Taking the initial BFS by any method discussed above, this method find the optimal solution to the transportation problem. The steps are given below :

(i) For each row consider a variable u; and for each column consider another variable v/

Find u_{i} and v_{j }such that

u_{i} + v_{j} = c_{ij} for every basic cells.

(ii) For every non-basic cells, calculate the net evaluations as follows :

C_{ij} = U_{i} + V_{j} - C_{ij}

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

If at least one c_{ij} > 0, select the variable having the largest positive net evaluation to enter the basis.

(iii) Let the variable x_{rc }enter the basis. Allocate an unknown quantity 9 to the cell (r, c).

Identify a loop that starts and ends in the cell (r, c).

Subtract and add 9 to the corner points of the loop clockwise/anticlockwise.

(iv) Assign a minimum value of 9 in such a way that one basic variable becomes zero and other basic variables remain non-negative. The basic cell which reduces to zero leaves the basis and the cell with e enters into the basis.

If more than one basic variables become zero due to the minimum value of 9, then only one basic cell leaves the basis and the solution is called degenerate.

(v) Go to step (i) until an optimal BFS has been obtained.

Note. In step (ii), if all c_{ij} < 0, then the optimal solution is unique. If at least one c_{ij} < 0, then we can obtain alternative solution. Assign e in that cell and repeat one iteration (from step (iii)).

**Example 4.** Consider the initial BFS by LCM of Example 2, find the optimal solution of the TP.

For non-basic cells · . c_{ij} = u_{i} + v_{j} - c_{ij}

C_{11} = - 5, C_{12} = - 2, C_{13} =- 6, C_{22}= 0, C_{23 }=- 3, C_{24 }= 2, C_{33} = 1, C_{41} =- 3, C_{44} = 0.

Since all q; are not non-positive, the current solution is not optimal.

Select the cell (2, 4) due to largest positive value 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 below :

Select e = min. (5, 15) = 5. The cell (3, 4) leaves the basis and the cell (2, 4) enters into the basis. Thus the current solution is updated.

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

c_{11} =- 3, c_{2} = 0, C_{13} =- 4, C_{22} = 0, C_{23 }=- 3, c_{33} =I, £:34 =- 2, c_{41} =- 3, c_{44} =- 2.

Since all c_{ij} are not non-positive, the current solution is not optimal.

Select the cell (3, 3) due to largest positive value and assign an unknown quantity e in that cell. Identify a loop and subtract and add e to the corner points of the loop which is shown below :

Select e = min. (5, 25) = 5. The cell (3, 2) leaves the basis and the cell (3, 3) enters into the basis. Thus the current solution is updated.

**Iteration 3.**

For non-basic cells : cij = ui + vj - cij

C_{11} =- 3, c_{12} = - 1, c_{13} =- 5, C_{12} = - 1, C_{23} =- 5, 2 =- I, c_{34} =- 2, c_{41 }=- 2, c_{44 }=- L

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

x_{14 }= 20, x_{21} = 10, x_{24 }= 5, x_{31} = 20, x_{33} = 5, x_{42} = 20, x_{43 }= 20.

The optimal transportation cost

= 1 x 20 + 2 x 10 + 3 x 5 + 3 x 20 + 2 x 5 + 3 x 20 + 1 x 20 = $ 205.

**Example 5.** Consider the initial BFS by VAM of Example 3, find the optimal solution of the TP **Solution.**

For non-basic cells c_{ij} = u_{i} + v_{j }– c_{ij}

C_{11} = - 3, c_{12} = o, c_{13} =- 4, C_{23} =- 3, c_{32} = 0, c_{33} = 1, c_{34} =- 2, c_{41} = - 3, c_{44} = - 2.

Since all c_{ij }are not non-positive, the current solution is not optimal.

Select the cell (3, 3) due to largest positive value 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 below :

Select e = min. (5, 25, 25) = 5. The cell (2, 2) leaves the basis and the cell (3, 3) enters into the basis. Thus the current solution is updated.

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

C_{11}=-3, G_{12}=-1, G_{13}=-5, C_{22}=-I, c_{23}=-5, c_{32}=-1, c_{34}=-2, c_{41}=-2, c_{44}=-1

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

x_{14 }= 20, x_{21} = 10, x_{24} = 5, x_{31} = 20, x_{33} = 5, x_{42} = 20, x_{43} = 20.

The optimal transportation cost= $ 205.

**Note.** To find optimal solution to a T.P., the number of iterations by uv-method is always more if we consider the initial BFS by NWC.