Find the sequence that minimizes the total elapsed time to complete the following jobs. Each job is processed in order \(AB\). Also find the idle time for machine \(B\).Jobs (Processing times in minutes)IIIIIIIVVVIVIIMachine A126511576Machine B789478…

Answer: Sequence III – V – II – VI – I – IV – VII; total elapsed time 55 minutes; idle time of B = 9 minutes

Johnson's rule: pick the smallest time; if it is on A place the job first, if on B place it last.

• Smallest 3 (VII on B): VII last.
• Next 4 (IV on B): IV second last.
• Next 5 (III and V, both on A): a tie on machine A, so the job with the larger time on B is placed first: III (9 min on B) first, V (7 min on B) second.
• Next 6 (II on A): II third.
• Next 7 (VI on A, I on B; V is already placed): one on each machine, so VI goes fourth (next from the start) and I fifth (next from the end).

Optimal sequence: III, V, II, VI, I, IV, VII (V, III, ... gives the same total time).

Total elapsed time \(=55\) minutes. Idle time of B \(=5+2+2=9\) minutes (check: \(55-46=9\)). Idle time of A \(=55-52=3\) minutes.