2G1503 Simulation and Modeling Exercise #1-2
Ali Ghodsi aligh@imit.kth.se
1
Linear Congruential Method
• Used to generate random numbers
• Recursive definition:
– Xn+1=a*Xn+c mod m
• If c=0 then
– Multiplicative LCM
• else
– Mixed LCM
2
Mixed LCG
• Let X0 = 27, a = 8, c = 47, and m = 100
• Exercise 1: Generate three numbers!
X0 = 27
X1 = 8*27 + 47 mod 100 = 63
X2 = 8*63 + 47 mod 100 = 51
X3 = 8*51 + 47 mod 100 = 55
3
Multiplicative LCG
• Let X0 = 847, a = 56, and m = 1000
• Exercise 2: Generate four numbers!
X0 = 847
X1 = 56*847 mod 1000 = 432
X2 = 56*432 mod 1000 = 192
X3 = 56*192 mod 1000 = 752
X4 = 56*752 mod 1000 = 112
4
Random Numbers and Integers
• The random integers from the LGC can
be converted to random numbers by
the formula:
Rn = Xn / m for all n
• Example:
– R1 = 847/1000 = 0.847
– R2 = 432/1000 = 0.432
– R3 = 192/1000 = 0.192
5
Testing for Uniformity
• Use the Kolmogorov-Smirnov test with
alpha = 0.05 to determine if the
hypothesis that the numbers are
uniformly distributed on the interval
[0,1] can be rejected.
• Numbers:
0.54, 0.73, 0.98, 0.11, 0.68
6
Sorting the numbers
N:
Numbers:
Sorted:
5
0.54, 0.73, 0.98, 0.11, 0.68
0.11, 0.54, 0.68, 0.73, 0.98
– Make N intervals and compute D+ and D-
7
Kolmogorov Smirnov Test of Uniformity
i
1
2
3
4
5
Ri
0.11 0.54 0.68 0.73 0.98
i/N
0.2
0.4
0.6
0.8
1.0
(i-1)/N
0.0
0.2
0.4
0.6
0.8
i/N–Ri
0.09 -
-
0.07 0.02
D+=0.07
Ri–(i-1)/N 0.11 0.34 0.28 0.13 0.18
D-=0.34
D=0.34
8
Null-hypothesis
• H0: The sample has a uniform
distribution
• H1: The sample does not have a
uniform distribution
9
Significance Level
• The significance level α gives the
probability that the null-hypothesis is
rejected even though it is true.
• P(H0 rejected|H0 is true)
10
Critical Values for KS-test
N
D0.10
D0.05
D0.01
1
0.950
0.975
0.995
…
…
…
..
5
0.510
0.565
0.669
…
…
…
…
11
Testing for Independence
Use the run-ups and run-downs test
with alpha = 0.05 to determine if the
hypothesis that the numbers are
independent holds.
12
Testing for Independence
0.34
0.83
0.96
0.47
0.79
0.90
0.76
0.99
0.30
0.71
0.25
0.79
0.77
0.17
0.23
0.89
0.64
0.67
0.82
0.19
0.87
0.70
0.56
0.56
0.82
0.44
0.81
0.41
0.05
0.93
0.12
0.94
0.52
0.45
0.65
0.21
0.74
0.73
0.31
0.37
0.46
0.22
0.99
0.78
0.39
0.67
0.74
0.02
0.05
0.42
13
Testing for Independence
+
0.83
0.96
0.47
0.79
0.90
0.76
0.99
0.30
0.71
0.25
0.79
0.77
0.17
0.23
0.89
0.64
0.67
0.82
0.19
0.87
0.70
0.56
0.56
0.82
0.44
0.81
0.41
0.05
0.93
0.12
0.94
0.52
0.45
0.65
0.21
0.74
0.73
0.31
0.37
0.46
0.22
0.99
0.78
0.39
0.67
0.74
0.02
0.05
0.42
14
Testing for Independence
+
0.83
0.96
0.47
0.79
0.76
0.99
0.30
0.71
0.25
0.79
0.77
0.17
0.23
0.89
0.64
0.67
0.82
0.19
0.87
0.70
0.56
0.56
0.82
0.44
0.81
0.41
0.05
0.93
0.12
0.94
0.52
0.45
0.65
0.21
0.74
0.73
0.31
0.37
0.46
0.22
0.99
0.78
0.39
0.67
0.74
0.02
0.05
0.42
15
Testing for Independence
+
0.83
0.96
0.47
0.79
0.76
0.99
0.30
0.71
+
0.79
0.77
0.17
0.23
0.89
0.64
0.67
0.82
0.19
0.87
0.70
0.56
0.56
0.82
0.44
0.81
0.41
0.05
0.93
0.12
0.94
0.52
0.45
0.65
0.21
0.74
0.73
0.31
0.37
0.46
0.22
0.99
0.78
0.39
0.67
0.74
0.02
0.05
0.42
16
Testing for Independence
+
+
-
+
-
+
+
-
+
+
+
+
+
+
+
-
+
+
-
+
+
+
+
+
+
+
+
+
+
+
17
Testing for Indepedence
• N=50 (numbers)
a=27 (runs)
• We know how to calculate σa, µa:
σa = (2N-1) / 3
=> 100-1 / 3 = 33
µ2a = (16N–29) / 90 => 800-29 / 90 = 8.57
µa = 8.570.5=2.93
• Converting to standard N(0,1):
Z0 = a-σa / µa = 27-33 / 2.93 = -2.05
18
Standard Normal Distribution
Acceptance region for hypothesis
of independence -Za/2 ≤ Z0 ≤ Za/2
α/2
α/2
-Zα / 2
fail to reject
Zα / 2
19
Critical Values for run-ups/run-downs
Zα
0.05
0.06
0.07
0.0
0.51994
0.52392
0.52790
…
…
…
..
1.9
0.97491
0.97500
0.97558
…
…
…
…
20
Testing the Hypothesis
• Z0.025 = 1.96 for N(0,1)
• Z0=-2.05 < -Z0.025 hence H0 is rejected
21