| Exam Board | Edexcel |
|---|---|
| Module | D2 (Decision Mathematics 2) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Game theory LP formulation |
| Difficulty | Standard +0.8 This is a standard D2 game theory question requiring LP formulation for player B's optimal strategy. While it follows a well-defined procedure taught in the syllabus (adding constant to make entries positive, defining probabilities as variables, setting up minimize objective with constraints), it requires careful systematic work across multiple parts with opportunity for sign errors. Slightly above average difficulty due to the procedural complexity and need for precision, but remains a textbook application rather than requiring novel insight. |
| Spec | 7.06a LP formulation: variables, constraints, objective function7.08a Pay-off matrix: zero-sum games7.08b Dominance: reduce pay-off matrix |
| \cline { 3 - 5 } \multicolumn{2}{c|}{} | \(B\) | |||
| \cline { 2 - 5 } \multicolumn{2}{c|}{} | I | II | III | |
| \multirow{3}{*}{\(A\)} | I | - 1 | 4 | - 3 |
| \cline { 2 - 5 } | II | - 3 | 7 | 1 |
| \cline { 2 - 5 } | III | 5 | - 2 | - 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Add 4 to all entries (since minimum entry is \(-3\)); \(v = V + 4\) | M1 | |
| New matrix stated with all entries positive; \(v = V+4\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(p_1, p_2, p_3\) = probability that \(B\) plays I, II, III respectively | B1 | Both variables defined |
| \(p_1 + p_2 + p_3 = 1\), \(p_i \geq 0\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Minimise \(v\) (or maximise \(V\)) | M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3p_1 + 8p_2 + p_3 \geq v\) | M1 | One correct constraint |
| \(p_1 + 11p_2 + 5p_3 \geq v\); \(5p_1 + 3p_2 + 3p_3 \geq v\); \(p_1+p_2+p_3=1\), \(p_i \geq 0\) | A1 | All constraints correct |
# Question 2:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Add 4 to all entries (since minimum entry is $-3$); $v = V + 4$ | M1 | |
| New matrix stated with all entries positive; $v = V+4$ | A1 | |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $p_1, p_2, p_3$ = probability that $B$ plays I, II, III respectively | B1 | Both variables defined |
| $p_1 + p_2 + p_3 = 1$, $p_i \geq 0$ | B1 | |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Minimise $v$ (or maximise $V$) | M1A1 | |
## Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3p_1 + 8p_2 + p_3 \geq v$ | M1 | One correct constraint |
| $p_1 + 11p_2 + 5p_3 \geq v$; $5p_1 + 3p_2 + 3p_3 \geq v$; $p_1+p_2+p_3=1$, $p_i \geq 0$ | A1 | All constraints correct |
---2. The payoff matrix for player $A$ in a two-person zero-sum game with value $V$ is shown below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\cline { 3 - 5 }
\multicolumn{2}{c|}{} & \multicolumn{3}{c|}{$B$} \\
\cline { 2 - 5 }
\multicolumn{2}{c|}{} & I & II & III \\
\hline
\multirow{3}{*}{$A$} & I & - 1 & 4 & - 3 \\
\cline { 2 - 5 }
& II & - 3 & 7 & 1 \\
\cline { 2 - 5 }
& III & 5 & - 2 & - 1 \\
\hline
\end{tabular}
\end{center}
Formulate this information as a linear programming problem, the solution to which will give the optimal strategy for player $B$.
\begin{enumerate}[label=(\alph*)]
\item Rewrite the matrix as necessary and state the new value of the game, $v$, in terms of $V$.
\item Define your decision variables.
\item Write down the objective function in terms of your decision variables.
\item Write down the constraints.
\end{enumerate}
\hfill \mbox{\textit{Edexcel D2 Q2 [8]}}