Find the sequence that minimizes the total elapsed time to complete the following jobs in the order AB. Find the total elapsed time and idle time for machine B :JobsIIIIIIIVVVIVIIMachine A716191014155Machine B121414101657

Answer: Sequence VII \(\to\) I \(\to\) IV \(\to\) V \(\to\) III \(\to\) II \(\to\) VI; total elapsed time = 91 units; idle time for B = 13 units.

Johnson's rule:

• Smallest time is 5: job VII on A (place first) and job VI on B (place last).
• Next is 7: job I on A, placed in the next position from the start (job VII's 7 on B is already used).
• Next is 10: job IV has 10 on both machines (tie for the same job), so it may be placed either in the next position from the start or from the end; we place it next from the start.
• Next is 14: job V on A goes next from the start; jobs II and III have 14 on B, so they fill the remaining positions from the end (either order).

Optimal sequence: VII, I, IV, V, III, II, VI (VII, I, IV, V, II, III, VI, or placing IV in the sixth position, e.g. VII, I, V, III, II, IV, VI, are alternatives with the same total time of 91 units).

Total elapsed time = 91 units.

Idle time for machine B = \(91-(12+14+14+10+16+5+7)=91-78=13\) units (5 at the start, then 2, 3, 2 and 1 while waiting for jobs V, III, II and VI).

(Idle time for machine A = \(91-86=5\) units, after it finishes at 86.)