an optimum solution of transportation problem result when net cost change values of all unoccupied cells are
Answers
Answer: hey mate
Here you go
Step-by-step explanation:
The net change in transportation cost as a result of this perturbation is called the evaluation of the empty cell in question.
Therefore,
Evaluation of cell (3,2) = Rs.100 x (8 x 1 – 5 x 1 + 2 x 1 – 5 x 1)
= Rs. (0 x 100)
= Rs.0.
Thus the total transportation cost increases by Rs.0 for each unit allocated to cell (3,2). Likewise, the net evaluation (also called opportunity cost) is calculated for every empty cell. For this the following simple procedure may be adopted.
Starting from the chosen empty cell, trace a path in the matrix consisting of a series of alternate horizontal and vertical lines. The path begins and terminals in the chosen cell. All other corners of the path lie in the cells for which allocations have been made. The path may skip over any number of occupied or vacant cells. Mark the corner of the path in the chosen vacant cell as positive and other corners of the path alternatively –ve, +ve, -ve and so on. Allocate I unit to the chosen cell; subtract and add 1 unit from the cells at the corners of the path, maintaining the row and column requirements. The net change in the total cost resulting from this adjustment is called the evaluation of the chosen empty cell. Evaluation of the various empty cells (in hundreds of rupees) are:
Cell (1, 3) = c13-c33+c31-c11 = 11-15+5 = -1,
Cell (1, 4) = c14-c34+c31-c11 = 7-9+5-2 = +1,
Cell (2, 1) = c21-c24+c34-c31 = 1-1+9-5 = +4,
Cell (2, 2) = c22-c24+c34-c31+c11-c12 = 0-1+9-5+2-3 = +2,
Cell (2, 3) = c23-c24+c34-c33 = 6-1+9-15 = -1,
Cell (3, 2) = c32-c31+c11-c12 = 8-5+2-3 = +2.
If any cell evaluation is negative, the cost can be reduced so that the solution under consideration can be improved i.e., it is not optimal. On the other hand, if all cell evaluations are positive or zero, the solution in question will be optimal. Since evaluations of cells (1, 3) and (2, 3) are –ve, initial basic feasible solution given in table 3.15 is not optimal.
Now in a transportation problem involving m rows and n columns, the total number of empty cells will be,
m.n – (m+n-1) = (m-1) (n-1).
Therefore, there are (m-1) (n-1) such cell evaluations which must be calculated and for large problems, the method can be quite inefficient. This method is named ‘stepping-stone’ since only occupied cells or ‘stepping stones’ are used in the evaluation of vacant cells.
Answer:
transportation cost solving transportation problem