1. updated 21 April2008 Linear Programs with Totally Unimodular Matrices

2. Basic Feasible Solutions Standard Form slide 1

3. Basic Feasible Solutions SolutionBasic VariablesNon-Basic VariablesIntersection BFS 1x = 2.25 & y = 3.75s 1 = s 2 = 0(1) and (2) BFS 2x = 6 & s 2 = 15y = s 1 = 0(1) and x-axis BFS 3y = 5 & s 1 = 1x = s 1 = 0(2) and y-axis BFS 4s 1 = 6 & s 2 = 45x = y = 0x-axis and y-axis slide 2

4. Vector-Matrix Representation slide 3

5. Example MCNFP 5 14 (1, 0,2) 3 2 0 -3 -2 (2, 0,2) (4, 1,3) (4, 0,3) (3, 2,5) slide 4

6. LP for Example MCNFP Min 3x 12 + 2 x 13 + x 23 + 4 x 24 + 4 x 34 s.t. x 12 + x 13 = 5{Node 1} x 23 + x 24 – x 12 = -2{Node 2} x 34 – x 13 - x 23 = 0 {Node 3} – x 24 - x 34 = -3 {Node 4} 2  x 12  5, 0  x 13  2, 0  x 23  2,1  x 24  3, 0  x 34  3, slide 5

7. Matrix Representation of Flow Balance Constraints slide 6

8. Solving for a Basic Feasible Solution slide 7

9. Cramer’s Rule Use determinants to solve x=A -1 b. Take the matrix A and replace column j with the vector b to form matrix B j. slide 8

10. Using Cramer’s Rule to Solve for x 12 slide 9

11. Total Unimodularity A square, integer matrix is unimodular if its determinant is 1 or -1. An integer matrix A is called totally unimodular (TU) if every square, nonsingular submatrix of A is unimodular. slide 10

12. Total Unimodularity A square, integer matrix is unimodular if its determinant is 1 or -1. An integer matrix A is called totally unimodular (TU) if every square, nonsingular submatrix of A is unimodular. slide 11

13. Sufficient Conditions for TU An integer matrix A is TU if 1.All entries are -1, 0 or 1 2.At most two non-zero entries appear in any column 3.The rows of A can be partitioned into two disjoint sets M 1 and M 2 such that If a column has two entries of the same sign, their rows are in different sets. If a column has two entries of different signs, their rows are in the same set. slide 12

14. The Matrix of Flow Balance Constraints Every column has exactly one +1 and exactly one -1. This satisfies conditions 1 and 2. Let the row partition be M 1 = {all rows} and M 2 = {}. This satisfies condition 3. Thus the flow balance constraint matrix is TU. slide 13

15. Using Cramer’s Rule to Solve for x 12 slide 14

16. Expansion by Minors: 4-by-4 Matrix slide 15

17. Expansion by Minors: 3-by-3 Matrix slide 16

18. Using Cramer’s Rule to Solve for x 12 slide 17

19. Using Cramer’s Rule to Solve for x 12 When we expand along minors, the determinants of the submatrices will be +1, -1, or 0. Therefore, the determinant will be an integer: (5)(+1, -1, or 0) + (-2) (+1, -1, or 0) + 0 + (-3) (+1, -1, or 0). slide 18

20. Using Cramer’s Rule to Solve for x 12 slide 19

21. TU Theorems Matrix A is TU if and only if A T is TU. Matrix A is TU if and only if [A, I] is TU. – I is the identity matrix. If the constraint matrix for an IP is TU, then its LP relaxation has an integral optimal solution. The BFSs of an MCNF LP are integer valued. slide 20