Topics

Formulating a linear programme

Covers AS91574

The three components

Every linear programme has exactly three parts, and writing them down separately is the first step of every answer.
Decision variables — the quantities you get to choose.
- Define them in words, with units: "Let $x$ be the number of tonnes of Blend A produced per week."
- Ambiguity here poisons everything downstream. "Let $x$ be Blend A" is not a definition — is it tonnes, batches, or dollars?
Constraints — the limits, written as linear inequalities.
- One per resource, plus non-negativity: $x \geq 0$ , $y \geq 0$ .
Objective function — the thing to maximise or minimise, written as a formula:

$P = 40x + 30y$

State whether you are maximising or minimising. A minimisation problem answered with a maximum is a complete answer to the wrong question.

Reading constraints out of a context

Each limited resource gives one constraint. Build it in three steps:
- How much of the resource does one unit of each product use?
- Multiply by the number of units.
- Compare with the total available, using the right inequality.
Worked through — "each tray of A needs 2 hours of oven time, each tray of B needs 1 hour, and there are 100 oven hours a week":

$2x + y \leq 100$

A table keeps this organised, and is worth drawing every time:

	Product A ( $x$ )	Product B ( $y$ )	Available
Oven hours	2	1	100
Labour hours	1	1	80
Storage (A only)	1	0	40
Profit each	$40	$30	maximise

The last row is the objective, not a constraint. Mixing them is a common error: profit is what you are optimising, not something you are limited to.

Getting the inequality direction right

Wording	Inequality
"at most", "no more than", "available", "capacity of"	$\leq$
"at least", "minimum of", "must supply", "contract requires"	$\geq$
"exactly", "must use all of"	$=$
"twice as many A as B"	$x = 2y$ — an equality constraint
"at least twice as many A as B"	$x \geq 2y$

Ratio constraints trip people up. "At least three times as many A as B" is $x \geq 3y$ . Test it with numbers: 9 of A and 2 of B satisfies $9 \geq 6$ ✓, and that is indeed more than three times as many.
Rearrange ratio constraints into standard form before graphing: $x \geq 3y$ becomes $x - 3y \geq 0$ , a line through the origin.

The assumptions built into the model

Linear programming assumes four things, and naming them is worth real marks at Excellence:
- Proportionality — doubling production exactly doubles resource use and profit. No bulk discounts, no economies of scale.
- Additivity — the products do not interfere with each other; total usage is just the sum.
- Divisibility — fractional amounts are allowed. False whenever the products are discrete items, which is why whole-number answers need separate treatment.
- Certainty — every coefficient is known exactly. Real prices and yields vary.
Say which assumption is shakiest in the given context. For a bakery, divisibility is fine for kilograms of dough and false for wedding cakes.

Writing the finished model

Set it out in a standard block, so a reader can check it at a glance:

$\textbf{Maximise } \quad P = 40x + 30y$

$\text{subject to} \quad \begin{cases} 2x + y \leq 100 \\ x + y \leq 80 \\ x \leq 40 \\ x \geq 0,\ y \geq 0 \end{cases}$

Check the model before graphing by testing one sensible plan. If making 10 of each is obviously possible in the context, it must satisfy every constraint — and if it does not, a constraint is wrong.

Tips and Tricks

Define the variables with units and a time period. "Number of tonnes per week" and "number of tonnes" lead to different constraints, and the difference will not show up until the answer is nonsense.

Draw the resource table before writing any inequality. It forces you to find the coefficient of each variable in each constraint, and it makes a missing constraint obvious.

Profit belongs in the objective, never in a constraint — unless the context genuinely imposes a profit target, in which case it is an extra $\geq$ constraint as well.

Test every ratio constraint with real numbers before you commit. Writing $3x \geq y$ instead of $x \geq 3y$ inverts the meaning and is very hard to spot on a graph.

Never omit the non-negativity constraints. They are part of the model, they bound the region, and assessors look for them.

State the assumptions when the context strains them. Whole aircraft, whole staff and whole cakes are not divisible, and a sentence saying so shows you understand what the model does.

Worked ExampleSetting up a full model

A small dairy company makes two cheeses. Each tray of Halloumi uses 2 hours of vat time, 1 hour of packing, and earns $40 profit. Each tray of Feta uses 1 hour of vat time, 1 hour of packing, and earns $30 profit. There are 100 vat hours and 80 packing hours available per week, and cold storage limits Halloumi to 40 trays a week. A supermarket contract requires at least 10 trays of Feta each week.

