CPSC 531:Input Modeling1 Instructor: Anirban Mahanti Office: ICT 745 Email: [email protected]...

CPSC 531:Input Modeling 1

CPSC 531:Input ModelingInstructor: Anirban MahantiOffice: ICT 745Email: [email protected] Location: TRB 101Lectures: TR 15:30 – 16:45 hoursClass web page:

http://pages.cpsc.ucalgary.ca/~mahanti/teaching/F05/CPSC531

Notes from “Discrete-event System Simulation” by Banks, Carson, Nelson, and Nicol, Prentice Hall, 2005, and

“Simulation Modeling and Analysis” by Law and Kelton, McGraw Hill, 2000.

http://pages.cpsc.ucalgary.ca/~mahanti/teaching/F05/CPSC531


Outline Quality of output depends on the input

models driving the simulation This module discusses the following:

Data collection from the real system Hypothesize probability distributions Choose parameters for the distributions Goodness of fit test – how well does the fitted

distribution model available data Selecting distributions in absence of data Models of arrival process (Poisson Process, Non-

stationary Poisson Process, Batch Arrivals)


Data Collection Plan ahead: begin by a practice or pre-observing

session, watch for unusual circumstances Analyze the data as it is being collected: check

adequacy Combine homogeneous data sets, e.g. successive

time periods, during the same time period on successive days

Be aware of data censoring: the quantity is not observed in its entirety, danger of leaving out long process times

Check for relationship between variables, e.g. build scatter diagram

Check for autocorrelation Collect input data, not performance data


Identifying Probability Distributions Several possible techniques (may use a

combination of these) Prior knowledge of the random variable’s role

• Inter-arrival times are exponential if arrivals occur one at a time, have constant mean rate, and are independent

• Service times are not normally distributed because service time can not be negative

• Product of many independent pieces may imply Lognormal

Use the physical basis of the distribution as guide Summary statistics Histograms


Distribution Guide Use the physical basis of the distribution as a guide,

for example: Binomial: # of successes in n trials Poisson: # of independent events that occur in a fixed

amount of time or space Normal: dist’n of a process that is the sum of a number of

component processes Exponential: time between independent events, or a

process time that is memoryless Weibull: time to failure for components Discrete or continuous uniform: models complete

uncertainty Triangular: a process for which only the minimum, most

likely, and maximum values are known Empirical: resamples from the actual data collected


Summary Statistics

LK00in 6.5 Table See

geometricor binomial negative candidate :1

Poisson candidate :1

binomial candidate:1

);(X /)(by / ratio lexis estimate ons,distributi discreteFor

1parameter shape with or Weibull gamma are candidates :1cv

lexponentia candidate :1 cv

1parameter shape with or Weibull gamma are candidates :1 cv

);(X /)(by / estimate ons,distributi continuousFor

close aremedian sample and )(X ifon distributi Symmetric

)(X :mean Sample

22

^

nnS

nnScvcv

n

n


Histograms A frequency distribution or histogram is useful

in determining the shape of a distribution The number of class intervals depends on:

The number of observations The dispersion of the data Suggested: the square root of the sample size

For continuous data: Corresponds to the probability density function of a

theoretical distribution For discrete data:

Corresponds to the probability mass function If few data points are available: combine

adjacent cells to eliminate the ragged appearance of the histogram


Histogram

0

0.05

0.1

0.15

0.2

0.1 0.6 1.1 1.6 2.1 2.6

Bin

Fre

qu

ency

Histogram for Continuous Data Sample n = 100

interarrival times of requests to a Web server during a 1-minute period (see Web page) Request arrival

approximately stationary – # of requests arriving in 10-second periods approx. equal

Sample mean = 0.534 s; median = 0.398; CV = 0.98

exponential distribution? Right-hand side shows two

histograms: top with interval or bin size of 0.1 s; bottom with bin size of 0.25 s.

Histogram

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.25 0.75 1.25 1.75 2.25 2.75

Bin

Fre

qu

ency


Histogram

0

0.04

0.08

0.12

0.16

0.2

0 1 2 3 4 5 6 7 8 9 10

x

h(x

)

Histogram for Discrete Data Sample n = 100 observations of the number of

items requested from a job shop per week over a long time period (# req., # observations): {(0,1), (1,3), (2,8), (3,14), (4,

18), (5,17), (6,16), (7,10), (8,8), (9,4), (10, 1)} Mean = 4.94, variance = 4.4, Lexis ratio = 0.9 Poisson distribution?


