Transportation Problem

Modified Distribution Method - Example

This example is the largest and the most involved you have read so far. So you must read the steps and the explanation mindfully.

Example

Consider the transportation problem presented in the following table.

Determine the optimal solution of the above problem.

Solution.

An initial basic feasible solution is obtained by Matrix Minimum Method and is shown in table 1.

Table 1

Distribution centre
		D1	D2	D3	D4	Supply
Plant	P1	19	30	50		7
	P2		30		60	10
	P3			60		18
Requirement		5	8	7	15

Initial basic feasible solution

12 X 7 + 70 X 3 + 40 X 7 + 40 X 2 + 10 X 8 + 20 X 8 = Rs. 894.

Calculating u_i and v_j using u_i + v_j = c_ij

Substituting u₁ = 0, we get

u₁ + v₄ = c₁₄ Þ 0 + v₄ = 12 or v₄ = 12
u₃ + v₄ = c₃₄ Þ u₃ + 12 = 20 or u₃ = 8
u₃ + v₂ = c₃₂ Þ 8 + v₂ = 10 or v₂ = 2
u₃ + v₁ = c₃₁ Þ 8 + v₁ = 40 or v₁ = 32
u₂ + v₁ = c₂₁ Þ u₂ + 32 = 70 or u₂ = 38
u₂ + v₃ = c₂₃ Þ 38 + v₃ = 40 or v₃ = 2

Table 2

Distribution centre
		D1	D2	D3	D4	Supply	ui
Plant	P1	19	30	50		7	0
	P2		30		60	10	38
	P3			60		18	8
Requirement		5	8	7	15
vj		32	2	2	12

Calculating opportunity cost using c_ij – ( u_i + v_j )

Table 3

Distribution centre
		D1	D2	D3	D4	Supply	ui
Plant	P1					7	0
	P2					10	38
	P3					18	8
Requirement		5	8	7	15
vj		32	2	2	12

Now choose the smallest (most) negative value from opportunity cost (i.e., –13) and draw a closed path from P1D1. The following table shows the closed path.

Table 4

Choose the smallest value with a negative position on the closed path(i.e., 2), it indicates the number of units that can be shipped to the entering cell. Now add this quantity to all the cells on the corner points of the closed path marked with plus signs and subtract it from those cells marked with minus signs. In this way, an unoccupied cell becomes an occupied cell.

Now again calculate the values for u_i & v_j and opportunity cost. The resulting matrix is shown below.

Table 5

Distribution centre
		D1	D2	D3	D4	Supply	ui
Plant	P1					7	0
	P2					10	51
	P3					18	8
Requirement		5	8	7	15
vj		19	2	–11	12

Choose the smallest (most) negative value from opportunity cost (i.e., –23). Now draw a closed path from P2D2 .

Table 6

Now again calculate the values for u_i & v_j and opportunity cost

Don't panic. The following table is the last table.

"Every complex human problem has an easy solution - neat & small, but wrong." - Vinay Chhabra & Manish Dewan

Table 7

Distribution centre
		D1	D2	D3	D4	Supply	ui
Plant	P1					7	0
	P2					10	28
	P3					18	8
Requirement		5	8	7	15
vj		19	2	12	12

Since all the current opportunity costs are non–negative, this is the optimal solution. The minimum transportation cost is: 19 X 5 + 12 X 2 + 30 X 3 + 40 X 7 + 10 X 5 + 20 X 13 = Rs. 799