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Transcript of 2 Random Numbers (1)
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Chapter 2
Random Numbers
1. Random-number generation
2. Random-variable generation
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Chapter 2
2
1. Objectives. Random numbers
Discuss how the random numbers are
generated.
Introduce the subsequent testing for
randomness:
Frequency test
Autocorrelation test.
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Chapter 2
3
Properties of random numbers
Two important statistical properties:
Uniformity
Independence.
Random Number, R i , must be independently drawn from auniform distribution with pdf:
Figure: pdf for random numbers
2
1
2)(
1
0
21
0
x xdxr E
12
1
4
1
3
1
2
1
3)(
21
0
31
0
22
xr E dx xr V
12
22
abr V
bar E
otherwise ,0
10 ,1)(
x x f
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Chapter 2
4
Generation of pseudo-random numbers
“Pseudo”, because generating numbers using a known
method removes the potential for true randomness.
Goal: To produce a sequence of numbers in [0,1] that
simulates, or imitates, the ideal properties of random numbers
(RN).
Important considerations in RN routines:
Fast
Portable to different computers
Have sufficiently long cycle
Replicable
Closely approximate the ideal statistical properties of uniformity
and independence.
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Chapter 2
5
Techniques to generate random numbers
Middle-square algorithm
Middle-product algorithm
Constant multiplier algorithm
Linear algorithm
Multiplicative congruential algorithm
Additive congruential algorithm
Congruential non-linear algorithms
Quadratic
Blumy Shub
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Chapter 2
6
Middle-square algorithm
1. Select a seed x0 with D digits (D > 3)
2. Rises to the square: (x0)2
3. Takes the four digits of the Center = x1
4. Rises to the square: (x1)2
5. Takes the four digits of the Center = x2
6. And so on. Then convert these values to r 1 between
0 and 1
If it is not possible to obtain the D digits in the Center,
add zeros to the left.
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Chapter 2
7
Middle-square algorithm
If x0 = 5735
Then,
Y0 = (5735)2 = 32890225 x1 = 8902 r 1 = 0.8902
Y1 = (8902)2
= 79245604 x2 = 2456 r 2 = 0.2456Y2 = (2456)2 = 06031936 x3 = 0319 r 3 = 0.0319
Y3 = (0319)2 = 101760 x4 = 0176 r 4 = 0.0176
Y4 = (0176)2 = 030976 x5 = 3097 r 5 = 0.3097
1. Small cycles
2. May lose the series (e.g.: X0 = 0012)
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Chapter 2
8
Middle-product algorithm
It requires 2 seeds of D digits
1. Select 2 seeds x0 and x1 with D > 3
2. Y0 = x0* x1 xi = digits of the center3. Yi = x1*xi xi+1 = digits of the center
If it is not possible to obtain the D digits in the center, add
zeros to the left.
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Chapter 2
9
Middle-product algorithm
Example:
Seeds x0 = 5015 y x1 = 5734
Y0 = (5015)(5734) = 28756010 x2 = 7560 r 1 = 0.756
Y1 = (5734)(7560) = 43349040 x3 = 3490 r 2 = 0.349
Y2 = (7560)(3490) = 26384400 x4 = 3844 r 3 = 0.3844
Y3 = (3490)(3844) = 13415560 x5 = 4155 r 4 = 0.4155
Y4 = (3844)(4155) = 15971820 x6 = 9718 r 5 = 0.9718
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Chapter 2
10
Constant multiplier algorithm
