CS计算机代考程序代写 CSC 445/545 – Summer 2021 – Network Simplex Examples III

CSC 445/545 – Summer 2021 – Network Simplex Examples III

CSC 445/545 – Summer 2021
Network Simplex Examples III

Bill Bird

Department of Computer Science
University of Victoria

July 11, 2021

University of Victoria – CSC 445/545 – Summer 2021 1

Examples

This lecture contains examples of exercises of the form ‘solve this minimum cost network
flow problem’. The term ‘solve’ means to determine if the problem is infeasible, unbounded
or has an optimal flow, and to find the optimal flow if one exists. Therefore, determining
that a problem instance is infeasible or unbounded is considered ‘solving’ that problem.

All of the examples in this lecture involve starting from a basis that is neither primal nor
dual feasible and applying a two phase method to solve the problem.

The examples here all use the network equivalent of the largest coefficient rule to select the
entering arc (primal pivoting) or leaving arc (dual pivoting) at each step.

University of Victoria – CSC 445/545 – Summer 2021 2

Example 1a (1)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

2

d

6
e

9

f14

g

-4

h

-2

i

2

-8

-9

-4

-8

2

10 4

5

-1

10

15

-11 -11

Task: Starting from the infeasible basis above, use a two phase method (dual first, primal
second) to solve the network flow problem.

University of Victoria – CSC 445/545 – Summer 2021 3

Example 1a (2)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

0

0

0

0

0

0

0

0

0

0

0

0

0

a

0

b

0

c

0

d

0
e

0

f0

g

0

h

0

i

0

-8

-9

-4

-8

2

10 4

5

0

0

0

0 0

Phase One: All edge costs are temporarily set to zero (shown in orange on the left). This
makes any basis dual feasible, so we can use the Dual Network Simplex Method to find an
optimal flow for the modified graph.

University of Victoria – CSC 445/545 – Summer 2021 4

Example 1a (3)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

0

0

0

0

0

0

0

0

0

0

0

0

0

a

0

b

0

c

0

d

0
e

0

f0

g

0

h

0

i

0

-8

-4

-8

2

10 4

5

0

0

0

0 0

-9

Iteration 1. Leaving arc is (b, c).

University of Victoria – CSC 445/545 – Summer 2021 5

Example 1a (4)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

0

0

0

0

0

0

0

0

0

0

0

0

0

a

0

b

0

c

0

d

0
e

0

f0

g

0

h

0

i

0

0

0 0

?

With (b, c) leaving, the entering arc is (d , e). The flow assigned to (d , e) will be 9 units.

University of Victoria – CSC 445/545 – Summer 2021 6

Example 1a (5)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

0

0

0

0

0

0

0

0

0

0

0

0

0

a

0

b

0

c

0

d

0
e

0

f0

g

0

h

0

i

0

-8

-13

1

9

2

10 4

5

0

0

0

0 0

End of Iteration 1. Total cost is 0.

University of Victoria – CSC 445/545 – Summer 2021 7

Example 1a (6)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

0

0

0

0

0

0

0

0

0

0

0

0

0

a

0

b

0

c

0

d

0
e

0

f0

g

0

h

0

i

0

-8

1

9

2

10 4

5

0

0

0

0 0

-13

Iteration 2. Leaving arc is (b, e).

University of Victoria – CSC 445/545 – Summer 2021 8

Example 1a (7)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

0

0

0

0

0

0

0

0

0

0

0

0

0

a

0

b

0

c

0

d

0
e

0

f0

g

0

h

0

i

0

0

0

0

?

With (b, e) leaving, the entering arc is (e, a). The flow assigned to (e, a) will be 13 units.

University of Victoria – CSC 445/545 – Summer 2021 9

Example 1a (8)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

0

0

0

0

0

0

0

0

0

0

0

0

0

a

0

b

0

c

0

d

0
e

0

f0

g

0

h

0

i

0

5

1

9

13

2

10 4

5

0

0

0

0 0

Optimal. Total cost is 0.

University of Victoria – CSC 445/545 – Summer 2021 10

Example 1a (9)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

0

0

0

0

0

0

0

0

0

0

0

0

0

a

0

b

0

c

0

d

0
e

0

f0

g

0

h

0

i

0

5

1

9

13

2

10 4

5

0

0

0

0 0

The optimal basis for the modified graph will be both primal and dual feasible. Since the
vertex supply values for the modified graph match those of the original graph, this basis will
be primal feasible for the original graph.

University of Victoria – CSC 445/545 – Summer 2021 11

Example 1a (10)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

-7

d

-3
e

-1

f4

g

-13

h

-11

i

-7

5

1

9

13

2

10 4

5

9

10

5

-10 -10

Phase Two: Starting from the feasible basis produced by Phase One, we can now run the
Primal Network Simplex Method.

University of Victoria – CSC 445/545 – Summer 2021 12

Example 1a (11)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

-7

d

-3
e

-1

f4

g

-13

h

-11

i

-7

5

1

9

13

2

10 4

5

9

10

5

-10-10

Iteration 1. Entering arc is (g , f ).

University of Victoria – CSC 445/545 – Summer 2021 13

Example 1a (12)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

-7

d

-3
e

-1

f4

g

-13

h

-11

i

-7

?

2

9

5

410

With (g , f ) entering, the leaving arc is (e, f ). The flow assigned to (g , f ) will be 2 units.

University of Victoria – CSC 445/545 – Summer 2021 14

Example 1a (13)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

-7

d

-3
e

-1

f-6

g

-13

h

-11

i

-7

5

1

7

13

8

2

2

3

9

10

10

-5

0

End of Iteration 1. Total cost is 86.

University of Victoria – CSC 445/545 – Summer 2021 15

Example 1a (14)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

-7

d

-3
e

-1

f-6

g

-13

h

-11

i

-7

5

1

7

13

8

2

2

3

9

10

10

0

-5

Iteration 2. Entering arc is (f , a).

University of Victoria – CSC 445/545 – Summer 2021 16

Example 1a (15)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

-7

d

-3
e

-1

f-6

g

-13

h

-11

i

-7

?

13

7

3

28

2

With (f , a) entering, the leaving arc is (h, i). The flow assigned to (f , a) will be 2 units.