Parameter Estimation

Next step after selecting a family of distributions If observations in a sample of size n are X1, X2, …,

Xn (discrete or continuous), the sample mean and variance are:

If the data are discrete and have been grouped in a frequency distribution:

where fj is the observed frequency of value Xj

1 1

2221

n

XnXS

n

XX

n

i i

n

i i

1 1

22

21

n

XnXfS

n

XfX

n

j jj

n

j jj


Parameter Estimation

When raw data are unavailable (data are grouped into class intervals), the approximate sample mean and variance are:

where fj is the observed frequency of in the jth class interval

mj is the midpoint of the jth interval, and c is the number of class intervals

A parameter is an unknown constant, but an estimator is a statistic.

1 1

22

21

n

XnmfS

n

XfX

n

j jj

c

j jj


Histogram

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.25 0.75 1.25 1.75 2.25 2.75

Bin

Fre

qu

en

cy

How Representative are the fits? Continuous data – plot

over the histogram and look for similarities

Discrete data – compare the observed frequency with the expected frequency

Try a Quantile-Quantile Plot

bttxF |)(|

^

Histogram

0

0.04

0.08

0.12

0.16

0.2

0 1 2 3 4 5 6 7 8 9 10

x

h(x)

Fitted Dist

Observed


Quantile-Quantile Plots

Q-Q plot is a useful tool for evaluating distribution fit If X is a random variable with cdf F, then the q-quantile of

X is the such that

When F has an inverse, = F-1(q)

Let {xi, i = 1,2, …., n} be a sample of data from X and {yj, j = 1,2, …, n} be the observations in ascending order:

where j is the ranking or order number

, for 0 1F( ) P(X ) q q

1 0.5is approximately -

j

j -y F

n



The plot of yj versus F-1( (j-0.5)/n) is Approximately a straight line if F is a member of an

appropriate family of distributions The line has slope 1 if F is a member of an appropriate

family of distributions with appropriate parameter values



Example: Check whether the door installation times follows a normal distribution [BCNN05] The observations are now ordered from smallest to

largest:

yj are plotted versus F-1( (j-0.5)/n) where F has a normal distribution with the sample mean (99.99 sec) and sample variance (0.28322 sec2)

j Value j Value j Value1 99.55 6 99.98 11 100.262 99.56 7 100.02 12 100.273 99.62 8 100.06 13 100.334 99.65 9 100.17 14 100.415 99.79 10 100.23 15 100.47


Quantile-Quantile Plots [BCNN05]

Example (continued): Check whether the door installation times follow a normal distribution.

Straight line, supporting the hypothesis of a

normal distribution

Superimposed density function of

the normal distribution


Quantile-Quantile Plots [BCNN05]

Consider the following while evaluating the linearity of a q-q plot: The observed values never fall exactly on a straight line The ordered values are ranked and hence not

independent, unlikely for the points to be scattered about the line

Variance of the extremes is higher than the middle. Linearity of the points in the middle of the plot is more important.

Q-Q plot can also be used to check homogeneity Check whether a single distribution can represent both

sample sets Plotting the order values of the two data samples against

each other


Goodness-of-Fit Tests [BCNN05]

Conduct hypothesis testing on input data distribution using: Kolmogorov-Smirnov (KS) test Chi-square test

No single correct distribution in a real application exists. If very little data are available, it is unlikely to reject any

candidate distributions If a lot of data are available, it is likely to reject all

candidate distributions


Chi-Square Test [BCNN05]

Compare histogram of the data to the shape of the candidate distribution function

Valid for large sample sizes when parameters are estimated by maximum likelihood

Arrange the n observations into a set of k class intervals or cells, the test statistics is:

which approximately follows the chi-square distribution with k-s-1 degrees of freedom, where s = # of parameters of the hypothesized distribution estimated by the sample statistics.

k

i i

ii

E

EO

1

220

)(

Observed Frequency

Expected Frequency

Ei = n*pi

where pi is the theoretical prob. of the ith interval.

