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CMA Intermediate · Operations Management and Strategic Management · Scheduling and Queuing Models

In Johnson's rule for sequencing n jobs through two machines (M1 then M2), which step correctly describes how the optimal sequence is built?

Johnson's rule selects the smallest processing time overall. If it lies on the first machine, the job is placed at the earliest free position; if on the second machine, at the latest free position. This reduces idle time on the second machine and minimises total elapsed time.

  1. ASelect the smallest processing time in the list; if it is on M1, place that job as early as possible, and if it is on M2, place it as late as possibleCorrect
  2. BSelect the largest processing time in the list; if it is on M1, place that job as late as possible, and if it is on M2, place it as early as possible
  3. CArrange jobs in decreasing order of their total processing time across both machines
  4. DArrange jobs in increasing order of their M2 processing time only

Explanation

Johnson's rule picks the minimum time among all jobs and both machines. A minimum on M1 sends the job to the front, so M2 starts early; a minimum on M2 sends it to the back, so M2 finishes quickly. Using the largest time reverses the logic and does not minimise makespan.

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