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

RuleOrder jobs byMinimises
SPT (shortest processing time)Increasing processing timeMean flow time, mean waiting time
EDD (earliest due date)Increasing due dateMaximum lateness (tardiness)
FCFSArrival orderNothing 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:

  1. Find the smallest time among all jobs on both machines.
  2. If it is on machine 1, put that job as early as possible; if on machine 2, as late as possible.
  3. 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

GATE-styleNAT · numerical answer

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 explanationAnswer: 10

SPT order 2, 4, 6, 8 gives completion times 2, 6, 12, 20; mean = 40/4 = 10 hours.

GATE-styleNAT · numerical answer

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 explanationAnswer: 13

The 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.

GATE-styleNAT · numerical answer

A c chart has an average of 9 defects per unit. What is its upper control limit?

Show answer and explanationAnswer: 18

UCL = c̄ + 3√c̄ = 9 + 3 × 3 = 18.

GATE-styleMCQ · one correct answer

To minimise the maximum lateness on a single machine, jobs should be sequenced by:

Show answer and explanationAnswer: B

EDD order minimises the maximum lateness; SPT minimises mean flow time.