MSIS 685: Linear Programming
Lecture 11
Minimal Network Flow Problem
11/24/98
Scriber: Yu Xia
Instructor: Farid Alizadeh
Transportation Problem
Example:
Problem:
a1  12 , a2  10 , a3  7
b1  9 , b2  6 , b3  8 , b4  6
c ij :
12
6
3 7
2
5
10 8 3
3 4 9
Solution:
1) Use Northwest corner method to find the initial solution:
9 3
12
3 7
10
x ij :
1 6 7
9
6 8 6
Dual:
12 6   u1
 10 8  u 2


4 9 u3
v1
v2
v3
v4
12
6
4
9
0
4
0
2) Change the base of primal:
9
3


3
6 6
 3
7
1 6
7
1 6
Dual:
12 6   u1
0
2
8  u 2  10
4 9 u 3  14
v1
12
v2
v3
v4
6 18 23
3) Change the primal base:
6 6

0 6
6


 9
3
7
1
1
6
7
This is a degenerate solution
The dual:
12 6  7 u1
0
2
8
u 2  10
4
u 3  14
v1
12
v2
v3
v4
6 18 7
4) The primal:
0 6  6
9
6 0 6
 9
1
7
The dual:
6 3 7 u1
2  8  u2
 4
v1
v2
u3
v3
v4
3 6 3
7
1
7
0
5
1
5) The primal:
9
6
0


6
5 1 6
 9 1
1
7
7
The dual:
6 3 7 u1
2 10
 u2
 4
u3
v1
v2
v3
v4
2 6 3
7
0
4
1
6) The primal:
5 1 6 

6 1 5
  9
9 1
1
7
7
The dual:
6 3 7 u1
2
3 u2
  4
u3
v1
v2
v3
0
4
1
v4
6 6 3
7
7) The primal:
6 1
5


9

1
7
5
  4
v2
v3
6 6 3
 4

The dual:
6 3 7 u1
2
3 u2
v1
6 6
u3
v4
7
0
4
1
6
2
8) The primal:
6
4
5
6

2
4 8
6  4
5 2

The dual:
6 3  u1
2
3 u2
5 3
u3
v1
v2
v3
9
9) The primal
4 8
2

v2
8 4
6  8
1 6

The dual:
3 7 u1
2
3 u2
5 3
u3
v1
0
6
3
v4
8 6 3
4
5
6
v3
6 4 3
0
4
1
v4
7
Conclusion:
Optimal base of the primal:
8 4
8
1 6
2
2
Minimal Cost Flow Problem
A graph can be represented by G  ( N , A) . N is the set of nodes, A is the set of arcs.
: Supplying nodes
: Transportation nodes
: Demanding nodes
: Arc
Transshipment Problem:
min  c ij x ij
( i , j ) A
s.t.

xik 
i
( i , k )A
x
kj
 bk (k  1,, m)
j
( k , j )A
xij  0
Definitions:
Walk: ( i1 ,i2 )  (i2 , i3 )  …  (ik 1 , ik )  (ik , ik 1 )
Path: A path is a walk with no vertices visited more than once.
Circuit: A circuit is a walk where i1  ik 1
Cycle: A cycle is a circuit, and it is a path if (ik , ik 1 ) is taken away.
In cycle number of edges equals number of vertices. In path number of edges
equals number of vertices minus 1.
Strongly Connected: A directed network G=(N, A) is strongly connected if
there exits a path between any pair of nodes.
Weakly Connected: A network is weakly connected if it is connected when
treated as an undirected network.
Forest: A forest is a network with no cycles if treated as undirected
Tree: A tree is a weakly connected forest
Degree of Node: Degree of Node (i) is the number of arcs with one end
pointing to i.
Indegree of Node: Indegree of Node (i) is the number of arcs with the form (k,
i).
Outdegree of Node: Outdegree of Node (i) is the number of arcs with the form
(i, k).
Degree of Node (i)= Indegree of Node (i)+ Outdegree of Node (i)
Lemma: A tree with n nodes has n-1 arcs.
Proof:
By induction:
1) The lemma is true for 1-node and 2-node trees.
2) Consider n-node tree:
Claim: Every tree that has more than one node has at least one node of
degree 1.
Use contradiction: There is a path P between node 1 and node 2. If
the degree of each node of a tree is not less than one, then it is
possible to find another path between node 1 and 2, and none of the
arcs in this path is identical to any arc in P.
Remove that degree 1 node and the arc, we obtain a tree has n-1 nodes.
Minimal cost flow problem can be presented as Ax  b
A  R mn : Incidence Matrix ( m : the number of nodes, n : the number of arcs)
Example:
1
1
2
2
3
5
4
3
4
1
 1
 1 1 1


A


1 1


1
1  1

m
For any feasible network, the condition  bk  0 must be satisfied.
k 1
If the rank of the incidence matrix is k, then the graph has m-k parts.
Spanning Tree: T is a spanning tree of G = (N, A ) , if T= (N, A / ) , A/  A , and T is a tree.
Lemma: For every (undirected) tree, there is a way to label nodes with 1, 2 … m
such that:
a) Node labeled 1 is arbitrary.
b) For each node labeled i (i  1) , there is exactly one edge {i, j} where j  i .
To find the label like this use:
Method 1: breadth-first search.
Example:
1)
1
2)
2
3
1
1
2
5
2
3)
3
4
1
1
2
3
4
 1 1

1
1  1


A
1


1



1 
Get rid of the first row, the incidence matrix is an upper triangular matrix and
no zero in diagonal.
Method 2:
Start from the leaf, and mark it m.
Take it and the associated arc away, and then mark m-1 to the leaf of the new
tree.
…
For a minimal cost flow problem, a basic feasible solution corresponds to a spanning
tree. One tree differs from another with one arc.