| Exam Board | Edexcel |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Three-variable constraint reduction |
| Difficulty | Standard +0.3 This is a standard linear programming question with straightforward constraint formulation and graphical solution. Parts (a)-(c) test basic translation to inequalities, (d)-(f) involve routine 2D graphical LP with objective lines, and (g) is simple substitution. The three-variable aspect is eliminated early by fixing z=150, making it easier than typical LP questions. |
| Spec | 7.06a LP formulation: variables, constraints, objective function7.06b Slack variables: converting inequalities to equations7.06c Working with constraints: algebra and ad hoc methods7.06d Graphical solution: feasible region, two variables7.06e Sensitivity analysis: effect of changing coefficients |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(5y \leq 3z\) | M1 A1 | a1M1: Correct method: \(5 \square 3z\) where \(\square\) is any inequality or equals. An exact equivalent answer (with or without integer coefficients) can score M1 or M1 for \(3y \leq 5z\) only. a1A1: CAO (or equivalent e.g. \(k(5y \leq 3z)\) where \(k\) is any positive integer only) |
| The total number of shirts must be at least 250. At most 20% of all the shirts should be small | B1 M1 A1 | b1B1: CAO oe e.g. the minimum number of shirts is 250 is fine for this mark (note that they must imply that the total number (and not one particular brand of shirt) is at least 250). b1M1: Three of 'at most', '20%', 'all' and 'small' (allow equivalents e.g. fifth or 0.2 for 20%) allow those who simply 'said' provided that 'it is clear that they aren't talking about one particular brand of shirt. b1A1: CAO (o.e. e.g. the number of small shirts is less than or equal to a fifth of the total number of shirts, the number of small shirts is at most 20% of all the shirts sold) – give bod of these that clearly imply 'all' provided that they aren't talking about only one particular brand. Do not allow statements which contain use of 0.2 or \(\frac{1}{4}\) for this mark, e.g. the number of small shirts is at most 0.2 of all the shirts is A0 |
| (Minimise) \(6x + 10y + 15z\) | B1 | c1B1: Expression correct (or \(600x + 1000y + 1500z\)) |
| \(z = 150 \Rightarrow x + y \geq 100\); \(y \leq 90\); \(4x – y \leq 150\) | M1 A1 A1 | d(i)1M1: Eliminating \(z\) from all their inequalities by using the substitution \(z = 150\) – accept unsimplified (e.g. \(x + y + 150 \geq 250, x + y \geq 100, 4x + y \leq 150\) ) and their \(5y \leq 3(150)\)). d(i)1A1: CAO e.g. \(x + y \geq 100, 4x \leq y + 150\) (oe) - all constraints must be correct with integer coefficients but allow positive multiples – ignore \(x \geq 0, y \geq 0\) but any other additional constraints is A0 – allow recovery in this part if \(y \leq 90\) (oe) seen in (d) even if their \(5y \leq 3z\) is incorrect in (a). |
| [Graph with 6 marks: one each for four lines, correct region labelled, feasible region correctly shown] | B1 B1 B1 B1 | In (d), lines must be long enough to define the correct feasible region and would pass through one small square of the points stated: \(x + y = 100\) must pass within one small square of its intersection with the axes – (0, 100) and (100, 0); \(y = 90\) must pass within one small square of its intersection with the y-axis and (60, 90); \(4x – y = 150\) must pass within one small square of (37.5, 0) and (60, 90). di1B1: Any one line correctly drawn. dii2B1: Any two lines correctly drawn. diii3B1: All three lines correctly drawn. div4B1: Region, \(\mathcal{R}\), correctly labelled – not just implied by shading – dependent on scoring the three previous B marks in this part. |
| Correct objective line; V correctly labelled | B1 B1 | e1B1: Drawing the correct objective line on the graph with gradient of –0.6. Line must be correct to within one small square if extended from axis to axis. If line is shorter than (0, 6) to (10, 0) then B0. e2B1: \(V\) correctly labelled – note that this mark is dependent on the correct feasible region in (d) (so must have scored at least B1B1B1B0 in (d)) and the previous B mark in (e) |
| 50 small and 50 medium shirts; Cost = £3050 | B1 B1 | f1B1: CAO – must be in context (50 small and 50 medium and not for \(x = y = 50\)) note that this mark is dependent on the correct feasible region in (d) (so must have scored at least B1B1B1B0 in (d)) and the first B mark (for a correct objective line) in (e). f2B1: CAO (3050) – units not required - note that this mark is dependent on the correct feasible region in (d) (so must have scored at least B1B1B1B0 in (d)) and the first B mark (for a correct objective line) in (e) |
| \(x = 50, y = 75 \Rightarrow z \geq 125\) therefore minimum number of large shirts is 125. This leads to a cost of £2925 which is less than the cost in (f) | M1 A1 A1 | g1M1: Substitute to obtain the correct value for \(z\) of 125 (accept \(z \geq 125\) or \(>\)) – if no method allow 125 seen but M0 if 125 found but a different value of \(z\) stated and subsequently used. g1A1: Correct cost (2925) and dependent on the final B mark in (f). |
| Total: 18 marks |
| Answer | Marks | Guidance |
|--------|-------|----------|
| $5y \leq 3z$ | M1 A1 | a1M1: Correct method: $5 \square 3z$ where $\square$ is any inequality or equals. An exact equivalent answer (with or without integer coefficients) can score M1 or M1 for $3y \leq 5z$ only. a1A1: CAO (or equivalent e.g. $k(5y \leq 3z)$ where $k$ is any positive integer only) |
