USA: +1-585-535-1023

UK: +44-208-133-5697

AUS: +61-280-07-5697

UV-Method/Modi Method

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 ui and vj such that

ui + vj = cij for every basic cells.

 

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

Cij = Ui + Vj - Cij

If all cij are non-positive, the current solution is optimal.

If at least one cij > 0, select the variable having the largest positive net evaluation to enter the basis.

 

(iii) Let the variable xrc 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 cij < 0, then the optimal solution is unique. If at least one cij < 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.

Solution. Iteration 1

For non-basic cells · . cij = ui + vj - cij

 

C11 = - 5, C12 = - 2, C13 =- 6, C22= 0, C23 =- 3, C24 = 2, C33 = 1, C41 =- 3, C44 = 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.

 

Iteration 2.

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

 

c11 =- 3, c2 = 0, C13 =- 4, C22 = 0, C23 =- 3, c33 =I, £:34 =- 2, c41 =- 3, c44 =- 2.

 

Since all cij 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

 

C11 =- 3, c12 = - 1, c13 =- 5, C12 = - 1, C23 =- 5, 2 =- I, c34 =- 2, c41 =- 2, c44 =- L

 

Since all cij are non-positive, the current solution is optimal. Thus the optimal solution is

 

x14 = 20, x21 = 10, x24 = 5, x31 = 20, x33 = 5, x42 = 20, x43 = 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.

Iteration 1.

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

C11 = - 3, c12 = o, c13 =- 4, C23 =- 3, c32 = 0, c33 = 1, c34 =- 2, c41 = - 3, c44 = - 2.

 

Since all cij 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.

 

Iteration 2.

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

 

C11=-3, G12=-1, G13=-5, C22=-I, c23=-5, c32=-1, c34=-2, c41=-2, c44=-1

 

Since all cij are non-positive, the current solution is optimal. Thus the optimal solution is

 

x14 = 20, x21 = 10, x24 = 5, x31 = 20, x33 = 5, x42 = 20, x43 = 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.