Maximization Case Of Transportation Model
a) By multiplying the matrix by –1, we can convert it into a maximisation problem.
b) Now in VAM we have to find the row opportunity cost and column opportunity costs.
c) In minimization problem, we use to subtract the smallest element in the row from next highest element in that row for finding row opportunity cost.
d) Similarly, we use to subtract smallest element in the column by next highest element in that column to get column opportunity cost.
e) Multiplied the matrix by –1 the highest element will become lowest element.
f) Subtract the lowest element from the next highest element as usual.
g) Otherwise, instead of multiplying by –1 simply find the difference between highest element and the next lowest element and take it as opportunity cost of that row or column.
h) in the given problem in the row A, the highest element is 4 and the next lowest element is 3 and hence the opportunity cost is 4 – 3 = 1. (Or smallest element is –4 and the next highest element is –3 and the opportunity cost is –3 – (– 4) = –3 4 = 1). Similarly, we can write all opportunity costs.
i) Once we find the opportunity costs, rest of the procedure is same. That is, we have to select highest opportunity cost and select the highest profit element in that row or column for allocation.
j) Obtain the basic feasible solution.
k) As usual the basic feasible solution must have m n – 1 allocation.
l) If the allocations are not equal to m n – 1, the problem degenerate. In that case, add ∈to an empty cell, which do not form loop with other loaded cells.
m) The optimality test by MODI method is followed. Here, once the opportunity costs of all the cells are positive, (as we have converted the maximistion problem into minimization problem) the solution is said to be optimal.
n) In the given problem as the opportunity costs of all empty cells are positive, the solution is optimal. And the optimal return to the company is Rs. 125/-.
