Topics

Whole-number answers and special cases

Covers AS91574

Why the corner may not be the answer

The graphical method treats the region as continuous. It will happily report an optimum of 22.5 trucks or 7.3 workers.
When the decision variables count discrete items, that answer is not a plan. The real optimum is a whole-number point in the region, and it is usually not at the corner.
This is the divisibility assumption failing, and handling it properly is a standard Merit/Excellence discriminator.

Finding the whole-number optimum

Never simply round. Rounding can land outside the feasible region, giving a plan that breaks a constraint.
The reliable procedure:
- Solve the continuous problem first and find the fractional corner.
- List the whole-number points near that corner — round each coordinate up and down, giving up to four candidates, and add the neighbours along each binding edge.
- Test every candidate in EVERY constraint. Discard any that fail.
- Evaluate the objective at the survivors and take the best.
Show the rejected candidates. " $(23, 5)$ rejected — exceeds the labour constraint by 2 hours" is evidence of method.
State the loss. "The whole-number optimum earns $2,580 against the continuous $2,600, so rounding costs $20 a week."

Why rounding down is not automatically safe

Rounding both coordinates down is always feasible when all constraints are of the $\leq$ type with positive coefficients — but it is often not optimal.
With a $\geq$ constraint present, rounding down can break it. A contract requiring at least 20 units is violated by rounding 20.4 down to 20… but rounding 20.4 down to 20 is fine, whereas rounding 19.6 down to 19 is not.
Test, do not assume. The three lines of checking are quicker than the reasoning about when rounding is safe.

No feasible solution

The region is empty: the constraints contradict each other.
Report it as the answer, and name the conflict: "The 500-tonne requirement cannot be met with 45 drivers, which allow at most 450 tonnes."
Quantify the shortfall and say what would have to change. That is what the client needs.

Unbounded solutions

The objective can be increased without limit within the region.
In a real context this is always a modelling error, not a finding. No business has infinite profit.
The response is to find the missing constraint — a demand ceiling, a budget, a capacity — and add it.

Redundant constraints

A redundant constraint never binds, because another is always tighter. Its boundary line does not touch the region.
It is not an error, and it should stay in the model — a redundant constraint today may bind tomorrow when another limit is relaxed.
Identifying redundancy is useful advice: "the storage limit never binds, so expanding storage cannot help."

Sensitivity: what the answer is robust to

Changing a constraint's limit moves the boundary line, sliding the optimal corner along the other binding edge. Recalculate the corner and re-evaluate.
The marginal value of a resource is the increase in the objective from one extra unit of it — worth computing for each binding constraint, since it says what the business should pay.
That value holds only while the same constraints stay binding. Beyond some point another constraint takes over and the marginal value drops, often to zero.
Changing a price rotates the objective line. The corner stays optimal while the objective's gradient lies between the gradients of the two edges meeting there — one inequality gives the whole range.

What to write in the conclusion

The plan — how many of each, in the context's units.
The objective value — with a dollar sign or the right unit.
The binding constraints — what is fully used.
The slack — what is left over, and therefore not worth buying.
One caveat — the assumption most likely to fail, or the range over which the answer is stable.

Tips and Tricks

Never round without re-checking every constraint. A rounded point that breaks a limit is worse than a fractional one, because it looks like a valid answer.

Test all four rounded neighbours, not just the obvious one. The best whole-number plan is often the one that rounds one variable up and the other down.

Say what rounding costs. The difference between the continuous and whole-number objective values is a real number the business can use.

Infeasible and unbounded are answers, not failures. Report which constraints conflict, or which constraint is missing, and quantify it.

Marginal values expire. State the range over which "one more hour is worth $20" holds, or the advice becomes wrong as soon as it is acted on at scale.

Finish with the binding and slack constraints. It converts a number into a decision, which is what the internal is really assessing.