University of Victoria – CSC 445/545 – Summer 2021 17

Example 1a (16)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

-7

d

-3
e

-1

f-1

g

-8

h

-6

i

-7

5

1

5

11

2

6

4

1

9

10

5

5

-5

End of Iteration 2. Total cost is 76.

University of Victoria – CSC 445/545 – Summer 2021 18

Example 1a (17)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

-7

d

-3
e

-1

f-1

g

-8

h

-6

i

-7

5

1

5

11

2

6

4

1

9

10

5

5

-5

Iteration 3. Entering arc is (i , f ).

University of Victoria – CSC 445/545 – Summer 2021 19

Example 1a (18)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

-7

d

-3
e

-1

f-1

g

-8

h

-6

i

-7

?

2

11

5

1

With (i , f ) entering, the leaving arc is (i , d). The flow assigned to (i , f ) will be 1 units.

University of Victoria – CSC 445/545 – Summer 2021 20

Example 1a (19)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

-7

d

-3
e

-1

f-1

g

-8

h

-6

i

-2

5

1

4

10

3

6

4 1

9

10

5

0

5

Optimal. Total cost is 71.

University of Victoria – CSC 445/545 – Summer 2021 21

Example 1b (1)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

2

d

6
e

9

f14

g

-4

h

-2

i

2

-8

-9

-4

-8

2

10 4

5

-1

10

15

-11 -11

Task: Starting from the infeasible basis above, use a two phase method (primal first, dual
second) to solve the network flow problem.

University of Victoria – CSC 445/545 – Summer 2021 22

Example 1b (2)

Supply/Cost Flow

a

0

b

0

c

0

d

0
e

0

f0

g

0

h

0

i

0

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

2

d

6
e

9

f14

g

-4

h

-2

i

2

0

0

0

0

0

0 0

0

-1

10

15

-11 -11

Phase One: All vertex supplies are temporarily set to zero (shown in orange on the left).
This makes any basis primal feasible, so we can use the Primal Network Simplex Method to
find an optimal flow for the modified graph.

University of Victoria – CSC 445/545 – Summer 2021 23

Example 1b (3)

Supply/Cost Flow

a

0

b

0

c

0

d

0
e

0

f0

g

0

h

0

i

0

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

2

d

6
e

9

f14

g

-4

h

-2

i

2

0

0

0

0

0

0 0

0

-1

10

15

-11-11

Iteration 1. Entering arc is (g , f ).

University of Victoria – CSC 445/545 – Summer 2021 24

Example 1b (4)

Supply/Cost Flow

a

0

b

0

c

0

d

0
e

0

f0

g

0

h

0

i

0

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

2

d

6
e

9

f14

g

-4

h

-2

i

2

?

0

0

0

0

0

00

With (g , f ) entering, the leaving arc is (b, e). The flow assigned to (g , f ) will be 0 units.

University of Victoria – CSC 445/545 – Summer 2021 25

Example 1b (5)

Supply/Cost Flow

a

0

b

0

c

0

d

0
e

0

f0

g

0

h

0

i

0

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

2

d

6
e

-2

f3

g

-4

h

-2

i

2

0

0

0

0

0

0

0

0

11

10

-1

4

0

End of Iteration 1. Total cost is 0.

University of Victoria – CSC 445/545 – Summer 2021 26

Example 1b (6)

Supply/Cost Flow

a

0

b

0

c

0

d

0
e

0

f0

g

0

h

0

i

0

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

2

d

6
e

-2

f3

g

-4

h

-2

i

2

0

0

0

0

0

0

0

0

11

10

4

0

-1

Iteration 2. Entering arc is (e, a).

University of Victoria – CSC 445/545 – Summer 2021 27

Example 1b (7)

Supply/Cost Flow

a

0

b

0

c

0

d

0
e

0

f0

g

0

h

0

i

0

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

2

d

6
e

-2

f3

g

-4

h

-2

i

2

?

0

0

0

0

00

0

0

With (e, a) entering, the leaving arc is (e, f ). The flow assigned to (e, a) will be 0 units.

University of Victoria – CSC 445/545 – Summer 2021 28

Example 1b (8)

Supply/Cost Flow

a

0

b

0

c

0

d

0
e

0

f0

g

0

h

0

i

0

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

2

d

6
e

-1

f3

g

-4

h

-2

i

2

0

0

0

0

0

0

0

0

10

9

1

4

0

Optimal. Total cost is 0.

University of Victoria – CSC 445/545 – Summer 2021 29

Example 1b (9)

Supply/Cost Flow

a

0

b

0

c

0

d

0
e

0

f0

g

0

h

0

i

0

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

2

d

6
e

-1

f3

g

-4

h

-2

i

2

0

0

0

0

0

0

0

0

10

9

1

4

0

The optimal basis for the modified graph will be both primal and dual feasible. Since the
vertex supply values for the modified graph match those of the original graph, this basis will
be dual feasible for the original graph.

University of Victoria – CSC 445/545 – Summer 2021 30

Example 1b (10)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

2

d

6
e

-1

f3

g

-4

h

-2

i

2

-2

-7

-6

6

8

2

2

3

10

9

1

4

0

Phase Two: Starting from the feasible basis produced by Phase One, we can now run the
Dual Network Simplex Method.

University of Victoria – CSC 445/545 – Summer 2021 31

Example 1b (11)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

2

d

6
e

-1

f3

g

-4

h

-2

i

2

-2

-6

6

8

2

2

3

10

9

1

4

0

-7

Iteration 1. Leaving arc is (b, c).

University of Victoria – CSC 445/545 – Summer 2021 32

Example 1b (12)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

2

d

6
e

-1

f3

g

-4

h

-2

i

2

1

9

4

?

With (b, c) leaving, the entering arc is (f , a). The flow assigned to (f , a) will be 7 units.

University of Victoria – CSC 445/545 – Summer 2021 33

Example 1b (13)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

-2

d

2
e

-1

f-1

g

-8

h

-6

i

-2

5

1

6

7

1

9

-5

-4

4

10

5

5

0

End of Iteration 1. Total cost is 51.

