Question

A toy manufacturing company produces five types of toys. Each toy has to go through three machines A, B, C in the order ABC. The time required in hours for each process is given in the following table.

Type	1	2	3	4	5
Machine A	16	20	12	14	22
Machine B	10	12	4	6	8
Machine C	8	18	16	12	10

Solve the problem for minimizing the total elapsed time.

Answer 1

Type	1	2	3	4	5
Machine A	16	20	12	14	22
Machine B	10	12	4	6	8
Machine C	8	18	16	12	10

Here min A = 12, max B = 12, min C = 8.

Since min A ≥ max B is satisfied, the problem can be converted into a two machine problem.

Let G and H be two fictitious machines such that G = A + B and H = B + C

Then the problem can be written as

Type	1	2	3	4	5
Machine G	26	32	16	20	30
Machine H	18	30	20	18	18

Observe that Min (G, H) = 16, corresponds to Type 3 on machine G.

∴ Type 3 is placed first in sequence.

3

Then the problem reduces to

Type	1	2	4	5
Machine G	26	32	20	30
Machine H	18	30	18	18

Now, Min (G, H) = 18, corresponds to Type 1, Type 4 and Type 5 on machine H.

∴ Type 1, 4 and 5 are placed either last, second last or third last and Type 2 on remaining in the sequence.

3

2

5

4

1

OR

3

2

5

1

4

OR

3

2

1

4

5

OR

3

2

1

5

4

OR

3

2

4

1

5

OR

3

2

4

5

1

We take the optimal sequence as,

3 – 2 – 5 – 4 – 1

Total elapsed time

Job	Machine A		Machine B		Machine C
	In	Out	In	Out	In	Out
3 (12, 4, 16)	0	12	12	16	16	32
2 (20, 12, 18)	12	32	32	44	44	62
5 (22, 8, 10)	32	54	54	62	62	72
4 (14, 6, 12)	54	68	68	74	74	86
1 (16, 10, 8)	68	84	84	94	94	102

∴ Total elapsed time = 102 hrs

Idle time for machine A = 102 – 84 = 18 hrs

Idle time for machine B = (102 – 94) + 12 + 16 + 10 + 6 + 10 = 62hrs

Idle time for machine C = 16 + 12 + 2 + 8 = 38 hrs.