Similar to Middle-product algorithm
1. Select a seed x0 (D > 3)
2. Select a constant "a“
3. Y0 = a*x0 xi = D digits of the center
4. Yi = a*xi xi+1= D digits of the center
5. Repeat until you get the desired random numbers
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Chapter 2
11
Constant multiplier algorithm
Example:
If: seed x0 = 9803 and constant a = 6965
Y0 = (6965)(9803) = 68277895 x1 = 2778 r 1 = 0.2778
Y1 = (6965)(2778) = 19348770 x2 = 3487 r 2 = 0.3487
Y2 = (6965)(3487) = 24286955 x3 = 2869 r 3 = 0.2869
Y3 = (6965)(2869) = 19982585 x4 = 9825 r 4 = 0.9825
Y4 = (6965)(9825) = 68431125 x5 = 4311 r 5 = 0.4311
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Chapter 2
12
Linear algorithm
To produce a sequence of integers, X 1, X 2 , … between 0 andm-1 by following a recursive relationship:
The selection of the values for a, c , m, and X 0 drastically affects
the statistical properties and the cycle length, must be > 0
"mod (m)" is the residue of the term (axi + c) /m
The random integers are being generated [0,m-1], and toconvert the integers to random numbers:
,...2,1,0 ,mod)(1 imcaX X ii
The
multiplier
The
increment
Themodulus
,...2,1 ,1
1
im
X R i
i
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Chapter 2
13
Linear algorithm
Example:
Use X 0 = 27 , a = 17 , c = 43 y m = 100
The values of x i and r i are:
X 1 = (17*27+43) mod 100 = 502 mod 100 = 2 r 1 = 2/99 = 0.02
X 2 = (17*2+43) mod 100 = 77 mod 100 = 77 r 2 = 77/99 = 0.77
X 3 = (17*77+43) mod 100 = 1352 mod,100 = 52 r 3 = 52/99 = 0.52
The maximum life period N is equal to m
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Chapter 2
14
Multiplicative congruential algorithm
It arises from the linear algorithm when c = 0. The recursiveequation is:
The values of a, m, and x 0 must be > 0 and integers
The conditions that must be fulfilled to achieve the
maximum period are:
m = 2g
g must be integer
a = 3 + 8 k or 5 + 8 k where k = 0, 1, 2, 3,...
X0 = should be an odd number
With these conditions we achieve N = m/4 = 2
g-2
,...2,1,0 ,mod1 imaX X ii
,...2,1 ,1
1
im
X r ii
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Chapter 2
15
Multiplicative congruential algorithm
Example: Use x 0 = 17 , k = 2 , g = 5
Then: a = 5+8(2) = 21 and m=2 5 = 32
x 1 = (21*17) mod 32 = 5 r 1 = 5/31 = 0.1612
x 2 = (21*5) mod 32 = 9 r 2 = 9/31 = 0.2903
x 3 = (21*9) mod 32 = 29 r 3 = 29/31 = 0.9354
x 4
= (21*29) mod 32 = 1 r 4
= 1/31 = 0.3225
x 5 = (21*1) mod 32 = 21 r 5 = 21/31 = 0.6774
x 6 = (21*21) mod 32 = 25 r 6 = 25/31 = 0.8064
x 7 = (21*25) mod 32 = 13 r 7 = 13/31 = 0.4193
x 8
= (21*13) mod 32 = 17 r 8
= 17/31 = 0.5483
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Chapter 2
16
Additive congruential algorithm
Requires a sequence of n integer numbers x1, x2, x3,…xn
to generate new random numbers starting at xn+1, xn+2,…
The recursive equation is:
1m
X r ii
N nnim x x x niii ,...,2,1 ,mod1
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Chapter 2
17
Additive congruential algorithm
Example:
x1 =65, x2 = 89, x3 = 98, x4 = 03, x5 = 69, m = 100
x 6 =(x 5 +x 1 )mod100 = (69+65)mod100 = 34 r 6 = 34/99 = 0.3434
x 7 =(x 6 +x 2 )mod100 = (34+89)mod100 = 23 r 7 = 23/99 = 0.2323
x 8 =(x 7 +x 3 )mod100 = (23+98)mod100 = 21 r 8 = 21/99 = 0.2121
x 9=(x 8 +x 4 )mod100 = (21+03)mod100 = 24 r 9= 24/99 = 0.2424
x 10 =(x 9+x 5 )mod100 = (24+69)mod100 = 93 r 10 = 93/99 = 0.9393
