Kramer’s (a.k.a Cramer’s) Rule
• Component j of
x = A-1b is
 aij
det B j

x j  det , B j   
A
a
 n1
a
12
1


a
b
n2


 
ann
b a
n
• Form Bj by replacing
column j of A with b.
1n
Total Unimodularity
• A square, integer matrix B is unimodular
(UM) if its determinant is 1 or -1.
• An integer matrix A is called totally
unimodular (TUM) if every square,
nonsingular submatrix of A is UM.
• From Cramer’s rule, it follows that if A is
TUM and b is an integer vector, then every
BFS of the constraint system Ax = b is
integer.
TUM Theorem
• An integer matrix A is TUM 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 MCNFP constraint matrices are TUM.
General Form of the MCNF
Problem
min
c x

x


l  x u
( i , j )A
st
ij
j|( i , j )A
ij
ij
ij

ij
ij
 x
 j|( j ,i )A
ji

b
i
i V
(i, j )  A
Flow Balance Constraint Matrix
x
2
1
Capacity Constraints
0 0 1 0 0
1 0 0 1 0
0 1 0 0 1
0 0 0 0 0
1 0 0 0 0
0 1 0 0 0
12
x
23
x
31
 1 0  1


A   1 1 0 
 0  1 1 
3
1
0

0
C
1
0

0
x
0
0 0 0 
0 0 0

1 0 0 
0 1 0 

0 0  1
0
0
Constraints in Standard Form
Ax  b
 x
u 


C  s    
l


e 

x0
Shortest Path Problems
• Defined on a Network
– Nodes, Arcs and Arc Costs
– Two Special Nodes
• Origin Node s
• Destination Node t
• A path from s to t is an alternating sequence
of nodes and arcs starting at s and ending at
t:
s,(s,v1),v1,(v1,v2),…,(vi,vj),vj,(vj,t),t
s=1, t=3
1
We Want a Minimum Length Path From s to t.
5
10
2
7
1
7
4
1,(1,2),2,(2,3),3
Length = 15
1,(1,2),2,(2,4),4,(4,3) Length = 13
1,(1,4),4,(4,3),3
Length = 14
3
Maximizing Rent Example
• Optimally Select Non-Overlapping Bids for
Arrive Day Depart Day Bid ($)
10 periods
1
2
2
1
5
7
2
4
2
3
7
1
3
8
11
4
5
1
4
6
6
5
6
3
5
9
7
7
8
4
7
9
5
8
10
3
Shortest Path Formulation
t
d10
-7
-2
d1
s
-2
d2
0
d3
-7
0
d4 -1 d5
d6
-3
-6
-1
-11
d9
-5
0 d7 -4 d8
-3
MCNF Formulation of Shortest
Path Problems
•
•
•
•
•
Origin Node s has a supply of 1
Destination Node t has a demand of 1
All other Nodes are Transshipment Nodes
Each Arc has Capacity 1
Tracing A Unit of Flow from s to t gives a
Path from s to t
Maximum Flow Problems
• Defined on a Network
–
–
–
–
Source Node s
Sink Node t
All Other Nodes are Transshipment Nodes
Arcs have Capacities, but no Costs
• Maximize the Flow from s to t
Example: Rerouting Airline
Passengers
Due to a mechanical problem, Fly-By-Night
Airlines had to cancel flight 162 - its only nonstop flight from San Francisco to New York. The
table below shows the number of seats available
on Fly-By-Night's other flights.
Flight
160
115
153
102
170
150
180
From
San Francisco
San Francisco
Denver
Denver
Houston
Atlanta
Chicago
To
Denver
Houston
Atlanta
Chicago
Atlanta
New York
New York
Number of seats
5
6
4
2
5
7
4
Formulate a maximum flow problem that will tell Fly-By-Night
how to reroute as many passengers from San Francisco to
New York as possible.
5
SF
D
2
C
4
4
6
H
(4,5) D
SF
(5,6)
H
NY
7
5
(2,2)
A
(flow, capacity)
C
(2,4)
(5,5)
A
(2,4)
NY
(7,7)
Max Flow from SF to NY
= 2+2+5=9
MCNF Formulation of Maximum
Flow Problems
• Let Arc Cost = 0 for all Arcs
• Add an infinite capacity arc from t to s
– Give this arc a cost of -1
Maximum-Flow Minimum-Cut
Theorem
5
SF
D
H
SF
C
4
6
5
2
5
A
D
C
4
6
H
5
A
• Removing arcs (D,C)
4
and (A,NY) cuts off
NY
SF from NY.
7
• The set of arcs{(D,C),
(A,NY)} is an s-t cut
4
with capacity 2+7=9.
NY • The value of a
maximum s-t flow =
the capacity of a
minimum s-t cut.