IE 1055 2025 Homework 10 Due April 5 2017 1 40 pts Consider the single facility rectilinear location problem There are six 6 existing facilities numbered 1 through 6 which will interact with the new facility The x y coordinates of the existing facilities are given as follows 1 2 2 2 4 5 3 4 8 4 8 5 5 12 8 and 6 12 2 The new facility will have equal interaction with each existing facility In answering the following parts when applicable you must clearly show the feasible region and in those cases where more than one solution exists you must show all points that are optimum In presenting your answers please use graph paper or engineering paper Make sure you still show all your calculations on a separate page a Determine the location and the objective function value of the minisum optimum b Determine the location and the objective function value of the minimax optimum c Is there a point which is optimum for both the minisum and the minimax objective Why or why not d Suppose the optimum minisum location s you found in part a is not usable Given the minisum objective determine the best location for the new facility among the following three points X 3 5 Y 4 7 and Z 10 5 e Suppose the maximum distance to any existing facility must be less than or equal to 9 rectilinear distance units For the minisum objective determine the optimum location and the objective function value at the optimum location f Suppose the distance to each existing facility must be greater than or equal to 3 rectilinear distance units For the minisum objective determine the optimum location and the objective function value at the optimum location g Suppose the distance to each existing facility must be less than or equal to 3 rectilinear distance units For the minisum objective determine the optimum location and the objective function value at the optimum location Answer 40 Pts a 7 pts Minisum Note that the total weight is 6 so the median value is 6 2 3 X X value 2 4 8 12 facility 1 2 3 4 5 6 cum wt 1 3 and 3 3 4 and 4 3 6 Y Y value facility 2 1 6 5 2 4 8 3 5 cum wt 2 2 3 4 4 3 6 THIS SOLUTION MAY ONLY BE USED FOR IE 1055 2025 SPRING 2017 Thus for x there is a not a single optimal point but all points from x 4 to 8 are optimal Since at x 4 not more than half the total weight is to the right and at x 8 not more than half the total weight is to the left For y y 5 by the median condition To determine the objective function pick the point 4 5 The objective 4 2 5 2 0 4 4 5 8 4 8 5 5 4 12 5 8 412 5 2 34 Thus the optimal solution is the line segment from 4 5 to 8 5 b 7 pts Assume gi 0 i Facility ai 1 2 2 4 3 4 4 8 5 12 6 12 bi 2 5 8 5 8 2 ai bi 4 9 12 13 20 14 ai bi 0 1 4 3 4 10 c1 4 c2 20 c3 10 c4 4 c5 max 20 4 4 10 16 The optimal solution is the line segment joining 4 10 4 10 16 which is 7 5 and 20 4 20 4 16 which is 8 4 The optimal objective function value c5 2 16 2 8 c 4 pts Yes because the two line segments intersect at 7 5 THIS SOLUTION MAY ONLY BE USED FOR IE 1055 2025 SPRING 2017 6 d 5 pts wi 6 i 1 The same minisum contour line contains X 3 5 Y 4 7 Z 10 5 Thus all 3 are equivalent Note slopes for each one are in the boxes THIS SOLUTION MAY ONLY BE USED FOR IE 1055 2025 SPRING 2017 e 6 pts The answer is the line segment connecting 6 5 and 8 5 First find the minimax contour line for k 9 c1 9 13 4 9 1 13 1 13 1 7 6 c2 9 20 9 11 2 13 5 13 5 9 4 c3 9 10 9 1 3 11 5 11 5 8 3 4 11 1 11 1 6 5 c4 9 4 9 5 Connecting these points yields a contour line This intersects with the minimum optimal solution over the line segment 6 5 to 8 5 and results in a minisum objective function value of 34 f 7 pts Note that the diamonds around each existing facility represent infeasible regions Using the results of part d and looking at the minsum contour lines the best candidates for the minisum optimal are 6 6 6 4 a point on the segment 11 5 8 8 or the segment 11 5 8 2 Since 11 5 is on both segments check this point and the points 6 6 6 4 11 5 11 5 5 2 1 4 0 11 4 5 8 11 12 5 8 11 8 0 1112 5 2 40 6 6 6 2 6 2 6 4 6 5 6 4 6 8 16 8 6 0 6 12 6 8 612 6 2 36 similarly for 6 4 we get 36 THIS SOLUTION MAY ONLY BE USED FOR IE 1055 2025 SPRING 2017 The optimality of 6 4 and 6 6 is confirmed by constructing the contour line containing points 6 4 6 6 there are no interior contour lines containing a feasible point g 4 pts Refer to the diagram for f Now the diamonds represent the feasible region relative to each facility Note that the intersection of all of the diamonds is clearly the empty set thus there is no feasible region for the problem 2 35 pts Four machines are located in a plant at points 0 30 20 10 40 50 and 50 30 These machines require maintenance at expected frequencies of 30 12 25 and 20 times per year respectively Because of the nature of the maintenance all machines must be maintained at the maintenance center A machine can be serviced by exactly one maintenance center Moreover the cost of transporting the machines to and from the maintenance center is 5 per unit of distance the 5 per unit distance covers the cost to and from so it is the two way cost not a one way cost including the cost of lost profits resulting from the machines being down The annual cost of owning and operating a maintenance center is 5000 Rectilinear travel is assumed a Determine the optimal location if there will be one maintenance center b Determine the optimal cost if there is one maintenance center c Calculate the number of possible ways to allocate the machines across two maintenance centers d List all possible machine to maintenance center assignment combinations for two maintenance centers indicating which machines are allocated to …