: If an operation is completed at time sand the buffer is not empty, the

probability that the next operation is completed during the interval

[s+ t, s+ t+ t] is ett+ o(t) t. is the service rate.


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Sample pathNumber of customers in the system as a function of time.

1

2

3

4

56

t

n


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State Space

0 1 2

n1 n n+1

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Performance EvaluationLet p(n, t) be the probability that there are n parts in the system at timet. Then,

p(n, t+ t) = p(n 1, t)t+ p(n + 1, t)t

+p(n, t)(1 (t+ t)) + o(t)

for n > 0

and

p(0, t+ t) = p(1, t)t+ p(0, t)(1 t) + o(t).


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Performance EvaluationOr,

dp(n, t)

dt= p(n 1, t) + p(n + 1, t) p(n, t)( + ),

n > 0dp(0, t)

dt= p(1, t) p(0, t).

If a steady state distribution exists, it satisfies

0 = p(n 1) + p(n + 1) p(n)( + ), n > 00 = p(1) p(0).

Why if?


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Performance EvaluationLet = /. These equations are satisfied by

p(n) = (1 )n, n 0if < 1. The average number of parts in the system is

n =n

np(n) =

1 =

.

From Littles law, the average delay experienced by a part is

W=1

.


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Performance Evaluation

Delay in a M/M/1 Queue

Arrival Rate

Delay

0 0.25 0.5 0.75 10

10

20

30

40

Define the utilization = /.

What happens if > 1?


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Performance Evaluation

W

=2=1

0

20

40

60

80

100

0 0.5 1 1.5 2

To increase capacity, increase . To decrease delay for a given ,

increase .


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Other Single-Stage ModelsThings get more complicated when:

There are multiple servers.

There is finite space for queueing. The arrival process is not Poisson. The service process is not exponential.

Closed formulas and approximations exist for some cases.


Continuous random variablesPhilosophical issues
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1. Mathematically, continuous and discrete random variables are verydifferent.

2. Quantitatively, however, some continuous models are very close to

some discrete models.3. Therefore, which kind of model to use for a given system is a matter

ofconvenience.

Example: The production process for small metal parts (nuts, bolts,

washers, etc.) might better be modeled as a continuous flow than a largenumber of discrete parts.


Continuous random variablesProbability density
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Low density

High density

The probability of a two-dimensional

random variable being in a small square isthe probability density times the area ofthe square. (Actually, it is more generalthan this.)


Continuous random variablesProbability density
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Continuous random variablesSpaces
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Continuous random variables can be defined in one, two, three, ..., infinite dimensional spaces;

in finite or infinite regions of the spaces. Continuous random variables can have

probability measures with the same dimensionality as the space; lower dimensionality than the space; a mix of dimensions.


Continuous random variablesDimensionality
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M1 B1 M2 B2 M3

M1

M2

M3

B1

x1

B2

x2


Continuous random variablesDimensionality
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M1 B1 M2 B2 M3

M1

M2

M3

B1

x1

B2

x2


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DimensionalityOnedimensional density

Twodimensional density

Zerodimensional density

(mass)

M1

2x

B M B M1 2 2 3

x1

Probability distributionof the amount ofmaterial in each of thetwo buffers.

2 852 Manufacturing Systems Analysis 97/128 Copyright 2010 Stanley B Gershwin

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Discrete approximation

M1

2x

B M B M1 2 2 3

x1

Probability distributionof the amount ofmaterial in each of thetwo buffers.

2 852 M f t i S te A l i 98/128 C i ht 2010 St le B Ge h i

Continuous random variablesExample
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Problem

Productionsurplus fromanunreliablemachine

Production rate = p

1 0

r

Production rate = 0

Demand rate = d<

r

r+ p

. (Why?)

Problem: producing more than has been demanded creates inventory and iswasteful. Producing less reduces revenue or customer goodwill. How can we

anticipate and respond to random failures to mitigate these effects?

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Solution

We propose a production policy. Later we show that it is a solution to anoptimization problem.Model:

0 < u


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SolutionSurplus, or inventory/backlog:

dx(t)

dt= u(t) d

Productionpolicy: Choose Z(the hedgingpoint) Then,

if = 1, ifx< Z, u= , ifx= Z, u= d, ifx> Z, u= 0;

if = 0, u= 0.

d t + Z

t

demand dt

hedging point Z

production

surplus x(t)

Production and Demand

Cumulative

How do we choose Z?

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Mathematical modelDefinitions:f(x, , t) is a probability density function.

f(x, , t)x = prob (x X(t) x+ xand the machine state is at time t).

prob (Z, , t) is a probability mass.

prob (Z, , t) = prob (x= Z

and the machine state is at time t).

Note that x> Z is transient.

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Mathematical model

State Space:

=1x

= ddx

dt

= d

=0dx

dt

x=Z

2 852M f i

S

A l i

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C i h

2010

S l

B

G h i

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Mathematical model

Transitions to = 1, [x, x+ x]; x< Z:

=1

x

=0

x=Z

repair

no failure

x

2 852M f i

S

A l i

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C i h

2010

S l

B

G h i

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Mathematical model


=1

x

=0

x=Z

failure

no repair

x

2 852M f i

S

A l i

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C i h

2010

S l

B

G h i


Mathematical model
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Mathematical model


x + d t

x( d) tx(t) =

x(t) =

x(t+ t)

=1

=0

f(x, 1, t+ t)x=

[f(x+ dt, 0, t)x]rt+ [f(x ( d)t, 1, t)x](1 pt)

+o(t)o(x)

