What Johnson's rule is
Johnson's rule is a sequencing method that orders n jobs passing through two machines, always in the same order (machine 1, then machine 2), so that the makespan is as short as possible. The makespan is the time from the start of the first job to the finish of the last one. For this problem the rule gives the optimal order with a simple procedure and no trial and error.
S. M. Johnson published it in 1954. The logic is straightforward: machine 2 cannot start a job until machine 1 has finished it. So jobs with a short time on M1 go first (M2 starts working early) and jobs with a short time on M2 go last (M2 finishes right after M1). Dispatching rules such as FIFO, SPT and EDD deal with queues and due dates; Johnson's rule is the right choice when the goal is to clear the batch as early as possible.
Visual map of Johnson's rule
The map summarizes the steps, the six-job example and the total-time calculation on the Gantt chart.
How to apply the rule, step by step
List each job's time on machine 1 (M1) and machine 2 (M2), then:
- Find the smallest time in the whole table, across M1 and M2.
- If it is on M1, put that job in the first free position of the sequence (counting from the start).
- If it is on M2, put the job in the last free position (counting from the end).
- Remove the job from the list and repeat until none are left. Ties can be broken arbitrarily; the result stays optimal.
An equivalent shortcut works well in a spreadsheet: split the jobs into two groups. Group 1 holds jobs with M1 ≤ M2, sorted by M1 time in ascending order. Group 2 holds jobs with M1 > M2, sorted by M2 time in descending order. The final sequence is group 1 followed by group 2.
Makespan = time at which M2 finishes the last job
To compute times: a job starts on M1 as soon as the previous job leaves M1; on M2 it starts at the later of its own finish on M1 and the previous job's finish on M2.
Worked example with six jobs
Fictional example: a machine shop processes six parts in two operations, lathe (M1) and grinder (M2). Times in minutes:
| Job | M1 (lathe) | M2 (grinder) |
|---|---|---|
| A | 5 | 2 |
| B | 3 | 6 |
| C | 8 | 4 |
| D | 7 | 7 |
| E | 2 | 5 |
| F | 6 | 3 |
Applying the rule:
- Smallest time in the table: 2, for E on M1 (A on M2 is also 2). Taking E, it goes to position 1.
- Without E, the smallest is 2, for A on M2: A goes to the last position (6th).
- B, C, D, F remain. The smallest is 3, for B on M1 (F on M2 is also 3). B goes to position 2.
- C, D, F remain. The smallest is 3, for F on M2: F goes to position 5.
- C (8; 4) and D (7; 7) remain. The smallest is 4, for C on M2: C takes position 4 and D position 3.
Johnson sequence: E – B – D – C – F – A. The Gantt chart in minutes:
| Job | M1 (start–end) | M2 (start–end) |
|---|---|---|
| E | 0–2 | 2–7 |
| B | 2–5 | 7–13 |
| D | 5–12 | 13–20 |
| C | 12–20 | 20–24 |
| F | 20–26 | 26–29 |
| A | 26–31 | 31–33 |
The makespan is 33 minutes. If the parts were run in arrival order (A – B – C – D – E – F), the total would be 38 minutes. That is 5 minutes, or 13.2% less, with no change in machines, method or staffing.
| Sequence | Makespan (min) | M2 utilization |
|---|---|---|
| A–B–C–D–E–F (arrival order) | 38 | 27 ÷ 38 = 71.1% |
| E–B–D–C–F–A (Johnson) | 33 | 27 ÷ 33 = 81.8% |
Sanity check: the sum of M1 times is 31 min and the smallest M2 time is 2 min (job A). No sequence can finish before 31 + 2 = 33 min, so 33 is truly the minimum. The M2 times add up to 27 min, so M2 sits idle for 6 min (0–2, 24–26 and 29–31).
What about three machines?
The original rule is exact for two machines. For three (M1, M2, M3) it stays optimal only in a special case: when the smallest M1 time is greater than or equal to the largest M2 time, or when the smallest M3 time is greater than or equal to the largest M2 time. Then M2 is never the bottleneck, and you build two fictitious machines:
- M' = time on M1 + time on M2;
- M'' = time on M2 + time on M3.
Apply Johnson to M' and M''. If the condition does not hold, the method only gives a good starting solution, and the general problem calls for enumeration, branch and bound or heuristics. For many machines with different routings (job shop), use dispatching rules, linear programming or finite-capacity scheduling software.
When to use Johnson's rule
- Two-stage lines with the same routing: machining and finishing, stamping and welding, printing and cutting, oven and packing.
- Batches released at once by the production order, with every item available at the start.
- End of shift or end of campaign: when freeing machines and labor as early as possible is what matters.
- A complement to production planning: after deciding what to build in the master production schedule, the rule orders the batch at the scheduling stage of production planning and control.
To show the result to the team, draw the Gantt chart of both machines, as in the table above. If the constraint sits on a single resource, see also the theory of constraints.
Common mistakes and limitations
- Using Johnson to meet due dates. It minimizes total time, not tardiness. With critical delivery dates, compare it with EDD.
- Forgetting that every job goes through M1 and then M2, with no reversed routing. If some jobs have the reverse routing, it is a different problem.
- Ignoring sequence-dependent setup times. The rule assumes setup is included in the processing time or independent of the order.
- Applying it when jobs arrive over time. The rule assumes all jobs are available at time zero.
- Not allowing for queues. If there is no buffer between the machines (blocking), the result changes.
- Miscalculating the Gantt chart. The most frequent slip is starting M2 before the job has left M1.
Frequently asked questions
What is Johnson's rule?
A method that sequences n jobs on two machines with the same routing to minimize the total completion time (makespan).
Does Johnson's rule always give the optimal solution?
Yes, for two machines in a flow shop with all jobs available at the start. For three or more machines, only under special conditions.
What is makespan?
The time between the start of the first job and the finish of the last. In the example, 33 minutes with Johnson and 38 in arrival order.
How do you handle ties in Johnson's rule?
Pick any of the tied jobs. The final makespan is the same, even if the sequence differs.
What is the difference between Johnson's rule and SPT?
SPT orders by shortest processing time and reduces average flow time. Johnson uses the times on both machines to shorten the total time of the batch.
Can I use it in a spreadsheet?
Yes. Split the jobs into two groups (M1 ≤ M2 and M1 > M2), sort the first by M1 ascending and the second by M2 descending, then build the Gantt with the MAX function.
Sources
- JOHNSON, S. M. Optimal two- and three-stage production schedules with setup times included. Naval Research Logistics Quarterly, v. 1, 1954.
- PINEDO, M. L. Scheduling: Theory, Algorithms, and Systems. Springer.
- BAKER, K. R. Introduction to Sequencing and Scheduling. Wiley, 1974.
- CHASE, R. B.; JACOBS, F. R.; AQUILANO, N. J. Operations Management for Competitive Advantage. McGraw-Hill.
Keep studying: trusted sources and links
- ASCM: Association for Supply Chain Management
- ABEPRO: Associação Brasileira de Engenharia de Produção (in Portuguese)
- Wikipedia: Scheduling (production processes)
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About the author
Lean Manufacturing & Behavioral Management Specialist
Over 20 years in the automotive and metalworking industries (GM and Dana), Lean Manufacturing practitioner since 2006. SENAI instructor and mentor in Brazil’s Brasil Mais Produtivo program, delivering consulting, training and audits for 50+ companies, combining quality, productivity and people development.
