modified distribution method of transportation ppt

THE MODI (MODIFIED DISTRIBUTION)

METHOD OF TRANSPORTATION PROBLEM

In this method, we have to add a new row above the table for cost factors and a new column to the left of the table.

We make use of Vi for cost factors of supply, and Wj for cost factors of demand. The key to the use of MODI is to utilize the occupied cells for each Vi and Wj values, and then use these values to calculate the net contribution of the vacant cells.

Let us consider the RSV problem. The initial table with the cost factors’ row and column is shown below:

W1 W2 W3

A B C

V1 1 4158

8 8158

V2 2 1616

24168

16184

V3 3 8 1636

24143 19

174 204 143

The initial distribution is the same as in the stepping stone method, where the Northwest rule is utilized. Cost factors which are Vi and Wj are added.

Let Cij = the shipping cost per unit to a particular occupied cell.

Vi + Wj = Cij

So each value of Vi and Wj (from occupied cells) is computed as follows:

V1 + W1 = 4 V2 + W2 = 24 V3 + W3 = 24V2 + W1 = 16 V3 + W2 = 16

We may assume a value for anyone of them, so we can solve for the rest. Suppose V1 = 0, then we have:

0 + W1 = 4 V2 + W2 = 24 V3 + W3 = 24 W1 = 4 12 + W2 = 24 4 + W3 = 24 W2 = 12 W3 = 20

V2 + W1 = 16 V3 + W2 = 16V2 + 4 = 16 V3 + 12 = 16

V1 = 4 V1 = 4

Table I (with the indicated cost factor values)

W1 = 4 W2 = 12 W3 = 20

A B C

V1 = 0 1 4158

8- 4

8- 12 158

V2 = 12 2 1616

24168

16- 16 184

V3 = 4 3 8 1636

24143 19

174 204 143

To test for improvement, evaluate the vacant cells for the net contribution to cost, by subtracting Vi and Wj from cost of vacant cells.

Vacant Cells:

1-B 8 – 0 – 12 = -4

1-C 8 – 0 – 20 = -12

2-C 16 – 12 – 20 = -16

3-A 8 – 4 – 4 = 0

The improvement shows that some units must be transferred to 2-C. As in the stepping stone method, chose the least entry from used cells having negative sign if taken as a rectangular order in the process of improvement.

The affected cells are:

2-C

24

168

16

16

36

24

143

After the transfer of 143 units to 2-C the entries become:

2-C

24

168

16

143

16

36

24

Table II

W1 = 4 W2 = 12 W3 = 4

A B C

V1 = 0 1 4158

8- 4

8158

V2 = 12 2 1616

2425

16143 184

V3 = 4 3 8 16179

2419

174 204 143

Finding the values of the cost factors by using the occupied cells:

Let V1 = 0

V1 + W1 = 4 V2 + W2 = 24 V2 + W3 = 16

0 + W1 = 4 12 + W2 = 24 12 + W3 = 16

W1 = 4 W2 =12 W3 = 4

V2 + W1 = 16 V3 + W2 = 16

V2 + 4 = 16 V3 + 12 = 16

V2 = 12 V3 = 4

Testing for Improvement (Table II)

Vacant Cells:

1-B 8 – 0 – 12 = -4

1-C 8 – 0 – 4 = 4

3-A 8 – 4 – 4 = 0

3-C 24 – 4 – 4 = 16

Improvement shows that some units may be transferred to 1-B. Affected cells are:

1-B

24

158

16

16

16

24

25

After the transfer we have:

1-B

24

158

16

25

16

16

24

Table III

W1 = 4 W2 = 8 W3 = 4

A B C

V1 = 0 1 4133

825

8158

V2 = 12 2 1641

24 16143 184

V3 =8 3 8- 4

16179

2419

174 204 143

Computing the cost factors through the used cells

If V1 = 0

V1 + W1 = 4 V2 + W1 = 16 V3 + W2 = 16

0 + W1 = 4 V2 + 4 = 16 V3 + 8 = 16

W1 = 4 V2 =12 V3 = 8

V1 + W2 = 8 V2 + W3 = 16

0 + W2 = 8 12 + W3 = 16

W2 = 8 W3 = 4

Testing for Improvement (Table III)

Vacant cells:

1-C 8 – 0 – 4 = 4

1-B 24 – 12 – 8 = 4

2-A 8 – 8 – 4 = -4

3-C 24 – 8 – 4 = 12The improvement shows that some units must be transferred to 3-

A. the rectangular position of used cells in relation to 3-A are 1-A, 1-B and 3-B. In the process of improvement as in the stepping stone method, add to 3-A, subtract from 1-A, add to 1-B and subtract from 3-B. the alternatives are to get from 1-A or from 3-B. the entries from 1-A and 3-B are 133 and 179, respectively. Since 133 is smaller than 179, 133 must be transferred to 3-A.

Involved entries are:

A B

124

133

16

25

216

41

24

38 16

179

After the transfer we have:

A B

124 16

158

216

41

24

38

133

16

46

Table IV

W1 = 0 W2 = 8 W3 = 0

A B C

V1 = 0 1 4 8158

8158

V2 = 16 2 1641

24 16143 184

V3 =8 3 8133

1646

2419

174 204 143

Cost: 158 x 8 = 1,264 41 x 16 = 656143 x 16 = 2,288133 x 8 = 1,064 46 x 16 = 736 Cost = P 6,008

Test for Improvement 

Vacant Cells: 1-A 4 – 0 – 0 = 4

1-C 8 – 0 – 0 = 82-B 24 – 16 – 8 =

03-C 24 – 8 – 0 =

16Since there is no negative in the improvement, the above table is optimum.(Formulate the decision)

Summary of MODI Method

A. Add the cost factors to get the cost of the occupied cells, by one value and solving for the rest.

B. Subtract the outside cost factors from the indicated cost of the vacant cells. The most negative result indicates the point of destination for transfer.