Operations Management and Strategic Management · Scheduling and Queuing Models

Scheduling Rules and Priority Dispatching: FCFS, SPT, EDD, Critical Ratio

Updated 10 October 2026 · Fact-checked

Priority dispatching rules decide which waiting job a machine takes next. FCFS follows arrival order, SPT picks the shortest processing time, EDD picks the earliest due date, and critical ratio ranks jobs by time left divided by work left. To solve, sequence the jobs, find completion times, then compute flow time and tardiness.

Understand Scheduling Rules and Priority Dispatching

A machine or work centre often has several jobs waiting. Sequencing means deciding the order in which they are processed. A priority rule (dispatching rule) is a simple, fixed way to pick the next job so the decision is fast and fair.

The four rules you must know are these. FCFS (first come, first served) takes jobs in arrival order. SPT (shortest processing time) takes the job with the smallest processing time first. EDD (earliest due date) takes the job whose due date is nearest. Critical ratio (CR) is a dynamic rule: it is recalculated each time you pick a job, using today's date.

You compare rules using performance measures. Flow time of a job is its completion time minus its arrival (release) time. If all jobs are available at time 0, flow time equals completion time. Lateness is completion time minus due date and can be negative. Tardiness is lateness, but never below zero: tardiness = max(0, lateness). Early jobs have zero tardiness, not negative tardiness.

No single rule wins on everything. For a single machine with all jobs available at time 0, SPT gives the lowest average flow time (and so the lowest average completion time and average number of jobs in the system). EDD gives the lowest maximum tardiness (maximum lateness). FCFS is simple and seen as fair but often performs worst. Critical ratio is useful when jobs have different remaining work and due dates, because it shows who is behind schedule.

Key rules to remember

Completion time: Completion time of a job = Completion time of previous job + Its processing time; For jobs all available at time 0 and a single machine with no idle time. Add up in the chosen sequence.
Flow time: Flow time = Completion time − Arrival (release) time; If all jobs are ready at time 0, flow time = completion time.
Average flow time: Average flow time = Σ Flow times ÷ Number of jobs; SPT minimises this on a single machine when all jobs are available together.
Lateness and tardiness: Lateness = Completion time − Due date; Tardiness = max(0, Lateness); Average tardiness = Σ Tardiness ÷ Number of jobs. EDD minimises maximum tardiness.
Critical ratio: CR = (Due date − Today's date) ÷ Remaining processing time; Use the same time unit for both. Lowest CR goes first. CR < 1 means behind schedule, CR = 1 on schedule, CR > 1 ahead. A negative CR means the job is already past its due date.
Average number of jobs in system: Average jobs in system = Σ Flow times ÷ Total time to finish all jobs; Total time is the completion time of the last job (the makespan).

How to solve Scheduling Rules and Priority Dispatching questions

Use this method for any sequencing question on a single machine. Keep the working in a table so you earn step marks.

1Read the question and note the rule(s) asked, the arrival time of each job, and whether due dates are given.
2List each job with its processing time and due date.
3Fix the sequence for each rule: FCFS by arrival order as given, SPT by smallest time first, EDD by earliest due date first. For critical ratio, calculate CR for every job and rank the lowest first.
4Build a table in that sequence: job, processing time, completion time (running total), due date.
5Compute flow time (completion − arrival) and tardiness = max(0, completion − due date) for each job.
6Total the columns and find the averages, the number of late jobs and the maximum tardiness.
7Compare the rules on each measure in a small summary table.
8Write a one or two line conclusion: which rule is best on which measure, and why.

Quickest way: Running-total table method

When to use it: Use for MCQs and for 14-mark numericals where all jobs are available at time 0 on one machine.

Write the sequence first, then the processing times beneath it.
Add running totals once. These are the completion times.
Subtract due dates from the totals. Cross out negatives (write 0) to get tardiness.
Add the totals and divide by the number of jobs.
Cross-check: the last completion time must equal the sum of all processing times, and it is the same for every rule.

Common mistakes in Scheduling Rules and Priority Dispatching

Treating negative lateness as negative tardiness and subtracting it in the total.
Students confuse lateness with tardiness.
Fix: Tardiness is never below zero. Replace every negative lateness with 0 before adding.
Using processing time instead of completion time when calculating average flow time.
Flow time sounds like time spent on the job only.
Fix: Flow time includes waiting. Use the running total (completion time) less arrival time.
Sorting EDD or SPT in the wrong direction.
Rushing and sorting by the wrong column.
Fix: Both rules go smallest first: smallest processing time for SPT, earliest due date for EDD. Write the sequence before computing anything.
Putting the highest critical ratio first.
Students assume a bigger number means more urgent.
Fix: A low CR means little time compared with the work left. The lowest CR, including negatives, is processed first.
Mixing units in the critical ratio, such as due date in days and remaining work in hours.
The question gives data in different units.
Fix: Convert both to the same unit before dividing. Use time remaining until due date, not the due date itself.
Claiming one rule is best on every measure.
Students memorise that SPT is best.
Fix: State the measure: SPT for average flow time, EDD for maximum tardiness. Say the result for the number of late jobs and average tardiness comes from your table.