x 11=(x 10 +x 6 )mod100 = (93+34)mod100 = 27 r 11= 27/99 = 0.2727
x 12 =(x 11+x 7 )mod100 = (27+23)mod100 = 50 r 12 = 50/99 = 0.5050
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Chapter 2
18
Congruential non-linear algorithms
Quadratic:
Conditions to achieve a maximum period N = m
m = 2g
g must be integer
a must be an even number
c must be an odd number(b-a)mod4 = 1
1
1
m
X r ii
N imcbxax xi ,...,3,2,1,0 mod2
1
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Chapter 2
19
Congruential non-linear algorithms (quadratic)
Example:
x0 = 13, m = 8, a = 26, b = 27, c = 27
x 1 = (26*132 +27*13+27)mod 8 = 4
x 2 = (26*42 +27*4+27)mod 8 = 7
x 3 = (26*7 2 +27*7+27)mod 8 = 2
x 4 = (26*2 2 +27*2+27)mod 8 = 1
x 5 = (26*12 +27*1+27)mod 8 = 0
x 6 = (26*0 2 +27*0+27)mod 8 = 3
x 7 = (26*32 +27*3+27)mod 8 =6
x 8 = (26*6 2 +27*6+27)mod 8 = 5
x 9 = (26*5 2 +27*5+27)mod 8 = 4
N imcbxax xi ,...,3,2,1,0 mod2
1
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Chapter 2
20
Congruential non-linear algorithms
Algorithm Blum, Blumy Shub
It occurs when Quadratic is:
a = 1
b = 0
c = 0
nim x x ii ,...,3,2,1,0 mod2
1
h
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Chapter 2
21
Characteristics of a good generator
Maximum Density Such that the values assumed by R i , i = 1,2,…, leave no large
gaps on [0,1]
Problem: Instead of continuous, each R i is discrete
Solution: a very large integer for modulus m
Maximum Period
To achieve maximum density and avoid cycling.
Achieve by: proper choice of a, c , m, and X 0 .
Most digital computers use a binary representation ofnumbers
Speed and efficiency depend on the modulus, m, to be (or close
to) 2b
a big number Rev. 1
Ch 2
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Chapter 2
22
Tests for Random Numbers
Two categories:
Testing for uniformity:
H 0 : R i ~ U[0,1]
H 1: R i ~ U[0,1]
Test for independence:H 0 : R i ~ independently
H 1: R i ~ independently
Mean test
H0: µri = 0.5H1: µri = 0.5
We can use the normal distribution with σ2 = 1/12
or σ = 0.2887 with a significance level α = 0.05
/
/
/
Ch t 2
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Chapter 2
23
Tests for Random Numbers
When to use these tests:
If a well-known simulation languages or random-number
generators is used, it is probably unnecessary to test
If the generator is not explicitly known or documented, e.g.,
spreadsheet programs, symbolic/numerical calculators, testsshould be applied to the sample numbers.
Types of tests:
Theoretical tests: evaluate the choices of m, a, and c without
actually generating any numbers
Empirical tests: applied to actual sequences of numbers
produced. Our emphasis.
Rev. 1
Ch t 2
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Chapter 2
24
Kolmogorov-Smirnov test [Uniformity Test]
Very useful for little data.
For discrete and continuous variables.
Procedure:
1. Arrange the r i from lowest to highest
2. Determine D+, D-
3. Determine D4. Determine the critical value of Dα, n
5. If D> Dα, n then r i are not uniform
Chapter 2
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Chapter 2
25
Kolmogorov-Smirnov test [Uniformity Test]
Example: Suppose that 5 generated numbers are 0.44,0.81, 0.14, 0.05, 0.93.
Step 1:
Step 2:
Step 3: D = max(D + , D - ) = 0.26
Step 4: Fora = 0.05 ,
D a
n = 0.565 > D
So not reject H 0 .
