PROYECTO PRODUCTOR
4. PROMOCIÓN Y LANZAMIENTO :
The performance of genetic algorithms is highly dependent on the genetic operators, through which the population can become increasingly adapted to the problem. In this section, we conduct a sensitivity analysis to examine the influence of different types of genetic operators on the solution and the computation effort.
The operators designed for the proposed algorithm can be classified into two main categories: mutation and crossover. To test how different types of operators affect the solution, five scenarios are designed for the sensitivity analysis. The numbers of each genetic operator used in each scenario are listed in Table 9.4.
Note that the first four operators in the table are mutation-based, while the remaining four are crossover-based. Also note that scenario 1 is the base case we adopted to solve the problem in the previous section.
The total number of operators in all scenarios is 32. Each of the scenarios emphasizes different types of operators. The first one has equal numbers of mutation-based and crossover-based operators. In the second scenario, there are three times more mutation-based than crossover-based operators. The third one consists only of mutation-based operators, without any crossover-based operators. Scenarios 4 and 5 switch the numbers of mutation-based and crossover-based operators in scenarios 2 and 3. Recall that there exists a high correlation between the final solution and the random seed. Hence, we run 50 replications for each scenario to generalize the results. Moreover, in order to reduce the random fluctuations among different scenarios, each scenario uses the same random seeds as those adopted in scenario 1. “Common Random Numbers” is a popular variance-reduction technique widely used in simulation analysis (Law, 1991). Program outputs are summarized in Table 9.5.
Table 9.4: The numbers of genetic operators in the scenarios for sensitivity analysis.
Mutation Crossover Scenario
Total no.
of operators
Uniform Straight Non-uniform
Whole non-uniform
Simple Two-point
Arith. Heuristic
1 32 4 4 4 4 4 4 4 4 2 32 6 6 6 6 2 2 2 2 3 32 8 8 8 8 0 0 0 0 4 32 2 2 2 2 6 6 6 6 5 32 0 0 0 0 8 8 8 8
Intelligent Road Design 177 Table 9.5: Program outputs for different scenarios.
Outputs
(million) Scenario 1 Scenario 2 Scenario 3 Scenario 4 Scenario 5 1st Smallest 101.598 1101.093 108.426 4101.392 114.877 2nd Smallest 101.719 2101.108 109.386 101.472 122.883 3rd Smallest 101.726 3101.371 110.074 101.514 125.433 4th Smallest 101.762 5101.398 110.315 101.578 126.384 5th Smallest 101.764 101.431 110.648 101.669 126.397 5th Largest 104.481 103.411 115.716 104.876 169.496 4th Largest 104.518 103.871 115.838 105.120 170.043 3rd Largest 104.606 104.154 115.871 105.284 171.992 2nd Largest 105.392 104.328 115.935 105.641 175.070 1st Largest 105.430 104.861 116.452 106.713 183.403
Mean 102.989 102.385 113.479 103.038 151.298
Stand Dev. 0.976 0.828 1.870 1.150 14.618
Table 9.5 lists the outputs of the best and worst 5 replications, as well as the sample mean and standard deviation for each scenario. The best solution among all outputs is found in scenario 2 (101.093), while the worst solution occurs in scenario 5 (183.403). The sample mean also indicates that scenario 2 has the lowest mean output and scenario 5 has the highest mean output. Note that four of the five best solutions among all replications are found to occur in scenario 2.
Moreover, the mean outputs of scenarios 1, 2, and 4 are very close. The row of standard deviations tells us that the outputs of scenario 2 are more concentrated, whereas the outputs from scenario 5 have high variations. It is also found that the outputs of scenario 5 (around 150) are almost 150% as those of the other four scenarios (around 100). Since scenario 5 does not employ any types of mutation operators, we can conclude that without introducing mutation operators the proposed GA does not yield satisfactory results.
In addition to the program outputs, the computation times for running each replication of each scenario are also recorded. Note that all scenarios are tested on a Pentium II 266 Personal Computer (PC). Table 9.6 indicates that the mean computation time for scenario 3 is the highest among all scenarios, while that for scenario 5 is the lowest. The standard deviations also show that the computation times for scenario 3 vary considerably, while those for scenario 5 are very similar. The most interesting result in the table is that the computation times for scenario 5 are extremely small compared with those for other scenarios.
Moreover, from both Table 9.4 and Table 9.6, it is found that as more mutation-based operators are used in the model, the computation times increase. That is because mutation-based operators involve extensive computations on real numbers. (In computing terminology, these computations are referred to as
“floating point” operations.) On the other hand, crossover-based operators are just swapping and reassigning gene values in the selected chromosomes, and do
178 Intelligent Road Design
Table 9.6: Computation times for different scenarios.
Computation
time (sec) Scenario 1 Scenario 2 Scenario 3 Scenario 4 Scenario 5
Min 70 91 101 48 7
Max 159 177 200 99 12
Mean 97.88 108.48 144.06 60.2 9.9
Stand Dev. 24.98133 22.96336 36.21276 14.34701 1.035098 not require complex calculations. Therefore, the computation times for crossover-based operators are relatively small.
Recall from Table 9.5 that the program outputs for the first four scenarios do not show a big difference in objective values. To visualize the variations, we plot the output of each replication in Figure 9.8. The diagram shows that scenarios 1, 2 and 4 yield very similar results. The outputs of scenario 3 are relatively high when compared with those resulting from other scenarios.
1 6 11 16 21 26 31 36 41 46
Scenario 1 Scenario 2
Scenario 3 Scenario 4 90
95 100 105 110 115 120
Objective value
Replication number
Scenario 1 Scenario 2 Scenario 3 Scenario 4
Figure 9.8: Program outputs for the first four scenarios.
For a more rigorous analysis, we make statistical inferences about the difference between the program outputs for each pair of scenarios. Since each scenario has the same number of replications and uses the same random seed, the program outputs can be considered as the observations from matched samples.
Let oik and ojk denote the outputs of the kth replication for scenario i and j.
Intelligent Road Design 179 Then the difference between oik and ojk , denoted by wk is treated as an observation from a single population. The value of wk is computed by:
jk ik
k o o
w = − , ∀k=1,...,no (9.2)
where no= number of observations.
We denote the mean output of the population of differences by µ . Then w
j i jk ik
jk ik k
w E w E o o E o E o µ µ
µ = ( )= ( − )= ( )− ( )= − (9.3)
where µ and i µ are the mean output of scenarios j i and j.
Since µ and i µ are usually unknown, the sample mean of j wk, denoted by w is employed to estimate µ . Note that w E(w)=E(w)=µw and thus w is an unbiased estimator of µ . Next, we calculate the variance of w w with the following equation:
o w
n w s s
2
2( )= (9.4)
where s2(w) = sample variance of w
2
sw = sample variance of w .
It can be proved (Neter, et al, 1982) that s2(w) is an unbiased estimator of population variance of w. With the estimators w and s2(w), we are able to test the following hypothesis:
0:
H µi−µj=0 or µ =i µj (9.5)
1:
H µi−µj≠0 or µ ≠i µj. (9.6) The test statistic for the preceding matched samples is:
) ( ) (
* 0
w s
w w s
t = w− = (9.7)
where t* = t distribution with degree of freedom no−1.
Following the above approach, we test the difference between the program outputs for each pair of scenarios. The results are given in Table 9.7. Note that the statistical test for the hypothesis stated in eqn (10.5) is a two-tailed test.