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Unbalanced Transportation Problems

Example 5.7. A departmental store wishes to purchase the following quantity of ladies dresses.

Dress type

A

B

C

D

Quantity

150

100

75

250

Tenders are submitted by three different manufacturers who undertake to supply not more than the quantity given below (all types of dresses combined)

Manufacturer

W

X

Y

Total quantity

350

250

150

The store estimates that profit per dress will vary with the manufacturers as shown in the matrix below.

How should orders be placed?

Dress

A

B

C

D

Manufacturer

W

2.75

3.50

4.25

2.25

X

3.30

3.25

4.50

1.75

Y

2.50

3.50

4.75

2.00

Sol. The problem can be written in the form of the following matrix.

Step 1. Matrix

A

B

C

D

Supply

W

2.75

3.50

4.25

2.25

300

Manufacturer

X

3.00

3.25

4.50

1.75

250

Y

2.50

3.50

4.75

2.00

150 (Total 700)

Demand

150

100

75

250

(Total 575)

Since the supply and demand are not equal,. it is not a balanced problem. Here total supply is 700 and total demand is 575, so surplus supplies are 125.

We have to create dummy destination (store). The cost associated with store will be taken zero as the surplus quantity manufactured remains in the factory and is not transported at all, so the new matrix is

A

B

C

D

E

Supply

W

2.75

3.50

4.25

2.25

0

300 (0.25)

Manufacturer

X

3.00

3.25

4.50

1.75

0

125

250

Y

2.50

3.50

4.75

2.00

0

150 (0.50)

Demand

150

(0.25)

100

(0.25)

75

(0.25)

250

(0.25)

125

(0)

Step 2. Using Vogel’s approximation method (VAM). Let us write the difference between the smallest and second cost in each column and each row and write it below the column or on right side of the rows respectively.

Row with greatest difference is row X as indicated with ← an arrow. In this row the least cost cell is XE. In this we can allot 125 since column E is fully satisfied this column is crossed out. Now the shrunken matrix is shown below:

A

B

C

D

W

2.75

3.50

4.25

2.25

300 (0.25)

Manufacturer

3.00

3.25

4.50

1.75

X

125

250 (1.25)

Y

2.50

3.50

4.50

2.00

150 (0.50)

Demand

150

(0.25)

100

(0.25)

75

(0.25)

250

(0.25)

 

Step 3. In This matrix maximum difference is in row X and the least cost cell is XD. We can allot 125 units to this cell and since this row is fully satisfied it is crossed out. The new matrix is as follows:

A

B

C

D

Supply

W

2.75

3.50

4.25

2.25

300(0.25)

Manufacturer

2.50

3.50

4.75

2.00

150 (0.50)

Y

125

150

(0.25)

100

(0.25)

75

(0.25)

250

(0.25)

 

Step 4. In this above matrix maximum difference is in row Y which is shown with an ← arrow. In this row least cost cell is YO and so we allot 125 units to this cell sine this satisfies column D so this column is crossed out and the resulting matrix is rewritten as follows.

A

B

C

W

2.75

125

3.50

100

4.25

75

300 (0.25)

150

(0.25)

100

(0.25)

75

(0.25)

 

Step 5. In this least cost cell is WA in which 125 can be allotted. Also in WB we can allot 100 units and in WC 75 can be allotted.

Step 6. The matrix with all allocation is shown below

A

B

C

D

E

W

2.75

125

3.50

100

4.25

75

2.25

0

300

X

3.00

3.25

4.50

1.75

125

0

125

250

Y

250

25

3.50

4.75

2.00

125

0

150

150

100

75

250

125

 

The cost of this solution is

Z = Rs. (12 ×2.75 + 100 × 3.50 + 75 × 4.25 × 1.75 + 25 × 2.50 + 125 × 2.00)

= Rs. (333.75 + 350 + 318.75 + 218.75 + 62.50 + 250)

= Rs, 1533.75

 

Example 5.8. Obtain the initial basic feasible solution to the following transportation problem by VAM.

D1

D2

D3

D4

D5

a1

A1

5

7

10

5

3

5

A2

8

6

9

12

14

10

A3

10

9

8

10

15

10

bj

3

3

10

5

4

25

 

Sol.

To

From

D1

D2

D3

D4

D5

a1

A1

5

7

10

5

1

3

4

5

A2

8

3

6

3

9

4

12

14

10

A3

10

9

8

6

10

4

15

10

bj

3

3

10

5

4

25

 

 

up

up

up

up

Up

2

0

-

-

-

2

2

2

1

3 ←

1

1

1

2

2

up

3

1

1

5

11 ↑

up

3

1

1

5 ↑

-

up

2

3 ↑

1

2

-

up

2 ↑

-

1

2

-

up

-

-

1

2

-

 

Transportation cost = 5 × 1 + 3 × 4 + 8 × 3 + 6 × 3 + 9 × 4 + 8 × 6 + 10 × 4=183 units