A toy manufacturing company produces five types of toys. Each toy has to go through three machines A, B, C in the order ABC. - Mathematics and Statistics

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Sum

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.

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

Find the total elapsed time and also find idle time for machine B.

Solution

 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 2 5 4 1

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, which corresponds to Type 1, Type, 4 and Type 5 on machine H.

∴ Type 1, 4 and 5 are placed either last, second, and last or third last and Type 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 = 62 hrs

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

Notes

Students can refer to the provided solution According to their question.

Concept: Types of Sequencing Problem
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Chapter 7: Assignment Problem and Sequencing - Part II [Page 129]
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