EMIS 8374:
Network Flows
“Easy” Integer Programming
Problems: Network Flow Problems
updated 4 April 2004
Basic Feasible Solutions
max
z  5x  8 y
x  y  6
5 x  9 y  45
s.t.
x,
max
s.t.

y
1
2 
0
Standard Form
z  5x  8 y
y

5x  9 y

x,
s,
x

y,
1
s
s
s
1
2
2

6
 45

0
slide 1
Basic Feasible Solutions
max
z  5x  8 y
s.t.
y

5x  9 y

x,
s,
x

y,
1
s
s
s
1
2
2

6
 45

0
Non-Basic Variables Basic Variables Intersection
BFS 1 S1 = S2 = 0
x = 2.25, y = 3.75 Constraints (1) and (2)
BFS 2 y = S1 = 0
x = 6, S2 = 15
Constraint (1) and x-axis
BFS 3 x = S2 = 0
y= 5, S1 = 1
Constraint (2) and y-axis
BFS 4 x = y = 0
S1 = 6, S2 = 45
x-axis and y-axis
slide 2
Vector-Matrix Representation
x
y s s
1 1 1 0 
A 

5 9 0 1 
1
B

x , y 
B

x , s 
B

2
y , s 
1
B  s , s 
1
2
1
AB  5
1
AB  5
1
AB  9
1
AB  0
2
1
9
0
1
1
0
0
1
6
b 
45
2.25
AB b  3.75
6
1
AB b  15
5
1
AB b  1
6
1
AB b  45
1
slide 3
LP Formulation of Shortest Path
Example
min
8x  10 x  4 x  9 x 5x
6 x  2 x  3x  4 x  5x 
x x
x  x x  x
x  x x  x
x  x x  x
x  x x  x
x  x
0  x 1
12
13
34
st
35
23
45
12
13
23
24
25
12
34
35
13
23
45
46
24
34
56
25
35
45
46
56
24
46
25
56

1
0
0
0
0
 1
ij
slide 4
Matrix Representation
x x x
12
13
23
x
24
x
25
x
34
x
35
x
45
x
46
x 
56
1 0
0
0
0
0
0
0
0
 1
 1
 1 0
 0
1
1
1 0
0
0
0
0


 
0
1
1 0
0
0
 0 1 1 0
 0
A 
, b   
0
0
0

1
0

1
0
1
1
0


 0
 0
 0
0
0
0 1 0 1 1 0
1


 
0
0
0
0
0
0
0
0

1

1


 1
Observation: The last row of the matrix is equal to –1 times
the sum of the other rows.
MCNF LPs always have one redundant row.
slide 5
Matrix Representation without
the constraint for node 6
x x x
12
13
23
x
24
x
25
x
34
x
35
x
45
1 0
0
0
0
0
0
 1
 1 0
1
1
1 0
0
0

A   0 1 1 0
0
1
1 0

0
0 1 0 1 0
1
 0
 0
0
0
0 1 0 1 1
x
46
x 
56
0 0
 1
0
0 0

 
0 0, b  0

 
1 0
0
0
0 1
A BFS: B = {x12, x13, x24, x35, x56}
slide 6
Solving for the BFS
Constraints after non-basic variables are removed:
x x
x x
x x
x
x x
 1
12
13
24
12
35
13
 0
35
 0
24
56
 0
 0
Solution: x24 = 0, x12 = 0, x13 = 1, x35 = 1, x56 = 1
slide 7
Solving for the BFS with Matrix
Algebra
x x
12
13
1
 1
 1 0

AB   0  1
0
 0
 0
0
x
24
x
35
0
0
1
0
0
1
1
0
0
1
x 
56
0
 1
0
0

 
0, b  0

 
0
0
0
1
0  x12
 1  
   x13
1
x B  AB b  0   x24
 1  x35
 1  x56
slide 8
Kramer’s (a.k.a Cramer’s) Rule
Component j of x = A-1b is
x
j

det B j
det A
 aij

, Bj   
a
 n1
a
12
1


a
b
n2


 
ann
b a
n
1n
Take the matrix A and replace
column j with the vector b.
slide 9
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.
• From Cramer’s rule, it follows that if A is TU and b is an
integer vector, then every BFS of the constraint system Ax
= b is integer.
• Examples:
– The matrix AB from the shortest path example is TU.
– The matrix A from the shortest path example is TU.
– The constraint matrix for any MCNF LP is TU.
slide 10
TU Theorem
• An integer matrix A is TU if
– All entries are -1, 0 or 1
– At most two non-zero entries appear in any column
– The rows of A can be partitioned into two disjoint sets
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.
• The matrix A is TU if and only if is AT TU.
• The matrix A is TU if and only if [A, I] is TU.
Where I is the identity matrix.
slide 11
MCNF LPs are TU
 x12

 1
 1

 0
 0

 1
 0

 0
 0

 0
x
13
x
23
x
24
1
0
0
0
1
1
1
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
x


0
0

1
1

0
0

0
0

1
34
b
 5
 
 0
 
 3
 5
 
 5
2
 
2
 3
 
 3
Flow Balance:
A is TU, so AT is TU.
Capacity
A 
M   .
 I 
M
A
T
T
A I .
I is TU.

T
 M is TU.
slide 12