3 - Simulation in Practice and Intro to Probability and Statistics

7/25/2019 3 - Simulation in Practice and Intro to Probability and Statistics

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Session #3 Agenda

Problem 2.4 Hoboken Police problem limited sourceShow that GPSS Solution gets same result (done last week)

Class Problems 4.2 ( Java City)Class Problem 4.7 (Ginos Pizza) ManualClass Problem 4.7 (Ginos Pizza) Automatic with VLOOKUP & Function

Class Problems 4.3 (Geophone) Simulation with ExcelManual Automatic without Outstanding Inventory Rule Automatic with Outstanding Inventory Rule (One Outstanding Order)

Class Problem 4.3 GPSSIntroduction to Probability & Statistics

Probability Mass and Density FunctionsCumulative Distribution Functions

Continuous Process GeneratorsInverse Transform Method


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Review Basic Single Server Model

M/M/1 Infinite Source, FCFS, ..Poisson Arrivals Result in Exponential Inter -arrivals (& Vice Versa)Text gives formulas for key statistics for several queuing casesWe derived mathematically P(n), and Lq for the Basic Single Server CaseWe showed that Poisson Arrivals have exponential inter-arrival times dist.Tonight we will introduce INVERSE transform method to derive CPG

Can use spreadsheets for simple queuing casesMore complex problems may require use of higher level simulation languagesWe looked at a GPSS model for Basic Single Server caseWe ran the GPSS model and plot the inter-arrival times, arrival rate distribution,and queue statistics.

What distribution did the arrival rate have when we used the exponential inter-arrival distribution at the generate block?

We then looked at problems with s servers and finite sources.We were given a formula for generating an exponentially distributed RV from auniform random numbers. Lets check it out with Excel now .


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Class Problem 4.2

The manager of Cafe Java is trying to determine whether to hireanother cashier for the morning coffee rush hour. Does she needone? Justify your answer using the simulation results obtainedfrom 5 customers (use the table below)! Use the following randomnumber streams:

Lets look at the spreadsheet


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Class Problem 4-7

Gino Petralli is an enterprising businessman. He is inthe process of starting up a Pizzeria. Specifically, heneeds to know if 10 tables in his dining area isenough. Further, he would like to ensure the averagetime a party must wait for a table is less than 2minutes. Do a manual simulation ofGinos Pizzeria to determineif the facility can meet Ginoscriteria with its current setup.

Assume there are 10 tables inthe pizzeria and at the beginningof the simulation, 8 of the tablesare occupied.


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Interarrivals... # in Group... Order Size... Consumption and conversationtime...

Interarrivals are exponentiallydistributed with a mean of 1/ = 10min/group. Thus, = 6 groups/hr.This results in the following processgenerator for interarrivals.

xr

hr ln ( )

( min/ )16

60

=4

.20

=3

.40

=2.20

=5

.10

=6

.10

#of Pizzas# in Group Ordered (x) p(x)

2 1 1.00 3 1 (or 2) .60 (.40) 4 2 1.00 5 2 (or 3) .20 (.80) 6 3 (or 4) .50 (.50)

Cooking Time...

The pizza oven is of the conveyor belt type. Itakes 10 m inutes to cook a pizza and pizzasust be spaced 2 minutes apart.

This process is uniformlydistributed (a,b) where a=10and b=50. The continuous process generator for auniform distribution is...

x = r 4(50 - 10) + 10

which generates eat times

Note: Also,assume a 5minute delay between the timea group is seateduntil the time its pizza(s) enters theoven.

Problem 4.7 (Ginos Pizza) )


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GINOs PIZZA Problem (4.7)

VLOOKUP & Function Approach for Queue LengthGPSS Solution


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Class Problem 4-3

The commercial off the shelve (COTS) ProgramManager (PM) for the Seawolf class attacksubmarine has asked you to exam their proposedinventory policy for passive sonar geophones(PSGs). The inventory policy must support the

Navys diverse missions. The propo sed inventorypolicy is to order 30 PSGs whenever the inventorydrops below 16 at the east coast depot.


