The+Answer+of+OR+Chapter+8

Chapter 8

8.1-1

Plant 1 2 3

Demand

8.1-5 (a) Plant

1 500 650 400 10

Distributer center 2 3 750 300 800 400 700 500 10 10

Supply

4

450 600 550 10

12 17 11

1 2 3 4(D) Demand

8.2-7

1 -8 -5 -6 M 4 2 -7 -2 -4 M 6

Cost per unit Customer 3 3(Extra) -5 -5 -1 -1 -3 -3 M 0 2 6

Supply

4

-2 -3 -5 0 6

6 8 4 6

Because row 4 has the largest number of allocation-3, let u4=0. For each basic variables, Cij?ui?vj?0, then

v4?v3?v5?0,u3?3,v4?3,u2?1,v1?4,u1?3

Source 1 2

1 8 1 5 (25)-

3 4(D)

6 -1

2 6 3 M M-1 3 (25)+ 0 (0)- 25 0

Destination 3 3 (20) 8 7 9 6 0 (0) 20 0

4 7 1 4 (5)+ 6 b(5)- 0 -3 10 3

5 5 2 7 6 8 5 0 (20) 20 0

supply 20 30

ui 3 1

30 20

3 0

0

+ -4

Demand 25 vj

4

Iteration 2

Source 1

1 8 5 5 25_ ○3

6 -1 + 0 ○0

Demand 25 vj

Destination

2 6 7 M M-1 3 ○25 0 4 25 -4

3 3 ○20 8 3 9 2 0 ○0 20 0

4

7 5 4 ○5+ 6 ○5- 0 1 10 -1

5 5 2 7 2 8 1 0 ○20 20 0

supply 20

ui 3

2 30 5

30 7

4(D) 20 0

0

Source 1

1

8 5 5 20 ○

3

6 ○5

4(D)

0 0 ○

Demand 25 vj

0

2

6 6 M M-2 3 ○25 0 3 25 -3

Destination 3 3 ○20 8 3 9 3 0 ○0 20 0

4

7 5 4 ○10 6 1 0 1 10 -1

5 5 2 7 2 8 2 0 ○20 20 0

supply 20

ui 3

2 30 5

30 6

20 0

Z=305

There is no negative coefficient of Cij?ui?vj for non basice variable. So it is optimal solution.

8.2-8:

(a) Northweat corner rule

The initial BF soluion is list as follow. Source

Destination 1

2 3 4 1

3

7

6

4

○3 ○2

2

2 4 3

2 ○

1 ○1 3

4

3

8 5 ○1 ○2Demand 3 3 2

2

Iteration 0: Source

Destination 1

2 3 4 1 3 7

6

4 ○3 ○2 0

1 2

2 4

3

2

2

○1- ○1+ 2 3

4 3 8 5 -1

-6 +

○1- ○2 Demand 3

3

2

2

supply

5

2

3

Z=48 supply 5 2 3 Z=48

ui

0

-3

2

vj 3 7 6 3

Iteration 1: Source

Destination 1

2 3 4 1

3

7

6

4 ○3 ○2- 0

-5 + 2

2 4 3

2

2

○0 ○2 -4 3

4 3 8 5 5

○1 6

○2-

+

Demand 3 3 2 2 vj

3

7

6

9

Iteration 2: Source

Destination 1

2 3 4 1

3

7

6 4 ○3 5

5 ○2 2

2 4

3

2

-3

○0- ○2 -4 + 3

4 3 8 5 0

○3+ 6

○0-

supply 5

2

3

Z=42

supply 5

2

3

ui

0

-3

-4

ui

0

2

1

Demand 3 vj

3

3 2

2 1

2 4

Z=32

Iteration 3: Source

Destination 1

1

3

2 7

3 6 1 3

4 4 ○2 2 ○0 5 4 2 2

Z=32 optimal solution 3

-1

2

0

5

2

supply ui

○3 1

2

2 1

3

4 4

Demand 3 vj

1

3 4 4

○0 ○2 3 3 ○

8 6 2 3

Because degeneration appears in iteration 2 and 3, the optimal solution is not unique.

(b) Vogel’s approximation method The initial BF solution is: (1) source

1

1 2

3 2

2 7 4

Destination

3 6 3 □

4 4 2

5 2

Supply Row

difference 1 0

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