Formulate the linear programme completely, and check the model with a trial plan.

Step 1 — Define the decision variables

Both the unit (trays) and the period (per week) are stated, so every constraint that follows is measured consistently.

Step 2 — Build the resource table

	Halloumi ( $x$ )	Feta ( $y$ )	Available per week
Vat hours	2	1	100
Packing hours	1	1	80
Cold storage	1	0	40 (Halloumi only)
Contract	0	1	at least 10 (Feta)
Profit per tray	$40	$30	maximise

Step 3 — Write the constraints

Vat time — 2 hours per Halloumi tray, 1 per Feta, at most 100 available:

$2x + y \leq 100 \qquad (1)$

Packing time — 1 hour each, at most 80:

$x + y \leq 80 \qquad (2)$

Cold storage — Halloumi only, at most 40 trays:

$x \leq 40 \qquad (3)$

Supermarket contract — at least 10 trays of Feta:

$y \geq 10 \qquad (4)$

Non-negativity:

$x \geq 0, \qquad y \geq 0 \qquad (5)$

Step 4 — Write the objective function

Profit is $40 per Halloumi tray and $30 per Feta tray:

$\textbf{Maximise} \qquad P = 40x + 30y$

Step 5 — Set out the complete model

$\textbf{Maximise } \ P = 40x + 30y$

$\text{subject to} \quad \begin{cases} 2x + y \leq 100 & \text{(vat)} \\ x + y \leq 80 & \text{(packing)} \\ x \leq 40 & \text{(storage)} \\ y \geq 10 & \text{(contract)} \\ x \geq 0,\ y \geq 0 \end{cases}$

Step 6 — Check the model with a trial plan

Try 20 trays of each, which should obviously be possible:

Constraint	Check	Result
Vat	$2(20) + 20 = 60 \leq 100$	✓
Packing	$20 + 20 = 40 \leq 80$	✓
Storage	$20 \leq 40$	✓
Contract	$20 \geq 10$	✓

Feasible, with profit $P = 40(20) + 30(20) = \$1{,}400$ .

Now try a plan that should FAIL — 50 trays of Halloumi:

Storage: $50 \leq 40$ ✗ Infeasible, correctly caught by constraint (3).

Step 7 — State the assumptions

Proportionality: 10 trays of Halloumi use exactly 20 vat hours and earn exactly $400 — no setup time per batch and no bulk savings. In a real dairy each vat run has a cleaning cost, so this is an approximation.
Divisibility: the model will permit answers such as 20.5 trays. Trays are discrete, so the optimal corner may need adjusting to whole numbers.
Certainty: profits of $40 and $30 are treated as fixed, though milk prices and demand both vary week to week.

Formulating a linear programme

Covers AS91574

The three components

Every linear programme has exactly three parts, and writing them down separately is the first step of every answer.
Decision variables — the quantities you get to choose.
- Define them in words, with units: "Let $x$ be the number of tonnes of Blend A produced per week."
- Ambiguity here poisons everything downstream. "Let $x$ be Blend A" is not a definition — is it tonnes, batches, or dollars?
Constraints — the limits, written as linear inequalities.
- One per resource, plus non-negativity: $x \geq 0$ , $y \geq 0$ .
Objective function — the thing to maximise or minimise, written as a formula:

$P = 40x + 30y$

State whether you are maximising or minimising. A minimisation problem answered with a maximum is a complete answer to the wrong question.

Reading constraints out of a context

Each limited resource gives one constraint. Build it in three steps:
- How much of the resource does one unit of each product use?
- Multiply by the number of units.
- Compare with the total available, using the right inequality.
Worked through — "each tray of A needs 2 hours of oven time, each tray of B needs 1 hour, and there are 100 oven hours a week":

$2x + y \leq 100$

A table keeps this organised, and is worth drawing every time:

	Product A ( $x$ )	Product B ( $y$ )	Available
Oven hours	2	1	100
Labour hours	1	1	80
Storage (A only)	1	0	40
Profit each	$40	$30	maximise

The last row is the objective, not a constraint. Mixing them is a common error: profit is what you are optimising, not something you are limited to.

Getting the inequality direction right

Wording	Inequality
"at most", "no more than", "available", "capacity of"	$\leq$
"at least", "minimum of", "must supply", "contract requires"	$\geq$
"exactly", "must use all of"	$=$
"twice as many A as B"	$x = 2y$ — an equality constraint
"at least twice as many A as B"	$x \geq 2y$

Ratio constraints trip people up. "At least three times as many A as B" is $x \geq 3y$ . Test it with numbers: 9 of A and 2 of B satisfies $9 \geq 6$ ✓, and that is indeed more than three times as many.
Rearrange ratio constraints into standard form before graphing: $x \geq 3y$ becomes $x - 3y \geq 0$ , a line through the origin.

