Vorlesung02 Statistical Models in Simulation

8/14/2019 Vorlesung02 Statistical Models in Simulation

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UNIVERSITT

D U I S B U R GE S S E N

Prof. Dr.-Ing. Bernd Noche

Rechnergesttzte Netzanalysenehem.:Simulation in Logist ics II

Probability and Statistics in

Simulation

Lecturer: Prof. Dr.-Ing. Bernd Noche

tul06.11.2013 1Rechnergesttzte Netzanalysen


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Perform statistic

analyses of the

simulation output data

Design the

simulation

experiments

Probabilityand

statistics

Model aprobabilistic

s stem

Generate randomsamples from the

in ut distribution

Validate the

simulation

Choose the

input probabilistic

mode s r u on

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. ueue ng ys em: n erarr va mes an e serv ce mes.

Exponential distribution, Weibull distribution, and gamma distribution,

normal distribution.

2. Inventory System: the number of units demanded per order or per time period,

the time between placing an order, and the lead time.

Geometric, Poisson, and negative binomial distribution provide a range of

distribution shapes that satisfy a variety of demand patterns. The lead time distribution can often be fitted fairly well by a gamma

distribution.

3. Reliability and Maintainability: time to failure, time to repair.

Time to failure has been modeled with exponential, gamma and Weibull

.

4. Limited Data: in many instances simulations begin before data collection has

been completed.

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, , .


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Random Variables and Probabilit Distributions

Random Variable (RV)A real number assigned to each outcome of an

experiment in the sample spaceCan only take a finite or a countable infinite set of

values

e.g., hit or miss {0 or 1}, Flip a coin of shooting abasketball, outcome of throwing a dart {1, 2, ,

,Simulate Monte Carlo: throw a dice {1, 2, 3, 4, 5,

6}, a pair of dices {2, 3, 4, , 10, 11, 12} Continuous Random Variable

Can take on a continuum of values (infinite)

06.11.2013 Rechnergesttzte Netzanalysen 4e.g., customer interarrival time


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Discrete Random Variables

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Restrictions/Conditions

0 p(xi) 1 for all i x = 1 certain outcome

Alternative representation for the probability distribution is the

cumulativedistribution function (CDF) (Verteilungsfunktion),

F(x) Definition: F(x) = P(X x) relative to probability mass function: F(x) =(xi x)p(xi) Properties of F(x):

(1) 0 F(x) 1(2) F( ) = 0(3) F( ) = 1



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Ex: S = {0, 1, 2, 3} four possible outcomes

p(0) = 1/8 p(1) = 3/8 p(2) = 3/8 p(3) = 1/8

(i = 0 to 3)p(xi) = 1/8 + 3/8 + 3/8 + 1/8 = 1

p(Xi) F(Xi)

1 13

1.000

0.875

1/2 1/2

2

0.500

1/4

10 2 3 Xi 10 2 3

1/4

Xi

0.1250



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Continuous Random Variables

Probability density function (pdf)(Dichtefunktion), f(x)

e n on: a = (from a to b) x x

Conditions: f(x) f x(1) f(x) 0 and

=rom o

xa b

P(a X b)

Cumulative distribution function (cdf), F(x)Definition: F(x) = (from to x)f(y) dy = P(X x)

variable X assuming a value less than or equal to x06.11.2013 8Rechnergesttzte Netzanalysen


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values between 0 and 1 and equal probability

f (X) F (X)

= =0.75

UNFRM (0,1)

. .0.5

.

X X0 0.25 10.5 0.75 1

pdf cdf



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Discrete and Continuous Random

Variables

Mixed Distribution

-

probability Value 2 - 1/3 p(2) = 1/3

=Between 1 and 2 - 1/3 probability = 1

F (X)f (X)

1 1/3 1/3 1 1.00 X

2

1/3 1/3 0.33

.

(1,2)

1

X X0 1 2 0 1 2


+ x-


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Ex ectation and Moments

Used to characterize robabilit distribution functions

The Expectation (expected value) of a random variable

, (all i) i iE[x] =(all x)x f(x)dx when x is continuous

n genera , can e a unc on o xE[xn] =(all i)xin p(xi) when x is discreteE xn = (all x)xn x x w en x is continuous

The expectation of xn

is defined as the nth

moment of aran om var a e

Expected value is a special case when n = 1, it is thus called



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A variant of the nth moment is the nth moment of a

random variable about the meanE[(xE[x])n]

Important: the second moment

about the mean E[(xE[x])2

] =2

=Var[x]

where

the variance (Varianz) of x Var[x] = measures of the spreadof probability distribution

