Answer: Sequence VII, I, IV, V, III, II, VI; idle time for A = 5 units, for B = 13 units
Johnson's rule: the minimum time is 5, for job VII on A (place first) and job VI on B (place last). Next minimum 7 is job I on A (place next from the start). Next 10 is job IV (equal on A and B; placed third). Next 14: job V on A (fourth) and jobs II and III on B (placed from the end, III before II). Optimal sequence:
The jobs in the time table are therefore, row by row: VII, I, IV, V, III, II, VI (check: VII takes 5 on A and 7 on B; I takes 7 and 12; II takes 16 and 14; VI takes 15 and 5).
Total elapsed time T = 91 units.
Idle time for machine A \(=T-\) (total working time of A) \(=91-86=\boxed{5}\) units (A is idle after 86).
Idle time for machine B \(=91-78=\boxed{13}\) units (B works \(12+14+14+10+16+5+7=78\); it is idle 5 at the start, then 34 to 36, 52 to 55, 69 to 71 and 85 to 86, i.e. \(5+2+3+2+1=13\)).
