Question

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`

Answer 1

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	25
Q	IV	21
R	I	19
S	III	34

Hence, total (minimum) processing cost = 25 + 21 + 19 + 34 = ₹ 99