= standard deviation (Standardabweichung) of the


ran om var a e


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Expectation and Moments

Other higher order of moments - measures of probability of

distributions

Skewness (Schiefe) - measures if the distribution is symmetric

Mode

Mean

Skewed Positively Skewed Negatively

Kurtosis (Wlbung) - measures flatness or peakedness

Peaked long thin tailsFlat (with short broad tails)



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For two random variables x and y:

Cov[x, y] = E[(xE[x]) (yE[y])]

Causal relationship

, =

Formally, p(y|x) = p(y) for discrete

y x = y or con nuous

Measure of dependencecorrelation coefficient,

[-1,1]

Var[x] Var[y]= Cov[x,y]



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Functions of Random Variables and

e r roper es

E[x + y] = E[x] + E[y] x + y is a RV

E[x + k] = E[x] + k x + k is a RV

w ere s an ar rary cons an

Properties for Variances Var[x + y] = Var[x] + Var[y] + 2Cov[x, y]

If x, y are independent, Var[x + y] = Var[x] + Var[y] Var[kx] = k2 Var[x]

Var[x + k] = Var[x]

06.11.2013 Rechnergesttzte Netzanalysen 15 Var[kx + ny] = k2

Var[x] + n2

Var[y] + 2kn Cov[x, y]


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Example :Constant100 3s

Constant

No uncertaintyegenera e case No variation (sd = 0)

r a s



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Consider an experiment consisting of n trials, eachcan e a success or a a ure.

Xj= 1 if the jth experiment is a success

Xj= 0 if the jth experiment is a failure

Jacob Bernoulli (16541705)The Bernoulli distribution (one trial):

p,

xj =1,j=1,2,...,n

pj xj =p xj = 1 p = q,

0,xj = 0,j=1,2,...,notherwise whereE(Xj) = p and V(Xj) = p (1p) = p q



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Let N = # trials, p = probability of each success trial

Example: N=7, p=0.5 X = # of success

X=5

X=4

X=4

...e c.

The mean,E(x) = p + p + + p = np

The variance,V(X) = pq + pq + + pq = npq


9


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pro uc on process manu ac ures mac nes on e average a

nonconforming. Every day a random sample of size 50 is taken from the process. If

the sample contains more than 2 nonconforming machines , the process will be.

Question: what is the probability that the process is stopped by the sampling

scheme?

Solution: P(X > 2) = 1 P( X2)

the probability P(X2) is calculated from:

P( X2) = 0.92

Thus, the probability that the production process is stopped on any day, based on

, . . .

the mean number of nonconforming and variance machines in a random sample

of size 50 is:


= np = . = , = npq = . . = .


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e geome r c s r u on s re a e o a sequence o ernou r a s; e ran om

variable of interest , X, is defined to be the number of trials to achieve the first

success. The distribution of X is given by

p(x) = qx1p, x = 1, 2,

the event {X = 0} occurs when there are x 1 failures followed by a success. Eachof the failures has an associated probability of q = 1 p, and each success has a

robabilit . Thus

P(FFFFS) = qx1p

the mean and variance are given by

= p

and

V(X) = q/P



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Forty percent o t e assem e mac ines are rejecte at t e

inspection station.

ues on: n e pro a y a e rs accep a e

machines is the third one inspected.

consider each inspection as a Bernoulli trial with q = 0.4 and p

= . ,

P(3) = 0.4(0.6) = 0.096

,machines the third one from any arbitrary starting point.



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Gives (probability of) the number of events that occur in

a given period

Formula looks quite complicated (and NOT discrete), but

it is discrete and using it is not that difficult

X {0,1,...,t}

otherwise0x!

e p(x)=

- x

Poisson

(Simon Denis, Fr. 17811840 ) Whereis the mean arrival rate. Note thatmust bepositive E(X) = V(X) =Das Bild kann zurzeitnicht angezeigtwerden.

Applications of Poisson Distribution

Discrete distribution, used to model the number of

independent events occuring per unit time,Eg. Batc s zes o customers an tems

If the time betweeen successive events is exponential,

en e num er o even s n a xe me n erva s

poisson.06.11.2013 Rechnergesttzte Netzanalysen 23


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r ng mac ne repa rman s ca e eac me ere s a ca or serv ce. e

number of calls per hour is known to occur in accordance with a Poisson

distribution with a mean of 2 per hour. Question: T e pro a i ity o t ree ca s in t e next our?

Solution: p(3) = e22/3! = (0.135)(8)/6 = 0.18

Question: determine the probability of two or more calls in

one hour eriod. P 2 or more = ?



