| Exam Board | Edexcel |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Simplex algorithm execution |
| Difficulty | Standard +0.8 This is a multi-part linear programming question requiring formulation of constraints from a word problem, then application of the Simplex algorithm. While D1 is standard A-level, LP formulation with three variables and interpreting real-world constraints (person-hours, regulatory requirements) requires careful reasoning beyond routine exercises. The Simplex algorithm execution adds computational complexity, making this moderately challenging. |
| Spec | 7.06a LP formulation: variables, constraints, objective function7.06b Slack variables: converting inequalities to equations7.07a Simplex tableau: initial setup in standard format |
| Answer | Marks |
|---|---|
| - \(x \geq 0, y \geq 0, z \geq 0\) | M2 A2 |
| (b) To change inequalities into equations | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Basic Var. | \(x\) | \(y\) |
| \(r\) | \(\frac{3}{2}\) | 0 |
| \(s\) | \(\frac{1}{2}\) | 0 |
| \(y\) | \(\frac{1}{4}\) | 1 |
| \(R\) | 1 | 0 |
| M3 A3 | ||
| (d) Final tableau as all values on the objective row are \(\geq 0\) | B1 | |
| (e) Centre provides 10 courses for adults (not pensioners) and gets £100 revenue per day | B2 | |
| (f) no. e.g. the slack variable \(s\) associated with this constraint is not zero so optimal solution without this constraint would be the same | B2 | (16) |
**(a)** Maximise $R = 4x + 10y + 2z$ given:
- $x - y \leq 5$
- $y + 2z \leq 0$
- $2x + 4y + z \leq 40$
- $x \geq 0, y \geq 0, z \geq 0$ | M2 A2 |
**(b)** To change inequalities into equations | B1 |
**(c)** Only one positive value so pivot row is 3rd row. Second tableau is:
| Basic Var. | $x$ | $y$ | $z$ | $r$ | $s$ | $t$ | Value |
|---|---|---|---|---|---|---|---|
| $r$ | $\frac{3}{2}$ | 0 | $\frac{1}{4}$ | 1 | 0 | $\frac{1}{4}$ | 15 |
| $s$ | $\frac{1}{2}$ | 0 | $\frac{9}{4}$ | 0 | 1 | $\frac{1}{4}$ | 10 |
| $y$ | $\frac{1}{4}$ | 1 | $\frac{1}{4}$ | 0 | 0 | $\frac{1}{4}$ | 10 |
| $R$ | 1 | 0 | $\frac{1}{2}$ | 0 | 0 | $\frac{3}{2}$ | 100 |
| M3 A3 |
**(d)** Final tableau as all values on the objective row are $\geq 0$ | B1 |
**(e)** Centre provides 10 courses for adults (not pensioners) and gets £100 revenue per day | B2 |
**(f)** no. e.g. the slack variable $s$ associated with this constraint is not zero so optimal solution without this constraint would be the same | B2 | (16) |
**Total: (75)**7. A fitness centre runs introductory courses aimed at the following groups of customers:
Pensioners, who will be charged $\pounds 4$ for a 2 -hour session.\\
Other adults, who will be charged $\pounds 10$ for a 4 -hour session.\\
Children, who will be charged $\pounds 2$ for a 1 -hour session.\\
Let the number of pensioners, other adults, and children be $x , y$ and $z$ respectively.\\
Regulations state that the number of pensioners, $x$, must be at most 5 more than the number of adults, $y$. There must also be at least twice as many adults, $y$, as there are children, $z$.
The centre is able to supervise up to 40 person-hours each day at the centre and wishes to maximise the revenue $( \pounds R )$ that can be earned each day from these sessions. You may assume that the places on any courses that the centre runs will be filled.\\
(a) Modelling this situation as a linear programming problem, write down the constraints and objective function in terms of $x , y$ and $z$.
Using the Simplex algorithm, the following initial tableau is obtained.
(d) $\_\_\_\_$\\
\hfill \mbox{\textit{Edexcel D1 Q7 [16]}}