University of Victoria – CSC 445/545 – Summer 2021 34

Example 1b (14)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

-2

d

2
e

-1

f-1

g

-8

h

-6

i

-2

5

1

6

7

1

9

-4

4

10

5

5

0

-5

Iteration 2. Leaving arc is (h, i).

University of Victoria – CSC 445/545 – Summer 2021 35

Example 1b (15)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

-2

d

2
e

-1

f-1

g

-8

h

-6

i

-2

4

5

0

?

With (h, i) leaving, the entering arc is (i , f ). The flow assigned to (i , f ) will be 5 units.

University of Victoria – CSC 445/545 – Summer 2021 36

Example 1b (16)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

-2

d

2
e

-1

f-1

g

-8

h

-6

i

-2

5

1

6

7

6

4

-4

5

4

10

5

5

0

End of Iteration 2. Total cost is 51.

University of Victoria – CSC 445/545 – Summer 2021 37

Example 1b (17)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

-2

d

2
e

-1

f-1

g

-8

h

-6

i

-2

5

1

6

7

6

4 5

4

10

5

5

0

-4

Iteration 3. Leaving arc is (i , d).

University of Victoria – CSC 445/545 – Summer 2021 38

Example 1b (18)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

-2

d

2
e

-1

f-1

g

-8

h

-6

i

-2

4

5

?

With (i , d) leaving, the entering arc is (d , e). The flow assigned to (d , e) will be 4 units.

University of Victoria – CSC 445/545 – Summer 2021 39

Example 1b (19)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

5

1

2

7

4

4

1

a

0

b

1

c

-7

d

-3
e

-1

f-1

g

-8

h

-6

i

-2

5

1

4

10

3

6

4 1

9

10

5

0

5

Optimal. Total cost is 71.

University of Victoria – CSC 445/545 – Summer 2021 40

Example 2a (1)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

-5

1

2

7

4

4

1

a

0

b

1

c

2

d

2
e

4

f-1

g

-8

h

-6

i

-2

4

-1

3

9

12

6

4 1

5

4

5

0

0

Task: Starting from the infeasible basis above, use a two phase method (dual first, primal
second) to solve the network flow problem.

University of Victoria – CSC 445/545 – Summer 2021 41

Example 2a (2)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

0

0

0

0

0

0

0

0

0

0

0

0

0

a

0

b

0

c

0

d

0
e

0

f0

g

0

h

0

i

0

4

-1

3

9

12

6

4 1

0

0

0

0

0

Phase One: All edge costs are temporarily set to zero (shown in orange on the left). This
makes any basis dual feasible, so we can use the Dual Network Simplex Method to find an
optimal flow for the modified graph.

University of Victoria – CSC 445/545 – Summer 2021 42

Example 2a (3)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

0

0

0

0

0

0

0

0

0

0

0

0

0

a

0

b

0

c

0

d

0
e

0

f0

g

0

h

0

i

0

4

3

9

12

6

4 1

0

0

0

0

0

-1

Iteration 1. Leaving arc is (b, c).

University of Victoria – CSC 445/545 – Summer 2021 43

Example 2a (4)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

0

0

0

0

0

0

0

0

0

0

0

0

0

a

0

b

0

c

0

d

0
e

0

f0

g

0

h

0

i

0

0

?

With (b, c) leaving, the entering arc is (c , d). The flow assigned to (c , d) will be 1 units.

University of Victoria – CSC 445/545 – Summer 2021 44

Example 2a (5)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

0

0

0

0

0

0

0

0

0

0

0

0

0

a

0

b

0

c

0

d

0
e

0

f0

g

0

h

0

i

0

5

1

4

10

13

6

4 1

0

00

0

0

Optimal. Total cost is 0.

University of Victoria – CSC 445/545 – Summer 2021 45

Example 2a (6)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

0

0

0

0

0

0

0

0

0

0

0

0

0

a

0

b

0

c

0

d

0
e

0

f0

g

0

h

0

i

0

5

1

4

10

13

6

4 1

0

00

0

0

The optimal basis for the modified graph will be both primal and dual feasible. Since the
vertex supply values for the modified graph match those of the original graph, this basis will
be primal feasible for the original graph.

University of Victoria – CSC 445/545 – Summer 2021 46

Example 2a (7)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

-5

1

2

7

4

4

1

a

0

b

1

c

-2

d

2
e

4

f-1

g

-8

h

-6

i

-2

5

1

4

10

13

6

4 1

4

55

0

0

Phase Two: Starting from the feasible basis produced by Phase One, we can now run the
Primal Network Simplex Method.

University of Victoria – CSC 445/545 – Summer 2021 47

Example 2a (8)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

-5

1

2

7

4

4

1

a

0

b

1

c

-2

d

2
e

4

f-1

g

-8

h

-6

i

-2

5

1

4

10

13

6

4 1

4

55

0

0

Optimal. Total cost is 21.

University of Victoria – CSC 445/545 – Summer 2021 48

Example 2b (1)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

-5

1

2

7

4

4

1

a

0

b

1

c

2

d

2
e

4

f-1

g

-8

h

-6

i

-2

4

-1

3

9

12

6

4 1

5

4

5

0

0

Task: Starting from the infeasible basis above, use a two phase method (primal first, dual
second) to solve the network flow problem.

University of Victoria – CSC 445/545 – Summer 2021 49

Example 2b (2)

Supply/Cost Flow

a

0

b

0

c

0

d

0
e

0

f0

g

0

h

0

i

0

1

1

8

4

2

1

-5

1

2

7

4

4

1

a

0

b

1

c

2

d

2
e

4

f-1

g

-8

h

-6

i

-2

0

0

0

0

0

0

0 0

5

4

5

0

0

Phase One: All vertex supplies are temporarily set to zero (shown in orange on the left).
This makes any basis primal feasible, so we can use the Primal Network Simplex Method to
find an optimal flow for the modified graph.

University of Victoria – CSC 445/545 – Summer 2021 50

Example 2b (3)

Supply/Cost Flow

a

0

b

0

c

0

d

0
e

0

f0

g

0

h

0

i

0

1

1

8

4

2

1

-5

1

2

7

4

4

1

a

0

b

1

c

2

d

2
e

4

f-1

g

-8

h

-6

i

-2

0

0

0

0

0

0

0 0

5

4

5

0

0

Optimal. Total cost is 0.

