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A job production unit has four jobs P, Q, R, and 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

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Question

A job production unit has four jobs P, Q, R, and 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:

Job Machines
(Processing cost in ₹)
I II III IV
P 31 25 33 29
Q 25 24 23 21
R 19 21 23 24
S 38 36 34 40

Find the optimal assignment to minimize the total processing cost.

Answer in Brief
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Solution

Step 1: Subtract smallest element of each row from every element of that row.

Jobs 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 2: Since each column contains element 0, column minimization will give us the some matrix.

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

  Jobs I II III IV  
← P 6 0 8 4 →
← Q 4 3 2 0 →
← R 0 2 4 5 →
← S 4 2 0 6 →

∵ No. of lines drawn (4) = order of the matrix (4), optimal solution has reached.

Step 4: Assignment

Jobs I II III IV
P 6 0 8 4
Q 4 3 2 0
R 0 2 4 5
S 4 2 0 6

Job Assignment:

Job Machine Min.cost
P II 25
Q IV 21
R I 19
S III 34
  Total Cost ₹ 99

Minimum Processing Cost = ₹ 99.

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2021-2022 (March) Set 1

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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 :

Job Machines
(Processing cost in ₹)
I II III IV
P 31 25 33 29
Q 25 24 23 21
R 19 21 23 24
S 38 36 34 40

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 :

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 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:

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 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 :

Job Machine Min.cost
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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