Solutions for Operations Research Final Exam

Transcription

1 Solutions for Operations Research Final Exam. (a) The buffer stock is B = i a i = a + a + a + a + a + a 6 + a 7 = =. And the transportation tableau corresponding to the transshipment problem is Destination Origin a i + B M M M M M 6 M M M 6 M b j + B (b) We solve the problem by using the transportation algorithm. The starting basic feasible solution produced by the northwest corner method is illustrated as follows. 6 The st BFS. u M-6 M- M- M- -M M M M M- M-6-6 M M -M -M- M+ M M 8-M 7-M M+ M M 6 M 6 M+ M+ v M-6 - -M- -M Computations for the st BFS.

2 Then x 6 enters the basis and x 7 leaves the basis and θ =. -θ +θ -θ +θ 6 +θ -θ Pivot from the st BFS to the nd BFS. The nd BFS is illustrated as follows. 6 The nd BFS. u M-6 - M+ M M M 6 7-M M- M M M-6 M-8 -M- M+ M M M- M+ M M 6 M M- M-8 M+ M+ v -M -M -M- -M Computations for the nd BFS. Then x enters the basis and x 6 leaves the basis and θ =. -θ +θ 6 +θ -θ Pivot from the nd BFS to the rd BFS.

3 The rd BFS is illustrated as follows. 6 The rd BFS. u -9 - M+ M M M M-7 M- M- M M M-6 - -M- M+ M M -M- M+ M M 6 M M- - M+ M+ v -M 7 -M- -M Computations for the rd BFS. Then x 7 enters the basis and x leaves the basis and θ =. -θ +θ 6 +θ -θ Pivot from the rd BFS to the th BFS. The th BFS is illustrated as follows. 6 The th BFS.

4 u M- M+ - M M M M- M- M-9 M M M-6 M-7 M M M+ M+ - M 6 M M-6 - M v 6 7 Computations for the th BFS. Then x 7 enters the basis and x 7 leaves the basis and θ =. -θ +θ 6 +θ -θ Pivot from the th BFS to the th BFS. The nd BFS is illustrated as follows. 6 The th BFS.

5 u M- M+ - M M M M- M- M-8 M M M-6 M-7 M M M+ M+ - M 6 M M- M+ v 6 - Computations for the th BFS. Since the reduced cost values for each nonbasic variable are nonnegative, the th BFS is optimal. The interpretation of the solution is: units from node to node 6, units from node to node, units from node to node 6, units from node to node 7, units from node to node, and units from node to node 7. Moreover, if the cost is the least, the case that some units are transported backward can not happen at any transhipment nodes. Since the number of transportation units at each transshipment node is less than B, an upper bound for the sum of net out of nodes. Therefore we set the availability at each transhipment node as a i + B, and the requirement at each transhipment node as b i + B. This interprets the role of B in this transhipment problem. (c) Using the least cost method, the transportation tableau corresponding to the transshipment problem is Destination Origin a i + B 8 6 M M M 6 6 M b j + B We solve the problem by using the transportation algorithm. The starting basic feasible solution produced by the northwest corner method is illustrated as follows.

6 6 The st BFS. u 6 -M+ M+ M M M M- -M- M+8 M M - -M- M+ M M - M+ M M 6 M M+ M+ M+ M+ v -M+ -M+ -M- -M Computations for the st BFS. Then x enters the basis and x leaves the basis and θ =. -θ +θ 6 +θ -θ Pivot from the st BFS to the nd BFS. The nd BFS is illustrated as follows. 6 The nd BFS. 6

7 u 6 6 M M M M- - 7 M M - M M - M+ M+ - M 6 M M+ V - Computations for the nd BFS. Then x 6 enters the basis and x 7 leaves the basis and θ =. θ +θ +θ θ 6 Pivot from the nd BFS to the rd BFS. The rd BFS is illustrated as follows. 6 The rd BFS. 7

8 u M M M M- - 8 M M M M M+ M+ - M 6 M M+ V - Computations for the rd BFS. Then x enters the basis and x 6 leaves the basis and θ =. θ +θ +θ θ 6 Pivot from the rd BFS to the th BFS. The th BFS is illustrated as follows. 6 The th BFS. u M M M M- M M - M M - M+ M+ - M 6 M M+ V 6 - Computations for the th BFS. 8

9 Then x 7 enters the basis and x leaves the basis and θ =. θ +θ +θ θ 6 +θ θ Pivot from the th BFS to the th BFS. The nd BFS is illustrated as follows. 6 The th BFS. u 7 - M M M M- M M M M M+ M+ - M 6 M M+ V 6 - Computations for the th BFS. Since the reduced cost values for each nonbasic variable are nonnegative, the th BFS is optimal. The interpretation of the solution is: units from node to node 6, units from node to node, units from node to node 6, units from node to node 7, units from node to node, and units from node to node 7. 9

