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# hw6 - HW 6 Solutions Fall 2010 Points possible 10...

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HW 6 Solutions Fall 2010 1 Points possible: 10 Statistical Comparisons: The table below shows the objective function value for the best solution in each trial for three different algorithms applied to the same problem: Trial SA GA GS 1 91.94 147.90 47.66 2 77.13 97.88 150.53 3 10.93 39.76 97.04 4 18.6 204.48 82.62 5 28.63 488.83 99.89 6 86.52 113.00 76.52 7 64.58 141.97 87.84 8 22.23 53.76 51.73 9 59.75 408.20 147.51 10 134.11 226.95 115.98 Mean 59.44 192.27 95.73 Std. Dev 39.52 148.35 34.90 (i). (1 points) Make a boxplot for the data provided above (Use the Matlab command BOXPLOT). Comment on your plot: How do the means and variances compare for each of the algorithms? Are there any outliers? Which algorithm performed the best in your opinion and why? Answers we hope students will make: SA has the lowest mean, smaller variability compared to GA, and no outliers GA appears to perform the worst in this case with the highest mean, high variability, and an outlier GS has the lowest variability and no outliers, but a higher mean than SA

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HW 6 Solutions Fall 2010 2 Other good observations about the boxplot Choose SA as the best since it has the lowest mean and relatively low variability (ii). ( 1 points) Plot empirical CDF’s for the data above using the plotting position formula provided in class (i/[n+1]). Comment on your plot: Which algorithm appears to perform the best? Is there any evidence of stochastic dominance ? Since this is a minimization problem we want to have the CDFs toward the left side of the plot area and want the CDF to be relatively vertical (low variability). From the empirical CDF, can see that the SA algorithm stochastically dominates (i.e. is the farthest to the left and does not intersect with the other CDFs). (iii). (3 points) Perform all pairwise comparisons (three in all) of mean objective function value of best solution using a two-sample t-test. State your hypothesis. Report your test statistic and p-values for each comparison. At = 0.05 what is your conclusion for each test? For all three pair-wise comparisons, define hypothesis as Null hypothesis H 0 : two algorithms are equal, i.e. μ 1 = μ 2 Alternative hypothesis H a : two algorithms are not equal, i.e. μ 1 ≠ μ 2 (two-tailed test) The test statistic is 1 2 2 2 1 2 X X t S S n n The t distribution has the degrees of freedom
HW 6 Solutions Fall 2010 3 2 2 2 1 2 2 2 2 2 1 2 .

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hw6 - HW 6 Solutions Fall 2010 Points possible 10...

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