Worked ExampleFrom the fractional corner to a real plan

A workshop makes benches and tables. Each bench uses 3 hours of assembly and 2 kg of finish; each table uses 5 hours of assembly and 2 kg of finish. There are 47 assembly hours and 24 kg of finish available. Benches earn $90 profit, tables $140. Find the best whole-number production plan.

Step 1 — Formulate the model

$\text{Maximise } P = 90b + 140t \ \text{ subject to } \begin{cases} 3b + 5t \leq 47 & \text{(assembly)} \\ 2b + 2t \leq 24 & \text{(finish)} \\ b \geq 0,\ t \geq 0 \end{cases}$

Simplify the finish constraint by dividing by 2:

$b + t \leq 12$

Step 2 — Find the corners of the continuous problem

Corner $(0,0)$ — the origin.

Corner on the $b$ -axis: $t = 0$ gives $3b \leq 47$ (so $b \leq 15.67$ ) and $b \leq 12$ . The tighter is 12, so the corner is $(12, 0)$ .

Corner on the $t$ -axis: $b = 0$ gives $5t \leq 47$ (so $t \leq 9.4$ ) and $t \leq 12$ . The tighter is 9.4, so the corner is $(0, 9.4)$ .

Corner where the two constraints meet. Solve:

$3b + 5t = 47 \qquad (1)$ $b + t = 12 \qquad (2)$

From (2), $b = 12 - t$ . Substitute into (1):

$3(12 - t) + 5t = 47$

$36 + 2t = 47$

$t = 5.5, \qquad b = 6.5$

$\text{Corner } = (6.5,\ 5.5)$

Step 3 — Solve the continuous problem

Corner	$P = 90b + 140t$	Value
$(0, 0)$	0	$0
$(12, 0)$	$1080 + 0$	$1,080
$(6.5, 5.5)$	$585 + 770$	$1,355
$(0, 9.4)$	$0 + 1316$	$1,316

Step 4 — List the whole-number candidates near the corner

Rounding each coordinate up and down gives four candidates, and it is worth adding two neighbours along the binding edges:

Candidate	Assembly $3b + 5t \leq 47$	Finish $b + t \leq 12$	Feasible?
$(6, 5)$	$18 + 25 = 43$ ✓	$11$ ✓	Yes
$(7, 5)$	$21 + 25 = 46$ ✓	$12$ ✓	Yes
$(6, 6)$	$18 + 30 = 48$ ✗	$12$ ✓	No — assembly exceeded by 1 hour
$(7, 6)$	$21 + 30 = 51$ ✗	$13$ ✗	No — both exceeded
$(4, 7)$	$12 + 35 = 47$ ✓	$11$ ✓	Yes
$(9, 4)$	$27 + 20 = 47$ ✓	$13$ ✗	No — finish exceeded

Step 5 — Evaluate the feasible candidates

$(6, 5)$ :

$P = 90(6) + 140(5) = 540 + 700 = \$1{,}240$

$(7, 5)$ :

$P = 90(7) + 140(5) = 630 + 700 = \$1{,}330$

$(4, 7)$ :

$P = 90(4) + 140(7) = 360 + 980 = \$1{,}340$

Step 6 — Select the whole-number optimum

Note that this is NOT the nearest lattice point to the corner. $(7,5)$ is closer to $(6.5, 5.5)$ but earns $10 less. The best integer answer is often not the nearest one, which is exactly why a list of candidates must be tested rather than a single rounding performed.

Step 7 — State the cost of integrality

$\$1{,}355 - \$1{,}340 = \$15$

Step 8 — Identify what binds at the chosen plan

At $(4, 7)$ :

Constraint	Value	Status
Assembly	$12 + 35 = 47 \leq 47$	BINDING — every hour used
Finish	$4 + 7 = 11 \leq 12$	slack — 1 kg spare

Step 9 — Advise the workshop

What extra assembly time is worth. With 50 hours instead of 47, the plan $(5, 7)$ becomes feasible: assembly $15 + 35 = 50$ ✓, finish $12 \leq 12$ ✓, earning $450 + 980 = \$1{,}430$ .

