A2 June 2019 Q4
4.
| Player B | ||||
|---|---|---|---|---|
| Option X | Option Y | Option Z | ||
| Player A | Option P | \(3\) | \(-2\) | \(0\) |
| Option Q | \(-4\) | \(4\) | \(-2\) | |
| Option R | \(1\) | \(2\) | \(-1\) | |
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 P, Q and R, choosing option P with probability \(p_1\), option Q with probability \(p_2\) and option R 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 programming problem for the game, writing the constraints as inequalities.
Maximise \(P = V\)
\[\begin{array}{ll} \text{subject to} & V \geqslant 3p_1 - 4p_2 + p_3 \\ & V \geqslant -2p_1 + 4p_2 + 2p_3 \\ & V \geqslant -2p_2 - p_3 \\ & p_1 + p_2 + p_3 \leqslant 1 \\ & p_1 \geqslant 0,\ p_2 \geqslant 0,\ p_3 \geqslant 0,\ V \geqslant 0 \end{array}\]The Simplex algorithm is used to solve the corrected linear programming problem.
The optimal values are \(p_1 = 0.6\), \(p_2 = 0\) and \(p_3 = 0.4\)
| Scheme | Marks | AO |
|---|---|---|
| Row minima: \(-2, -4, -1\) max is \(-1\) | 1M1 | 1.2 |
| Column maxima: \(3, 4, 0\) min is 0 | ||
| Row maximin \((-1) \neq\) Column minimax (0) so not stable | 1A1 | 2.4 |
| (2) |
Notes
1M1: attempt at row minima and column maxima – condone one error
1A1: correct reasoning that the game is not stable (accept “\(-1 \neq 0\)” + statement) – dependent on correct row minima and column maxima
| Scheme | Marks | AO |
|---|---|---|
| As the value of \(V\) must be non-negative the coefficients of the three inequalities involving \(V\) must be non-negative so add at least 4 to each value \(\left[\text{e.g. adding 5 gives } \begin{pmatrix} 8 & 3 & 5 \\ 1 & 9 & 3 \\ 6 & 7 & 4 \end{pmatrix}\right]\) | 1B1 | 2.3 |
| Furthermore, as \(V\) is the minimum that A can expect to win, the constraints should be \(V \leqslant \ldots\) | 2B1 | 2.3 |
| e.g. ‘adding 5’ \(V \leqslant 8p_1 + p_2 + 6p_3\) \(V \leqslant 3p_1 + 9p_2 + 7p_3\) \(V \leqslant 5p_1 + 3p_2 + 4p_3\) e.g. ‘adding 4’ \(V \leqslant 7p_1 + 5p_3\) \(V \leqslant 2p_1 + 8p_2 + 6p_3\) \(V \leqslant 4p_1 + 2p_2 + 3p_3\) | 3B1 | 1.1b |
| (3) |
Notes
1B1: Indicates that coefficients are incorrect because \(V\) must be non-negative. Must convey both underlined aspects. Condone ‘positive’ for ‘non-negative’
2B1: Indicates that inequality signs are the wrong way because \(V\) is the minimum (so expected winnings are \(\geqslant V\)). Must convey both aspects but give bod.
3B1: CAO (all three inequalities). Give this mark for correct equations with slack variables provided 2B1 has been awarded.
| Scheme | Marks | AO | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
‘adding 5’
| 1B1 1M1 1A1 | 1.2 3.3 1.1b | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| (3) |
Notes
1B1: All row and column labels correct for Simplex tableau
1M1: Setting up the Simplex model - any two of my ‘\(r\)’, ‘\(s\)’ or ‘\(t\)’ rows correct. Or a completely correct answer with either one column or one row missing – condone lack of basic variable column. Should follow from changed constraints of the correct form from b). So, constraints must have been of the form \(V \leqslant ap_1 + bp_2 + cp_3\ (a, b, c \geqslant 0)\) o.e.
1A1: CAO on numerical values.
Note: The B mark is for labelling the simplex tableau correctly; the M and A marks are for values only.
| Scheme | Marks | AO |
|---|---|---|
| Substitute \(p\) values to obtain \(V \leqslant 4.6\) | 1M1 | 3.4 |
| Value of the game to player A \(= 4.6 - 5 = -0.4\) | 1A1 | 2.2a |
| (2) |
Notes
1M1: substitutes their \(p\) values into all three expressions for the upper bound of \(V\). Condone use of an equals sign here. But not ‘\(V \geqslant \ldots\)’
May see one of:
- \(V \leqslant 3(0.6) - 4(0) + 0.4 = 2.2\ \{= \tfrac{11}{5}\};\ V \leqslant -2(0.6) - 4(0) + 2(0.4) = -0.4\ \{= -\tfrac{2}{5}\};\ V \leqslant -2(0) - (0.4) = -0.4\ \{= -\tfrac{2}{5}\}\)
- \(V \leqslant 8(0.6) + (0) + 6(0.4) = 7.2\ \{= \tfrac{36}{5}\};\ V \leqslant 3(0.6) + 9(0) + 7(0.4) = 4.6\ \{= \tfrac{23}{5}\};\ V \leqslant 5(0.6) + 3(0) + 4(0.4) = 4.6\ \{= \tfrac{23}{5}\}\)
- \(V \leqslant 7(0.6) + 5(0.4) = 6.2\ \{= \tfrac{31}{5}\};\ V \leqslant 2(0.6) + 8(0) + 6(0.4) = 3.6\ \{= \tfrac{18}{5}\};\ V \leqslant 4(0.6) + 2(0) + 3(0.4) = 3.6\ \{= \tfrac{18}{5}\}\)
1A1: CAO for the value of the game to player A
| Scheme | Marks | AO |
|---|---|---|
| Player B’s choices are options Y and Z only | 1B1 | 3.4 |
| Either \(2q = 0.4\) or \(-2q + 1(1 - q) = 0.4\) (where \(q\) is the probability Player B plays their option Y and \((1 - q)\) is the corresponding probability for option Z) | 1M1 | 3.1a |
| \(q = 0.2\) | 1A1 | 1.1b |
| Player B should never play their option X, they should play their option Y with probability 0.2 and option Z with probability 0.8 | 2A1ft | 3.2a |
| (4) | ||
| (14 marks) |
Notes
1B1: CAO – uses the model to determine that Player B only plays Y and Z
1M1: A correct equation for B (where value of game to B = – 1 x their value of game to A)
1A1: CAO for \(q\) (the probability that B plays Y)
2A1ft: Correct optimal strategy in context (not just in terms of \(q\)) following through their \(q\)
Alternative for (e)
1B1: CAO – uses the model to determine that Player B only plays Y and Z
1M1: Formulates two correct expressions for the expected value of the game to B and finds the intersection: \(2q = -2q + 1(1 - q)\ (= 1 - 3q)\)
1A1: As above
2A1ft: As above