| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Discrete (Further AS Paper 2 Discrete) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Dynamic Programming |
| Type | Zero-sum game stable solution |
| Difficulty | Challenging +1.2 This is a game theory question requiring systematic checking of saddle points across different values of x. While it involves multiple cases and careful verification, the technique is standard for Further Maths students: find row minima and column maxima, then check for equality. The constraint that x is an integer significantly limits the search space, making this a methodical rather than insightful problem. |
| Spec | 7.08c Pure strategies: play-safe strategies and stable solutions7.08d Nash equilibrium: identification and interpretation |
| Mayon | |||
| \(\mathbf{M_1}\) | \(\mathbf{M_2}\) | \(\mathbf{M_3}\) | |
| \(\mathbf{B_1}\) | \(-2\) | \(-1\) | \(1\) |
| Bilal \quad \(\mathbf{B_2}\) | \(4\) | \(-3\) | \(1\) |
| \(\mathbf{B_3}\) | \(-1\) | \(x\) | \(0\) |
| Answer | Marks | Guidance |
|---|---|---|
| 10(a) | Finds at least 4 of the row minima | |
| and column maxima | 3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 4, x or –1, 1 | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| max(row minima) = –1 or x | 2.2a | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| min(column maxima) = –1 or x | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| = min(column maxima) | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| solution | 2.1 | R1 |
| Subtotal | 6 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 10(b) | States –1 | 1.1b |
| Answer | Marks |
|---|---|
| Subtotal | 1 |
| Question total | 7 |
| Question Paper total | 40 |
Question 10:
--- 10(a) ---
10(a) | Finds at least 4 of the row minima
and column maxima | 3.1a | M1 | Row minima = –2 , –3, –1 or x
Column maxima = 4 , –1 or x, 1
max(row minima) = –1 or x
min(column maxima) = –1 or x
For a stable solution,
max(row minima) = min(column
maxima)
x must equal –1 to ensure that
max(row minima) = –1
min(column maxima) = –1
Therefore a stable solution occurs
only when x = –1
Obtains row minima of
–2, –3, –1 or x
or
Obtains column maxima of
4, x or –1, 1 | 1.1b | A1
Obtains
max(row minima) = –1 or x | 2.2a | A1
Obtains
min(column maxima) = –1 or x | 1.1b | A1
States or uses max(row minima)
= min(column maxima) | 1.1b | B1
Completes fully reasoned
argument to show that x = –1 is
the only value that gives a stable
solution | 2.1 | R1
Subtotal | 6
Q | Marking instructions | AO | Marks | Typical solution
--- 10(b) ---
10(b) | States –1 | 1.1b | B1 | The value of the game for Bilal
is –1
Subtotal | 1
Question total | 7
Question Paper total | 40Bilal and Mayon play a zero-sum game.
The game is represented by the following pay-off matrix for Bilal, where $x$ is an integer.
\begin{tabular}{c|c|c|c|}
& \multicolumn{3}{c}{Mayon} \\
& $\mathbf{M_1}$ & $\mathbf{M_2}$ & $\mathbf{M_3}$ \\
\hline
$\mathbf{B_1}$ & $-2$ & $-1$ & $1$ \\
\hline
Bilal \quad $\mathbf{B_2}$ & $4$ & $-3$ & $1$ \\
\hline
$\mathbf{B_3}$ & $-1$ & $x$ & $0$ \\
\hline
\end{tabular}
The game has a stable solution.
\begin{enumerate}[label=10 (\alph*)]
\item Show that there is only one possible value for $x$
Fully justify your answer.
[6 marks]
\item State the value of the game for Bilal.
[1 mark]
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Discrete 2024 Q10 [7]}}