The assumptions built into the model

Linear programming assumes four things, and naming them is worth real marks at Excellence:
- Proportionality — doubling production exactly doubles resource use and profit. No bulk discounts, no economies of scale.
- Additivity — the products do not interfere with each other; total usage is just the sum.
- Divisibility — fractional amounts are allowed. False whenever the products are discrete items, which is why whole-number answers need separate treatment.
- Certainty — every coefficient is known exactly. Real prices and yields vary.
Say which assumption is shakiest in the given context. For a bakery, divisibility is fine for kilograms of dough and false for wedding cakes.

Writing the finished model

Set it out in a standard block, so a reader can check it at a glance:

$\textbf{Maximise } \quad P = 40x + 30y$

$\text{subject to} \quad \begin{cases} 2x + y \leq 100 \\ x + y \leq 80 \\ x \leq 40 \\ x \geq 0,\ y \geq 0 \end{cases}$

Check the model before graphing by testing one sensible plan. If making 10 of each is obviously possible in the context, it must satisfy every constraint — and if it does not, a constraint is wrong.

Tips and Tricks

Draw the resource table before writing any inequality. It forces you to find the coefficient of each variable in each constraint, and it makes a missing constraint obvious.

Profit belongs in the objective, never in a constraint — unless the context genuinely imposes a profit target, in which case it is an extra $\geq$ constraint as well.

Test every ratio constraint with real numbers before you commit. Writing $3x \geq y$ instead of $x \geq 3y$ inverts the meaning and is very hard to spot on a graph.

Never omit the non-negativity constraints. They are part of the model, they bound the region, and assessors look for them.

State the assumptions when the context strains them. Whole aircraft, whole staff and whole cakes are not divisible, and a sentence saying so shows you understand what the model does.

Worked ExampleSetting up a full model

Formulate the linear programme completely, and check the model with a trial plan.

Step 1 — Define the decision variables

Both the unit (trays) and the period (per week) are stated, so every constraint that follows is measured consistently.

Step 2 — Build the resource table

	Halloumi ( $x$ )	Feta ( $y$ )	Available per week
Vat hours	2	1	100
Packing hours	1	1	80
Cold storage	1	0	40 (Halloumi only)
Contract	0	1	at least 10 (Feta)
Profit per tray	$40	$30	maximise

Step 3 — Write the constraints

Vat time — 2 hours per Halloumi tray, 1 per Feta, at most 100 available:

$2x + y \leq 100 \qquad (1)$

Packing time — 1 hour each, at most 80:

$x + y \leq 80 \qquad (2)$

Cold storage — Halloumi only, at most 40 trays:

$x \leq 40 \qquad (3)$

Supermarket contract — at least 10 trays of Feta:

$y \geq 10 \qquad (4)$

Non-negativity:

$x \geq 0, \qquad y \geq 0 \qquad (5)$

Step 4 — Write the objective function

Profit is $40 per Halloumi tray and $30 per Feta tray:

$\textbf{Maximise} \qquad P = 40x + 30y$

Step 5 — Set out the complete model

$\textbf{Maximise } \ P = 40x + 30y$

Step 6 — Check the model with a trial plan

Try 20 trays of each, which should obviously be possible:

Constraint	Check	Result
Vat	$2(20) + 20 = 60 \leq 100$	✓
Packing	$20 + 20 = 40 \leq 80$	✓
Storage	$20 \leq 40$	✓
Contract	$20 \geq 10$	✓

Feasible, with profit $P = 40(20) + 30(20) = \$1{,}400$ .

Now try a plan that should FAIL — 50 trays of Halloumi:

Storage: $50 \leq 40$ ✗ Infeasible, correctly caught by constraint (3).

Step 7 — State the assumptions

Proportionality: 10 trays of Halloumi use exactly 20 vat hours and earn exactly $400 — no setup time per batch and no bulk savings. In a real dairy each vat run has a cleaning cost, so this is an approximation.
Divisibility: the model will permit answers such as 20.5 trays. Trays are discrete, so the optimal corner may need adjusting to whole numbers.
Certainty: profits of $40 and $30 are treated as fixed, though milk prices and demand both vary week to week.