A toy manufacturing company produces five types of toys. Each toy has to go through three machines A, B and C in the order ABC. The time required in hours for each process is given in the following table :Type12345Machine A1620121422Machine B1012468M…

Answer: Optimal sequence 3 – 2 – 1 – 4 – 5; total elapsed time = 102 hours; idle time for machine B = 62 hours.

Min A = 12 and max B = 12, so min A \(\ge\) max B and the problem can be converted to two fictitious machines G = A + B and H = B + C.

Johnson's rule: the smallest value is 16 (G, type 3), so type 3 goes first. Next smallest is 18 in H for types 1, 4 and 5 (a tie), so these go at the end; we take type 5 last, then 4, then 1. Type 2 fills the remaining place. Optimal sequence: 3 – 2 – 1 – 4 – 5. (Other orders of the tied types 1, 4, 5 give the same total time.)

Total elapsed time T = 102 hours.

Idle time for machine B = T − (total processing time on B) \(= 102 - (10 + 12 + 4 + 6 + 8) = 102 - 40 = 62\) hours. (Break-up: 12 h before the first job, then 16 + 4 + 4 + 16 h between jobs, and 10 h after the last job until T: \(12 + 16 + 4 + 4 + 16 + 10 = 62\).)