University of Victoria – CSC 445/545 – Summer 2021 51

Example 2b (4)

Supply/Cost Flow

a

0

b

0

c

0

d

0
e

0

f0

g

0

h

0

i

0

1

1

8

4

2

1

-5

1

2

7

4

4

1

a

0

b

1

c

2

d

2
e

4

f-1

g

-8

h

-6

i

-2

0

0

0

0

0

0

0 0

5

4

5

0

0

The optimal basis for the modified graph will be both primal and dual feasible. Since the
vertex supply values for the modified graph match those of the original graph, this basis will
be dual feasible for the original graph.

University of Victoria – CSC 445/545 – Summer 2021 52

Example 2b (5)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

-5

1

2

7

4

4

1

a

0

b

1

c

2

d

2
e

4

f-1

g

-8

h

-6

i

-2

4

-1

3

9

12

6

4 1

5

4

5

0

0

Phase Two: Starting from the feasible basis produced by Phase One, we can now run the
Dual Network Simplex Method.

University of Victoria – CSC 445/545 – Summer 2021 53

Example 2b (6)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

-5

1

2

7

4

4

1

a

0

b

1

c

2

d

2
e

4

f-1

g

-8

h

-6

i

-2

4

3

9

12

6

4 1

5

4

5

0

0

-1

Iteration 1. Leaving arc is (b, c).

University of Victoria – CSC 445/545 – Summer 2021 54

Example 2b (7)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

-5

1

2

7

4

4

1

a

0

b

1

c

2

d

2
e

4

f-1

g

-8

h

-6

i

-2

4

?

With (b, c) leaving, the entering arc is (c, d). The flow assigned to (c , d) will be 1 units.

University of Victoria – CSC 445/545 – Summer 2021 55

Example 2b (8)

Supply/Cost Flow

a

-6

b

-5

c

1

d

3
e

6

f-2

g

10

h

-6

i

1

1

1

8

4

2

1

-5

1

2

7

4

4

1

a

0

b

1

c

-2

d

2
e

4

f-1

g

-8

h

-6

i

-2

5

1

4

10

13

6

4 1

4

55

0

0

Optimal. Total cost is 21.

University of Victoria – CSC 445/545 – Summer 2021 56

Example 3a (1)

Supply/Cost Flow

a

-3

b

-5

c

-3

d

4

e10 f 3

g

-2

h

-4

i

-1

j

1

5

10

10

2 10

9

512 11

2

5

2

1

2 2

1

a

0

b

-10

c

-12

d

-2

e-12 f -17

g

-19

h

-17

i

-15

j

-18

-4 1 -4

7

7

-3

-2 -6

1

24

12

25

2011

1

5

Task: Starting from the infeasible basis above, use a two phase method (dual first, primal
second) to solve the network flow problem.

University of Victoria – CSC 445/545 – Summer 2021 57

Example 3a (2)

Supply/Cost Flow

a

-3

b

-5

c

-3

d

4

e10 f 3

g

-2

h

-4

i

-1

j

1

0

0

0

0 0

0

00 0

0

0

0

0

0 0

0

a

0

b

0

c

0

d

0

e0 f 0

g

0

h

0

i

0

j

0

-4 1 -4

7

7

-3

-2 -6

1

0

0

0

00

0

0

Phase One: All edge costs are temporarily set to zero (shown in orange on the left). This
makes any basis dual feasible, so we can use the Dual Network Simplex Method to find an
optimal flow for the modified graph.

University of Victoria – CSC 445/545 – Summer 2021 58

Example 3a (3)

Supply/Cost Flow

a

-3

b

-5

c

-3

d

4

e10 f 3

g

-2

h

-4

i

-1

j

1

0

0

0

0 0

0

00 0

0

0

0

0

0 0

0

a

0

b

0

c

0

d

0

e0 f 0

g

0

h

0

i

0

j

0

-4 1 -4

7

7

-3

-2

1

0

0

0

00

0

0-6

Iteration 1. Leaving arc is (h, i).

University of Victoria – CSC 445/545 – Summer 2021 59

Example 3a (4)

Supply/Cost Flow

a

-3

b

-5

c

-3

d

4

e10 f 3

g

-2

h

-4

i

-1

j

1

0

0

0

0 0

0

00 0

0

0

0

0

0 0

0

a

0

b

0

c

0

d

0

e0 f 0

g

0

h

0

i

0

j

0

0

0

?

With (h, i) leaving, the entering arc is (a, g). The flow assigned to (a, g) will be 6 units.

University of Victoria – CSC 445/545 – Summer 2021 60

Example 3a (5)

Supply/Cost Flow

a

-3

b

-5

c

-3

d

4

e10 f 3

g

-2

h

-4

i

-1

j

1

0

0

0

0 0

0

00 0

0

0

0

0

0 0

0

a

0

b

0

c

0

d

0

e0 f 0

g

0

h

0

i

0

j

0

6

-4 1 -4

13

1

3

4

1

0

0

00

0

0 0

End of Iteration 1. Total cost is 0.

University of Victoria – CSC 445/545 – Summer 2021 61

Example 3a (6)

Supply/Cost Flow

a

-3

b

-5

c

-3

d

4

e10 f 3

g

-2

h

-4

i

-1

j

1

0

0

0

0 0

0

00 0

0

0

0

0

0 0

0

a

0

b

0

c

0

d

0

e0 f 0

g

0

h

0

i

0

j

0

6

1 -4

13

1

3

4

1

0

0

00

0

0 0

-4

Iteration 2. Leaving arc is (b, a).

University of Victoria – CSC 445/545 – Summer 2021 62

Example 3a (7)

Supply/Cost Flow

a

-3

b

-5

c

-3

d

4

e10 f 3

g

-2

h

-4

i

-1

j

1

0

0

0

0 0

0

00 0

0

0

0

0

0 0

0

a

0

b

0

c

0

d

0

e0 f 0

g

0

h

0

i

0

j

0

0

0

00

?

