Answer in Operations Research for Andinet #283667

1. Given the following transportation problem:

D1 D2 D3 D4 SS

S1 5 10 7 10 50

S2 9 10 3 3 40

S3 6 4 2 2 60

DD 40 30 10 70

Where D1, D2, D3, and D4 are destinations and the values in the shaded cells are costs;

A. Determine the basic feasible solution through NWCM, LCM and VAM

B. Determine the optimal solution through Stepping Stone and MODI Method

A.

NWCM:

Step-1:

Select the upper left corner cell of the transportation matrix and allocate min(s1, d1).

Step-2:

a. Subtract this value from supply and demand of respective row and column.

b. If the supply is 0, then cross (strike) that row and move down to the next cell.

c. If the demand is 0, then cross (strike) that column and move right to the next cell.

d. If supply and demand both are 0, then cross (strike) both row & column and move diagonally to the next cell.

Step-3:Repeat this steps until all supply and demand values are 0.

The minimum total transportation cost = 5×40+10×10+10×20+3×10+3×10+2×60=680

LCM:

Step-1:Select the cell having minimum unit cost c_ij and allocate as much as possible, i.e. min(s_i,d_j).Step-2:a. Subtract this min value from supply s_i and demand d_j.

b. If the supply s_i is 0, then cross (strike) that row and If the demand d_j is 0 then cross (strike) that column.

c. If min unit cost cell is not unique, then select the cell where maximum allocation can be possible

Step-3:Repeat this steps for all uncrossed (unstriked) rows and columns until all supply and demand values are 0.

The minimum total transportation cost = 5×40+10×10+10×20+3×10+3×10+2×60=680

VAM:

Step-1:Find the cells having smallest and next to smallest cost in each row and write the difference (called penalty) along the side of the table in row penalty.

Step-2:Find the cells having smallest and next to smallest cost in each column and write the difference (called penalty) along the side of the table in each column penalty.

Step-3:Select the row or column with the maximum penalty and find cell that has least cost in selected row or column. Allocate as much as possible in this cell.

If there is a tie in the values of penalties then select the cell where maximum allocation can be possible

Step-4:Adjust the supply & demand and cross out (strike out) the satisfied row or column.

Step-5:Repeat this steps until all supply and demand values are 0.

The minimum total transportation cost = 5×40+7×10+3×40+4×30+2×30=570

B.

Stepping Stone:

Step-1:

Find an initial basic feasible solution using any one of the three methods NWCM, LCM or VAM.

Step-2:

1. Draw a closed path (or loop) from an unoccupied cell. The right angle turn in this path is allowed only at occupied cells and at the original unoccupied cell. Mark (+) and (-) sign alternatively at each corner, starting from the original unoccupied cell.

2. Add the transportation costs of each cell traced in the closed path. This is called net cost change.

3. Repeat this for all other unoccupied cells.

Step-3:

1. If all the net cost change are ≥0, an optimal solution has been reached. Now stop this procedure.

2. If not then select the unoccupied cell having the highest negative net cost change and draw a closed path.Step-4:1. Select minimum allocated value among all negative position (-) on closed path

2. Assign this value to selected unoccupied cell (So unoccupied cell becomes occupied cell).

3. Add this value to the other occupied cells marked with (+) sign.

4. Subtract this value to the other occupied cells marked with (-) sign.

Step-5:

Repeat Step-2 to step-4 until optimal solution is obtained. This procedure stops when all net cost change ≥0 for unoccupied cells.

The minimum total transportation cost = 5×40+7×10+3×40+4×30+2×30=570

MODI:

Step-1:

Find an initial basic feasible solution using any one of the three methods NWCM, LCM or VAM.

Step-2:

Find u_i and v_j for rows and columns. To start

a. assign 0 to u_i or v_j where maximum number of allocation in a row or column respectively.

b. Calculate other u_i‘s and v_j‘s using c_ij=u_i+v_j, for all occupied cells.

Step-3:For all unoccupied cells, calculate d_ij=c_ij–(u_i+v_j), .

Step-4:Check the sign of d_ij

a. If d_ij>0, then current basic feasible solution is optimal and stop this procedure.

b. If d_ij=0 then alternative solution exists, with different set allocation and same transportation cost. Now stop this procedure.

b. If d_ij<0, then the given solution is not an optimal solution and further improvement in the solution is possible.

Step-5:Select the unoccupied cell with the largest negative value of d_ij, and included in the next solution.

Step-6:Draw a closed path (or loop) from the unoccupied cell (selected in the previous step). The right angle turn in this path is allowed only at occupied cells and at the original unoccupied cell. Mark (+) and (-) sign alternatively at each corner, starting from the original unoccupied cell.

Step-7:

1. Select the minimum value from cells marked with (-) sign of the closed path.

2. Assign this value to selected unoccupied cell (So unoccupied cell becomes occupied cell).

3. Add this value to the other occupied cells marked with (+) sign.

4. Subtract this value to the other occupied cells marked with (-) sign.

Step-8:Repeat Step-2 to step-7 until optimal solution is obtained. This procedure stops when all d_ij≥0 for unoccupied cells.

The minimum total transportation cost = 5×40+7×10+3×40+4×30+2×30=570