10 . (a) Suppose that B and B both determine the same BFS x. Then x has zeros in the n m columns not in B; it also must have zeros in the columns in B B. Hence it is degenerate. (b) The converse is not true. A counterexample is In this counterexample, and min x + x [ ][ ] x s.t. = x x, x. n = m = n m =. [ ] Therefore the vertex (, ) which contains a zero is degenerate. Moreover, it corresponds to the unique basis [ ] [ ] {, }.. Relax the integer constraint, the optimum solution is (, 8 ), the intersection of x x + x = and x + x + x =. Staring with this, we have x = x x x = 8 + x x. Observe the equation x = 8 + x x, we have x x = 8 x x 8. Actually, x x because of the integer constraint. Substituting x with x + x, we have x ( x + x ),

11 or x. Then we have the first cut, x =. After cutting by x =, the optimal solution is (, ), already an integer solution. The problem is solved. And the graph for feasible domain and the cut are illustrated in the following graph. feasible domain xy x = x +x = x -x = -x +x = x -7-8 the first cut -9. For each constraints in AX + s b, - - we multiply respective to each constraint and have new constraints with integer coefficients. If there exists any non-original variables, we can always substitute them with the combination of original variables by the constraints.. (a) The network for the numerical example is as follows.

12 (b) The solution to the maximum flow problem in (a) is as follows (c) Let S = {,, 6, } and S = {,,, 7, 8, 9,, }.

13 Then we have the minimum cut K = {(6, ), (, ), (.), (, )}. Follow the instruction, we use the st, the nd, the th rows and the st column to cover all zeros in the reduced matrix. (d) Consider the assignment problem corresponding to zero cells: max x + x + x + x + x + x + x + x + x s.t. x ij = or x, x, x, x, x, x, x, x, x. x x x x x x x x x It is obvious that only one of {x, x } can be. Similarly, only one of {x, x, x } can be. Only one of {x, x } can be, etc. Therefore, the optimal values of this LP equals the number of independent zero cells. Consider the dual problem of the above LP: min u + u + u + u + u + v + v + v + v + v s.t. u + v u + v u + v u + v u + v u u + v u + v v u + v The optimal value of the dual LP equals the number of lines required (least) to cover all zero cells. It concludes that the maximum number of independent zero cells in a reduced cost assignment matrix is equal to the minimum number of lines to cover all zeros in the matrix since the primal optimum equals the dual optimum.

14 6. (a) min i j c ij x ij s.t. 6 x ij =, j =,..., 6 i= 6 x ij =, i =,..., 6 j= u i u j + 6x ij, i j 6 u i R, i =,...,6 x ij {, }. (b) Step : From each row, we find the row minimum and subtract it from all entries on that row. M M 6 M 6 M M 9 M Step : From each column, we find the column minimum and subtract it from all entries on that column. M M 9 M 9 M 6 7 M 8 M Step : We draw lines across rows and columns in such a way that all zeros are covered and that the minimum number of lines have been used(in this case lines across the th row, the 6th row, the th column, and the 6th column). M 7 M 9 M 9 M 6 M M Step : A test for optimality If the number of lines just drawn equals the number of rows of the cost matrix,

15 we are done. If the number of lines is less than the number of rows of the cost matrix, we go to step. Now the number of lines is, which is less than 6, the number of rows of the cost matrix. Step : We find the smallest entry which is not covered by the lines, which in this case is the (, )-entry, and subtract it from each entry not covered by the lines. We also add it to each entry which is covered by a vertical and horizontal line. Now we can go back to Step. Step *: Draw lines across zeros (the st, the th, and the 6th columns, the th, and the 6th rows). M 6 M 8 M 9 M M 6 M Step *: A test for optimality If the number of lines just drawn equals the number of rows of the cost matrix, we are done. If the number of lines is less than the number of rows of the cost matrix, we go to step. Now the number of lines is, which is less than 6, the number of rows of the cost matrix. Step *: We find the smallest entry which is not covered by the lines, which in this case is the (, )-entry, and subtract it from each entry not covered by the lines. We also add it to each entry which is covered by a vertical and horizontal line. Now we can go back to Step. Step **: We draw lines across rows and columns in such a way that all zeros are covered and that the minimum number of lines have been used. And the number of lines just drawn equals the number of rows of the cost matrix, we are done. M 6 9 M 6 8 M 9 M M 7 9 M The answer to the assignment problem is assigning to, to, to, to, to 6, and 6 to.

16 (c) Consider the subtour {(, ), (, 6), (6, )}. This can be eliminated by Tucker s formulation because we will get a contradiction, u u + 6 () u u () u 6 u + 6 () () + () + () 8. = (d) The cost of the solution in (c) is =. This value,, can be a lower bound on the minimum cost. And we branch by adding one more constraint x st = or x st =, respectively, to the constraints on the candidate problem, two candidate subproblem are generated. 6