STATISTICAL MODELS IN SIMULATION - CECS -...

2/08/20072/08/2007 ENGN8101 Modelling and OptimizationENGN8101 Modelling and Optimization 11

STATISTICAL MODELS IN SIMULATION

How is probability and variation related to modelling of system performance?

Random variable– any system variable, X, that can

take different values- continuous or discrete

i.e. X has a range of values


How are the values within this range distributed?

- Described using Probability FunctionsProbability Functions

Probability functions – used to define probabilities of events associated with a random variable

Often mathematical functions, or graphical in nature


Discrete variables

X – described by a probability mass function P(x) – not P(X)

why? So P(x) = P(X = x)

Cumulative distribution function: F(x) = P(X≤x)

F(x) = step function bounded by 0 and 1

Xx ⊂


QA properties is considering making an offer to purchase an apartment building. Management has subjectively assessed a probability distribution for x, the purchase price: x P(x) $148000 .20 $150000 .40 $152000 .40

a) Find the cumulative distribution function F(x). What is F(152000)? b) What is the probability that the apartment house can be purchased for $150000 or less?

Problem to be solved:


Continuous variables

Here – need a probability density function (pdf)

pdf = P(x ≤ X ≤ x + dx)/dx

e.g.

More generally:

F(x) = P(X ≤ x)

∫∞−

=x

dxxf )(

1)( == ∫∞

∞−

dxxf


Probability Density FunctionProbability Density Function


Descriptive parameters

Mean value = Expected value

Discrete:

Continuous:

Variance:

Discrete:

Continuous

∫

∑∞

∞−

==

==

dxxfxXE

xPxXEiall

ii

)()(

)()(

μ

μ

222

222

)()(])[(

)()(])[(

σμμ

σμμ

=−=−

=−=−

∫

∑∞

∞−

dxxfxXE

xpxXE

i

xallii

i


Median (xm) P(X>xm) = 0.50

Mode: occurs where the density function has its peak

Skewness: θ = E([X-μ]3)/σ3

What would the 4th moment describe?

CASE STUDY 8CASE STUDY 8The number of signals arriving at a satellite monitoring system in any hour is defined by a random variable X. The probability mass function of X is believed to be

a) compute the constant c b) compute the mean and standard deviation of X c) compute the probability that the number of signals detected by the system in

any hour is less than or equal to 2

elsewhere0)(

4and3,2,1,0for1

)( 2

=

=+

=

xp

xx

cxp


a) For all x, P(x) = 1

So

Giving c = 0.538

b) Mean =

= 0.772

Similarly – variance = 1.094 giving σ = 1.05

c) P(X≤2) desired = P(X=0)+ P(X=1)+ P(X=2)= 0.538(1 +0.5 + 0.2) = 0.915

1116

119

114

111

110

1=⎥⎦⎤

⎢⎣⎡

++

++

++

++

+c

1 2 3 4 52 2 2 2 21 2 3 4 5

1 1 1 1 1( )1 1 1 1 1

i

i iall x

x P x c x x x x xx x x x x

⎡ ⎤= + + + +⎢ ⎥+ + + + +⎣ ⎦

∑


Glazer’s Winton Woods apartment complex has 80 two-bedroom apartments. The number of apartment air conditioner units that must be replaced during the summer season has the probability distribution shown below:

Air conditioner Probability Replaced 0 0.30 1 0.35 2 0.20 3 0.10 4 0.05

1)What is the expected number of airconditioner units that will be replaced during a summer season?

2)What is the variance in the number of air conditioner replacements?

3) What is the standard deviation?