M f i

S

A l i

/

C i h

S l

B

G h i

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Mathematical model

Or,

f(x, 1, t+ t) =o(t)o(x)

x

+f(x+ dt, 0, t)rt+ f(x ( d)t, 1, t)(1 pt)

In steady state,

f(x, 1) =o(t)o(x)

x

+f(x+ dt, 0)rt+ f(x ( d)t, 1)(1 pt)

S

/

C

S

G

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Mathematical model

Expand in Taylor series:f(x, 1) =

f(x, 0) + df(x, 0)

dxdt

rt

+

f(x, 1)

df(x, 1)

dx( d)t

(1 pt)

+o(t)o(x)

x

/

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Mathematical model

Multiply out:

f(x, 1) = f(x, 0)rt+df(x, 0)

dx(d)(r)t2

+f(x, 1) df(x, 1)

dx( d)t

f(x, 1)ptdf(x, 1)

dx( d)pt2

+o(t)o(x)

x

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Mathematical model

Subtract f(x, 1) from both sides and move one of the terms:

df(x, 1)

dx( d)t=

o(t)o(x)

x

+f(x, 0)rt+df(x, 0)

dx(d)(r)t2

f(x, 1)pt df(x, 1)dx

( d)pt2

2.852ManufacturingSystemsAnalysis Copyright 2010StanleyB.Gershwin.

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Mathematical model

Divide through by t:

df(x, 1)

dx( d) =

o(t)o(x)

tx

+f(x, 0)r+df(x, 0)

dx(d)(r)t

f(x, 1)p df(x, 1)dx

( d)pt


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Mathematical model

Take the limit as t 0:

df(x, 1)

dx( d) = f(x, 0)r f(x, 1)p



Mathematical model
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x + d tx(t) =

x( d) tx(t) =

x(t+ t)

=1

=0

failure

no repair

f(x, 0, t+ t)x=

[f(x+ dt, 0, t)x](1 rt) + [f(x ( d)t, 1, t)x]pt

+o(t)o(x)


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Mathematical model

By following essentially the same steps as for the transitions to = 1, [x, x+ x]; x< Z, we have

df(x, 0)dx

d= f(x, 0)r f(x, 1)p

Note:

df(x, 1)

dx ( d) =df(x, 0)

dx d

Why?


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Mathematical model

Transitions to = 1, x= Z:

1p t

1p t

Z ( d) t

=1

=0

no failure

x=Z

no failure

P(Z, 1) = P(Z, 1)(1 pt)

+ prob (Z ( d)t< X< Z, = 1)(1 pt)+o(t).


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Mathematical model

Or,

P(Z, 1) = P(Z, 1) P(Z, 1)pt

+f(Z ( d)t, 1)( d)t(1 pt) + o(t),

or,P(Z, 1)pt= o(t)+

+f(Z, 1) df(Z, 1)

dx( d)t

( d)t(1 pt),


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Mathematical model

Or,

P(Z, 1)pt= f(Z, 1)( d)t+ o(t)or,

P(Z, 1)p= f(Z, 1)( d)


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Mathematical model

=1

x

= ddx

dt

= d

=0dx

dt

x=Z

P(Z, 0) = 0. Why?


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Mathematical model

Transitions to = 0,Z ( d)t< x< Z:

p t 1r t

Z d t

=1

failure

=0

x=Z

no repair

prob (Z dt< X< Z, 0) = f(Z, 0)dt+ o(t)= P(Z, 1)pt+ o(t)


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Mathematical model

Or,

f(Z, 0)d= P(Z, 1)p= f(Z, 1)( d)


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Mathematical model

df

dx(x, 0)d= f(x, 0)r f(x, 1)p

df(x, 1)dx

( d) = f(x, 0)r f(x, 1)p

f(Z, 1)( d) = f(Z, 0)d

0 = pP(Z, 1) + f(Z, 1)( d)

1 = P(Z, 1) +Z

[f(x, 0) + f(x, 1)] dx


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SolutionSolution of equations:

f(x, 0) = Aebx

f(x, 1) = A ddebx

P(Z, 1) = A dpebZ

P(Z, 0) = 0

whereb= r

d p

d

and A is chosen so that normalization is satisfied.


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SolutionDensity Function -- Controlled Machine

-40 -20 0 200

1E-2

0.02

0.03

x


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Observations

1. Meanings of b:Mathematical:In order for the solution on the previous slide to make sense, b> 0.Otherwise, the normalization integral cannot be evaluated.



Ob i
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Observations

Intuitive:

The average duration of an up period is 1/p. The rate that x increases(while x< Z) while the machine is up is d. Therefore, the average

increase ofxduring an up period while x< Z is ( d)/p.

The average duration of a down period is 1/r. The rate that xdecreaseswhile the machine is down is d. Therefore, the average decrease ofxduringan down period is d/r.

In order to guarantee that xdoes not move toward , we must have( d)/p> d/r.


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Observations

If ( d)/p> d/r,

thenp

d 0.

That is, we must have b> 0 so that there is enough capacity for x to increase on

the average when x< Z.


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Observations

Also, note that b> 0 =r

d>

p

d=

r( d) > pd=

r rd> pd=

r > rd+ pd=

r

r+ p> d

which we assumed.



Observations
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2. Let C= AebZ. Then

f(x, 0) = Ceb(Zx)

f(x, 1) = C ddeb(Zx)

P(Z, 1) = Cdp

P(Z, 0) = 0

That is, the probability distribution really depends on Z x. IfZ ischanged, the distribution shifts without changing its shape.


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2.852 Manufacturing Systems AnalysisSpring 2010

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