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Class Problem 4.3 ManualProcess Generator 1: DEMAND (given)

x = 5/6 ln(r1) exp inter-arrival time with lambda = 6/5 orders per day

i.e 1/lambda - 5/6 days per order Inter-arrival time avg. (IAT avg.)

Process Generator 2: Number of PSGs in Order (Given)

Numberof PSGs Frequency

1 6

2 5

3 9

4 30

5 25

6 25

Continued ____


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Class Problem 4.3

Number of PSGs P(x) Cum Prob RN Range1 0.06 0.06 0


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Simulation Class Problem 4.3

r1 Days Between Clock Time r2demand r3

Le

adTime

Inve

ntor y

PlaceOrder

Orders Level

0 0.904 6 0.782 3.078 15 Yes

0.659 0.347526454 0.347526454 0.398 4 0.872 11

0.9 0.08780043 0.435326883 0.808 6 0.084 5

Orderarriv 3.07376

0.024 3.108084541 3.543411424 0.647 5 0.393 30

0.219 1.265569624 4.808981048 0.695 5 0.211 25

0.411 0.740968387 5.549949435 0.65 5 0.132 20

0.237 1.199745948 6.749695384 0.667 5 0.843 2.1349 15 yes

0.599 0.427078067 7.176773451 0.04 1 0.921 14

0.826 0.159300421 7.336073872 0.516 5 0.909 9

Order

arrives 39 8.88455


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Problem 4.3 Excel Solution

EXCEL MANUALEXCEL Automatic Without No Outstanding Order RuleEXCEL Automatic With Outstanding Order Rule


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Geophone Problem with GPSS

Run GPSS For Geophone Problem Policy 2 (1 OutstandingInventory Order at a time)


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Department of Systems Engineeringand Engineering ManagementCharles V. Schaefer Jr. School ofEngineeringStevens Institute of TechnologyHoboken, New Jersey 07030

Introduction toProbability and Statistics

Part I

0

0 .1

0 .2

0 .3

0 .4

-4 -2 0 2 4

x

f (

x )

N (0 ,1 )

a b x

f(x)

1 b - a

f(x)dx = b - a

b - a1

a

b


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Lesson Objectives

Lesson ObjectivesProbability & Statistics Review PDFs, CDFs, etc. Role of Continuous Process Generators Inverse Transform Method


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Role of Statistics in Modeling andSimulation

Uniform


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Definitions

Population Sample

Mythical - ExactWorld

Random - RealWorld

Probability

Statistics


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Process Generators

Continuous Process Generators

In many simulations, it is morerealistic and practical to usecontinuous random variables

- Computational it is more efficient- More representative of the

real world

Discrete Process Generators

Easier to explain andunderstand


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Discrete Process Generators

0

2

4

6

8

10

12

Frequency

b. Histogram

0 0.5 1.00.250

0.25

0.50

0.75

1

a. Sorted Observations

12

9

3

6

TimeBetween

TruckArrivals,

Hours

0

0.4

0.8

1.0

Cumulative

Probability

c. Cumulative Distribution Function

0 0.5 1.00.25

0.75

0.75

0.2

0.6

Uniform Time BetweenRandom Truck Arrivals

Variable (hours)

0 < r < 0.40 0.250.40 < r < 0.70 0.500.70 < r < 0.80 0.750.80 < r < 1.00 1.00

d. Process Generator

Time Between Truck Arrivals, Hours

12/30

21/3024/30

30/30

Time Between Truck Arrivals, Hours

10 20 30


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Role of Statistics in Modeling andSimulation

Frequency tables or histograms suppress much detailIdealized mathematical representations are needed

real world is not discrete mathematically efficient

The curve describing the shape of the distribution is calleda frequency curve

0

0.1

0.2

0.3

0.4

5 6 7 8 9

Interarrival Time, Students/Hour

f (

x )

