This matrix is not TU since the submatrix shown below has determinant of

Transcription

1 EMIS 8373: Integer Programming [Homework 5 Solutions] 1 Problem 1 Problem 1 on page 50 of the Wolsey text book. (a) A 1 = This matrix is not TU since the submatrix shown below has determinant of 2. A 1 =

2 EMIS 8373: Integer Programming [Homework 5 Solutions] 2 (b) A 2 = Partition the rows such that M 1 = {1, 2} and M 2 = {3, 4, 5}. A 2 =

3 EMIS 8373: Integer Programming [Homework 5 Solutions] 3 For column 1, 2 i=1 a i1 5 i=3 a i1 = 1 ( 1) = 0. For column j {2, 3, 4, 7}, 2 i=1 a ij 5 i=3 a ij = 1 1 = 0. For column j {5, 6}, 2 i=1 a ij 5 i=3 a ij = 0 0 = 0. Thus, A 2 is TU by Proposition 3.2

4 EMIS 8373: Integer Programming [Homework 5 Solutions] 4 Is this matrix TU?

5 EMIS 8373: Integer Programming [Homework 5 Solutions] 5 Problem 2 (i) For each pair of employees (i, j) let c ij be a compatibility score for putting employees i and j in the same room. If we let x ij = 1 if and only if employees i and j are assigned to the same room. A valid IP formulation for the general case where there are n employees is max s.t. n 1 i=1 n j=i+1 c ij x ij x ji + x ij = 1 j<i j>i x ij {0, 1}. i = 1, 2,..., n Notice the constraint matrix for this IP is a 0-1 matrix such that each column has exactly two non-zero elements. This matrix appears to be TU. It turns out, however, that it is not in all cases.

6 EMIS 8373: Integer Programming [Homework 5 Solutions] 6 For example, suppose that the compatibility score is either 0 or 1 and that employees 1, 2, and 3, are compatible with each other. In this case the formulation would contain the constraints listed below. x 12 + x = 1 (Employee 1) x 12 + x = 1 (Employee 2) x 13 + x = 1 (Employee 3) Since the constraints above are part of the formulation, the constraint matrix would contain the submatrix shown below which has a determinant of Thus, the matrix is not TU

7 EMIS 8373: Integer Programming [Homework 5 Solutions] 7 This problem can also be formulated as matching problem. Let G = (V, E) be a graph with one node for each employee and edge (i, j) in E if employees i and j are compatible. A rooming assignment is possible if and only the maximum cardinality matching in G has size V 2. Recall from our discussion of the solutions to Homework 4 that the matching problem in a non-bipartite graph cannot be solved as an LP.

8 EMIS 8373: Integer Programming [Homework 5 Solutions] 8 (ii) This can be formulated as a minimum-cost network flow problem. Specifically, it is an assignment problem. Nodes: The network has nodes E1, E2,... En for the English children, each with a supply of 1, and nodes F 1, F 2,... F n for the French children each with a demand of 1. Arcs: For each potential pair (i, j) there is a unit-capacity arc (Ei, F j) with cost w ij where w ij is the willingness of that pair to form a team. Since this is a network flow problem, the constraint matrix is TU which means that we can solve it as a linear program and the optimal values of the flow variables will all be in {0, 1}. If there is flow on arc (i, j), that indicates English child i is teamed with French child j.

9 EMIS 8373: Integer Programming [Homework 5 Solutions] 9 Problem 3 (a) This is a minimum spanning tree (MST) problem which can be solved with Kruskal s algorithm. The minimum length of a spanning three in the given graph is 93. The edges of a MST are shown below. (1,2) (3,4) (3,6) (6,10) (7,11) (9,12) (1,4) (3,5) (5,8) (7,8) (8,9) (12,13)

10 EMIS 8373: Integer Programming [Homework 5 Solutions] 10 (b) We can use the following integer programming model to solve this problem. # AMPL model for the minimum cost tree problem. set V; # nodes in the network set E within {V,V}; # undirected edges in the network param c {E} default 0; # cost of edge (i,j) param b {V} default 0; # supply/demand for node i # Let U be the subset of nodes that are to be # connected by the tree. # Pick a root node r in U and let b[r] = card(u) - 1; # Let b[i] = -1 for every node in U\{r} and # let b[i] = 0 for every node in V\U.

11 EMIS 8373: Integer Programming [Homework 5 Solutions] 11 param M := sum{i in V: b[i] > 0} b[i]; # Maximum amount of flow that would be on any arc. var y {E} binary; # y[i,j] = 1 if edge (i,j) is in the tree var x {i in V, j in V: (i,j) in E or (j,i) in E} >= 0; # x[i,j] represents the "flow" on edge (i,j) minimize cost: sum{(i,j) in E} c[i,j] * y[i,j]; subject to flow_balance {i in V}: sum{j in V: (i,j) in E or (j,i) in E} x[i,j] - sum{j in V: (i,j) in E or (j,i) in E} x[j,i] = b[i]; subject to fixed_cost {(i,j) in E}: M * y[i,j] >= (x[i,j] + x[j,i]);

12 EMIS 8373: Integer Programming [Homework 5 Solutions] 12 This model is essentially a MCNFP model, but with a fixed (rather than variable) cost for putting flow on an arc. The following script uses the model to solve the MST problem from part (a) and then the suggested problems from part (b). # prob3_run.txt model tree_model.txt; data smu_data.txt; # set the supply/demand parameters for the MST problem let {i in V} b[i] := -1; let b[1] := (card(v) -1); solve; display {(i,j) in E: (x[i,j] + x[j,i]) > 0};

13 EMIS 8373: Integer Programming [Homework 5 Solutions] 13 # set the supply/demand parameters for the MST problem # where we only need to connect the odd-numbered nodes set U; let U := {1, 3, 5, 7, 9, 11, 13}; let b[1] := (card(u) -1); let {i in U diff {1}} b[i] := -1; let {i in V diff U} b[i] := 0; solve; display {(i,j) in E: (x[i,j] + x[j,i]) > 0}; # set the supply/demand parameters for the MST problem # where we only need to connect the prime-numbered nodes let U := {1, 2, 3, 5, 7, 11, 13}; let b[1] := (card(u) -1); let {i in U diff {1}} b[i] := -1; let {i in V diff U} b[i] := 0; solve; display {(i,j) in E: (x[i,j] + x[j,i]) > 0};

14 EMIS 8373: Integer Programming [Homework 5 Solutions] 14 % ampl < prob3_run.txt ILOG AMPL 8.000, licensed to "southern methodist-dallas, tx". AMPL Version (OSF1 V4.0) CPLEX 8.0.0: optimal integer solution; objective MIP simplex iterations 358 branch-and-bound nodes set {(i,j) in E: x[i,j] + x[j,i] > 0} := (1,2) (3,4) (3,6) (6,10) (7,11) (9,12) (1,4) (3,5) (5,8) (7,8) (8,9) (12,13); CPLEX 8.0.0: optimal integer solution; objective MIP simplex iterations 23 branch-and-bound nodes set {(i,j) in E: x[i,j] + x[j,i] > 0} := (1,4) (3,4) (3,5) (5,8) (7,8) (7,11) (8,9) (

15 EMIS 8373: Integer Programming [Homework 5 Solutions] 15 CPLEX 8.0.0: optimal integer solution; objective MIP simplex iterations 12 branch-and-bound nodes set {(i,j) in E: x[i,j] + x[j,i] > 0} := (1,2) (2,3) (3,5) (5,8) (7,8) (7,11) (8,13);