linear programs with totally unimodular matrices
DESCRIPTION
Linear Programs with Totally Unimodular Matrices. updated 19 March 2009. Basic Feasible Solutions. Standard Form. Basic Feasible Solutions. Vector-Matrix Representation. Example MCNFP. -2. (3, 2,5). (4, 1,3). 2. 5. -3. 1. 4. (1, 0,2). (4, 0,3). (2, 0,2). 3. 0. - PowerPoint PPT PresentationTRANSCRIPT
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updated 19 March 2009
Linear Programs with Totally Unimodular Matrices
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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
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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
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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
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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)
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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,
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Matrix Representation of Flow Balance Constraints
3025
11000101100110100011
34
24
23
13
12
xxxxx
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Solving for a Basic Feasible Solution
3025
1100101001010011
34
24
13
12
xxxx
3025
1100101001010011
1
34
24
13
12
xxxx
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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.
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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
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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
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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
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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.
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The Matrix of Flow Balance Constraints
3025
11000101100110100011
34
24
23
13
12
xxxxx
• Every column has exactly one +1 and exactly one -1.• Conditions 1 and 2 are satisfied.• Let the row partition be M1 = {all rows} and M2 = {}.• Condition 3 is satisfied.• Therefore, the flow balance constraint matrix is TU.
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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.
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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
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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
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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
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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).
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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.
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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