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CVEN 5393 Spring 2013 Homework 8 Solution Revelle, Whitlatch and Wright Civil and Environmental Systems Engineering – 2 nd Edition Problem 5.3

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a) Plot feasible region in decision space 4x 1 - 12x 2 <= -6 -4x 1 + 6x 2 <= 12 4x 1 + 2x 2 >= 8 x 1 + x 2 <= 9 4x 2 <= 16

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b) Plot the corresponding feasible region in objective space. For each extreme point indicate if it is a noninferior or dominated solution Z2Z2 Z1Z1 Min Z 1 Max Z 2 X1X2Z1Z2 A1.501.004.0011.50dominated B0.752.504.0019.75noninferior C3.004.0010.0037.00noninferior D5.004.0014.0043.00noninferior E6.382.6315.3837.50dominated

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C. Use the constraint method (graphically) to generate an approximation of the non- inferior set having 6 noninferior solutions evenly spaced along the Z1 axis. SolutionX1X2Z 1 = 2x 1 + x 2 minZ 2 = 3x 1 + 7x 2 max Z1 Optima1.5.75 1.0 2.5 4.0 11.50 19.75 ABAB Z2 Optima5.04.014.043.00D Construct a payoff table. Each row represents the solution of one individual objective function, and shows the values for the other objects at that point. If alternate optima are detected, compare the values of other objectives to select the noninferior point. Every point in the payoff table is a noninferior solution. The table gives the entire range of values each objective can have on the noninferior set. Need 6 noninferior solutions evenly spaced along the Z1 axis Max is 14; Min is 4.0 Other 4 points are 6, 8, 10, 12 Create constrained problem by selecting one of the objectives to optimize and moving the other(s) into the constraint set with the addition of a right hand side coefficient for each. 4 additional constraints to add to the problem, one at a time and find optimal solution for Z2: 2X1 + X2 <= 6 2X1 + X2 <= 8 2X1 + X2 <= 10 2X1 + X2 <= 12

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Add each additional constraint to the set and find the optimal solution for the remaining objective. Each solution is a noninferior solution of the multiobjective problem. 4x 1 - 12x 2 <= -6 -4x 1 + 6x 2 <= 12 4x 1 + 2x 2 >= 8 x 1 + x 2 <= 9 4x 2 <= 16 2x 1 + x 2 <= 6 Max 3X 1 + 7X 2 --------------------------------------------------------------------------------------------------------------------0--------------------------------------------------------------------------------------------------------------------0 2x 1 + x 2 <= 8 2x 1 + x 2 <= 10 2x 1 + x 2 <= 6 ---------- -------- -------- -------- X1X2Z1Z2 0.752.504.0019.75 1.503.006.0025.50 2.253.508.0031.25 3.004.0010.0037.00 4.00 12.0040.00 5.004.0014.0043.00 Noninferior Solutions 2x 1 + x 2 <= 12 2x 1 + x 2 <= 14 New constraint set: 2X1 + X2 <= 6 2X1 + X2 <= 8 2X1 + X2 <= 10 2X1 + X2 <= 12

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c) The selection of the Z1 points gave the same solution for the pareto front as we got in a) and b). If we had not used as many Z1 points we may have gotten some approximation errors. Z2Z2 Z1Z1 Min Z 1 Max Z 2 X1X2Z1Z2 A1.501.004.0011.50dominated B0.752.504.0019.75noninferior C3.004.0010.0037.00noninferior D5.004.0014.0043.00noninferior E6.382.6315.3837.50dominated X1X2Z1Z2 0.752.504.0019.75 1.503.006.0025.50 2.253.508.0031.25 3.004.0010.0037.00 4.00 12.0040.00 5.004.0014.0043.00

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d.) Use the weighting method (graphically) to generate an approximiation of the non- inferior set having 6 noninferior solutions evenly spaced along the Z 2 axis. What is range of Z 2 point? We know this from the payoff table. Z 2 ranges from 19.75 to 43.0. Six evenly space values are: Z 2 = 19.75; 24.40; 29.05; 33.70; 38.35; 43.0 To solve: use the weighting method to identify noninferior points in objective space. Then interpolate between these to find the value of noninferior points at the designated Z 2 values. Note that since Z 1 is a minimization objective and Z 2 is a max objective, we must make Z 1 also a max objective by taking the negative of it. We arbitrarily select a set of weights and compute the Grand Objective: grad ient w1w1 w2w2 w 1 *(-)Z 1 + w 2 *Z 2 Grand objective 11.001.0 * (-2x1 – x2) + 0*(3x1+7x2) Max -2x 1 – x 2 20.80.20.8 * (-2x1 – x2) + 0.2*(3x1+7x2) Max -x 1 + 0.6x 2 30.60.40.6 * (-2x1 – x2) + 0.4*(3x1+7x2) Max 2.2x 2 40.40.60.4 * (-2x1 – x2) + 0.6*(3x1+7x2) Max x 1 + 3.8x 2 50.20.80.2 * (-2x1 – x2) + 0.8*(3x1+7x2) Max 2.0x 1 + 5.4x 2 601.00 * (-2x1 – x2) + 1.0*(3x1+7x2) Max 3x 1 + 7x 2

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d) Solve single grand objectives 4x 1 - 12x 2 <= -6 -4x 1 + 6x 2 <= 12 4x 1 + 2x 2 >= 8 x 1 + x 2 <= 9 4x 2 <= 16 GrObjective 1-2x 1 – x 2 2-x 1 + 0.6x 2 32.2x 2 4x 1 + 3.8x 2 52.0x 1 + 5.4x 2 63x 1 + 7x 2 Solution A,B B C,D D D D x 1 x 2 Z 1 Z 2 0.75 2.5 4.0 19.75 3.0 4.0 10.0 37.0 5.0 4.0 14.0 43.0

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b) Use linear interpolation to find the noninferior points at Z 2 = 19.75; 24.40; 29.05; 33.70; 38.35; 43.0 Z2Z2 Z1Z1 Z2Z1 19.754.0 24.405.62 29.057.64 33.708.85 38.3510.9 43.0014.0 10.9, 38.35

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Things to keep in mind The “classical” methods of solving multi-objective optimization problems are based on creating a set of single objective optimization problems, each of which identifies a non-inferior solution. The set forms a pareto front or surface (if more than 2-d). Other solutions can be inferred by interpolation if the variables are continuous. If you have a plot of extreme points in objective space, you can identify the noninferior extreme points by the relative values of the objectives or by the “corner” rule. The NE corner rule depends on sign of objectives The payoff table identifies the range of values that each objective can have.

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