Common probabilistic functions

A random variable – often takes values that follow a probability ‘trend’

If they follow a numerical pattern – can be modelled easily using distribution functions

DISCRETE

Hypergeometric

Binomial

Poisson

CONTINUOUS

Gaussian

Exponential

Weibull


HYPERGEOMTRIC

Applicable to sampling a population without subsequent replacement of sample

For D non-conforming samples in a population N,

The probability of getting x non-conforming items in a sample of size n is

)!(!!meanswhere)(

xDxD

xD

nN

xnDN

xD

xP−⎟⎟

⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

⎟⎟⎠

⎞⎜⎜⎝

⎛

=

⎟⎠⎞

⎜⎝⎛

−−

⎟⎠⎞

⎜⎝⎛ −==

11VarianceMean

NnN

ND

NnD

NnD


CASE STUDY 9CASE STUDY 9

Here, N = 20, D = 5, n = 4, x = 3

Probability of 3 non-conformers:

031.0

420

115

35

)3( =

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

=P

A batch of 20 transistors is known to contain 5 non-conforming ones. If an inspector randomly samples 4 items, find the probability of picking out 3 non-conforming ones


BINOMIALSeries of independent trials – each trial gives ‘yes’ or ‘no’

Probability of success = p = constant for any trial

Probability of x successes in n trials:

Mean = np, variance = np(1-p)

Uses:Sampling without replacement from large populations, orSampling with replacement from small populations

As N →∞ Hypergeometric → Binomial

xnppxn

xP −−⎟⎟⎠

⎞⎜⎜⎝

⎛= )1()( 3



Here, n = 5, x = 2 and p = 0.05

(if success is defined as finding a dud filtered signal)

Additionally, mean and variance of the distribution:

021.0)95.0()05.0(25

)2( 32 =⎟⎟⎠

⎞⎜⎜⎝

⎛==XP

25(0.05) 0.25 (1 ) 5(0.05)(0.95) 0.2375np and np pμ = = = σ = − = =

A signal filtering device is known to be 95% successful. If a random sample of 5 filtered signals is chosen, find the probability that 2 of them have not been correctly processed.


At a particular university it has been found that 20% of the students withdraw without completing the introductory statistics course. Assume that 20 students have registered for the course this semester.

1) What is the probability that two or fewer will withdraw?

2) What is the probability that exactly four will withdraw?

3) What is the expected number of withdrawals?



POISSON RANDOM VARIABLEPOISSON RANDOM VARIABLE

Named after Simeon D. Poisson (1781Named after Simeon D. Poisson (1781--1840)1840)Originated as an approximation to binomialOriginated as an approximation to binomialUsed extensively in stochastic modelingUsed extensively in stochastic modelingExamples include:Examples include:

Number of phone calls received, number of messages Number of phone calls received, number of messages arriving at a sending node, number of radioactive arriving at a sending node, number of radioactive disintegration, number of misprints found a printed disintegration, number of misprints found a printed page, number of defects found on sheet of processed page, number of defects found on sheet of processed metal, number of blood cells counts, etc.metal, number of blood cells counts, etc.


POISSON

Models the number of occurrences of an event over timeor space, or volume…

Events = random and independent

Uses: number of non-conformities in a productnumber of machine breakdowns per month

here λ = average no. of events over specified time period

mean = variance = λ

If n →∞ and p → 0, then Poisson → binomial

!)(

xexp

xλλ−=



Here, 1 unit is now 40m2

so λ is now 6

We need: P(X≤2) = P(X=0) + P(X=1) + P(X=2)

The mean and variance of this distribution are both 6

062.0!26

!16

!06 260606

=++=−−− eee

It is estimated that the average number of surface defects in 20m2 of paper produced by a process is 3. What is the probability of finding no more than 2 defects in 40m2 of paper through random selection?

Poisson approximation will be good as long as p≤0.05 and n>20.


A certain restaurant has a reputation for good food. Restaurant management boasts that on a Saturday night groups of customers arrive at the rate of 15 groups every half hour.

1) What is the probability that 5 minutes will pass with no customers arriving?

2) What is the probability that 8 groups of customers will arrive in 10 minutes?

3) What is the probability that more than 5 groups of customers will arrive in a 10-minute period of time?



GAUSSIAN (normal)

The most important continuous distribution in The most important continuous distribution in probability and statisticsprobability and statisticsThe story of the outcome of normal is really the story The story of the outcome of normal is really the story of the development of statistics as a science.of the development of statistics as a science.Gauss discovered this while incorporating the method Gauss discovered this while incorporating the method of least squares for reducing the errors in fitting of least squares for reducing the errors in fitting curves for astronomical observations.curves for astronomical observations.


