The idea, step by step
More people than jobs
When there are more workers than jobs, add enough dummy job columns to make a square table. A worker assigned to a dummy job is idle. If being idle has no cost, enter zero in that dummy column; if there is a genuine idle or stand-by cost, enter that stated cost instead.
In the four-worker/three-job table, adding one zero-cost dummy job permits one idle worker. A→1, B→2 and C→3 cost 4+3+2=$9, with D idle. The dummy does not invent a fourth real job, and the answer must state who is not used.
| Worker | Job 1 | Job 2 | Job 3 |
|---|---|---|---|
| A | 4 | 8 | 7 |
| B | 6 | 3 | 8 |
| C | 5 | 7 | 2 |
| D | 9 | 4 | 6 |
More jobs than people
A dummy worker represents an unfilled task, not an imaginary person who completes it for free. If every real job is compulsory and each worker can perform only one, having too few workers makes the original model infeasible. You need another worker, permission for multiple jobs, or a revised schedule.
If leaving a job unfilled is allowed, use its stated penalty in the dummy row. In the two-worker table, assigning A→1, B→3 and the dummy→2 costs 6+3+2=$11. Job 2 remains uncompleted, with its $2 penalty; the result must not be reported as all jobs being completed.
| Resource | Job 1 | Job 2 | Job 3 |
|---|---|---|---|
| A | 6 | 4 | 8 |
| B | 5 | 7 | 3 |
| Unfilled-job penalty | 10 | 2 | 7 |
A common trap
Adding a zero-cost dummy worker and claiming all compulsory jobs have been done.
Instead: A dummy worker marks an unfilled job. Free omission must be an explicit allowed assumption, not a way to create extra capacity.
See the thinking
One worker stays idle
Solve the four-worker/three-job problem with free idling.
| Worker | Job 1 | Job 2 | Job 3 |
|---|---|---|---|
| A | 4 | 8 | 7 |
| B | 6 | 3 | 8 |
| C | 5 | 7 | 2 |
| D | 9 | 4 | 6 |
- Add a fourth column of zeros for an idle worker.
- A→1, B→2, C→3 and D→dummy is feasible.
- The cost is 4+3+2+0=9.
- The three real-job column minima 4,3,2 are attained, so no lower total is possible.
Penalties change which job is omitted
Use the two-worker table including unfilled-job penalties. Compare A→2, B→3, omit 1 with A→1, B→3, omit 2.
| Resource | Job 1 | Job 2 | Job 3 |
|---|---|---|---|
| A | 6 | 4 | 8 |
| B | 5 | 7 | 3 |
| Unfilled-job penalty | 10 | 2 | 7 |
- First option costs 4+3+10=17.
- Second option costs 6+3+2=11.
- The apparently cheaper real work in the first option is outweighed by its omission penalty.
Check your understanding
Five workers must cover three jobs, with each worker doing at most one. How many dummy job columns balance the assignment?
Show answer
Two.
The 5×3 table needs two more columns to become 5×5, representing two idle workers.
Your turn
6 questions · 15 marks · Work at your own pace.
Read the questions online
- Add the correct type and number of dummy entries to a 4-worker/3-job table with free idling.2 marks
- Solve the displayed four-worker problem and state who is idle.3 marks
Costs ($) for four workers and three required jobs. A worker may remain idle at no cost. Worker Job 1 Job 2 Job 3 A 4 8 7 B 6 3 8 C 5 7 2 D 9 4 6 - Two workers must each do exactly one job, but there are three compulsory jobs. Can a dummy row alone make all three jobs physically complete?2 marks
- Using the penalty table, find the total for A→3, B→1 and job 2 left unfilled.2 marks
Costs ($) for two workers and three jobs, with stated unfilled-job penalties. Resource Job 1 Job 2 Job 3 A 6 4 8 B 5 7 3 Unfilled-job penalty 10 2 7 - Using the penalty table, compare all six assignments and give the minimum.4 marks
Costs ($) for two workers and three jobs, with stated unfilled-job penalties. Resource Job 1 Job 2 Job 3 A 6 4 8 B 5 7 3 Unfilled-job penalty 10 2 7 - Four workers cover three jobs. Idle-worker costs are A:$2, B:$1, C:$4 and D:$3. What should the dummy job column contain?2 marks
Read the worked solutions
Equivalent correct methods earn the corresponding marks. Include units where asked.
- Add the correct type and number of dummy entries to a 4-worker/3-job table with free idling.2 marks
There is one more worker than required jobs.
The dummy job means one person is idle.
One dummy job column, all zeros. - Solve the displayed four-worker problem and state who is idle.3 marks
Costs ($) for four workers and three required jobs. A worker may remain idle at no cost. Worker Job 1 Job 2 Job 3 A 4 8 7 B 6 3 8 C 5 7 2 D 9 4 6 The real costs are 4,3 and 2.
The dummy assignment costs 0.
Cost $9; A→1, B→2, C→3, D idle. - Two workers must each do exactly one job, but there are three compulsory jobs. Can a dummy row alone make all three jobs physically complete?2 marks
A dummy is a modelling device, not extra labour.
The original capacity restriction makes completion of all jobs infeasible.
No. - Using the penalty table, find the total for A→3, B→1 and job 2 left unfilled.2 marks
Costs ($) for two workers and three jobs, with stated unfilled-job penalties. Resource Job 1 Job 2 Job 3 A 6 4 8 B 5 7 3 Unfilled-job penalty 10 2 7 A/job 3 costs 8; B/job 1 costs 5; job 2 penalty is 2.
8+5+2=15.
$15. - Using the penalty table, compare all six assignments and give the minimum.4 marks
Costs ($) for two workers and three jobs, with stated unfilled-job penalties. Resource Job 1 Job 2 Job 3 A 6 4 8 B 5 7 3 Unfilled-job penalty 10 2 7 Totals for A/B/dummy job sequences (1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1) are 20,11,16,17,15,25.
Choose the smallest feasible total and identify the unfilled job.
Minimum $11: A→1, B→3, leave job 2 unfilled. - Four workers cover three jobs. Idle-worker costs are A:$2, B:$1, C:$4 and D:$3. What should the dummy job column contain?2 marks
The dummy column represents the actual cost of leaving that particular worker idle.
It is not automatically zero.
2,1,4,3 in rows A, B, C, D.
Curriculum reference & lesson scope
SACE Board of South Australia
Stage 2 General Mathematics: discrete models, project scheduling and assignment. Activity-on-arrow conventions are used for project networks; technology may verify calculations but does not replace interpreting the model.
Official curriculum source ↗