| Exam Board | OCR MEI |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Formulation from word problem |
| Difficulty | Moderate -0.3 This is a standard D1 linear programming formulation question with straightforward constraints. Students must interpret cutting plans, write waste expressions, and formulate/solve a 2-variable LP graphically—all routine techniques for this module. The multi-part structure guides students through each step, making it slightly easier than average despite requiring multiple skills. |
| Spec | 7.06a LP formulation: variables, constraints, objective function7.06b Slack variables: converting inequalities to equations7.06d Graphical solution: feasible region, two variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Diagram showing \(z\) m width with 3 strips of 32cm and waste of 4cm | M1 | 3 widths + waste |
| A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.06x + 0.21y + 0.04z\) (m²) | M1 | 3 areas |
| \(A1\sqrt{}\) | any units OK; ignore scaling |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2x + y > 20\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y + 3z > 24\) | \(B1\sqrt{}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Use of \(z = 20 - x - y\) | M1 | |
| Minimise \(0.02x + 0.17y\) (constant of 0.8 not needed but OK if there) | \(A1\sqrt{}\) | "minimise" not needed – given |
| s.t. \(2x + y > 20\) | \(A1\sqrt{}\) | |
| \(-3x - 2y > -36\) or \(3x + 2y < 36\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Line \(2x + y = 20\) drawn correctly | B1 | line (cao) |
| Line \(3x + 2y = 36\) drawn correctly | B1 | line (cao) |
| Shading – follow two negative gradient lines making a triangle with base on x-axis | B1 | shading |
| Objective valued at \((10, 0)\) and at \((4, 12)\) giving values 212 or 292 or 2.12 or 2.92 | M1 | objective valued at \((10,0)\) and at \((4,12)\) |
| \(z = 10\); Cut 10m according to plan \(x\) and 10m according to plan \(z\) | M1 | \(z=10\) |
| 20m of material with width 47cm; 30m of material with width 32cm | A1 | both |
| \(1\text{m}^2\) of waste | A1 | waste |
| Min waste at \((10, 0)\) |
# Question 4:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Diagram showing $z$ m width with 3 strips of 32cm and waste of 4cm | M1 | 3 widths + waste |
| | A1 | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.06x + 0.21y + 0.04z$ (m²) | M1 | 3 areas |
| | $A1\sqrt{}$ | any units OK; ignore scaling |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2x + y > 20$ | B1 | |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y + 3z > 24$ | $B1\sqrt{}$ | |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Use of $z = 20 - x - y$ | M1 | |
| Minimise $0.02x + 0.17y$ (constant of 0.8 not needed but OK if there) | $A1\sqrt{}$ | "minimise" not needed – given |
| s.t. $2x + y > 20$ | $A1\sqrt{}$ | |
| $-3x - 2y > -36$ or $3x + 2y < 36$ | | |
## Part (vi)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Line $2x + y = 20$ drawn correctly | B1 | line (cao) |
| Line $3x + 2y = 36$ drawn correctly | B1 | line (cao) |
| Shading – follow two negative gradient lines making a triangle with base on x-axis | B1 | shading |
| Objective valued at $(10, 0)$ and at $(4, 12)$ giving values 212 or 292 or 2.12 or 2.92 | M1 | objective valued at $(10,0)$ and at $(4,12)$ |
| $z = 10$; Cut 10m according to plan $x$ and 10m according to plan $z$ | M1 | $z=10$ |
| 20m of material with width 47cm; 30m of material with width 32cm | A1 | both |
| $1\text{m}^2$ of waste | A1 | waste |
| Min waste at $(10, 0)$ | | |4 Two products are to be made from material that is supplied in a single roll, 20 m long and 1 m wide. The two products require widths of 47 cm and 32 cm respectively.
Two ways of cutting lengths of material are shown in the plans below.\\
\includegraphics[max width=\textwidth, alt={}, center]{e88abde1-8769-4a3c-b115-031cea08d9a6-5_408_1538_520_269}\\
\includegraphics[max width=\textwidth, alt={}, center]{e88abde1-8769-4a3c-b115-031cea08d9a6-5_403_1533_952_274}\\
(i) Given that there should be no unnecessary waste, draw one other cutting plan that might be used for a cut of length $z$ metres.\\
(ii) Write down an expression for the total area that is wasted in terms of $x , y$ and $z$.
All of the roll is to be cut, so $x + y + z = 20$.\\
There needs to be a total length of at least 20 metres of the material for the first product, the one requiring width 47 cm .\\
(iii) Write this as a linear constraint on the variables.
There needs to be a total length of at least 24 metres of the material for the second product, the one requiring width 32 cm .\\
(iv) Write this as a linear constraint on the variables.\\
(v) Formulate an LP in terms of $x$ and $y$ to minimise the area that is wasted. You will need to use the relationship $x + y + z = 20$, together with your answers to parts (ii), (iii) and (iv).\\
(vi) Solve your LP graphically, and interpret the solution.
\hfill \mbox{\textit{OCR MEI D1 2016 Q4 [16]}}