| The total number of shirts must be at least 250. At most 20% of all the shirts should be small | B1 M1 A1 | b1B1: CAO oe e.g. the minimum number of shirts is 250 is fine for this mark (note that they must imply that the total number (and not one particular brand of shirt) is **at least** 250). b1M1: Three of 'at most', '20%', 'all' and 'small' (allow equivalents e.g. fifth or 0.2 for 20%) allow those who simply 'said' provided that 'it is clear that they aren't talking about one particular brand of shirt. b1A1: CAO (o.e. e.g. the number of small shirts is less than or equal to a fifth of the total number of shirts, the number of small shirts is at most 20% of all the shirts sold) – give bod of these that clearly imply 'all' provided that they aren't talking about only one particular brand. Do not allow statements which contain use of 0.2 or $\frac{1}{4}$ for this mark, e.g. the number of small shirts is at most 0.2 of all the shirts is A0 |
| (Minimise) $6x + 10y + 15z$ | B1 | c1B1: Expression correct (or $600x + 1000y + 1500z$) |
| $z = 150 \Rightarrow x + y \geq 100$; $y \leq 90$; $4x – y \leq 150$ | M1 A1 A1 | d(i)1M1: Eliminating $z$ from all **their** inequalities by using the substitution $z = 150$ – accept unsimplified (e.g. $x + y + 150 \geq 250, x + y \geq 100, 4x + y \leq 150$ ) and their $5y \leq 3(150)$). d(i)1A1: CAO e.g. $x + y \geq 100, 4x \leq y + 150$ (oe) - all constraints must be correct with integer coefficients but allow positive multiples – ignore $x \geq 0, y \geq 0$ but any other additional constraints is A0 – allow recovery in this part if $y \leq 90$ (oe) seen in (d) even if their $5y \leq 3z$ is incorrect in (a). |
| [Graph with 6 marks: one each for four lines, correct region labelled, feasible region correctly shown] | B1 B1 B1 B1 | **In (d), lines must be long enough to define the correct feasible region and would pass through one small square of the points stated:** $x + y = 100$ must pass within one small square of its intersection with the axes – (0, 100) and (100, 0); $y = 90$ must pass within one small square of its intersection with the y-axis and (60, 90); $4x – y = 150$ must pass within one small square of (37.5, 0) and (60, 90). **di1B1:** Any one line correctly drawn. **dii2B1:** Any two lines correctly drawn. **diii3B1:** All three lines correctly drawn. **div4B1:** Region, $\mathcal{R}$, correctly labelled – not just implied by shading – dependent on scoring the three previous B marks in this part. |
| Correct objective line; V correctly labelled | B1 B1 | e1B1: Drawing the correct objective line on the graph with gradient of –0.6. Line must be correct to within one small square if extended from axis to axis. If line is shorter than (0, 6) to (10, 0) then B0. e2B1: $V$ correctly labelled – note that this mark is dependent on the correct feasible region in (d) (so must have scored at least B1B1B1B0 in (d)) and the previous B mark in (e) |
| 50 small and 50 medium shirts; Cost = £3050 | B1 B1 | f1B1: CAO – must be in context (50 small and 50 medium and not for $x = y = 50$) note that this mark is dependent on the correct feasible region in (d) (so must have scored at least B1B1B1B0 in (d)) and the first B mark (for a correct objective line) in (e). f2B1: CAO (3050) – units not required - note that this mark is dependent on the correct feasible region in (d) (so must have scored at least B1B1B1B0 in (d)) and the first B mark (for a correct objective line) in (e) |
| $x = 50, y = 75 \Rightarrow z \geq 125$ therefore minimum number of large shirts is 125. This leads to a cost of £2925 which is less than the cost in (f) | M1 A1 A1 | g1M1: Substitute to obtain the correct value for $z$ of 125 (accept $z \geq 125$ or $>$) – if no method allow 125 seen but M0 if 125 found but a different value of $z$ stated and subsequently used. g1A1: Correct cost (2925) and dependent on the final B mark in (f). |
| **Total: 18 marks** | | |
---5. A clothing shop sells a particular brand of shirt, which comes in three different sizes, small, medium and large.
Each month the manager of the shop orders $x$ small shirts, $y$ medium shirts and $z$ large shirts.\\
The manager forms constraints on the number of each size of shirts he will have to order.\\
One constraint is that for every 3 medium shirts he will order at least 5 large shirts.
\begin{enumerate}[label=(\alph*)]
\item Write down an inequality, with integer coefficients, to model this constraint.
Two further constraints are
$$x + y + z \geqslant 250 \text { and } x \leqslant 0.2 ( x + y + z )$$
\item Use these two constraints to write down statements, in context, that describe the number of different sizes of shirt the manager will order.
The cost of each small shirt is $\pounds 6$, the cost of each medium shirt is $\pounds 10$ and the cost of each large shirt is $\pounds 15$
The manager must minimise the total cost of all the shirts he will order.
\item Write down the objective function.
Initially, the manager decides to order exactly 150 large shirts.
\item \begin{enumerate}[label=(\roman*)]
\item Rewrite the constraints, as simplified inequalities with integer coefficients, in terms of $x$ and $y$ only.
\item Represent these constraints on Diagram 1 in the answer book. Hence determine, and label, the feasible region $R$.
\end{enumerate}\item Use the objective line method to find the optimal vertex, $V$, of the feasible region. You must make your objective line clear and label $V$.
\item Write down the number of each size of shirt the manager should order. Calculate the total cost of this order.
Later, the manager decides to order exactly 50 small shirts and exactly 75 medium shirts instead of 150 large shirts.
\item Find the minimum number of large shirts the manager should order and show that this leads to a lower cost than the cost found in (f).
\end{enumerate}
\hfill \mbox{\textit{Edexcel D1 2019 Q5 [18]}}