A caveat. At 50 hours the finish constraint becomes binding as well, so beyond that point extra assembly hours are worth nothing until more finish is bought too. The $30 an hour applies for the next three hours only, and the advice must say so.

Whole-number answers and special cases

Covers AS91574

Why the corner may not be the answer

The graphical method treats the region as continuous. It will happily report an optimum of 22.5 trucks or 7.3 workers.
When the decision variables count discrete items, that answer is not a plan. The real optimum is a whole-number point in the region, and it is usually not at the corner.
This is the divisibility assumption failing, and handling it properly is a standard Merit/Excellence discriminator.

Finding the whole-number optimum

Never simply round. Rounding can land outside the feasible region, giving a plan that breaks a constraint.
The reliable procedure:
- Solve the continuous problem first and find the fractional corner.
- List the whole-number points near that corner — round each coordinate up and down, giving up to four candidates, and add the neighbours along each binding edge.
- Test every candidate in EVERY constraint. Discard any that fail.
- Evaluate the objective at the survivors and take the best.
Show the rejected candidates. " $(23, 5)$ rejected — exceeds the labour constraint by 2 hours" is evidence of method.
State the loss. "The whole-number optimum earns $2,580 against the continuous $2,600, so rounding costs $20 a week."

Why rounding down is not automatically safe

Rounding both coordinates down is always feasible when all constraints are of the $\leq$ type with positive coefficients — but it is often not optimal.
With a $\geq$ constraint present, rounding down can break it. A contract requiring at least 20 units is violated by rounding 20.4 down to 20… but rounding 20.4 down to 20 is fine, whereas rounding 19.6 down to 19 is not.
Test, do not assume. The three lines of checking are quicker than the reasoning about when rounding is safe.

No feasible solution

The region is empty: the constraints contradict each other.
Report it as the answer, and name the conflict: "The 500-tonne requirement cannot be met with 45 drivers, which allow at most 450 tonnes."
Quantify the shortfall and say what would have to change. That is what the client needs.

Unbounded solutions

The objective can be increased without limit within the region.
In a real context this is always a modelling error, not a finding. No business has infinite profit.
The response is to find the missing constraint — a demand ceiling, a budget, a capacity — and add it.

Redundant constraints

A redundant constraint never binds, because another is always tighter. Its boundary line does not touch the region.
It is not an error, and it should stay in the model — a redundant constraint today may bind tomorrow when another limit is relaxed.
Identifying redundancy is useful advice: "the storage limit never binds, so expanding storage cannot help."

Sensitivity: what the answer is robust to

Changing a constraint's limit moves the boundary line, sliding the optimal corner along the other binding edge. Recalculate the corner and re-evaluate.
The marginal value of a resource is the increase in the objective from one extra unit of it — worth computing for each binding constraint, since it says what the business should pay.
That value holds only while the same constraints stay binding. Beyond some point another constraint takes over and the marginal value drops, often to zero.
Changing a price rotates the objective line. The corner stays optimal while the objective's gradient lies between the gradients of the two edges meeting there — one inequality gives the whole range.

What to write in the conclusion

The plan — how many of each, in the context's units.
The objective value — with a dollar sign or the right unit.
The binding constraints — what is fully used.
The slack — what is left over, and therefore not worth buying.
One caveat — the assumption most likely to fail, or the range over which the answer is stable.

Tips and Tricks

Never round without re-checking every constraint. A rounded point that breaks a limit is worse than a fractional one, because it looks like a valid answer.

Test all four rounded neighbours, not just the obvious one. The best whole-number plan is often the one that rounds one variable up and the other down.

Say what rounding costs. The difference between the continuous and whole-number objective values is a real number the business can use.

