Understand the idea.
What the rows and columns mean
An assignment table records the value of pairing a row item with a column item. In a complete square assignment, each person does one job and each job receives one person, so selected entries must occupy distinct rows and columns.
Read the caption before calculating: cost is usually minimised, while profit is usually maximised. Check units and unavailable pairings. A forbidden pairing is excluded; giving it a zero would instead make it look particularly attractive in a cost problem.
| Technician | Job 1 | Job 2 | Job 3 |
|---|---|---|---|
| Ali | 9 | 2 | 7 |
| Bea | 6 | 4 | 3 |
| Chen | 5 | 8 | 1 |
Whole assignments, not isolated bargains
Individual bargains can conflict. Choosing the cheapest entry in each row is valid only if the selected columns are distinct. For three people there are complete allocations, so every feasible total can be compared directly.
In the technician table, Ali→2, Bea→1 and Chen→3 costs dollars. Bea's own cheapest job is 3, but assigning her there as well as Chen would leave job 1 uncovered.
See the method in action.
Evaluate a complete allocation
Using the technician table, calculate the cost of Ali→3, Bea→2 and Chen→1.
| Technician | Job 1 | Job 2 | Job 3 |
|---|---|---|---|
| Ali | 9 | 2 | 7 |
| Bea | 6 | 4 | 3 |
| Chen | 5 | 8 | 1 |
Worked solution
- 1
Check that every job is used once.
Ali, Bea and Chen receive different jobs, so this is a feasible complete assignment.
- 2
Read the three original entries.
Each cost is found at the intersection of the named technician's row and assigned job's column.
- 3
Add the selected costs.
The assignment objective is the sum of costs for these three pairings.
$16.
Reject an infeasible selection
A student chooses Ali→2, Bea→3 and Chen→3. Why is the proposed $6 total not a valid solution?
Worked solution
- 1
Check the proposed arithmetic.
The entries have been added correctly, but a correct sum alone does not establish feasibility.
- 2
Count how often each job is selected.
Bea and Chen both select job 3, while nobody selects job 1.
- 3
Reject the allocation as infeasible.
The one-person-per-job rule is broken, so six dollars cannot be reported as the cost of an admissible complete assignment.
It violates the one-person-per-job constraint, so the $6 sum is not an admissible assignment cost.
Try it yourself.
Check your understanding
For a four-person/four-job table, how many entries are selected in a complete feasible assignment?
Show answer & reasoning
Four: one per row and one per column.
Four entries are selected. Each of the four people supplies one pairing, and distinct job columns ensure that the four required jobs are all covered exactly once.
Keep practising
6 worksheet questions · 16 marks
Read the questions online
- Use the technician table to find the cost of Ali→1, Bea→2 and Chen→3.2 marks
Cost ($) of assigning one technician to each job. Technician Job 1 Job 2 Job 3 Ali 9 2 7 Bea 6 4 3 Chen 5 8 1 - List all six possible job sequences for Ali, Bea and Chen.3 marks
- Evaluate all six technician assignments and identify the minimum.4 marks
Cost ($) of assigning one technician to each job. Technician Job 1 Job 2 Job 3 Ali 9 2 7 Bea 6 4 3 Chen 5 8 1 - A table records the time each driver needs for a separate delivery. Explain one assumption needed before minimising the sum of the times.2 marks
- A proposed allocation uses one entry in each column but two entries in one row. Is it feasible for a standard one-to-one square assignment?2 marks
- If every selected cost in a four-person assignment is increased by $3, how much does the total increase, and does this change which complete assignment is cheapest?3 marks
Read the worked solutions
Equivalent correct methods earn the corresponding marks. Include units where asked.
- Use the technician table to find the cost of Ali→1, Bea→2 and Chen→3.2 marks
Cost ($) of assigning one technician to each job. Technician Job 1 Job 2 Job 3 Ali 9 2 7 Bea 6 4 3 Chen 5 8 1 Read 9,4 and 1.
.
$14. - List all six possible job sequences for Ali, Bea and Chen.3 marks
Choose one of three jobs for Ali, then one of the remaining two for Bea.
Chen receives the remaining job.
, , , , , . - Evaluate all six technician assignments and identify the minimum.4 marks
Cost ($) of assigning one technician to each job. Technician Job 1 Job 2 Job 3 Ali 9 2 7 Bea 6 4 3 Chen 5 8 1 Use the six sequences from the preceding question.
Add the three original entries for each sequence.
The smallest total is 9.
Totals $14, $20, $9, $10, $21, $16 respectively; minimum $9 with Ali→2, Bea→1, Chen→3. - A table records the time each driver needs for a separate delivery. Explain one assumption needed before minimising the sum of the times.2 marks
The sum and the maximum are different objective functions.
The correct model must match the operational goal.
For example, the objective really is total driver-hours, not the time when the last simultaneous delivery finishes. - A proposed allocation uses one entry in each column but two entries in one row. Is it feasible for a standard one-to-one square assignment?2 marks
One person would be assigned two jobs.
Another row/person would necessarily be unassigned.
No. - If every selected cost in a four-person assignment is increased by $3, how much does the total increase, and does this change which complete assignment is cheapest?3 marks
Each complete assignment contains four entries.
Every possible total gains the same .
The total increases by $12; the ranking of complete assignments is unchanged.
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 ↗