GAUSSIAN (normal)

Most widely used distribution for continuous random variables

- “Natures’ Distribution” –

For a population mean = μ and variance = σ2

Probability density function for x =

i.e. open-endedbell curve

∞<<∞−⎥⎦

⎤⎢⎣

⎡ −−= xxxf 2

2

2 2)(exp

21)(

σμ

πσ


Graphs of various normal PDFGraphs of various normal PDF


Often standardized such that σ2 = 1 and μ =0

Here - Z = standardized random variable and

Alternatively

Note

and due to symmetry

∞<<∞−⎥⎦

⎤⎢⎣

⎡−= xzzf

2exp

21)(

2

π

∫∞−

⎥⎦⎤

⎢⎣⎡−=≤=Φ

z

dzzzZPz 2

21exp

21)()(π

)(1)(5.0)0(

zz Φ−=−Φ=Φ


Impossible to document every σ/μcombination

∴standardization required for easy tabulation and reference

Tabulated values –

give areas under the curve –hence – probabilities

σμ−

=Xz


a)

P(X>103.3) = P(X>1.65)

From tables: P(z≤1.65) = 0.9505

So P(z>1.65) = 1-0.9505 = 0.0495

4.95% of the parts will be above 103.3mm

65.12

1003.10311 =

−=

−=

σμXz

CASE STUDY 12CASE STUDY 12The length of a machined part is known to have a normal distribution with a mean of 100mm and a standard deviation of 2mm

a) What proportion of the parts can be expected to be over 103.3mm in length? b) What proportion will be between 98.5mm and 102.0mm?


b)

We needP(98.5≤X≤102.0)

now

From tables: 1.00 → 0.8413 and -0.75 → 0.2266

∴answer = 0.8413 – 0.2266 = 0.6142

61.47% of the output lies in the specified range

75.02

1005.9800.12

1000.10221 −=

−==

−= zz


EXPONENTIAL

Main use – reliability analysis

e.g. time to failure of a system entity

Here λ = failure rate

pdf:

i.e. failure most likely at t=0 (switching on)

Mean = 1/λ variance = 1/λ2

xexf λλ −=)(


Most important facet – memorylessmemoryless distribution

i.e. no reliance on what has occurred before

e.g. Markov chains in simulation and modelling- also memoryless- define probability of state change

Solution – very simple: failure rate = 1/500

P(X≤200) = 1 – e-(1/500)200 = 1 – e-0.04 = 0.330


It is known that the battery for a video game has an average life of 500 hours. The failures of batteries are known to be random, independent and exponentially distributed. What is the probability of a battery failing within 200 hours?


WEIBULL

Main use – reliability and failure analysis

γ = location parameter α = scale parameterβ = shape parameter

Very generic!

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −

−⎟⎠⎞

⎜⎝⎛ −

=− ββ

αααβ yxyxxf exp)(

1


Joint PDFJoint PDF

So far we saw one random variable at a time. So far we saw one random variable at a time. However, in practice, we often see situations where However, in practice, we often see situations where more than one variable at a time need to be studied.more than one variable at a time need to be studied.For example, tensile strength (X) and diameter(Y) of For example, tensile strength (X) and diameter(Y) of a beam are of interest.a beam are of interest.Diameter (X) and thickness(Y) of an injectionDiameter (X) and thickness(Y) of an injection--molded molded disk are of interest.disk are of interest.


Joint PDF (ContJoint PDF (Cont’’d)d)X and Y are continuousX and Y are continuous

f(x,y) f(x,y) dxdx dydy = P( x < X < = P( x < X < x+dxx+dx, y < Y < , y < Y < y+dyy+dy) is ) is the probability that the random variables X will the probability that the random variables X will take values in (x, take values in (x, x+dxx+dx) and Y will take values in ) and Y will take values in ((y,y+dyy,y+dy).).f(x,y) f(x,y) > 0 for all x and y and> 0 for all x and y and ∫∫

∞

∞−

∞

∞−=1),( dydxyxf

∫∫=<<<<d

c

b

adydxyxfdYcbXaP ),(),(


Measures of Joint PDFMeasures of Joint PDF

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