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Normal

BinomialBernoulli

Poisson



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the interval (a,b),U(a,b), if its pdf and cdf are:

0,

f(x)= ba

, axb

otherwise

0,x F x =

x a

ax b

The uniform distribution plays a vital role

a1,b a

xb

in simulation. Random numbers, uniformly

distributed between 0 and 1, provide the

means to generate random events.

E(X) = (a+b)/2

V(X) = (ba)2/12



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terminal MArea beginning at 6:50 P.M. until

7:50 P.M. A certain passenger does not know

(uniformly distributed) between 7:00 P.M.

and 7:30 P.M. every Wednesday evening.

than 5 minutes for a bus?

X is uniform random variable on (0,30).

The desired probability is given by

F(15)F(5) + F(30)F(20) = 15/305/30 + 120/30 = 2/3


8


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exponential

x x

CDF F(x) = f(y)dy = e

y

dy=1 ex

E[X]=1 ;

Var(X)= 1 ..

2

0 0

Applications of Exponential Distribution:Used to model time between independent

, .

model service times that are highly variable.

Inappropriate for modeling process delay

Exponential

Memoryless property: P{X > s + t|X > t}= P{X > s} t,s 0.

.



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Example of the use of the memoryless property

of Exponential Distribution

A queueing system has two servers. The

service times are assumed to be exponentiallydistributed (with the same parameter). Upon

occupied () but there are no other waiting.



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e e o a g u s g ven y , w c s sa o ave an exponen a

distribution with mean 2 years. Its pdf is shown below:

Question: whats the probability that the life of the light bulb is between 2 and 3

years? Solution:

Question:What is the probability that it lasts for another year if it has already.



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Agner Krarup Erlang

The Erlang distribution is the distribution of the sum of k independent identically

distributed random variables each having an exponential distribution. Events

which occur independently with some average rate are modeled with a Poisson

process. The waiting times between k occurrences of the event are Erlang

distributed.

PDF:

the shapek, which is a nonnegative integer, and the rate, which is anon

negative real number.



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we ng mac ne as wo we ng ea s. e we ng ea s w sw c e

current to the second welding head if the first fails. The life expectancy of the

welding heads is exponentially distributed with average life 1000 hours. Question: W at is t e pro a i ity t at t e mac ine sti wor s a ter 90 days.

The probability that the system will still operate at least x hours is called the

reliability function R(x), where

R(x) = 1 F(x)

Solution: the total system lifetime is given k = 2 welding head, and = 1/1000.

F 2160 = 0.636

therefore, the chances are about 36% that the machine still works after 90 days.



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normal

E[X]=, Var(X)=2.Johann Carl Friedrich Gauss

Normal

anon ca : mean = , = . f( -x)=f(+x); the pdf is symmetric about . The maximum value of the pdfoccurs at x = .

.

Shape determined by its standard deviation .06.11.2013 Rechnergesttzte Netzanalysen 34


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rans orma on o var a es: e = ,



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e t me requ re to oa an oceango ng vesse , , s str ute as

N(12,4).

Question: What is the robabilit that the vessel is loaded in less than 10hours?

Solution:



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The random variable X withprobability density function

xex x = x

1 Waloddi Weibull

18 June 188712 October 1979

for x > 0

s a e u ran om var a e w

scale parameter > 0 and shapeparameter > 0.By inspecting the probabilitydensity function, it is seen thatw en = , e e udistribution is identical to theexponential distribution.

The effect of the Weibull shape parameter on thepdf.

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Weibul distribution is often used to modelthetime until failureof many different physicalsystems.

Distribution parameters provide a great deal offlexibility to model systems in which the numberof failures:

Increases wit time e.g., earing wear .

Decreases with time (some semiconductors). Remains constant (failures caused by external

s oc s to t e system .

with < 1exhibit a failure rate that decreases withtime,with = 1have a constant failure rate(consistent with the exponential distribution) andpopulations with > 1have a failure rate that

The effect of on the Weibull failure rate function..


8



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9



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T e time to ai ure or a e ectronic evice is nown ave a

Weibull distribution with= 1/3, and = 200 hours (time.

What is the probability that it fails before 2000 hours?

Solution:

= = = F(2000) = 1 exp[(2000/200) ] = 0.884



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A distribution whose parameters are the

observed values in a sample of data.May be used when it is impossible or unnecessary

particular parametric distribution

values in the sample.

sa van age: samp e m g no cover e en rerange of possible values.



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The world that the simulation analyst sees is

probabilistic, not deterministic. Reviewed several important probability

probability distributions in a simulation

context.

Difference between discrete continuous andempirical distributions.


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Vorlesung02 Statistical Models in Simulation