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Maharashtra State BoardHSC Commerce (English Medium) 12th Standard Board Exam
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A job production unit has four jobs A, B, C, D which can be manufactured on each of the four machines P, Q, R and S. The processing cost of each job is given in the following table - Mathematics and Statistics

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Question

A job production unit has four jobs A, B, C, D which can be manufactured on each of the four machines P, Q, R and S. The processing cost of each job is given in the following table:

         Jobs

 

 

                          Machines

P

Q

R

S

                Processing Cost (Rs.)

 

A

31

25

33

29

B

25

24

23

21

C

19

21

23

24

D

38

36

34

40

 How should the jobs be assigned to the four machines so that the total processing cost is minimum?

Sum

Solution

We can express the matrix form

`[(31,25,33,29), (25,24,23,21), (19,21,23,24), (38,36,34,40)]`

Subtracting the smallest element in each row from every element of it,

`[(6,0,8,4), (4,3,2,0), (0,2,4,5), (4,2,0,6)]`

Subtracting the smallest element In each column from very element of it.

`[(6,0,8,4), (4,3,2,0), (0,2,4,5), (4,2,0,6)]`

All the zeros of the above matrix are covered with minimum number of lines as below :

No. of lines = No. of rows/columns

Assignment of jobs :

A → Q, B → S, C → P, D → R

Minimum cost = 25 + 21 + 19 + 34

                        = Rs 99

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Assignment Problem
  Is there an error in this question or solution?
Q 6.2.3
2014-2015 (March)

APPEARS IN

2014-2015 (March) (with solutions)
Q 6.2.3 | 4 marks

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A job production unit has four jobs P, Q, R, S which can be manufactured on each of the four machines I, II, III and IV. The processing cost of each job for each machine is given in the following table :

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Complete the following activity to find the optimal assignment to minimize the total processing cost.

Solution:

Step 1: Subtract the smallest element in each row from every element of it. New assignment matrix is obtained as follows :

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I II III IV
P 6 0 8 4
Q 4 3 2 0
R 0 2 4 5
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Step 2: Subtract the smallest element in each column from every element of it. New assignment matrix is obtained as above, because each column in it contains one zero.

Step 3: Draw minimum number of vertical and horizontal lines to cover all zeros:

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(Processing cost in ₹)
I II III IV
P 6 0 8 4
Q 4 3 2 0
R 0 2 4 5
S 4 2 0 6

Step 4: From step 3, as the minimum number of straight lines required to cover all zeros in the assignment matrix equals the number of rows/columns. Optimal solution has reached.

Examine the rows one by one starting with the first row with exactly one zero is found. Mark the zero by enclosing it in (`square`), indicating assignment of the job. Cross all the zeros in the same column. This step is shown in the following table :

Job Machines
(Processing cost in ₹)
I II III IV
P 6 0 8 4
Q 4 3 2 0
R 0 2 4 5
S 4 2 0 6

Step 5: It is observed that all the zeros are assigned and each row and each column contains exactly one assignment. Hence, the optimal (minimum) assignment schedule is :

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P II `square`
Q `square` 21
R I `square`
S III 34

Hence, total (minimum) processing cost = 25 + 21 + 19 + 34 = ₹`square`


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