Sort R (i) Ascending
D+ = max {i/N – R (i) }
D- = max {R (i) - (i-1)/N}
R (i) 0.05 0.14 0.44 0.81 0.93
i/N 0.20 0.40 0.60 0.80 1.00
i/N – R (i) 0.15 0.26 0.16 - 0.07R (i) – (i-1)/N 0.05 - 0.04 0.21 0.13
Chapter 2
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Chapter 2
26
Chi-square test [Uniformity Test]
The Chi-square test uses the statistical sample:
Approximately the chi-square distribution with n-1 degrees of
freedom is tabulated in Table A.6
For a uniform distribution, the expected number Ei, in each class
is:
Valid only for large samples, example: N> = 50
If X02>X2
(α; n-1) reject the hypothesis of uniformity
n
i i
ii
E
E O
1
22
0
)(
nobservatioof #totaltheis N where,n N E i
n is the # of classes
Oi is the observed # of
the class i th
E i is the expected # of
the class i th
Chapter 2
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Chapter 2
27
Chi-square test [Uniformity Test]
Example: Use Chi-square with a = 0.05 to test that the data shownare uniformly distributed
The test uses n = 10 intervals of equal length:
[0, 0.1), [0.1, 0.2), …, [0.9, 1.0)
0.34 0.90 0.25 0.89 0.87 0.44 0.12 0.21 0.46 0.67
0.83 0.76 0.79 0.64 0.70 0.81 0.94 0.74 0.22 0.74
0.96 0.99 0.77 0.67 0.56 0.41 0.52 0.73 0.99 0.02
0.47 0.30 0.17 0.82 0.56 0.05 0.45 0.31 0.78 0.05
0.79 0.71 0.23 0.19 0.82 0.93 0.65 0.37 0.39 0.42
0.99 0.17 0.99 0.46 0.05 0.66 0.10 0.42 0.18 0.49
0.37 0.51 0.54 0.01 0.81 0.28 0.69 0.34 0.75 0.49
0.72 0.43 0.56 0.97 0.30 0.94 0.96 0.58 0.73 0.05
0.06 0.39 0.84 0.24 0.40 0.64 0.40 0.19 0.79 0.62
0.18 0.26 0.97 0.88 0.64 0.47 0.60 0.11 0.29 0.78
Chapter 2
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Chapter 2
28
Chi-square test [Uniformity Test]
The X02 value = 3.4
Compared to the value of Table A.6, X20.05, 9 = 16.9
Then H0 is not rejected.
Intervalo Oi Ei Oi-Ei (Oi-Ei)2
(Oi-Ei)2
Ei
1 8 10 -2 4 0.4
2 8 10 -2 4 0.4
3 10 10 0 0 0.0
4 9 10 -1 1 0.1
5 12 10 2 4 0.4
6 8 10 -2 4 0.4
7 10 10 0 0 0.0
8 14 10 4 16 1.6
9 10 10 0 0 0.0
10 11
100
10
100
1
0
1 0.1
3.4
Chapter 2
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Chapter 2
29
Tests for Autocorrelation [Test of Independence]
It is a test between each m numbers, starting with thenumber i The auto-correlation r im between the numbers R i , R i+m, R i+2m,
R i+(M+1)m
M is the largest integer such that, , where N is
the total number of values in the sequence Hypothesis:
If the values are not correlated: For large values of M, the distribution estimation r im, is
approximately a normal distribution.
N )m(M i 1
tindependen NOTarenumbers ,0 :
,0 :
1
0 tindependenarenumbers
im
im
H
H
r
r
Chapter 2
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Chapter 2
30
Tests for Autocorrelation [Test of Independence]
The statistical test is:
Z 0 is distributed normally with mean = 0 and variance = 1, and:
If r im > 0, the subsequence has positive autocorrelation
High random numbers tend to be followed by high ones, and vice versa.
If r im < 0, the subsequence has negative autocorrelation
Low random numbers tend to be followed by high ones, and vice versa.
im
im Z r
r
ˆ
0ˆ
ˆ
)(M
M σ
. R R M
ρ
im ρ
M
k
)m(k ikmiim
112
713ˆ
2501
1ˆ
0
1
Chapter 2
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Chapter 2
31
Example [Test for autocorrelation]
Test if the 3rd, 8th, 13th and so on, are auto-correlated.Use a = 0.05
Then i = 3, m = 5, N = 30
Then M = 4
0.12 0.01 0.23 0.28 0.89 0.31 0.64 0.28 0.83 0.93
0.99 0.15 0.33 0.35 0.91 0.41 0.60 0.27 0.75 0.88
0.68 0.49 0.05 0.43 0.95 0.58 0.19 0.36 0.69 0.87
30 28
305143
1
N )m(M i
Chapter 2
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p
32
Example [Test for autocorrelation]
From Table A.3, z 0.025 = 1.96. then the hypothesis is not
rejected.
