| Exam Board | OCR MEI |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2011 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Graphical optimization with objective line |
| Difficulty | Moderate -0.8 This is a standard linear programming question with straightforward constraint formulation and graphical solution. Parts (i)-(ii) involve translating word problems into inequalities using given variables (routine skill), part (iii) is basic graph sketching, and parts (iv)-(v) require solving two simultaneous linear equations and interpreting the result. All steps are mechanical applications of D1 techniques with no novel problem-solving required, making it easier than average A-level maths. |
| Spec | 7.06a LP formulation: variables, constraints, objective function7.06b Slack variables: converting inequalities to equations7.06d Graphical solution: feasible region, two variables7.06e Sensitivity analysis: effect of changing coefficients |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Thousands of litres of A in stock \(= 2\) | B1 | |
| \(b \geq -4\) | B1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(5(a+2) + 6(b+4) \geq 61\) | M1 A1 | |
| \((a+2) + (b+4) \leq 12\) giving \(a + b \leq 6\) | M1 A1 | watch for fluke |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| lines (with graph showing vertices at \(b=6\), \(a=6\), point \((5.4, 4.5)\), and \((9,-3)\)) | B4 | \(\sqrt{}\) their negative gradient stock line |
| shading | B1 | \(\sqrt{}\) shape \(= \bigtriangledown\) or \(\bigtriangledown\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Increase stock levels of A by 9000 litres | B1 | Give the marks for \(9000, -3000\), or equivalent \(\pm 200\) litres on both |
| Reduce stock levels of B by 3000 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| New stock levels are 11000 of A and 1000 of B | B1 | \(\sqrt{}\) (iv) SC correct answer from nowhere OK |
| \(5 \times 11000 + 6 \times 1000 = 61000\) | B1 | Allow comment only for the "fully stocked" B1 |
| \(11000 + 1000 = 12000\) | B1 |
# Question 6:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Thousands of litres of A in stock $= 2$ | B1 | |
| $b \geq -4$ | B1 | cao |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $5(a+2) + 6(b+4) \geq 61$ | M1 A1 | |
| $(a+2) + (b+4) \leq 12$ giving $a + b \leq 6$ | M1 A1 | watch for fluke |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| lines (with graph showing vertices at $b=6$, $a=6$, point $(5.4, 4.5)$, and $(9,-3)$) | B4 | $\sqrt{}$ their negative gradient stock line |
| shading | B1 | $\sqrt{}$ shape $= \bigtriangledown$ or $\bigtriangledown$ |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Increase stock levels of A by 9000 litres | B1 | Give the marks for $9000, -3000$, or equivalent $\pm 200$ litres on both |
| Reduce stock levels of B by 3000 | B1 | |
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| New stock levels are 11000 of A and 1000 of B | B1 | $\sqrt{}$ (iv) SC correct answer from nowhere OK |
| $5 \times 11000 + 6 \times 1000 = 61000$ | B1 | Allow comment only for the "fully stocked" B1 |
| $11000 + 1000 = 12000$ | B1 | |6 A manufacturing company holds stocks of two liquid chemicals. The company needs to update its stock levels.
The company has 2000 litres of chemical A and 4000 litres of chemical B currently in stock. Its storage facility allows for no more than a combined total of 12000 litres of the two chemicals.
Chemical A is valued at $\pounds 5$ per litre and chemical B is valued at $\pounds 6$ per litre. The company intends to hold stocks of these two chemicals with a total value of at least $\pounds 61000$.
Let $a$ be the increase in the stock level of A, in thousands of litres ( $a$ can be negative).\\
Let $b$ be the increase in the stock level of B , in thousands of litres ( $b$ can be negative).\\
(i) Explain why $a \geqslant - 2$, and produce a similar inequality for $b$.\\
(ii) Explain why the value constraint can be written as $5 a + 6 b \geqslant 27$, and produce, in similar form, the storage constraint.\\
(iii) Illustrate all four inequalities graphically.\\
(iv) Find the policy which will give a stock value of exactly $\pounds 61000$, and will use all 12000 litres of available storage space.\\
(v) Interpret your solution in terms of stock levels, and verify that the new stock levels do satisfy both the value constraint and the storage constraint.
\hfill \mbox{\textit{OCR MEI D1 2011 Q6 [16]}}