| Exam Board | OCR |
|---|---|
| Module | Further Discrete AS (Further Discrete AS) |
| Year | 2018 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Game theory LP formulation |
| Difficulty | Standard +0.3 This is a straightforward game theory question requiring standard techniques: finding play-safe strategies by examining row/column minima and maxima, checking stability conditions, and applying dominance concepts. While it involves multiple parts, each step follows routine procedures taught in Further Discrete modules with no novel problem-solving required. |
| Spec | 7.08a Pay-off matrix: zero-sum games7.08b Dominance: reduce pay-off matrix7.08d Nash equilibrium: identification and interpretation |
| P | Q | R | |
| X | \(( 1,4 )\) | \(( 5,3 )\) | \(( 2,6 )\) |
| Y | \(( 5,2 )\) | \(( 1,3 )\) | \(( 0,1 )\) |
| Z | \(( 4,3 )\) | \(( 3,1 )\) | \(( 2,1 )\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Worst pay-off row added: \(\min(P,Q,R)\) for rows \(= (2,1,1)\) | M1 | Using worst pay-off in column(s); Worst result in column P (for player on columns) is 2 |
| \(2 > 1\) so P is better than Q and R | A1 [2] | Maximin (in column P); Explaining why P (as given) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| e.g. If player on columns plays safe (P) then player on rows does best by playing Y \((1, 5, 4)\) | M1 | Play-safe for player on rows is Z \((1, 0, 2)\) |
| But then player on columns does better by changing from P to Q | A1 [2] | \((Y, P)\) has a better pay-off for the player on rows than \((Z, P)\); Or cell \((P,Z) \neq (2,2)\), i.e. the play-safe values |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(A = \{y : y \leq 4\}\) | B1 [1] | Or equivalent; Allow e.g. \(A \leq 4\); Allow \(y < 4\) or \(y \leq 3\); But not just a single specific value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| In row Z, column P has the best pay-off for the player on columns | B1 [1] | Row Z: \(3 > 1\) (for player on cols); For the player on cols \(P > Q\) in row X (or Z) and \(P > R\) in row Z |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Worst pay-off row added: $\min(P,Q,R)$ for rows $= (2,1,1)$ | M1 | Using worst pay-off in column(s); Worst result in column P (for player on columns) is 2 |
| $2 > 1$ so P is better than Q and R | A1 [2] | Maximin (in column P); Explaining why P (as given) |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| e.g. If player on columns plays safe (P) then player on rows does best by playing Y $(1, 5, 4)$ | M1 | Play-safe for player on rows is Z $(1, 0, 2)$ |
| But then player on columns does better by changing from P to Q | A1 [2] | $(Y, P)$ has a better pay-off for the player on rows than $(Z, P)$; Or cell $(P,Z) \neq (2,2)$, i.e. the play-safe values |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $A = \{y : y \leq 4\}$ | B1 [1] | Or equivalent; Allow e.g. $A \leq 4$; Allow $y < 4$ or $y \leq 3$; But not just a single specific value |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| In row Z, column P has the best pay-off for the player on columns | B1 [1] | Row Z: $3 > 1$ (for player on cols); For the player on cols $P > Q$ in row X (or Z) and $P > R$ in row Z |
---3 In the pay-off matrix below, the entry in each cell is of the form $( r , c )$, where $r$ is the pay-off for the player on rows and $c$ is the pay-off for the player on columns when they play that cell.
\begin{center}
\begin{tabular}{ c | c c c }
& P & Q & R \\
\hline
X & $( 1,4 )$ & $( 5,3 )$ & $( 2,6 )$ \\
Y & $( 5,2 )$ & $( 1,3 )$ & $( 0,1 )$ \\
Z & $( 4,3 )$ & $( 3,1 )$ & $( 2,1 )$ \\
\end{tabular}
\end{center}
(i) Show that the play-safe strategy for the player on columns is P .\\
(ii) Demonstrate that the game is not stable.
The pay-off for the cell in row Y , column P is changed from $( 5,2 )$ to $( y , p )$, where $y$ and $p$ are real numbers.\\
(iii) What is the largest set of values $A$, so that if $y \in A$ then row Y is dominated by another row?\\
(iv) Explain why column P can never be redundant because of dominance.
\hfill \mbox{\textit{OCR Further Discrete AS 2018 Q3 [6]}}