516.11280.0
1945.0
128.01412
7)4(13ˆ
1945.0
250)36.0)(05.0()05.0)(27.0(
)27.0)(33.0()33.0)(28.0()28.0)(23.0(
14
1ˆ
0
35
35
Z
)( σ
. ρ
ρ
Chapter 2
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p
33
Other tests of Independence
Test runs up and down Test runs up and below the mean
Test poker
Test series
Test holes
-----HOMEWORK 3 at SIDWeb-----
Bonus: (3 points for the partial exam)
Write a computer program that generates random numbers (4 digits)using the Multiplicative Congruential Algorithm. Allow user to entervalues x0, k, g
Give an example printed and the program file
Chapter 2
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p
34
Understand how to generate random variables
from a specific distribution as inputs of a
simulation model.
Illustrate some used techniques for generating
random variables.
Inverse-transform technique
Acceptance- rejection technique
2. Objectives. Random Variables
Chapter 2
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Inverse-transform technique
It can be used to generate random variables from anexponential distribution, uniform, Weibull, triangular and
empirical distribution.
Concept:
For cdf function: r = F(x)
Generate r (0,1)
Find x:
x = F-1(r)
r 1
x1
r = F(x)
Chapter 2
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Exponential Distribution [inverse-transform]
The exponential cdf:
To generate X 1, X 2 , X 3 …
r = F(x)
r = 1 – e-l x for x 0
Xi = F-1(r i)
= -(1/l ln(1-r i)
= -(1/l) ln(r i)
Figure: Inverse-transform
technique for exp (l
= 1)
)1ln(1
)1ln(
1
r X
r X
r e X
l
l
l
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Table of Distributions [inverse-transform]
Distribution Cumulative Inverse-Transform
Uniform
Weibull
Triangular
Normal
b xa
aba x x F
0 1
xe x F
x
a
21 221
10 2
2
2
x x x F
x x x F
ii r aba x
a 1
1ln r x
121 122
210 2
r r x
r r x
z x
Chapter 2
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Exponential Distribution [inverse-transform]
Example: Generate 200 variables Xi with exponential distribution(l = 1)
The histograms of r i and xi are:
Uniform distribution Exponential distribution
i 1 2 3 4 5
Ri 0.1306 0.0422 0.6597 0.7965 0.7696
Xi 0.1400 0.0431 1.078 1.592 1.468
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Uniform Distribution [Example]
The temperature of an ovenbehaves uniformly within the
range of 95 to 100 ° C.
Modeling the behavior of the
temperature.
ii r aba x
Measurements r i Temperature oC
1 0.48 95+5*0.48=97.40
2 0.82 99.103 0.69 98.45
4 0.67 98.35
5 0.00 95.00
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Exponential Distribution [Example]
Historical data service time a Bankbehaves exponentially with an
average of 3 min / customer.
Simulate the behavior of this
random variable
Customer r i Service Time(min)
1 0.36 3.06
2 0.17 5.313 0.97 0.09
4 0.50 2.07
5 0.21 0.70
ii r x 1ln3
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Poisson Distribution [Example]
The number of pieces entering to a production systemfollows a Poisson distribution with mean 2 pcs/hour.
