Updated 21 April2008 Linear Programs with Totally Unimodular Matrices.
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Transcript of Updated 21 April2008 Linear Programs with Totally Unimodular Matrices.
updated 21 April2008
Linear Programs with Totally Unimodular Matrices
Basic Feasible Solutions
0,
)2(4595
)1(6s.t.
85max
yx
yx
yx
yxz
0,,,
4595
6s.t.
85max
21
2
1
ssyx
syx
syx
yxz
Standard Form
slide 2
Basic Feasible Solutions
0,,,
4595
6s.t.
85max
21
2
1
ssyx
syx
syx
yxz
Solution Basic Variables Non-Basic Variables Intersection
BFS 1 x = 2.25 & y = 3.75 s1 = s2 = 0 (1) and (2)
BFS 2 x = 6 & s2 = 15 y = s1 = 0 (1) and x-axis
BFS 3 y = 5 & s1 = 1 x = s1 = 0 (2) and y-axis
BFS 4 s1 = 6 & s2 = 45 x = y = 0 x-axis and y-axis
slide 3
Vector-Matrix Representation
45
6
1095
011121
b
ssyx
A
45
6,
10
01,,
1
5,
09
11,,
15
6,
15
01,,
75.3
25.2,
95
11,,
121
11
12
1
bAAssB
bAAsyB
bAAsxB
bAAyxB
BB
BB
BB
BB
slide 4
Example MCNFP
5 1 4(1, 0,2)
3
2
0
-3
-2
(2, 0,2)
(4, 1,3)
(4, 0,3)
(3, 2,5)
slide 5
LP for Example MCNFP
Min 3x12 + 2 x13 + x23 + 4 x24 + 4 x34 s.t. x12 + x13 = 5 {Node 1} x23 + x24 – x12 = -2 {Node 2}
x34 – x13 - x23 = 0 {Node 3} – x24 - x34 = -3 {Node 4}
2 x12 5, 0 x13 2, 0 x23 2, 1 x24 3,
0 x34 3,
slide 6
Matrix Representation of Flow Balance Constraints
3025
11000101100110100011
34
24
23
13
12
xxxxx
slide 7
Solving for a Basic Feasible Solution
3025
1100101001010011
34
24
13
12
xxxx
3025
1100101001010011
1
34
24
13
12
xxxx
slide 8
Cramer’s Rule
Use determinants to solve x=A-1b.
abaa
abaaB
AB
xnnnnn
nij
j
j
j
21
1112
,
Take the matrix A and replace column j with the vector b to form matrix Bj.
slide 9
Using Cramer’s Rule to Solve for x12
A
Bx
)2,1(
12
1100
1010
0101
0011
1103
1010
0102
0015
integer?an Is )2,1(B
?1Does Aslide 10
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.
TUTUNot 1100101001010011
1111
slide 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.
1100
1010
0101
0011
111
01
slide 12
Sufficient Conditions for TU
An integer matrix A is TU if1. All entries are -1, 0 or 12. At most two non-zero entries appear in any column3. The rows of A can be partitioned into two disjoint
sets M1 and M2 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 13
The Matrix of Flow Balance Constraints
3025
11000101100110100011
34
24
23
13
12
xxxxx
• Every column has exactly one +1 and exactly one -1.• This satisfies conditions 1 and 2.
• Let the row partition be M1 = {all rows} and M2 = {}.• This satisfies condition 3.
• Thus the flow balance constraint matrix is TU.
slide 14
Using Cramer’s Rule to Solve for x12
A
Bx
)2,1(
12
1100
1010
0101
0011
1103
1010
0102
0015
integer?an Is )2,1(B?1Does A Yes.
slide 15
Expansion by Minors: 4-by-4 Matrix
aaaaaaaaa
aaaaaaaaaa
a
aaaaaaaaa
aaaaaaaaaa
a
aaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaa
342414
332313
322212
41
442414
432313
422212
31
443414
433313
423212
21
443424
433323
423222
11
44342414
43332313
42322212
41312111
44434241
34333231
24232221
14131211
slide 16
Expansion by Minors: 3-by-3 Matrix
aaaaa
aaaaa
aaaaa
aaaaaaa
aaaaaaaa
aaaaaaaaa
3122322113
3123332112
3223332211
3231
222113
3331
232112
3332
232211
333231
232221
131211
slide 17
Using Cramer’s Rule to Solve for x12
A
Bx
)2,1(
12
1100
1010
0101
0011
1103
1010
0102
0015
integer?an Is )2,1(B
?1Does A Yes.slide 18
Using Cramer’s Rule to Solve for x12
1100101001013025
1103101001020015
• 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 19
Using Cramer’s Rule to Solve for x12
A
Bx
)2,1(
12
1100
1010
0101
0011
1103
1010
0102
0015
integer?an Is )2,1(B
?1Does A Yes.
Yes.
slide 20
TU Theorems
• Matrix A is TU if and only if AT 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 21
