A2 June 2024 Q7
7.
| Player B | ||||
|---|---|---|---|---|
| Option X | Option Y | Option Z | ||
| Player A | Option R | \(3\) | \(2\) | \(-3\) |
| Option S | \(4\) | \(-2\) | \(1\) | |
| Option T | \(-1\) | \(3\) | \(6\) | |
A two person zero-sum game is represented by the pay-off matrix for player A, shown above.
Player A intends to make a random choice between options R, S and T, choosing option R with probability \(p_1\), option S with probability \(p_2\) and option T with probability \(p_3\)
Player A wants to find the optimal values of \(p_1\), \(p_2\) and \(p_3\) using the Simplex algorithm.
Player A formulates the following objective function for the corresponding linear programme.
\[\text{Maximise } P = V \qquad \text{where } V = \text{the value of the game} + 3\]After several iterations of the Simplex algorithm, a possible final tableau is
| b.v. | \(V\) | \(p_1\) | \(p_2\) | \(p_3\) | \(r\) | \(s\) | \(t\) | \(u\) | Value |
|---|---|---|---|---|---|---|---|---|---|
| \(p_3\) | \(0\) | \(0\) | \(0\) | \(1\) | \(\dfrac{1}{10}\) | \(-\dfrac{3}{80}\) | \(-\dfrac{1}{16}\) | \(\dfrac{33}{80}\) | \(\dfrac{33}{80}\) |
| \(p_2\) | \(0\) | \(0\) | \(1\) | \(0\) | \(-\dfrac{1}{10}\) | \(\dfrac{13}{80}\) | \(-\dfrac{1}{16}\) | \(\dfrac{17}{80}\) | \(\dfrac{17}{80}\) |
| \(V\) | \(1\) | \(0\) | \(0\) | \(0\) | \(\dfrac{1}{2}\) | \(\dfrac{5}{16}\) | \(\dfrac{3}{16}\) | \(\dfrac{73}{16}\) | \(\dfrac{73}{16}\) |
| \(p_1\) | \(0\) | \(1\) | \(0\) | \(0\) | \(0\) | \(-\dfrac{1}{8}\) | \(\dfrac{1}{8}\) | \(\dfrac{3}{8}\) | \(\dfrac{3}{8}\) |
| \(P\) | \(0\) | \(0\) | \(0\) | \(0\) | \(\dfrac{1}{2}\) | \(\dfrac{5}{16}\) | \(\dfrac{3}{16}\) | \(\dfrac{73}{16}\) | \(\dfrac{73}{16}\) |
Player B intends to make a random choice between options X, Y and Z.
| Scheme | Marks | AO |
|---|---|---|
| Row minima: \(-3, -2, -1\) (max is \(-1\)) Column maxima: 4, 3, 6 (min is 3) | M1 | 1.1b |
| Row maximin \((-1) \neq\) Column minimax (3) (so not stable) | A1 | 2.4 |
| (2) |
Notes
M1: Attempt to calculate row minima and column maxima – condone one error (note row max are 3, 4, 6 and column min are -1, -2, -3 so we must see where values come from)
A1: Correct reasoning that the game is not stable (accept \(-1 \neq 3\)) – dependent on correct row maximin and column minimax
| Scheme | Marks | AO | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| \(\begin{pmatrix} 3 & 2 & -3 \\ 4 & -2 & 1 \\ -1 & 3 & 6 \end{pmatrix} \rightarrow \begin{pmatrix} 6 & 5 & 0 \\ 7 & 1 & 4 \\ 2 & 6 & 9 \end{pmatrix}\) | B1 | 1.1b | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| \(V - 6p_1 - 7p_2 - 2p_3 + r = 0\) \(V - 5p_1 - p_2 - 6p_3 + s = 0\) \(V - 4p_2 - 9p_3 + t = 0\) \(p_1 + p_2 + p_3 + u = 1\) \((P - V = 0)\) | M1 A1 | 2.1 2.5 | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| M1 A1 | 3.3 2.2a | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| (5) |
Notes
(b) Note – a fully correct tableau implies all marks in (b)
B1: Correct augmentation – possibly implied by later working in tableau
M1: At least three equations in \(V, p_1, p_2, p_3\) and at least one dummy variable seen (must be using columns)
A1: CAO for all four equations (possibly implied by later working in tableau)
M1: Any two (numerical in nature) row correct (ignore labelling of b.v. column)
A1: CAO
| Scheme | Marks | AO |
|---|---|---|
| A should play R with probability \(\dfrac{3}{8}\), option S with probability \(\dfrac{17}{80}\) and option T with probability \(\dfrac{33}{80}\) | B1 | 3.2a |
| Value of the game to player A is \(\dfrac{73}{16} - 3\) | M1 | 1.1b |
| So, value of the game to player B is \(-\dfrac{25}{16}\) | A1 | 2.2a |
| (3) |
Notes
B1: Correct optimal strategy in context (dependent on both M marks in (b))
M1: For \(\pm\left(\dfrac{73}{16} \pm 3\right)\)
A1: CAO
| Scheme | Marks | AO |
|---|---|---|
| \(6q_1 + 5q_2 = 4.5625\) \(7q_1 + q_2 + 4q_3 = 4.5625\) \(2q_1 + 6q_2 + 9q_3 = 4.5625\) \(q_1 + q_2 + q_3 = 1\) or \(3q_1 + 2q_2 - 3q_3 = 1.5625\) \(4q_1 - 2q_2 + q_3 = 1.5625\) \(-q_1 + 3q_2 + 6q_3 = 1.5625\) \(q_1 + q_2 + q_3 = 1\) | M1 A1 | 2.1 1.1b |
| B should play X with probability \(\dfrac{1}{2}\), option Y with probability \(\dfrac{5}{16}\) and option Z with probability \(\dfrac{3}{16}\) | A1 | 3.2a |
| (3) | ||
| (13 marks) |
Notes
M1: Attempt to set up at least three equations in \(q_1, q_2, q_3\) using the value of the game from (c)
A1: CAO (for any three of the four correct equations)
A1: CAO in context (must have at least three correct equations)