Suggested Minimum = 5


Chi-Square Test Null hypothesis – observations come from a specified

distribution cannot be rejected at a significance of α if:

Comments: Errors in cells with small Ei’s affect the test statistics more

than cells with large Ei’s.

Minimum size of Ei debated: [BCNN05] recommends a value of 3 or more; if not combine adjacent cells.

Test designed for discrete distributions and large sample sizes only. For continuous distributions, Chi-Square test is only an approximation (i.e., level of significance holds only for n->∞).

2]1,1[

20 sk Obtained from a table


Chi-Square Test Example 1: 500 random numbers generated using a random

number generator; observations categorized into cells at intervals of 0.1, between 0 and 1. At level of significance of 0.1, are these numbers IID U(0,1)?

Interval Oi Ei [(Oi-Ei)^2]/Ei1 50 50 02 48 50 0.083 49 50 0.024 42 50 1.285 52 50 0.086 45 50 0.57 63 50 3.388 54 50 0.329 50 50 0

10 47 50 0.18500 5.84

0.10. of level cesignificanat accepted Hypothesis

;68.14 table thefrom ;85.5 2]9,9.0[

20


Chi-Square test [BCNN05]

Example 2: Vehicle ArrivalH0: the random variable is Poisson distributed.

H1: the random variable is not Poisson distributed.

Degree of freedom is k-s-1 = 7-1-1 = 5, hence, the hypothesis is rejected at the 0.05 level of significance.

!

)(

x

en

xnpEx

i

x i Observed Frequency, Oi Expected Frequency, Ei (Oi - Ei)2/Ei

0 12 2.61 10 9.62 19 17.4 0.153 17 21.1 0.84 19 19.2 4.415 6 14.0 2.576 7 8.5 0.267 5 4.48 5 2.09 3 0.8

10 3 0.3> 11 1 0.1

100 100.0 27.68

7.87

11.62 Combined because of min Ei

1.1168.27 25,05.0

20


Chi-Square Test

If the distribution tested is continuous:

where ai-1 and ai are the endpoints of the ith class interval

and f(x) is the assumed pdf, F(x) is the assumed cdf. Recommended number of class intervals (k):

Caution: Different grouping of data (i.e., k) can affect the hypothesis testing result.

)()( )( 11

ii

a

ai aFaFdxxf pi

i

Sample Size, n Number of Class Intervals, k

20 Do not use the chi-square test

50 5 to 10

100 10 to 20

> 100 n1/2 to n/5


Kolmogorov-Smirnov (KS) Test Difference between observed CDF F0(x) and

expected CDF Fe(x) should be small; formalizes the idea behind the Q-Q plot.

Step 1: Rank observations from smallest to largest:Y1 ≤ Y2 ≤ Y3 ≤ … ≤ Yn

Step 2: Define Fe(x) = (#i: Yi ≤ x)/n Step 3: Compute K as follows:

}1

)(),({max

|)()(|max

1 n

jYFYF

n

jK

xFxFK

jejenj

oex


KS Test Example: Test if given population is exponential with

parameter β = 0.01; that is Fe(x) = 1 – e–βx; K[0.9,15] = 1.0298.

KS Test for exponential distributionbeta 0.01 N 15Y_j j (j/n)-F(Yj) F(Yj)-(j-1)/n

5 1 0.017896 0.0487716 2 0.075098 -0.008436 3 0.141765 -0.0751

17 4 0.110331 -0.0436625 5 0.112134 -0.0454739 6 0.077057 -0.0103960 7 0.015478 0.05118861 8 0.076684 -0.0100272 9 0.086752 -0.0200974 10 0.143781 -0.07711

104 11 0.086788 -0.02012150 12 0.02313 0.043537170 13 0.04935 0.017316195 14 0.075607 -0.00894229 15 0.101266 -0.0346

MAX 0.143781 0.051188


KS Test

KS test suitable for small samples, continuous as well as discrete distributions

KS test, unlike the Chi-Square test, uses each observation in the given sample without grouping data into cells (intervals).

KS test is exact provided all parameters of the expected distribution are known.