With (b, a) leaving, the entering arc is (e, c). The flow assigned to (e, c) will be 4 units.

University of Victoria – CSC 445/545 – Summer 2021 63

Example 3a (8)

Supply/Cost Flow

a

-3

b

-5

c

-3

d

4

e10 f 3

g

-2

h

-4

i

-1

j

1

0

0

0

0 0

0

00 0

0

0

0

0

0 0

0

a

0

b

0

c

0

d

0

e0 f 0

g

0

h

0

i

0

j

0

6

5 -4

9 4

1

3

4

1

0

0

0

0

0

0 0

End of Iteration 2. Total cost is 0.

University of Victoria – CSC 445/545 – Summer 2021 64

Example 3a (9)

Supply/Cost Flow

a

-3

b

-5

c

-3

d

4

e10 f 3

g

-2

h

-4

i

-1

j

1

0

0

0

0 0

0

00 0

0

0

0

0

0 0

0

a

0

b

0

c

0

d

0

e0 f 0

g

0

h

0

i

0

j

0

6

5

9 4

1

3

4

1

0

0

0

0

0

0 0

-4

Iteration 3. Leaving arc is (c , d).

University of Victoria – CSC 445/545 – Summer 2021 65

Example 3a (10)

Supply/Cost Flow

a

-3

b

-5

c

-3

d

4

e10 f 3

g

-2

h

-4

i

-1

j

1

0

0

0

0 0

0

00 0

0

0

0

0

0 0

0

a

0

b

0

c

0

d

0

e0 f 0

g

0

h

0

i

0

j

0

0

0

?

With (c , d) leaving, the entering arc is (d , f ). The flow assigned to (d , f ) will be 4 units.

University of Victoria – CSC 445/545 – Summer 2021 66

Example 3a (11)

Supply/Cost Flow

a

-3

b

-5

c

-3

d

4

e10 f 3

g

-2

h

-4

i

-1

j

1

0

0

0

0 0

0

00 0

0

0

0

0

0 0

0

a

0

b

0

c

0

d

0

e0 f 0

g

0

h

0

i

0

j

0

6

5

49 8

1

7

4

1

0

0

0

0

0

0 0

Optimal. Total cost is 0.

University of Victoria – CSC 445/545 – Summer 2021 67

Example 3a (12)

Supply/Cost Flow

a

-3

b

-5

c

-3

d

4

e10 f 3

g

-2

h

-4

i

-1

j

1

0

0

0

0 0

0

00 0

0

0

0

0

0 0

0

a

0

b

0

c

0

d

0

e0 f 0

g

0

h

0

i

0

j

0

6

5

49 8

1

7

4

1

0

0

0

0

0

0 0

The optimal basis for the modified graph will be both primal and dual feasible. Since the
vertex supply values for the modified graph match those of the original graph, this basis will
be primal feasible for the original graph.

University of Victoria – CSC 445/545 – Summer 2021 68

Example 3a (13)

Supply/Cost Flow

a

-3

b

-5

c

-3

d

4

e10 f 3

g

-2

h

-4

i

-1

j

1

5

10

10

2 10

9

512 11

2

5

2

1

2 2

1

a

0

b

1

c

-1

d

-22

e-12 f -17

g

5

h

7

i

-15

j

-18

6

5

49 8

1

7

4

1

11

23

31

5

25

24 5

Phase Two: Starting from the feasible basis produced by Phase One, we can now run the
Primal Network Simplex Method.

University of Victoria – CSC 445/545 – Summer 2021 69

Example 3a (14)

Supply/Cost Flow

a

-3

b

-5

c

-3

d

4

e10 f 3

g

-2

h

-4

i

-1

j

1

5

10

10

2 10

9

512 11

2

5

2

1

2 2

1

a

0

b

1

c

-1

d

-22

e-12 f -17

g

5

h

7

i

-15

j

-18

6

5

49 8

1

7

4

1

11

23

31

5

25

24 5

Optimal. Total cost is 302.

University of Victoria – CSC 445/545 – Summer 2021 70

Example 3b (1)

Supply/Cost Flow

a

-3

b

-5

c

-3

d

4

e10 f 3

g

-2

h

-4

i

-1

j

1

5

10

10

2 10

9

512 11

2

5

2

1

2 2

1

a

0

b

-10

c

-12

d

-2

e-12 f -17

g

-19

h

-17

i

-15

j

-18

-4 1 -4

7

7

-3

-2 -6

1

24

12

25

2011

1

5

Task: Starting from the infeasible basis above, use a two phase method (primal first, dual
second) to solve the network flow problem.

University of Victoria – CSC 445/545 – Summer 2021 71

Example 3b (2)

Supply/Cost Flow

a

0

b

0

c

0

d

0

e0 f 0

g

0

h

0

i

0

j

0

5

10

10

2 10

9

512 11

2

5

2

1

2 2

1

a

0

b

-10

c

-12

d

-2

e-12 f -17

g

-19

h

-17

i

-15

j

-18

0 0 0

0

0

0

0 0

0

24

12

25

2011

1

5

Phase One: All vertex supplies are temporarily set to zero (shown in orange on the left).
This makes any basis primal feasible, so we can use the Primal Network Simplex Method to
find an optimal flow for the modified graph.

University of Victoria – CSC 445/545 – Summer 2021 72

Example 3b (3)

Supply/Cost Flow

a

0

b

0

c

0

d

0

e0 f 0

g

0

h

0

i

0

j

0

5

10

10

2 10

9

512 11

2

5

2

1

2 2

1

a

0

b

-10

c

-12

d

-2

e-12 f -17

g

-19

h

-17

i

-15

j

-18

0 0 0

0

0

0

0 0

0

24

12

25

2011

1

5

Optimal. Total cost is 0.

University of Victoria – CSC 445/545 – Summer 2021 73

Example 3b (4)

Supply/Cost Flow

a

0

b

0

c

0

d

0

e0 f 0

g

0

h

0

i

0

j

0

5

10

10

2 10

9

512 11

2

5

2

1

2 2

1

a

0

b

-10

c

-12

d

-2

e-12 f -17

g

-19

h

-17

i

-15

j

-18

0 0 0

0

0

0

0 0

0

24

12

25

2011

1

5

The optimal basis for the modified graph will be both primal and dual feasible. Since the
vertex supply values for the modified graph match those of the original graph, this basis will
be dual feasible for the original graph.

