| Exam Board | OCR |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Marks | 42 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Game theory LP formulation |
| Difficulty | Challenging +1.2 This is a standard game theory question requiring formulation and solution of an LP for a 2×3 zero-sum game. While it involves multiple steps (formulating constraints, solving the LP, interpreting results), the procedure is algorithmic and follows a well-defined template taught in D2. The matrix is small and the simplex method application is routine, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 7.08e Mixed strategies: optimal strategy using equations or graphical method7.08f Mixed strategies via LP: reformulate as linear programming problem |
| Answer | Marks | Guidance |
|---|---|---|
| B I | B II | B III |
| A I | 6 | 1 |
| A II | 0 | 9 |
| Answer | Marks | Guidance |
|---|---|---|
| Maximise \(P - v = 0\) | A1 | |
| (b) Tableau 1: Initial tableau with P, \(x_1\), \(x_2\), \(x_3\), r, s columns; taking 10 as pivot | M1 | |
| Tableau 2: After pivoting; taking \(8\frac{3}{5}\) as pivot | M1 A2 | |
| Tableau 3: After further pivoting | M1 A1 | |
| Tableau 4: Final optimal tableau with \(x_1 = 0\), \(x_2 = \frac{3}{67}\), \(x_3 = \frac{8}{67}\), \(P = \frac{1}{67}\), \(v = \frac{67}{11}\) | A1 | |
| \(p_1 = 0\), \(p_2 = \frac{67}{11} \times \frac{3}{67} = \frac{3}{11}\), \(p_3 = \frac{67}{11} \times \frac{8}{67} = \frac{8}{11}\) | M1 | |
| B should not play I, should play II \(\frac{3}{11}\) of time and III \(\frac{8}{11}\) of time | A1 | |
| Value of original game = \(\frac{67}{11} - 6 = \frac{1}{11}\) | A1 | (21) |
**(a)** Payoff matrix after changing signs and adding 4:
| | B I | B II | B III |
|---|---|---|---|
| A I | 6 | 1 | 5 |
| A II | 0 | 9 | 2 |
Set up constraints:
- From A I: $6p_1 + 1p_2 + 5p_3 \leq v$ so $-6p_1 - 1p_2 - 5p_3 + s = 0$
- From A II: $0p_1 + 9p_2 + 2p_3 \leq v$ so $-9p_2 - 2p_3 + t = 0$
- Also $p_1 + p_2 + p_3 + r = 1$
Maximise $P - v = 0$ | A1 | |
**(b)** Tableau 1: Initial tableau with P, $x_1$, $x_2$, $x_3$, r, s columns; taking 10 as pivot | M1 | |
Tableau 2: After pivoting; taking $8\frac{3}{5}$ as pivot | M1 A2 | |
Tableau 3: After further pivoting | M1 A1 | |
Tableau 4: Final optimal tableau with $x_1 = 0$, $x_2 = \frac{3}{67}$, $x_3 = \frac{8}{67}$, $P = \frac{1}{67}$, $v = \frac{67}{11}$ | A1 | |
$p_1 = 0$, $p_2 = \frac{67}{11} \times \frac{3}{67} = \frac{3}{11}$, $p_3 = \frac{67}{11} \times \frac{8}{67} = \frac{8}{11}$ | M1 | |
B should not play I, should play II $\frac{3}{11}$ of time and III $\frac{8}{11}$ of time | A1 | |
Value of original game = $\frac{67}{11} - 6 = \frac{1}{11}$ | A1 | (21) |
# Total
(60)The payoff matrix for player $A$ in a two-person zero-sum game is shown below.
\begin{array}{c|c|c|c|c}
& & \multicolumn{3}{c}{B} \\
& & \text{I} & \text{II} & \text{III} \\
\hline
\multirow{2}{*}{A} & \text{I} & -2 & 3 & -1 \\
& \text{II} & 4 & -5 & 2 \\
\end{array}
\begin{enumerate}[label=(\alph*)]
\item Formulate this information as a linear programming problem, the solution to which will give the optimal strategy for player $B$. [7 marks]
\item By solving this linear programming problem, find the optimal strategy for player $B$ and the value of the game. [14 marks]
\end{enumerate}
[21 marks]
\hfill \mbox{\textit{OCR D2 Q6 [42]}}