Transcript AsefCh01bSolutions

Problem
A project has three sequential tasks. The first task follows
Exponential distribution with parameter of 5 days. The second task
follows Uniform distribution with parameters of 5 and 10 days, the
last task follows Normal distribution with parameters of 9 and 2
days. The numbers on the next page were generated using RAND()
function in excel. Starting from the first row and column, moving
through the first row first, and not using any given element twice,
generate a random instance of duration of this project.
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Problem
0.1
0.5
0.7
0.9
0.8
0.1
0.3
0.2
0.4
0.1
0.4
0.4
0.0
0.2
0.6
By not using any element twice we mean not using a specific
element – in a specific row and column twice.
For example you can not use the number in row 1 and column 1
twice. Obviously you can use all the three 0.1s as soon as you
reach their corresponding box in the table.
End of the book Table provide you with uniform [0,1] random
numbers
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Continuous Probability Distributions
Uniform Distribution
a
Exponential Distribution
b

Normal Distribution
s

Uniform Probability Density Distribution
b
a
Uniform Probability Density Function
f (x) = 1/(b - a) for a < x < b
= 0 elsewhere
a = smallest value the variable can assume
b = largest value the variable can assume
µ = (a + b)/2
s 2 = (b - a)2/12
Continuous Probability Distributions
a
x1 x2
P(x1 ≤ x≤ x2)

b
a
x1

P(x≤ x1)
P(x≥ x1)= 1- P(x<x1)
a
x1

P(x≥ x1)
b
P(x =x1)= 0
b
Generate a Uniform Random Number
We want to select a student, randomly, to come
to the white board.
How?
Simulation is based on random number
generation
Crystal Ball is a simulation software
AutoMod is another simulation software
Excel is also a simulation tool
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Random Variable: Uniform Distribution
The second task follows Uniform distribution with parameters of 5 and
10 days. The random number we have is the second number in the first
row, and that is 0.1. For uniform distribution
X  a  (b  a) Rand ()
X  5  0.1(10  5)  5.5
The second activity takes 5.5 days
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Exponential Probability Distribution
f ( x)  e
F (x )
 x
.4
.3
P(x < x1 ) = area from 0 to x1
P( x  x1 )  1  e
.2
.1
 x1
x
1
2
3
4
5 6
7
8
Parameters: Mean and StDev.
How can I generate an exponential random
variable?
9 10
Random Variable: Exponential
Distribution
The first task follows exponential distribution with parameters of 5
days. The density function of exponential distribution is
f ( x)  e
 x
I do not know how to use a random number between 0 and 1 to generate
a random X from this distribution. But I know given a specific X, the
probability of the exponential random variable to be less than or equal
to that X is computed as
P( x  X )  1  e
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 X
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Random Variable: Exponential
Distribution
I also know that P(x≤ X) is a number in the closed range of [0,1]. In
addition I know that I have some tools – for example excel- to generate
a uniform random number in the range of [0,1]. Therefore, I can
generate a random probability and associate it with an exponential
random variable. Then, I can claim that I have generated a random
variable from exponential distribution.
Rand () 1  e
 X

e
 X
 1  Rand ()
Furthermore, I know that 1-rand() is between 0 and 1.
Therefore, for simplicity, I can replace 1-rand() with rand(). Thus
 X
e  Rand ()   X  ln[ Rand ()]

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Random Variable: Exponential
Distribution
X 
1

ln[ Rand ()]
Our Exponential distribution has a parameter of 5 days.
Do not forget: lambda is rate, 1/lambda is time.
Our 0 to 1 random number is 0.1. Therefore our exponential random
variable has a value of
X 
1

ln[ Rand ()] 
X  5 ln( 0.1)
X = 11.51 days. The first activity takes 11.51 days
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Random Variable: Normal
The last task follows Normal distribution with parameters of 9 and 2
days. The density function of exponential distribution is
1
(x μ)2 / 2 σ 2
f(x) 
e
2π σ
I do not know how to use a random number between 0 and 1 to generate
a random X from this distribution. However, I know that there is a
standard normal variable with parameters of 0 and 1. Its density
function is
1 z2/ 2
f(z) 
e
2π
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I also know that x= x + zsx
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Random Variable: Standard Normal
The first task follows exponential distribution with parameters of 5
days. The density function of exponential distribution is
1 z2/ 2
f(z) 
e
2π
I do not know how to use a random number between 0 and 1 to generate
a random z from this distribution. But I know that P(z≤ Z) is a number
in the range of [0,1].
P( z  Z )  rand ()
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
P( z  Z )  0.4
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Given Probability, Find z
Given a 40% Probability
The table will give you z
Z =- 0.25
Random Variable: Normal Distribution
We go to the end of the book. And find (after a little bit of
manipulation) as shown on the next page z = -0.25
Then to generate our random variable we simple transform z into x
X= 9 -0.25*2 = 8.5
The first task duration: 11.5
The second task duration: 5.5
The third task duration: 8.5
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Port Automation Model
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