University of Victoria – CSC 445/545 – Summer 2021 74

Example 3b (5)

Supply/Cost Flow

a

-3

b

-5

c

-3

d

4

e10 f 3

g

-2

h

-4

i

-1

j

1

5

10

10

2 10

9

512 11

2

5

2

1

2 2

1

a

0

b

-10

c

-12

d

-2

e-12 f -17

g

-19

h

-17

i

-15

j

-18

-4 1 -4

7

7

-3

-2 -6

1

24

12

25

2011

1

5

Phase Two: Starting from the feasible basis produced by Phase One, we can now run the
Dual Network Simplex Method.

University of Victoria – CSC 445/545 – Summer 2021 75

Example 3b (6)

Supply/Cost Flow

a

-3

b

-5

c

-3

d

4

e10 f 3

g

-2

h

-4

i

-1

j

1

5

10

10

2 10

9

512 11

2

5

2

1

2 2

1

a

0

b

-10

c

-12

d

-2

e-12 f -17

g

-19

h

-17

i

-15

j

-18

-4 1 -4

7

7

-3

-2

1

24

12

25

2011

1

5-6

Iteration 1. Leaving arc is (h, i).

University of Victoria – CSC 445/545 – Summer 2021 76

Example 3b (7)

Supply/Cost Flow

a

-3

b

-5

c

-3

d

4

e10 f 3

g

-2

h

-4

i

-1

j

1

5

10

10

2 10

9

512 11

2

5

2

1

2 2

1

a

0

b

-10

c

-12

d

-2

e-12 f -17

g

-19

h

-17

i

-15

j

-18

1

24

?

With (h, i) leaving, the entering arc is (a, g). The flow assigned to (a, g) will be 6 units.

University of Victoria – CSC 445/545 – Summer 2021 77

Example 3b (8)

Supply/Cost Flow

a

-3

b

-5

c

-3

d

4

e10 f 3

g

-2

h

-4

i

-1

j

1

5

10

10

2 10

9

512 11

2

5

2

1

2 2

1

a

0

b

-10

c

-12

d

-2

e-12 f -17

g

5

h

7

i

-15

j

-18

6

-4 1 -4

13

1

3

4

1

12

25

2011

25

24 5

End of Iteration 1. Total cost is 134.

University of Victoria – CSC 445/545 – Summer 2021 78

Example 3b (9)

Supply/Cost Flow

a

-3

b

-5

c

-3

d

4

e10 f 3

g

-2

h

-4

i

-1

j

1

5

10

10

2 10

9

512 11

2

5

2

1

2 2

1

a

0

b

-10

c

-12

d

-2

e-12 f -17

g

5

h

7

i

-15

j

-18

6

1 -4

13

1

3

4

1

12

25

2011

25

24 5

-4

Iteration 2. Leaving arc is (b, a).

University of Victoria – CSC 445/545 – Summer 2021 79

Example 3b (10)

Supply/Cost Flow

a

-3

b

-5

c

-3

d

4

e10 f 3

g

-2

h

-4

i

-1

j

1

5

10

10

2 10

9

512 11

2

5

2

1

2 2

1

a

0

b

-10

c

-12

d

-2

e-12 f -17

g

5

h

7

i

-15

j

-18

12

25

2011

?

With (b, a) leaving, the entering arc is (e, c). The flow assigned to (e, c) will be 4 units.

University of Victoria – CSC 445/545 – Summer 2021 80

Example 3b (11)

Supply/Cost Flow

a

-3

b

-5

c

-3

d

4

e10 f 3

g

-2

h

-4

i

-1

j

1

5

10

10

2 10

9

512 11

2

5

2

1

2 2

1

a

0

b

1

c

-1

d

9

e-12 f -17

g

5

h

7

i

-15

j

-18

6

5 -4

9 4

1

3

4

1

11

23

36

31

25

24 5

End of Iteration 2. Total cost is 178.

University of Victoria – CSC 445/545 – Summer 2021 81

Example 3b (12)

Supply/Cost Flow

a

-3

b

-5

c

-3

d

4

e10 f 3

g

-2

h

-4

i

-1

j

1

5

10

10

2 10

9

512 11

2

5

2

1

2 2

1

a

0

b

1

c

-1

d

9

e-12 f -17

g

5

h

7

i

-15

j

-18

6

5

9 4

1

3

4

1

11

23

36

31

25

24 5

-4

Iteration 3. Leaving arc is (c , d).

University of Victoria – CSC 445/545 – Summer 2021 82

Example 3b (13)

Supply/Cost Flow

a

-3

b

-5

c

-3

d

4

e10 f 3

g

-2

h

-4

i

-1

j

1

5

10

10

2 10

9

512 11

2

5

2

1

2 2

1

a

0

b

1

c

-1

d

9

e-12 f -17

g

5

h

7

i

-15

j

-18

36

31

?

With (c , d) leaving, the entering arc is (d , f ). The flow assigned to (d , f ) will be 4 units.

University of Victoria – CSC 445/545 – Summer 2021 83

Example 3b (14)

Supply/Cost Flow

a

-3

b

-5

c

-3

d

4

e10 f 3

g

-2

h

-4

i

-1

j

1

5

10

10

2 10

9

512 11

2

5

2

1

2 2

1

a

0

b

1

c

-1

d

-22

e-12 f -17

g

5

h

7

i

-15

j

-18

6

5

49 8

1

7

4

1

11

23

31

5

25

24 5

Optimal. Total cost is 302.

University of Victoria – CSC 445/545 – Summer 2021 84

Example 4a (1)

Supply/Cost Flow

a

0

b

0

c-6

d

-6

e

-2

f

9

g

5

48

28

10

7

65

7

38

15

56

48108

24

33

19

a

0

b

-7

c48

d

28

e

10

f

-56

g

-32

6

6

2

0

14

-5

10

-10

73

33

-14 8

-23

(This is the motivating example from Chapter 14 of Vanderbei)

University of Victoria – CSC 445/545 – Summer 2021 85

Example 4a (2)

Supply/Cost Flow

a

0

b

0

c-6

d

-6

e

-2

f

9

g

5

48

28

10

7

65

7

38

15

56

48108

24

33

19

a

0

b

-7

c48

d

28

e

10

f

-56

g

-32

6

6

2

0

14

-5

10

-10

73

33

-14 8

-23

Task: Starting from the infeasible basis above, use a two phase method (dual first, primal
second) to solve the network flow problem.