Area UnderCurve Must Be

Equal to 1


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Probability & Statistics Review


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Random Variable

Probability mass functionDiscrete

P(X = x i) = p(x i)

p(x i) = 1


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Random Variable

Probability density functionContinuousf(x) = e x x > 0 (example)P(X = a) = 0

- f(x) dx = 1P(a < x < b) = a b f(x) dx


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Probability Density Function

In mathematics, a probability density function (PDF)serves to represent a probability distribution in terms ofintegrals

If a probability distribution has density f(x): The infinitesimal interval [x, x + dx] has probability f(x) dx It can be seen as a "smoothed out" version of a histogram

24


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( ) 0

( ) 1

p x fo r a ll x

p x

( ) 0

( ) ( ) ( )

( ) ( ) 1

b

a

P x a

P a x b p x dx area under p x

P x p x dx

a b

p(x)

x


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Random Variable

Cumulative distribution function (CDF)

F(X) = P(x


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Cumulative Distribution Function

The cumulative distribution function (CDF) for a randomvariable X is the probability that the random variable is lessthan or equal to a specific value x

x

PDFdt CDF

0

27


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Random Variable

Expected value

= E(x)= xi p (x i)= x f(x) dx


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Probability Density FunctionExample


0

0.001

0.002

0.0030.004

0.005

0.006

0.007

0.008

0.009

1 15 29 43 57 71 85 99 113 127 141 155 169 183 197 211 225 239 253 267 281 295

29


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Cumulative Distribution FunctionExample


0

0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

1 15 29 43 57 71 85 99 113 127 141 155 169 183 197 211 225 239 253 267 281 295

Cumulative Distribution Function

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1 18 35 52 69 86 103 120 137 154 171 188 205 222 239 256 273 290

0.999843268881617

x

PDFd t CDF

0

PDF = f(t)

30


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Some UsefulContinuous

Probability Distributions

Uniform Distribution

Triangular Distribution

Normal Distribution

Exponential Distribution

31


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Continuous Distributions

)/(1)( ab x f

Discrete analogy of the continuous uniform distribution

Characterizes a random variable for which all integer outcomes between someminimum value and maximum value are equally likely

The density function is

if

and F(X) =

Mean = (a+b)/2

Uniform Distribution

b xa

a b x

f(x)

1b - a

f(x)dx =b - ab - a

1a

b

32


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Continuous Distributions

Triangular Distribution

33

Defined by three parameters: the minimum, a; maximum, b; and most likely, c

The density function is

2(x - a) if a


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Random Variable

Variance

))((

))((2

2

22

22

2

)(

]2[

])[()(

x p x x

x E x

ii xi p

E

x x E

x E xV

i


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Random Variable

Standard deviation

Sums of R.V.

when x 1 and x 2 are independent random variables

)()( X V X SD

)()()(

)()()(

2

2

21

2

1

2211

2211

xV a xV aY V

x E a x E a y E

xa xaY


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Random Variable

n

X X SampleMean

i

11

)(222

2 n

n

n

X X anceSampleVari x xS

ii


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AS IN THE DISCRETE METHOD, WEFIRST GENERATE A U(0,1) RANDOMNUMBER AND THEN TRANSFORM ITINTO A RANDOM VARIATE FROM THE

SPECIFIED DISTRIBUTION.

Two methods to transform:- Inverse Transformation Method- Acceptance-Rejection Method


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The ITM is generally used fordistributions whose cumulativedistribution function can be obtainedin closed form.