Worked examples

Example 1

Five jobs are waiting at a machine at time 0 (days). In arrival order: A (processing 6, due 10), B (2, due 4), C (8, due 20), D (3, due 7), E (4, due 12). Sequence the jobs by FCFS, SPT and EDD. Compare average flow time, average tardiness, number of late jobs and maximum tardiness.

Show the solution

FCFS sequence: A, B, C, D, E. Completion times: 6, 8, 16, 19, 23. Sum = 72, average flow time = 72 ÷ 5 = 14.4 days.
FCFS tardiness: A 6−10 → 0; B 8−4 = 4; C 16−20 → 0; D 19−7 = 12; E 23−12 = 11. Total = 27, average = 5.4 days, late jobs = 3, maximum = 12.
SPT sequence: B(2), D(3), E(4), A(6), C(8). Completion times: 2, 5, 9, 15, 23. Sum = 54, average flow time = 10.8 days.
SPT tardiness: B 2−4 → 0; D 5−7 → 0; E 9−12 → 0; A 15−10 = 5; C 23−20 = 3. Total = 8, average = 1.6 days, late jobs = 2, maximum = 5.
EDD sequence by due date: B(4), D(7), A(10), E(12), C(20). Completion times: 2, 5, 11, 15, 23. Sum = 56, average flow time = 11.2 days.
EDD tardiness: B 0; D 0; A 11−10 = 1; E 15−12 = 3; C 23−20 = 3. Total = 7, average = 1.4 days, late jobs = 3, maximum = 3.
Check: every rule finishes at day 23, which equals 6+2+8+3+4.

Answer: FCFS: average flow time 14.4 days, average tardiness 5.4 days, 3 late jobs, maximum tardiness 12. SPT: 10.8 days, 1.6 days, 2 late jobs, maximum 5. EDD: 11.2 days, 1.4 days, 3 late jobs, maximum 3. SPT gives the lowest average flow time and fewest late jobs; EDD gives the lowest maximum and average tardiness here; FCFS is worst on every measure.

Example 2

Today is day 10. Five jobs wait at a work centre. P: due day 25, remaining processing 10 days. Q: due day 18, remaining 8 days. R: due day 14, remaining 7 days. S: due day 30, remaining 12 days. T: due day 9, remaining 3 days. Use the critical ratio rule to rank the jobs and comment on each.

Show the solution

Formula: CR = (Due date − Today) ÷ Remaining processing time.
P: (25 − 10) ÷ 10 = 15 ÷ 10 = 1.50.
Q: (18 − 10) ÷ 8 = 8 ÷ 8 = 1.00.
R: (14 − 10) ÷ 7 = 4 ÷ 7 = 0.57 (approx.).
S: (30 − 10) ÷ 12 = 20 ÷ 12 = 1.67 (approx.).
T: (9 − 10) ÷ 3 = −1 ÷ 3 = −0.33 (approx.).
Rank from lowest to highest: T (−0.33), R (0.57), Q (1.00), P (1.50), S (1.67).

Answer: Sequence: T, R, Q, P, S. T is already past its due date (negative CR) and goes first. R is behind schedule (CR below 1). Q is exactly on schedule (CR = 1). P and S are ahead of schedule (CR above 1), with S having the most slack. Recalculate the ratios whenever the next job is chosen, because today's date changes.

Exam tips

Draw the table first: sequence, processing time, completion time, due date, tardiness. Even a wrong final figure earns method marks.
In MCQs, check the measure asked. Average flow time points to SPT as the best; maximum tardiness points to EDD as the best.
Always show the check that the last completion time equals the sum of processing times.
For critical ratio, say what the number means: below 1 behind, 1 on time, above 1 ahead. Examiners reward the interpretation.
End a comparison question with a short recommendation based on the business need, such as customer deadlines or machine utilisation.

Practice questions from Scheduling and Queuing Models

Scheduling Rules and Priority Dispatching: frequently asked questions

Which scheduling rule is best, FCFS, SPT or EDD?

It depends on the measure. For a single machine with all jobs ready at time 0, SPT minimises average flow time and EDD minimises maximum tardiness. FCFS is simple and seen as fair but is rarely the best on any measure.

What is the difference between lateness and tardiness?

Lateness is completion time minus due date and can be negative if a job finishes early. Tardiness is the larger of zero and lateness, so early jobs count as zero tardiness.

How do I calculate the critical ratio?

Divide the time remaining until the due date by the processing time still needed. CR = (Due date − Today) ÷ Remaining processing time. The job with the lowest ratio is scheduled first.

Does the sequence change the time at which all jobs finish?

No. On a single machine with no idle time, the last job finishes at the total of all processing times whichever rule you use. The rules change individual completion times, so flow time and tardiness differ.