Simulate the arrival behavior of the pieces to the system.
x P(x) F(x) Random digits
0 0.1353 0.1353 0-0.1353
1 0.2706 0.4060 0.1353-0.4060
2 0.2706 0.6766 0.4060-0.6766
3 0.1804 0.8571 0.6766-0.8571
4 0.0902 0.9473 0.8571-0.9473
5 0.0369 0.9834 0.9473-0.9834
6 0.0120 0.9954 0.9834-0.9954
7 0.0034 0.9989 0.9954-0.9989
8 0.0008 0.9997 0.9989-0.9997
9 0.0001 0.9999 0.9997-0.9999
10 0.00003 0.9999 0.9999-0.9999
Arrival
Time
r i Pieces/hour
1 0.6754 2
2 0.0234 0
3 0.7892 3
4 0.5134 2
5 0.3331 1
x
i
i
x
i
e x F
x
x
e x p
0 !)(
,...1,0 ,
!a
a
a
a
Chapter 2
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Convolution Method
In some probability distributions, the random variable
can be generated by adding other random variables
Erlang Distribution
The sum of exponential variables of equal means
(1/λ)
Y = x1+x2+…+xk
Y = (-1/kλ)ln(1-r 1)+ (-1/kλ)ln(1-r 2)+…+ (-1/kλ)ln(1-r k)
Where:
Y = Er i= (-1/kλ)lnπ(1-r i) The factor productoria
Chapter 2
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Convolution Method
The processing time of certain machine follows a 3-erlang distribution with mean 1/λ = 8 min/piece
Y = -(8/3)[lnπ(1-r i)] = -(8/3)ln[(1-r 1) (1-r 2) (1-r 3)]
Piece 1-r 1 1-r 2 1-r 3 Process Time
1 0.28 0.52 0.64 6.328
2 0.96 0.37 0.83 3.257
3 0. 04 0.12 0.03 23.588
4 0.35 0.44 0.50 6.837
5 0.77 0.09 0.21 11.279
Chapter 2
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Empirical continuous distribution [inverse-transform]
When theoretical distribution is not applicable To collect empirical data:
Resample the observed data
Interpolate between observed data points to fill in the gaps
For a small and large sample set (size n):
Arrange the data from smallest to largest
Assign the probability 1/n to each interval
Where 1 / n is the probability
(n)(2)(1) xxx
(i)1)-(i xxx
ni Ra x R F X ii )1()(ˆ
)1(1
n
x x
nin
x xa
iiii
i
/1/)1(/1
)1()()1()(
Chapter 2
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Empirical continuous distribution [inverse-transform]
Example: Suppose that the observed data of 100 repair times of a machineare:
i
Interval
(Hours) Frecuency
Relative
Frecuency
Cumulative
Frecuencia, c i Slot, a i
1 0.25 ≤ x ≤ 0.5 31 0.31 0.31 0.81
2 0.5 ≤ x ≤ 1.0 10 0.10 0.41 5.0
3 1.0 ≤ x ≤ 1.5 25 0.25 0.66 2.0
4 1.5 ≤ x ≤ 2.0 34 0.34 1.00 1.47
Consider r 1 = 0.83:
c3 = 0.67 < r 1 < c4 = 1.00
X1 = x(4-1) + a4(r 1 – c(4-1))
= 1.5 + 1.47(0.83-0.66)
= 1.75
Chapter 2
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Empirical Discrete Distribution
Example: Suppose the number of shipments, x, on theloading dock of IHW company is either 0 , 1, or 2
Data - Probability distribution:
Method - Given R, the generation
scheme becomes:
0.18.0
8.05.0
5.0
,2
,1
,0
R
R
R
xConsider R1 = 0.73:
F(xi-1) < R <= F(xi)
F(x0) < 0.73 <= F(x1)
Hence, x1 = 1
x p(x) F(x )
0 0.50 0.50
1 0.30 0.80
2 0.20 1.00
Rev. 1
Chapter 2
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Acceptance-Rejection technique
Useful particularly when inverse cdf does not exist in closedform, a.k.a. thinning
To generate random variables X ~ U (1/4, 1) we must:
This procedure is only for uniform distributions
Acceptance-rejection technique is also applied to Poisson,nonstationary Poisson, and gamma distributions
-----HOMEWORK 4 at SIDWeb-----
procedure :
Step 1. Generate R ~ U[0,1]Step 2a. If R >= ¼, accept X=R.
Step 2b. If R < ¼, reject R,back to step 1
Generate R
Condition
output R’
si
no