modified distribution method of transportation ppt
TRANSCRIPT
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.