| Exam Board | Edexcel |
|---|---|
| Module | FD1 AS (Further Decision 1 AS) |
| Year | 2020 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Reverse engineering objective from solution |
| Difficulty | Challenging +1.2 This requires reading coordinates from a graph, determining constraint inequalities from boundary lines, and reverse-engineering an objective function from a given optimal value. While it involves multiple steps (finding gradients, intercepts, and solving for objective coefficients), these are all standard linear programming techniques with no novel problem-solving required. It's moderately harder than average due to the reverse-engineering aspect and need for careful algebraic manipulation. |
| Spec | 7.06d Graphical solution: feasible region, two variables |
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| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Line through \((0, 12)\) and \((6, 0)\) is \(2x + y = 12\) | M1 | 1.1b – Correct method for finding the equation of one of the three lines |
| Line through \((0, 12)\) and \((10, 0)\) is \(6x + 5y = 60\) | ||
| Line through \((7, 2)\) and \((9, 8)\) is \(3x - y = 19\) | ||
| \(2x + y \geq 12\) | A1 | 3.4 – CAO (with correct inequality sign from shading); allow a positive multiple but must have integer coefficients |
| \(6x + 5y \leq 60\) | A1 | 1.1b – CAO; allow a positive multiple but must have integer coefficients |
| \(3x - y \leq 19\) | A1 | 1.1b – CAO; allow a positive multiple but must have integer coefficients |
| Solving correct two equations to find \(V\) | M1 | 1.1b – Attempt to find \(V\) by solving correct pair of simultaneous equations; implied by correct exact coordinates |
| \(V\left(\dfrac{155}{21}, \dfrac{22}{7}\right)\) | A1 | 2.2a – Correct deduction of exact coordinates for \(V\) |
| \(P = k(5x + 3y)\) and substituting \(P = 556\) and their \(V\) | M1dep | 3.4 – Uses model to write suitable objective and substitutes; dependent on previous M mark. Or: forming both equations \(\dfrac{155}{21}x + \dfrac{22}{7}y = 556\) and \(3x - 5y = 0\) |
| Maximise | B1 | 2.5 – Allow 'max'; independent of all other marks |
| \(P = 60x + 36y\) | A1 | 2.2a – Correct objective function; cannot be awarded for \(5x + 3y\) |
## Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Line through $(0, 12)$ and $(6, 0)$ is $2x + y = 12$ | M1 | 1.1b – Correct method for finding the equation of one of the three lines |
| Line through $(0, 12)$ and $(10, 0)$ is $6x + 5y = 60$ | | |
| Line through $(7, 2)$ and $(9, 8)$ is $3x - y = 19$ | | |
| $2x + y \geq 12$ | A1 | 3.4 – CAO (with correct inequality sign from shading); allow a positive multiple but must have integer coefficients |
| $6x + 5y \leq 60$ | A1 | 1.1b – CAO; allow a positive multiple but must have integer coefficients |
| $3x - y \leq 19$ | A1 | 1.1b – CAO; allow a positive multiple but must have integer coefficients |
| Solving correct two equations to find $V$ | M1 | 1.1b – Attempt to find $V$ by solving correct pair of simultaneous equations; implied by correct exact coordinates |
| $V\left(\dfrac{155}{21}, \dfrac{22}{7}\right)$ | A1 | 2.2a – Correct deduction of exact coordinates for $V$ |
| $P = k(5x + 3y)$ and substituting $P = 556$ and their $V$ | M1dep | 3.4 – Uses model to write suitable objective and substitutes; dependent on previous M mark. Or: forming both equations $\dfrac{155}{21}x + \dfrac{22}{7}y = 556$ and $3x - 5y = 0$ |
| Maximise | B1 | 2.5 – Allow 'max'; independent of all other marks |
| $P = 60x + 36y$ | A1 | 2.2a – Correct objective function; cannot be awarded for $5x + 3y$ |
**Complete LP formulation:**
Maximise $P = 60x + 36y$
Subject to $2x + y \geq 12$
$\quad\quad\quad\quad\; 6x + 5y \leq 60$
$\quad\quad\quad\quad\; 3x - y \leq 19$
**Total: 9 marks**
**Note:** If A0A0A0 then award A1A0A0 only for one 'correct' strict inequality and/or non-integer coefficients e.g. $x + 0.5y > 6$4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a2a6e659-aab5-4eec-9af4-ca6ab895f1c8-05_1472_1320_233_376}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows the constraints of a linear programming problem in $x$ and $y$, where $R$ is the feasible region. Figure 3 also shows an objective line for the problem and the optimal vertex, which is labelled as $V$.
The value of the objective at $V$ is 556\\
Express the linear programming problem in algebraic form. List the constraints as simplified inequalities with integer coefficients and determine the objective.
Please check the examination details below before entering your candidate information\\
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\includegraphics[max width=\textwidth, alt={}, center]{a2a6e659-aab5-4eec-9af4-ca6ab895f1c8-09_122_433_356_991}\\
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\section*{Thursday 14 May 2020}
Afternoon\\
Paper Reference 8FMO/27
\section*{Further Mathematics}
Advanced Subsidiary\\
Further Mathematics options\\
27: Decision Mathematics 1\\
(Part of options D, F, H and K)
\section*{Answer Book}
Do not return the question paper with the answer book.\\
1.\\
$\begin{array} { l l l l l l l l l l } 3.7 & 2.5 & 5.4 & 1.9 & 2.7 & 3.2 & 3.1 & 2.7 & 4.2 & 2.0 \end{array}$
\begin{enumerate}
\item (a)
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | c | }
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Activity & \begin{tabular}{ c }
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preceding \\
activities \\
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\end{tabular}
\end{center}
\begin{center}
\begin{tabular}{ | c | c | }
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Activity & \begin{tabular}{ c }
Immediately \\
preceding \\
activities \\
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E & \\
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\end{center}
\begin{center}
\begin{tabular}{ | c | c | }
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Immediately \\
preceding \\
activities \\
\end{tabular} \\
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I & \\
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\end{center}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a2a6e659-aab5-4eec-9af4-ca6ab895f1c8-12_734_1646_925_196}
\captionsetup{labelformat=empty}
\caption{Diagram 1}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a2a6e659-aab5-4eec-9af4-ca6ab895f1c8-13_1116_1475_979_296}
\captionsetup{labelformat=empty}
\caption{Grid 1}
\end{center}
\end{figure}
3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a2a6e659-aab5-4eec-9af4-ca6ab895f1c8-14_716_1467_255_299}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
[The weight of the network is $5 x + 246$ ]
4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a2a6e659-aab5-4eec-9af4-ca6ab895f1c8-18_1470_1319_255_388}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
\hfill \mbox{\textit{Edexcel FD1 AS 2020 Q4 [9]}}