| Exam Board | Edexcel |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Game theory LP formulation |
| Difficulty | Challenging +1.2 This is a standard D2 game theory question requiring systematic checking for saddle points and then formulating/solving a 2×2 LP to find mixed strategies. While it involves multiple steps (checking row minima/column maxima, setting up equations for optimal mixed strategy), the procedures are algorithmic and well-practiced in D2. It's moderately harder than average A-level due to being Further Maths content, but routine within that specification. |
| Spec | 7.08a Pay-off matrix: zero-sum games7.08c Pure strategies: play-safe strategies and stable solutions7.08e Mixed strategies: optimal strategy using equations or graphical method |
| B plays 1 | B plays 2 | B plays 3 | B plays 4 | |
| A plays 1 | 2 | - 1 | 1 | - 3 |
| A plays 2 | - 3 | 2 | - 2 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Scheme | Marks |
| (a) | Row mins [-3, -3] Column max [2, 2, 1, 1] Row maximin (-3) \(\neq\) column minmax (1) so not stable Column 4 dominates column 2 so delete column 2 or if B plays 2 A's expected winnings are \(-p + 2(1-p) = 2-3p\) | M1 A1 B1 |
| (b) | Let A play 1 with probability p and 2 with probability \(1-p\) If B plays 1 A's expected winnings are \(2p – 3(1-p) = 5p – 3\) If B plays 3 A's expected winnings are \(p – 2(1-p) = 3p – 2\) If B plays 4 A's expected winnings are \(-3p + (1-p) = 1 – 4p\) | M1 A1 M1 A1 |
| Solving three simultaneous equations and stating incorrect p is M0. | M1 A1 | b3DM1: Finding their correct optimal point, must have three (or four) lines and set up an equation to find \(0 \leq p \leq 1\). Dependent on previous M mark. Must have at least three intersection points. Solving all three simultaneous equations and stating incorrect p is M0. b3A1: CAO (must have scored all marks except b2B1 (define p mark) in this part). b4A1: CAO |
| \(5p – 3 = 1 – 4p\) \(9p = 4\) \(p = \frac{4}{9}\) A should play row 1 with probability \(\frac{4}{9}\) and row 2 with probability \(\frac{5}{9}\) | M1 A1 A1 | |
| 11 marks |
| Part | Scheme | Marks | Guidance |
|------|--------|-------|----------|
| (a) | Row mins [-3, -3] Column max [2, 2, 1, 1] Row maximin (-3) $\neq$ column minmax (1) so not stable Column 4 dominates column 2 so delete column 2 or if B plays 2 A's expected winnings are $-p + 2(1-p) = 2-3p$ | M1 A1 B1 | a1M1: Finding row minimums and column maximums – condone one error. a1A1: CAO states -3 $\neq$ 1 (or row (maximin) $\neq$ col (minmax)) and draws conclusion. b1B1: CAO Col 4 dominates Col 2 (maybe implied by later working) or correctly stating the expression for A's expected winnings if B plays 2 ($2 - 3p$). |
| (b) | Let A play 1 with probability p and 2 with probability $1-p$ If B plays 1 A's expected winnings are $2p – 3(1-p) = 5p – 3$ If B plays 3 A's expected winnings are $p – 2(1-p) = 3p – 2$ If B plays 4 A's expected winnings are $-3p + (1-p) = 1 – 4p$ | M1 A1 M1 A1 | b1M1: Setting up three probability equations, implicit definition of p. b1A1: CAO (condone incorrect simplification). b2M1: Either attempt at three lines (correct slant direction and relative intersection with 'axes') or four lines if no earlier domination, accept $p > 1$ or $p < 0$ here. Must be functions of p. b2A1: CAO $0 \leq p \leq 1$, scaling correct and clear (or 1 line = 1), condone lack of labels. Rulers used. |
| | Solving three simultaneous equations and stating incorrect p is M0. | M1 A1 | b3DM1: Finding their correct optimal point, must have three (or four) lines and set up an equation to find $0 \leq p \leq 1$. Dependent on previous M mark. Must have at least three intersection points. Solving all three simultaneous equations and stating incorrect p is M0. b3A1: CAO (must have scored all marks except b2B1 (define p mark) in this part). b4A1: CAO |
| | $5p – 3 = 1 – 4p$ $9p = 4$ $p = \frac{4}{9}$ A should play row 1 with probability $\frac{4}{9}$ and row 2 with probability $\frac{5}{9}$ | M1 A1 A1 | |
| | | **11 marks** | |
---4. A two-person zero-sum game is represented by the following pay-off matrix for player A.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | }
\hline
& B plays 1 & B plays 2 & B plays 3 & B plays 4 \\
\hline
A plays 1 & 2 & - 1 & 1 & - 3 \\
\hline
A plays 2 & - 3 & 2 & - 2 & 1 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Verify that there is no stable solution to this game.
\item Find the best strategy for player A.
\end{enumerate}
\hfill \mbox{\textit{Edexcel D2 2014 Q4 [11]}}