University of Victoria – CSC 445/545 – Summer 2021 86

Example 4a (3)

Supply/Cost Flow

a

0

b

0

c-6

d

-6

e

-2

f

9

g

5

0

0

0

0

0

0

0

0

0

00

0

0

0

a

0

b

0

c0

d

0

e

0

f

0

g

0

6

6

2

0

14

-5

0

0

0

0

00 0

0

Phase One: All edge costs are temporarily set to zero (shown in orange on the left). This
makes any basis dual feasible, so we can use the Dual Network Simplex Method to find an
optimal flow for the modified graph.

University of Victoria – CSC 445/545 – Summer 2021 87

Example 4a (4)

Supply/Cost Flow

a

0

b

0

c-6

d

-6

e

-2

f

9

g

5

0

0

0

0

0

0

0

0

0

00

0

0

0

a

0

b

0

c0

d

0

e

0

f

0

g

0

6

6

2

0

14

0

0

0

0

00 0

0

-5

Iteration 1. Leaving arc is (f , g).

University of Victoria – CSC 445/545 – Summer 2021 88

Example 4a (5)

Supply/Cost Flow

a

0

b

0

c-6

d

-6

e

-2

f

9

g

5

0

0

0

0

0

0

0

0

0

00

0

0

0

a

0

b

0

c0

d

0

e

0

f

0

g

0

0

0

?

With (f , g) leaving, the entering arc is (g , b). The flow assigned to (g , b) will be 5 units.

University of Victoria – CSC 445/545 – Summer 2021 89

Example 4a (6)

Supply/Cost Flow

a

0

b

0

c-6

d

-6

e

-2

f

9

g

5

0

0

0

0

0

0

0

0

0

00

0

0

0

a

0

b

0

c0

d

0

e

0

f

0

g

0

6

6

2

5

9

5

0

0

0

0

00

0

0

Optimal. Total cost is 0.

University of Victoria – CSC 445/545 – Summer 2021 90

Example 4a (7)

Supply/Cost Flow

a

0

b

0

c-6

d

-6

e

-2

f

9

g

5

0

0

0

0

0

0

0

0

0

00

0

0

0

a

0

b

0

c0

d

0

e

0

f

0

g

0

6

6

2

5

9

5

0

0

0

0

00

0

0

The optimal basis for the modified graph will be both primal and dual feasible. Since the
vertex supply values for the modified graph match those of the original graph, this basis will
be primal feasible for the original graph.

University of Victoria – CSC 445/545 – Summer 2021 91

Example 4a (8)

Supply/Cost Flow

a

0

b

0

c-6

d

-6

e

-2

f

9

g

5

48

28

10

7

65

7

38

15

56

48108

24

33

19

a

0

b

-7

c48

d

28

e

10

f

-56

g

-40

6

6

2

5

9

5

10

-10

73

33

-14

8

-31

Phase Two: Starting from the feasible basis produced by Phase One, we can now run the
Primal Network Simplex Method.

University of Victoria – CSC 445/545 – Summer 2021 92

Example 4a (9)

Supply/Cost Flow

a

0

b

0

c-6

d

-6

e

-2

f

9

g

5

48

28

10

7

65

7

38

15

56

48108

24

33

19

a

0

b

-7

c48

d

28

e

10

f

-56

g

-40

6

6

2

5

9

5

10

-10

73

33

-14

8

-31

Iteration 1. Entering arc is (g , e).

University of Victoria – CSC 445/545 – Summer 2021 93

Example 4a (10)

Supply/Cost Flow

a

0

b

0

c-6

d

-6

e

-2

f

9

g

5

48

28

10

7

65

7

38

15

56

48108

24

33

19

a

0

b

-7

c48

d

28

e

10

f

-56

g

-40

?

2

5

5

With (g , e) entering, the leaving arc is (a, e). The flow assigned to (g , e) will be 2 units.

University of Victoria – CSC 445/545 – Summer 2021 94

Example 4a (11)

Supply/Cost Flow

a

0

b

0

c-6

d

-6

e

-2

f

9

g

5

48

28

10

7

65

7

38

15

56

48108

24

33

19

a

0

b

-7

c48

d

28

e

-21

f

-56

g

-40

6

6

3

9

3

2

31

10

21

73

64

-14

8

End of Iteration 1. Total cost is 1118.

University of Victoria – CSC 445/545 – Summer 2021 95

Example 4a (12)

Supply/Cost Flow

a

0

b

0

c-6

d

-6

e

-2

f

9

g

5

48

28

10

7

65

7

38

15

56

48108

24

33

19

a

0

b

-7

c48

d

28

e

-21

f

-56

g

-40

6

6

3

9

3

2

31

10

21

73

64

4

8

-1

Iteration 2. Entering arc is (f , b).

University of Victoria – CSC 445/545 – Summer 2021 96

Example 4a (13)

Supply/Cost Flow

a

0

b

0

c-6

d

-6

e

-2

f

9

g

5

48

28

10

7

65

7

38

15

56

48108

24

33

19

a

0

b

-7

c48

d

28

e

-21

f

-56

g

-40

?

3

9

With (f , b) entering, the leaving arc is (f , a). The flow assigned to (f , b) will be 9 units.

University of Victoria – CSC 445/545 – Summer 2021 97

Example 4a (14)

Supply/Cost Flow

a

0

b

0

c-6

d

-6

e

-2

f

9

g

5

48

28

10

7

65

7

38

15

56

48108

24

33

19

a

0

b

-7

c48

d

28

e

-21

f

-55

g

-40

6

6

12

9 3

2

31

10

21

73

64

1

5

9

Optimal. Total cost is 1109.

