The idea, step by step
What the rows and columns mean
An assignment table gives the cost, time or profit for pairing each row item with each column item. In a standard square problem, one person is allocated to each job and each person does exactly one job. A feasible assignment therefore selects one entry in every row and every column.
First identify the quantity being optimised and its units. A cost table is normally minimised; a profit table is normally maximised. Do not assume that numbers represent costs just because they appear in a matrix. Unavailable pairings must be stated rather than silently treated as zero cost.
| Technician | Job 1 | Job 2 | Job 3 |
|---|---|---|---|
| Ali | 9 | 2 | 7 |
| Bea | 6 | 4 | 3 |
| Chen | 5 | 8 | 1 |
Whole assignments, not isolated bargains
The smallest cost in each row may lie in the same column, but the same job cannot be allocated twice. For three people there are 3!=6 complete assignments, small enough to enumerate. For larger tables, an algorithm avoids a rapidly growing search.
Using the technician table, Ali→2, Bea→1 and Chen→3 costs 2+6+1=$9. Choosing Bea→3 because it is her cheapest job would conflict with Chen→3. The objective is the lowest total for a feasible complete allocation, not necessarily each person's individually cheapest job.
A common trap
Choosing the lowest number in each row and adding them without checking columns.
Instead: Each required column must be chosen exactly once. Compare complete feasible assignments.
See the thinking
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 |
- Read Ali/job 3 as $7, Bea/job 2 as $4 and Chen/job 1 as $5.
- The jobs are distinct, so the assignment is feasible.
- Add the original costs: 7+4+5=16.
Reject an infeasible selection
A student chooses Ali→2, Bea→3 and Chen→3. Why is the proposed $6 total not a valid solution?
- The selected costs are 2,3 and 1.
- Job 3 is assigned to both Bea and Chen.
- Job 1 is not assigned to anyone.
Check your understanding
For a four-person/four-job table, how many entries are selected in a complete feasible assignment?
Show answer
Four: one per row and one per column.
An assignment contains one pairing for each person, and the job constraint prevents repeated columns.
Your turn
6 questions · 16 marks · Work at your own pace.
Read the questions online
- Use the technician table to find the cost of Ali→1, Bea→2 and Chen→3.2 marks
Cost ($) 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.
9+4+1=14.
$14. - List all six possible job sequences for Ali, Bea and Chen.3 marks
Choose one of three jobs for Ali, then one of the remaining two for Bea.
Chen receives the remaining job.
(1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1). - Evaluate all six technician assignments and identify the minimum.4 marks
Cost ($) 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 4×3=$12.
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 ↗