Scheduling, Work Study and Quality Control
Sequencing rules (SPT, EDD), Johnson's rule for two machines, work study and standard time, break-even analysis, control charts and acceptance sampling.
This lesson covers short-term scheduling, work study (how long a job should take), break-even analysis, and statistical quality control.
Sequencing jobs on one machine
| Rule | Order jobs by | Minimises |
|---|---|---|
| SPT (shortest processing time) | Increasing processing time | Mean flow time, mean waiting time |
| EDD (earliest due date) | Increasing due date | Maximum lateness (tardiness) |
| FCFS | Arrival order | Nothing in particular; it is simple and fair |
Johnson's rule (n jobs, 2 machines)
Every job goes through machine 1 then machine 2. To minimise the makespan:
- Find the smallest time among all jobs on both machines.
- If it is on machine 1, put that job as early as possible; if on machine 2, as late as possible.
- Remove the job and repeat.
Work study
- Method study finds a better way to do the job (process charts, therbligs).
- Time study: observed time × rating factor = normal time. Standard time = normal time × (1 + allowance fraction).
- Work sampling: many random observations estimate the fraction of time spent on each activity.
Break-even analysis
With fixed cost F, selling price p and variable cost v per unit:
- Break-even quantity = F/(p − v). (p − v) is the contribution per unit.
- Profit at quantity Q = Q(p − v) − F.
Statistical quality control
- Control charts for variables: X̄ chart (mean) and R chart (range). Control limits for X̄: X̿ ± A₂R̄.
- Control charts for attributes: p chart (fraction defective): p̄ ± 3√(p̄(1 − p̄)/n); c chart (defects per unit): c̄ ± 3√c̄.
- A point outside the limits means an assignable cause to investigate.
- Acceptance sampling: inspect a sample of n and accept the lot if defects ≤ c. The OC curve shows the probability of accepting lots of each quality. Producer's risk α (rejecting a good lot), consumer's risk β (accepting a bad lot).
Worked example
Observed time 2.0 min, rating 110%, allowances 10%.
- Normal time = 2.0 × 1.10 = 2.2 min. Standard time = 2.2 × 1.10 = 2.42 min.
Practice
Jobs with processing times 6, 2, 8 and 4 hours are done in SPT order. What is the mean flow time in hours?
Show answer and explanation
Answer: 10SPT order 2, 4, 6, 8 gives completion times 2, 6, 12, 20; mean = 40/4 = 10 hours.
Three jobs pass through machines M1 then M2 with times A (3, 6), B (5, 2), C (1, 4). What is the minimum makespan using Johnson's rule?
Show answer and explanation
Answer: 13The smallest time is C on M1 (1), so C goes first; next is B on M2 (2), so B goes last; the order is C, A, B. M1 finishes the jobs at 1, 4 and 9. M2 finishes C at 5, A at max(5, 4) + 6 = 11 and B at max(11, 9) + 2 = 13.
A c chart has an average of 9 defects per unit. What is its upper control limit?
Show answer and explanation
Answer: 18UCL = c̄ + 3√c̄ = 9 + 3 × 3 = 18.
To minimise the maximum lateness on a single machine, jobs should be sequenced by:
Show answer and explanation
Answer: BEDD order minimises the maximum lateness; SPT minimises mean flow time.