STEP 1: Obtain CDF from PDF(interject X into the equation)

STEP 2: Set F(x) = r

STEP 3: Solve for x(as a function of r)

We did this last week for the exponential distribution of arrival times


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IVT for Uniform Distribution

[ . ]

( ) / ( ) / ( ) ] / ( ) / ( ) 1

( ) 0

( )

'

( ) ( ) / ( ) / ( )] ( ) / ( )

'

( ) ( ) / ( )

b

a

X X

X a

a

U N IF O R M a b

Check this is a pdf

p x dx b a x b a b b a a b a

and p x

p x is th us a pdf

Let s find CD F

F t f t dt d t b a t b a x a b a

Let s find CP G

r F x x a b a

( ) ( )

( )

r b a x a

x a r b a

Converts uniform random numbers from U(0,1) to values of the random with distribution U(a,b)

h=1/(b-a)

p(x)

a b


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Upper Ramp Example

2

2 2

2 2 2 2 2 2

2 2 2 2 2 2 2 2 2

( ) 2 ( ) / ( )

( ) 2 ( ) / ( ) 2 / ( ) ( )

2 / ( ) [ / 2 ] (2 / ( ) )[ / 2 / 2 ]

(1 / ( ) )[ 2 2 ] (2 / ( ) [ 2 ] ( ) / ( )

1, ( ) 0, ( )

b b

a a

ba

p x x a b a

p x dx x a dx b a b a x a dx

b a x ax b a b ab a a

b a b ab a a b a b ab a b a b a

p x is G E p x is a p

robabi li ty dis tr ibution

1

0

2

2

2 / ( )

( , 2 / ( ))

( , 0 )

(2 / ( )) / ( ) 2 / ( )

( ) (2 / ( ) )( )

h b a

y b b a

y a

m b a b a b a

p x b a x a

ab

2/(b-a) (b,2/(b-a))

P(x)

2

2 2 2

2 2 2 2

2 2

2 2 2

2

( ) 2 ( ) / ( )

( ) ( ) 2 ( ) / ( ) 2 / ( ) [ / 2 ]

2 / ( ) [ / 2 / 2 ]

/ ( )[( / 2 ) / 2 ]

2 ( ) / 2 ( )

( ) 2 ( ) /

2 ( ) /

x x xa

f t t a b a

F x f t dt t a b a b a t at

b a x ax a a r

r h b a x ax a

b a r h x ax a x a

x a b a r h

x a b a r h

Find CDF CPG:

Show p(x) is a pdf:

=CDFubstituting h=2/(b-a):

=CPG

0 1r

0 1 for r


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Distribution Diagram PDF CDF

Process

Generator

Uniform

Ramp - up

a b

f(x ) =

1

b - a a x b

0 Otherwise

F (x ) =

x - a

b - a

a x b

0 Otherwise

x = a + ( b - a ) r

0 r 1

a b

f (x ) =2 (x - a )( b - a )

a x b

0 Otherw ise

2 F(x ) =

( x - a )

( b - a ) a x b

0 Otherwise

2

2


10

)(2

r

h

abr a x


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Ramp-Up

&Ramp -Down

a m b

f (x ) =

2 (x - a )

( b - a )( m - a ) a x m

2 ( b - x )

( b - a ) ( b - m ) m x b

0 Otherwise

f (x ) =

( x - a )

( b - a )( m - a ) a x m

( b - x )

( b - a ) ( b - m ) m x b

0 Otherwise

2

2

1

x = a+ r(b-a)( )

0 r ( ) / ( )

x = b - (1 -r)(b -a )( b m )

( ) / ( ) r 1

m a

m a b a

m a b a

Distribution Diagram PDF CDF ProcessGenerator

A

A=1/2base*height=(m-a)*(2/b-a)/2= (m-a)/(b-a)


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Ramp - Down

Exponential

Distribution Diagram PDF CDFProcess

Generator

a b

f (x ) =

2 (b - x )

(b - a ) a x b

0 Otherwise

2

F (x ) =

( b - x )

(b - a ) a x b

0 Otherwise

2

21

x = b - (b - a)

r 1

2 ( )1

0

r

f (x ) = e x 0, 0

0 Otherwise

- x

F (x ) =

1 - e x 0

0 Otherwise

- x

x = -1

ln r

0 r 1


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Questions?

Documents

3 - Simulation in Practice and Intro to Probability and Statistics