Maximization Transportation Problem

Maximization Transportation Problem

There are certain types of transportation problem where the objective function is to be maximized instead of minimized. These kinds of problems can be solved by converting the maximization problem into minimization problem. The conversion of maximization into minimization is done by subtracting the unit costs from the highest unit cost of the table.
The maximization of transportation problem is illustrated with the following Example.
Example:
A company has three factories located in three cities viz. X, Y, Z. These factories supplies consignments to four dealers viz. A, B, C and D. The dealers are spread all over the country. The production capacity of these factories is 310, 100 and 290 units per month respectively. The net return per unit product is given in the following table:
Dealers

Factory A B C D Capacity
X 6 6 6 4 310
Y 4 2 4 5 100
Z 5 6 7 8 290
Requirement 150 130 120 300 700

Determine a suitable allocation to maximize the total return.
Solution:
This is a maximization problem. Hence first we have to convert this in to minimization problem. The conversion of maximization into minimization is done by subtracting the unit cost of the table from the highest unit cost. Look the table, here 8 is the highest unit cost. So, subtract all the unit cost from the 8, and then we get the revised minimization transportation table, which is given below.
Dealers

Factory A B C D Capacity
X 2 2 2 4 310 = a1
Y 4 6 4 3 100 = a2
Z 3 2 1 0 290 =a3
Requirement 150 =b1 130 =b2 120 =b3 300 =b4 700

Now we can solve the problem as a minimization problem.
First we have to find out the basic feasible solution, we will use the (LCM).
2
150 2
130 2
30 4
X 310
4
X 6
X 4
90 3
10 100
3
X 2
X 1
X 0
290 290
150 130 120 300 700








The basic feasible solution by ( LCM ) is : x11=150, x12=130, x13=30, x23=90, x24=10 and x34=290.
Once the basic feasible solution is found, next we have to determine the optimum solution. Here we use MODI (Modified Distribution Method) method. By using this method we obtain
u1+v1=2 , u1+v2=2 , u1+v3=2 , u2+v3=4 , u2+v4=3 , u3+v4=0.
Taking u1=0 arbitrarily we obtain
u1=0, v1=2, v2=2 and v3=2, u2=2, v4=1 , u3= - 1.
By verifying the condition of optimality, we get:
Cell (1,4) = 4 – 0 – 1 = 3
Cell (2,1) = 4 – 2 – 2 = 0
Cell (2,2) = 6 – 2 – 2 = 2
Cell (3,1) = 3 – (–1) – 2 = 2
Cell (3,2) = 2 – (–1) – 2 = 1
Cell (3,3) = 1 – (–1) – 2 = 0
So, the optimality condition is satisfied.
x11=150, x12=130, x13=30, x23=90, x24=10 and x34=290.Thus, the maximum return is:
6*150 + 6*130 + 6*30 + 4*90 + 5*10 + 8*290 = 4590.