University of Victoria – CSC 445/545 – Summer 2021 98

Example 4b (1)

Supply/Cost Flow

a

0

b

0

c-6

d

-6

e

-2

f

9

g

5

48

28

10

7

65

7

38

15

56

48108

24

33

19

a

0

b

-7

c48

d

28

e

10

f

-56

g

-32

6

6

2

0

14

-5

10

-10

73

33

-14 8

-23

Task: Starting from the infeasible basis above, use a two phase method (primal first, dual
second) to solve the network flow problem.

University of Victoria – CSC 445/545 – Summer 2021 99

Example 4b (2)

Supply/Cost Flow

a

0

b

0

c0

d

0

e

0

f

0

g

0

48

28

10

7

65

7

38

15

56

48108

24

33

19

a

0

b

-7

c48

d

28

e

10

f

-56

g

-32

0

0

0

0

0

0

10

-10

73

33

-14 8

-23

Phase One: All vertex supplies are temporarily set to zero (shown in orange on the left).
This makes any basis primal feasible, so we can use the Primal Network Simplex Method to
find an optimal flow for the modified graph.

University of Victoria – CSC 445/545 – Summer 2021 100

Example 4b (3)

Supply/Cost Flow

a

0

b

0

c0

d

0

e

0

f

0

g

0

48

28

10

7

65

7

38

15

56

48108

24

33

19

a

0

b

-7

c48

d

28

e

10

f

-56

g

-32

0

0

0

0

0

0

10

-10

73

33

-14 8

-23

Iteration 1. Entering arc is (g , e).

University of Victoria – CSC 445/545 – Summer 2021 101

Example 4b (4)

Supply/Cost Flow

a

0

b

0

c0

d

0

e

0

f

0

g

0

48

28

10

7

65

7

38

15

56

48108

24

33

19

a

0

b

-7

c48

d

28

e

10

f

-56

g

-32

?

0

0

0

With (g , e) entering, the leaving arc is (a, e). The flow assigned to (g , e) will be 0 units.

University of Victoria – CSC 445/545 – Summer 2021 102

Example 4b (5)

Supply/Cost Flow

a

0

b

0

c0

d

0

e

0

f

0

g

0

48

28

10

7

65

7

38

15

56

48108

24

33

19

a

0

b

-7

c48

d

28

e

-13

f

-56

g

-32

0

0

0

0

0

0

23

10

13

73

56

-14 8

End of Iteration 1. Total cost is 0.

University of Victoria – CSC 445/545 – Summer 2021 103

Example 4b (6)

Supply/Cost Flow

a

0

b

0

c0

d

0

e

0

f

0

g

0

48

28

10

7

65

7

38

15

56

48108

24

33

19

a

0

b

-7

c48

d

28

e

-13

f

-56

g

-32

0

0

0

0

0

0

23

10

13

73

56

4 8-1

Iteration 2. Entering arc is (f , b).

University of Victoria – CSC 445/545 – Summer 2021 104

Example 4b (7)

Supply/Cost Flow

a

0

b

0

c0

d

0

e

0

f

0

g

0

48

28

10

7

65

7

38

15

56

48108

24

33

19

a

0

b

-7

c48

d

28

e

-13

f

-56

g

-32

?

0

0

With (f , b) entering, the leaving arc is (f , a). The flow assigned to (f , b) will be 0 units.

University of Victoria – CSC 445/545 – Summer 2021 105

Example 4b (8)

Supply/Cost Flow

a

0

b

0

c0

d

0

e

0

f

0

g

0

48

28

10

7

65

7

38

15

56

48108

24

33

19

a

0

b

-7

c48

d

28

e

-12

f

-55

g

-31

0

0

0

0

0

0

22

10

12

73

55

1

5 9

Optimal. Total cost is 0.

University of Victoria – CSC 445/545 – Summer 2021 106

Example 4b (9)

Supply/Cost Flow

a

0

b

0

c0

d

0

e

0

f

0

g

0

48

28

10

7

65

7

38

15

56

48108

24

33

19

a

0

b

-7

c48

d

28

e

-12

f

-55

g

-31

0

0

0

0

0

0

22

10

12

73

55

1

5 9

The optimal basis for the modified graph will be both primal and dual feasible. Since the
vertex supply values for the modified graph match those of the original graph, this basis will
be dual feasible for the original graph.

University of Victoria – CSC 445/545 – Summer 2021 107

Example 4b (10)

Supply/Cost Flow

a

0

b

0

c-6

d

-6

e

-2

f

9

g

5

48

28

10

7

65

7

38

15

56

48108

24

33

19

a

0

b

-7

c48

d

28

e

-12

f

-55

g

-31

6

6

12

12

-3

2

22

10

12

73

55

1

5 9

Phase Two: Starting from the feasible basis produced by Phase One, we can now run the
Dual Network Simplex Method.

University of Victoria – CSC 445/545 – Summer 2021 108

Example 4b (11)

Supply/Cost Flow

a

0

b

0

c-6

d

-6

e

-2

f

9

g

5

48

28

10

7

65

7

38

15

56

48108

24

33

19

a

0

b

-7

c48

d

28

e

-12

f

-55

g

-31

6

6

12

12

2

22

10

12

73

55

1

5 9

-3

Iteration 1. Leaving arc is (f , g).

University of Victoria – CSC 445/545 – Summer 2021 109

Example 4b (12)

Supply/Cost Flow

a

0

b

0

c-6

d

-6

e

-2

f

9

g

5

48

28

10

7

65

7

38

15

56

48108

24

33

19

a

0

b

-7

c48

d

28

e

-12

f

-55

g

-31

22

12

55

9

?

With (f , g) leaving, the entering arc is (g , b). The flow assigned to (g , b) will be 3 units.

University of Victoria – CSC 445/545 – Summer 2021 110

Example 4b (13)

Supply/Cost Flow

a

0

b

0

c-6

d

-6

e

-2

f

9

g

5

48

28

10

7

65

7

38

15

56

48108

24

33

19

a

0

b

-7

c48

d

28

e

-21

f

-55

g

-40

6

6

12

9 3

2

31

10

21

73

64

1

5

9

Optimal. Total cost is 1109.

University of Victoria – CSC 445/545 – Summer 2021 111