A marketing manager has 5 salesman and 5 sales districs. Considering the capabilities of the salesman and the nature of the districts, the marketing manager estimates that sales per month in thousand rupees for each salesman in each district would be as follows. Find the best allocation to maximize the sales :
District
1 2 3 4 5
P 36 36 31 21 46
Salesman Q 38 26 12 21 23
R 46 44 36 56 46
S 16 15 16 14 21
T 54 56 55 51 49
Answers
Answer:
The above is the answer.
Maximum Sales would be Rs.19,100
GIVEN: Salaries of 5 Salesmen and 5 Sales districts.
TO FIND: Maximum Sales
SOLUTION:
As we know,
This is a maximization problem and has to be converted into a minimization problem by subtracting all the elements from the largest element of the sales table, Here the largest element is 41. Hence, the equivalent sales table for this problem would be obtained by reducing all the elements from 41 and rewriting it.
A series of following steps will be used:
Step 1: Subtract the smallest element of each row from every element of the corresponding row.
Step 2: Subtract the smallest element of each column from every element of the corresponding column, we get the reduced matrix.
Step 3: Starting with row one, we make an assignment in a single zero and cross out all other zeros in the column marked.
Step 4: Draw the minimum number of horizontal and vertical lines which cover all the zeroes.
Step 5. The least uncovered element 4 is subtracted from all the uncovered elements and added to the intersection of the elements, we get the following reduced matrix.
Optimum assignment is
i.e.
1 -> B, 2 -> A, 3 -> E, 4 -> C, 5 -> D
38 + 40 + 37 + 41 + 35 = 191 (in hundred rupees).
Maximum sales would be Rs 19,100.
#SPJ3
