A2 October 2020 Q6
6.
| Player B | ||||
|---|---|---|---|---|
| Option X | Option Y | Option Z | ||
| Player A | Option Q | \(1\) | \(5\) | \(3\) |
| Option R | \(4\) | \(-3\) | \(1\) | |
| Option S | \(2\) | \(-4\) | \(-2\) | |
| Option T | \(3\) | \(-2\) | \(0\) | |
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 Q, R and T, choosing option Q with probability \(p_1\), option R 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 linear programme, writing the constraints as inequalities.
Maximise \(P = V\), where \(V =\) the value of original game \(+ 3\)
\[\begin{array}{ll} \text{subject to} & V \leqslant 4p_1 + 7p_2 + 6p_3 \\ & V \leqslant 8p_1 + p_3 \\ & V \leqslant 6p_1 + 4p_2 + 3p_3 \\ & p_1 + p_2 + p_3 \leqslant 1 \\ & p_1 \geqslant 0,\ p_2 \geqslant 0,\ p_3 \geqslant 0,\ V \geqslant 0 \end{array}\]The Simplex algorithm is used to solve the linear programming problem.
Given that the optimal value of \(p_1 = \dfrac{7}{11}\) and the optimal value of \(p_3 = 0\)
Player B intends to make a random choice between options X, Y and Z, choosing option X with probability \(q_1\), option Y with probability \(q_2\) and option Z with probability \(q_3\)
| Scheme | Marks | AO |
|---|---|---|
| Option R (or option T) dominates option S | B1 | 1.2 |
| Because e.g. \(4 > 2\) and \(-3 > -4\) and \(1 > -2\) | B1 | 2.4 |
| (2) |
Notes
B1: correct statement – must include the word ‘dominate’ (note that T dominates S too)
B1: correct inequalities – must be clear that all inequalities must hold
| Scheme | Marks | AO |
|---|---|---|
| Row minima: \(1, -3, -2\) max is 1 | M1 | 1.1b |
| Column maxima: \(4, 5, 3\) min is 3 | A1 | 1.1b |
| Row maximin (1) \(\neq\) Column minimax (3) so not stable | A1 | 2.4 |
| (3) |
Notes
M1: attempt at row minima and column maxima – condone one error
A1: correct max(row min) and min(col max)
A1: correct reasoning that the game is not stable (accept \(1 \neq 3\) + statement)
| Scheme | Marks | AO |
|---|---|---|
| \(V\) is less than or equal to each of these three expressions since we need to find the maximum value of the worst possible augmented expected pay-off for each value of \(p\) | B1 | 2.3 |
| (1) |
Notes
B1: an understanding that for each value of \(p\) we are seeking the minimum possible output
| Scheme | Marks | AO |
|---|---|---|
| It is necessary to use an inequality because it enables the Simplex algorithm to pivot on a row that will increase the value of \(P\) | B1 | 3.5a |
| (1) |
Notes
B1: as a minimum accept an answer that implies that an inequality is required so that we can apply the Simplex algorithm
| Scheme | Marks | AO |
|---|---|---|
| \(p_2 = \dfrac{4}{11}\) | B1 | 1.1b |
| Substitute \(p\) values to obtain \(V \leqslant \dfrac{56}{11}, \dfrac{56}{11}, \dfrac{58}{11}\) | M1 | 3.4 |
| Value of the game to player A \(= \dfrac{56}{11} - 3 = \dfrac{23}{11}\) | A1 | 2.2a |
| (3) |
Notes
B1: cao
M1: substitute their \(p\) values into all three expressions for the upper bound of \(V\)
A1: cao for the value of the game to player A
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{aligned} q_1 + 5q_2 + 3q_3 &= \dfrac{23}{11} \\ 4q_1 - 3q_2 + q_3 &= \dfrac{23}{11} \\ q_1 + q_2 + q_3 &= 1 \end{aligned}\) | M1 A1ft A1 | 3.1a 1.1b 1.1b |
| Player B should play option X with probability \(\dfrac{8}{11}\), option Y with probability \(\dfrac{3}{11}\) and never play option Z | A1 | 3.2a |
| (4) | ||
| (14 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 (e)
A1ft: Two correct ft “their” \(V\)
A1: cao (for exactly three equations correct)
A1: cao in context