Selecting Model without Data If data is not available, some possible sources to

obtain information about the process are: Engineering data: often product or process has

performance ratings provided by the manufacturer or company rules specify time or production standards.

Expert option: people who are experienced with the process or similar processes, often, they can provide optimistic, pessimistic and most-likely times, and they may know the variability as well.

Physical or conventional limitations: physical limits on performance, limits or bounds that narrow the range of the input process.

The nature of the process.

The uniform, triangular, and beta distributions are often used as input models. [ see LK00 for details]


Models of Arrival Processes Poisson Process Non-stationary Poisson Batch Arrivals


Poisson Process

Definition: Let N(t) denote the number of arrivals that occur in time interval [0,t].

The stochastic process {N(t), t>=0} is a Poisson process with mean rate if: N(0) = 0 Arrivals occur one at a time {N(t), t>=0} has stationary increments – number of arrivals

in a given interval depends only on the length of the interval, not its location

{N(t), t>=0} has independent increments – number of arrivals in disjoint time intervals are independent.

And …

0}2)({

lim

}1)({lim

h

hNPh

hNP

oh

oh


Stationary & Independent

Memoryless

Poisson Process: Interarrival Times

Consider the interarrival times of a Possion process (A1, A2, …), where Ai is the elapsed time between arrival i and arrival i+1

The 1st arrival occurs after time t iff there are no arrivals in the interval [0,t], hence:

P{A1 > t} = P{N(t) = 0} = e-t

P{A1 <= t} = 1 – e-t [cdf of exp()] Interarrival times, A1, A2, …, are exponentially distributed and

independent with mean 1/Arrival counts ~ Poisson()

Interarrival time ~ Exp(1/)


Splitting: Suppose each event of a Poisson process can be classified

as Type I, with probability p and Type II, with probability 1-p.

N(t) = N1(t) + N2(t), where N1(t) and N2(t) are both Poisson processes with rates p and (1-p)

Pooling: Suppose two Poisson processes are pooled together N1(t) + N2(t) = N(t), where N(t) is a Poisson processes

with rates 1 + 2

Poisson Process: Splitting and Pooling

N(t) ~ Poisson()

N1(t) ~ Poisson[p]

N2(t) ~ Poisson[(1-p)]

p

(1-p)

N(t) ~ Poisson(12)

N1(t) ~ Poisson[]

N2(t) ~ Poisson[]

1

2


Non-stationary Poisson Process (NSPP) Poisson Process without the stationary increments,

characterized by (t), the arrival rate at time t. The expected number of arrivals by time t, (t):

Relating stationary Poisson process n(t) with rate and NSPP N(t) with rate (t): Let arrival times of a stationary process with rate = 1 be

t1, t2, …, and arrival times of a NSPP with rate (t) be T1, T2, …, we know: ti = (Ti)

Ti = (ti)

tλ(s)dsΛ(t)

0


Example: Suppose arrivals to a Post Office have rates 2 per minute from 8 am until 12 pm, and then 0.5 per minute until 4 pm.

Let t = 0 correspond to 8 am, NSPP N(t) has rate function:

Expected number of arrivals by time t:

Hence, the probability distribution of the number of arrivals between 11 am and 2 pm.

P[N(6) – N(3) = k] = P[N((6)) – N((3)) = k]= P[N(9) – N(6) = k]= e(9-6)(9-6)k/k! = e3(3)k/k!

84 ,5.0

40 ,2)(

t

tt

84 ,6

25.02

40 ,2)( 4

0 4

tt

dsds

ttt t

Non-stationary Poisson Process (NSPP)


Batch Process Let N(t) be the number of batches that have

arrived by time t. If interarrival times of batches are IID exponential

random variables, {N(t), t≥0} can be modeled as a Poisson process.

Let X(t) = total number of individual customers to arrive by time t; let Bi = number of customers in the ith batch; then

If Bi’s are IID random variables independent of {N(t) t≥0}, and if {N(t), t≥0} is a Poisson process, then the stochastic process {X(t), t≥0} is a Compound Poisson process

0,)()(

1

tBtX

tN

ii

CPSC 531:Input Modeling1 Instructor: Anirban Mahanti Office: ICT 745 Email: [email protected]...

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