Infeasible and unbounded are answers, not failures. Report which constraints conflict, or which constraint is missing, and quantify it.

Marginal values expire. State the range over which "one more hour is worth $20" holds, or the advice becomes wrong as soon as it is acted on at scale.

Finish with the binding and slack constraints. It converts a number into a decision, which is what the internal is really assessing.

Worked ExampleFrom the fractional corner to a real plan

Step 1 — Formulate the model

$\text{Maximise } P = 90b + 140t \ \text{ subject to } \begin{cases} 3b + 5t \leq 47 & \text{(assembly)} \\ 2b + 2t \leq 24 & \text{(finish)} \\ b \geq 0,\ t \geq 0 \end{cases}$

Simplify the finish constraint by dividing by 2:

$b + t \leq 12$

Step 2 — Find the corners of the continuous problem

Corner $(0,0)$ — the origin.

Corner on the $b$ -axis: $t = 0$ gives $3b \leq 47$ (so $b \leq 15.67$ ) and $b \leq 12$ . The tighter is 12, so the corner is $(12, 0)$ .

Corner on the $t$ -axis: $b = 0$ gives $5t \leq 47$ (so $t \leq 9.4$ ) and $t \leq 12$ . The tighter is 9.4, so the corner is $(0, 9.4)$ .

Corner where the two constraints meet. Solve:

$3b + 5t = 47 \qquad (1)$ $b + t = 12 \qquad (2)$

From (2), $b = 12 - t$ . Substitute into (1):

$3(12 - t) + 5t = 47$

$36 + 2t = 47$

$t = 5.5, \qquad b = 6.5$

$\text{Corner } = (6.5,\ 5.5)$

Step 3 — Solve the continuous problem

Corner	$P = 90b + 140t$	Value
$(0, 0)$	0	$0
$(12, 0)$	$1080 + 0$	$1,080
$(6.5, 5.5)$	$585 + 770$	$1,355
$(0, 9.4)$	$0 + 1316$	$1,316

Step 4 — List the whole-number candidates near the corner

Rounding each coordinate up and down gives four candidates, and it is worth adding two neighbours along the binding edges:

Candidate	Assembly $3b + 5t \leq 47$	Finish $b + t \leq 12$	Feasible?
$(6, 5)$	$18 + 25 = 43$ ✓	$11$ ✓	Yes
$(7, 5)$	$21 + 25 = 46$ ✓	$12$ ✓	Yes
$(6, 6)$	$18 + 30 = 48$ ✗	$12$ ✓	No — assembly exceeded by 1 hour
$(7, 6)$	$21 + 30 = 51$ ✗	$13$ ✗	No — both exceeded
$(4, 7)$	$12 + 35 = 47$ ✓	$11$ ✓	Yes
$(9, 4)$	$27 + 20 = 47$ ✓	$13$ ✗	No — finish exceeded

Step 5 — Evaluate the feasible candidates

$(6, 5)$ :

$P = 90(6) + 140(5) = 540 + 700 = \$1{,}240$

$(7, 5)$ :

$P = 90(7) + 140(5) = 630 + 700 = \$1{,}330$

$(4, 7)$ :

$P = 90(4) + 140(7) = 360 + 980 = \$1{,}340$

Step 6 — Select the whole-number optimum

Step 7 — State the cost of integrality

$\$1{,}355 - \$1{,}340 = \$15$

Step 8 — Identify what binds at the chosen plan

At $(4, 7)$ :

Constraint	Value	Status
Assembly	$12 + 35 = 47 \leq 47$	BINDING — every hour used
Finish	$4 + 7 = 11 \leq 12$	slack — 1 kg spare

Step 9 — Advise the workshop

What extra assembly time is worth. With 50 hours instead of 47, the plan $(5, 7)$ becomes feasible: assembly $15 + 35 = 50$ ✓, finish $12 \leq 12$ ✓, earning $450 + 980 = \$1{,}430$ .