Instead of unit cost in transportation table, unit profit is considered then the objective of the T.P. changes to maximize the total profits subject to supply and demand restrictions. Then this problem is called 'max-type' T.P.

To obtain optimal solution, we consider

Loss = - Profit

and convert the max type transportation matrix to a loss matrix. Then all the methods described in the previous sections can be applied. Thus the optimal BFS obtained for the loss matrix will be the optimal BFS for the max-type T.P.

**Example 7.** A company has three plants at locations A, B and C, which supply to four markets

D, E, F and G. Monthly plant capacities are 500, 800 and 900 units respectively. Monthly demands of the markets are 600, 700, 400 and 500 units respectively. Unit profits (in rupees) due to transportation are given below :

Determine an optimal distribution for the company in order to maximize the total transportation profits.

**Solution.** The given problem is balanced max type T.P. All profits are converted to losses by multiplying - 1.

The initial BFS by LCM is given below

To find optimal solution let us apply uv-method.

** **

For non-basic cells : c_{ij} = U_{i} + v_{j }-c_{ij}

S_{2} =- 4, C_{13} = - 3, S_{4} = 3, c_{22} =- 4, c_{31} = - 1, c_{33} = - L

Since all c_{ij} are not non-positive, the current solution is not optimal Select the cell (1, 4) due to largest positive value and assign an unknown quantity 9 in that cell Identify a loop and subtract and add 9 to the corner points of the loop which is shown above.

Select θ= min. (500, 300) = 300. The cell (2, 4) leaves the basis and the cell (1, 4) enters into the .basis. Thus the current solution is updated.

For non-basic cells,

C_{12} =- 7, c_{13} =-3, C_{12} =- 7, C_{24} =- 3, C_{31} = 2, C_{33 }= 2.

Since all the c_{ij }are not non-positive, the current solution is not optimal. There is a tie in largest positive values. Let us select the cell (3, 1) and assign an unknown quantity e in that cell. Identify a loop and subtract and add e to the comer points of the loop which is shown above.

Selectθ= min. (200, 200) = 200. Since only one cell will leave the basis, let the cell (3, 3) leaves the basis and assign a zero in the cell (1, 1). The cell (3, I) enters into the basis. Thus the current solution is updated.

For non-basic cells,

C_{12 }=- 5, c_{13} = - 3, C_{12} =- 5, C_{24 }=-3, c_{33} = 0, C_{34} = - 4.

Since all the c_{ij} are non-positive, the current solution is optimal.

Thus the optimal solution, which is degenerate, is

x_{11} = 0, x_{14} =500, x_{21} = 400, x_{23 }= 400, x_{31} = 200, x_{32 }= 700.

The maximum transportation profit

=0 + 3000 + 2800 + 2000 + 1200 + 5600 = $14600.

Since c_{33} = 0, this indicates that there exists an alternative optimal solution. Assign an unknown quantity e in the cell (3, 3). Identify a loop and subtract and add e to the comer points of the loop which is shown below :

Select θ = min. (200, 400) = 200. The cell (3, I) leaves the basis and the cell (3, 3) enters into the basis.

** **

For non- basic cells ,

C_{12} = -5, c_{13} =- 3, C_{22 }=- 5, c_{24} =- 3, C_{31} = 0, c_{34} = - 2.

Since all the c_{ij} are non-positive, the current solution is optimal. Thus the alternative optimal solution is

X_{11 }= 0, x_{14} = 500, x_{21} = 600, x_{23} = 200, x_{32} = 700, x_{33} = 200.

and the maximum transportation profit is $ 14,600.