| Exam Board | Edexcel |
|---|---|
| Module | FD1 AS (Further Decision 1 AS) |
| Year | 2018 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Three-variable constraint reduction |
| Difficulty | Standard +0.3 This is a straightforward linear programming question requiring interpretation of constraints, translation to inequalities, and graphical solution of a 2D feasible region. The constraint reduction from 3 to 2 variables is given (not derived), and the graphical method is standard AS-level technique with clear corner points. Slightly easier than average due to the scaffolding provided. |
| 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 | Mark | Guidance |
| \(x\) is the number of cabinets produced in week 1, \(y\) is the number of cabinets produced in week 2, \(z\) is the number of cabinets produced in week 3 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x + y \leq z\) | B1 | |
| \(z \leq 2y\) | B1 | |
| \(y + z \leq 125\) | ||
| \((x, y, z \geq 0)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Objective is \(P = 250x + 275y + 200(150 - x - y)\) | M1 | |
| \(P = 50x + 75y \ (+30000)\) | A1 | |
| Objective line drawn or at least two vertices tested | M1 | |
| Optimal point \(\left(25, \dfrac{125}{3}\right)\) | A1 | |
| Consideration of integer coordinates around the optimal vertex | M1 | |
| Correct integer coordinate \((25, 42)\) | A1 | |
| The production schedule is 25 cabinets in week 1, 42 cabinets in week 2 and 83 cabinets in week 3 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Total cost of production is £34 400 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Define variables as "number of..." (must include phrase like 'number of', 'amount of', 'quantity of') | B1 | Must contain appropriate counting phrase at least once |
| Answer | Marks | Guidance |
|---|---|---|
| Any one correct constraint/inequality | B1 | Accept strict inequalities |
| All three correct constraints/inequalities | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Attempt to derive new objective function using \(x + y + z = 150\) to eliminate \(z\), or calculate all three values of \(z\) | M1 | Using \(x + y + z = 150\) |
| Correct objective in terms of \(x\) and \(y\) only or all three correct \(z\) values \(\left(\frac{250}{3}, 75, 75\right)\) | A1 | |
| Objective line drawn consistent with objective function (or reciprocal) or testing two correct vertices to at least 1 d.p. in objective function involving \(x\) and \(y\) only or testing two correct vertices in \(250x + 275y + 200z\) | M1 | |
| Correct optimal point \(\left(25, \frac{125}{3}\right)\) or \(\left(25, \frac{125}{3}, \frac{250}{3}\right)\) — accept 41.6 or 41.7 or better | A1 | At least 1 d.p. if not exact |
| Consider integer point(s) e.g. \((25, 41)\) etc. around the optimal vertex | M1 | Must have attempted point testing of vertices of FR or objective line |
| Correct integer coordinate \((25, 42)\) stated and clear rejection of \((26, 41)\) by checking \(x + 3y \geq 150\) or testing \((27, 41)\) in correct objective function | A1 | |
| Conclusion stated in context (not in terms of \(x\), \(y\) and \(z\)) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| £34,400 | B1 | Condone lack of units |
## Question 4:
### Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $x$ is the number of cabinets produced in week 1, $y$ is the number of cabinets produced in week 2, $z$ is the number of cabinets produced in week 3 | B1 | |
### Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $x + y \leq z$ | B1 | |
| $z \leq 2y$ | B1 | |
| $y + z \leq 125$ | | |
| $(x, y, z \geq 0)$ | | |
### Part (c)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Objective is $P = 250x + 275y + 200(150 - x - y)$ | M1 | |
| $P = 50x + 75y \ (+30000)$ | A1 | |
| Objective line drawn or at least two vertices tested | M1 | |
| Optimal point $\left(25, \dfrac{125}{3}\right)$ | A1 | |
| Consideration of integer coordinates around the optimal vertex | M1 | |
| Correct integer coordinate $(25, 42)$ | A1 | |
| The production schedule is 25 cabinets in week 1, 42 cabinets in week 2 and 83 cabinets in week 3 | B1 | |
### Part (c)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Total cost of production is £34 400 | B1 | |
## Question (a):
| Define variables as "number of..." (must include phrase like 'number of', 'amount of', 'quantity of') | B1 | Must contain appropriate counting phrase at least once |
---
## Question (b):
| Any one correct constraint/inequality | B1 | Accept strict inequalities |
| All three correct constraints/inequalities | B1 | |
**Note:** Vertices of the Feasible Region are $\left(25, \frac{125}{3}\right)$, $(25, 50)$, $\left(\frac{75}{2}, \frac{75}{2}\right)$
---
## Question (c)(i):
| Attempt to derive new objective function using $x + y + z = 150$ to eliminate $z$, **or** calculate all three values of $z$ | M1 | Using $x + y + z = 150$ |
| Correct objective in terms of $x$ and $y$ only **or** all three correct $z$ values $\left(\frac{250}{3}, 75, 75\right)$ | A1 | |
| Objective line drawn consistent with objective function (or reciprocal) **or** testing two correct vertices to at least 1 d.p. in objective function involving $x$ and $y$ only **or** testing two correct vertices in $250x + 275y + 200z$ | M1 | |
| Correct optimal point $\left(25, \frac{125}{3}\right)$ **or** $\left(25, \frac{125}{3}, \frac{250}{3}\right)$ — accept 41.6 or 41.7 or better | A1 | At least 1 d.p. if not exact |
| Consider integer point(s) e.g. $(25, 41)$ etc. around the optimal vertex | M1 | Must have attempted point testing of vertices of FR or objective line |
| Correct integer coordinate $(25, 42)$ stated **and** clear rejection of $(26, 41)$ by checking $x + 3y \geq 150$ **or** testing $(27, 41)$ in correct objective function | A1 | |
| Conclusion stated in context (not in terms of $x$, $y$ and $z$) | B1 | |
---
## Question (c)(ii):
| £34,400 | B1 | Condone lack of units |4. The manager of a factory is planning the production schedule for the next three weeks for a range of cabinets. The following constraints apply to the production schedule.
\begin{itemize}
\item The total number of cabinets produced in week 3 cannot be fewer than the total number produced in weeks 1 and 2
\item At most twice as many cabinets must be produced in week 3 as in week 2
\item The number of cabinets produced in weeks 2 and 3 must, in total, be at most 125
\end{itemize}
The production cost for each cabinet produced in weeks 1,2 and 3 is $\pounds 250 , \pounds 275$ and $\pounds 200$ respectively.\\
The factory manager decides to formulate a linear programming problem to find a production schedule that minimises the total cost of production.
The objective is to minimise $250 x + 275 y + 200 z$
\begin{enumerate}[label=(\alph*)]
\item Explain what the variables $x , y$ and $z$ represent.
\item Write down the constraints of the linear programming problem in terms of $x , y$ and $z$.
Due to demand, exactly 150 cabinets must be produced during these three weeks. This reduces the constraints to
$$\begin{gathered}
x + y \leqslant 75 \\
x + 3 y \geqslant 150 \\
x \geqslant 25 \\
y \geqslant 0
\end{gathered}$$
which are shown in Diagram 1 in the answer book.\\
Given that the manager does not want any cabinets left unfinished at the end of a week,
\item \begin{enumerate}[label=(\roman*)]
\item use a graphical approach to solve the linear programming problem and hence determine the production schedule which minimises the cost of production. You should make your method and working clear.
\item Find the minimum total cost of the production schedule.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel FD1 AS 2018 Q4 [11]}}