A2 June 2022 Q7
7.
| Player B | |||||
|---|---|---|---|---|---|
| Option W | Option X | Option Y | Option Z | ||
| Player A | Option Q | \(4\) | \(3\) | \(-1\) | \(-2\) |
| Option R | \(-3\) | \(5\) | \(-4\) | \(k\) | |
| Option S | \(-1\) | \(6\) | \(3\) | \(-3\) | |
A two person zero-sum game is represented by the pay-off matrix for player A shown above. It is given that \(k\) is an integer.
Given that Z is the play-safe option for player B,
Player A intends to make a random choice between options Q, R and S, choosing option Q with probability \(p_1\), option R with probability \(p_2\) and option S with probability \(p_3\)
Player A wants to find the optimal values of \(p_1\), \(p_2\) and \(p_3\) using the Simplex algorithm.
Given that \(k > -4\), player A formulates the following objective function for the corresponding linear program.
Maximise \(P = V\), where \(V =\) the value of the original game \(+\, 4\)
The Simplex algorithm is used to solve the linear programming problem. It is given that in the final Simplex tableau the optimal value of \(p_1 = \dfrac{7}{37}\), the optimal value of \(p_2 = \dfrac{17}{37}\) and all the slack variables are zero.
| Scheme | Marks | AO |
|---|---|---|
| Row minima are \(-2\), \(\min(k, -4)\) and \(-3\) | M1 | 1.1b |
| e.g., If the row minimum for option R is \(-4\) then the play safe is Q (as \(-2\) is greater than \(-4\) and \(-3\)) If the row minimum for option R is \(k\) then \(-4 > k\) and the play safe is still Q | A1 | 2.3 |
| (2) |
Notes
M1: Attempt to calculate row minima. Condone ‘\(k\) or \(-4\)’ for \(\min(k, -4)\).
A1: Correct argument/conditions for why the play-safe for player A is always their option Q
| Scheme | Marks | AO |
|---|---|---|
| Column maxima are 4, 6, 3 and \(\max(k, -2)\) | B1 | 1.1b |
| Since the play-safe is option Z, \(k < 3\) or \(k \leqslant 2\) | B1 | 2.2a |
| (2) |
Notes
B1: Attempt to calculate column maxima. Condone ‘\(k\) or \(-2\)’ for \(\max(k, -2)\)
B1: CAO ignore lower limit figure, if given
| Scheme | Marks | AO |
|---|---|---|
| e.g. Option Y dominates option X | B1 | 1.2 |
| Because e.g. \(-1 < 3\), \(-4 < 5\) and \(3 < 6\) | B1 | 2.4 |
| (2) |
Notes
B1: Correct statement – must include the word ‘dominate’. Also e.g. option Z dominates option X
B1: Correct inequalities – must be clear that all three inequalities must hold
| Scheme | Marks | AO | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| (i) \(\begin{pmatrix} 4 & -1 & -2 \\ -3 & -4 & k \\ -1 & 3 & -3 \end{pmatrix} \to \begin{pmatrix} 8 & 3 & 2 \\ 1 & 0 & k+4 \\ 3 & 7 & 1 \end{pmatrix}\) | B1 | 1.1b | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| \(V - 8p_1 - p_2 - 3p_3 + r = 0\) \(V - 3p_1 - 7p_3 + s = 0\) \(V - 2p_1 - (k + 4)p_2 - p_3 + t = 0\) | M1 A1 | 3.3 2.5 | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| \(p_1 + p_2 + p_3 + u = 1\) | B1 | 3.3 | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
(ii) e.g.
| B1 M1 A1 | 1.2 3.3 1.1b | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| (7) |
Notes
(d)(i) B1: Correct augmentation – possibly implied by later working, X column may be included
M1: At least three equations in \(V, p_1, p_2, p_3\) and at least one dummy variable seen
A1: CAO (ignore extra probability equation, if seen, see note below)
B1: Correct probability equation
(ii) B1: Correct row and column labels for Simplex tableau
M1: Any one (numerical in nature) row correct
A1: CAO
Special case
(d) If augmentation with +5 maximum marks possible B0 M1 A1 B1 B1 M1 A0
Note: (d)(i) A1 The extra probability equation you may see is \(V - 7p_1 - 9p_2 - 10p_3 + \text{slack} = 0\)
| Scheme | Marks | AO |
|---|---|---|
| \(p_3 = \dfrac{13}{37}\) | B1 | 1.1b |
| \(V - 8\left(\dfrac{7}{37}\right) - \left(\dfrac{17}{37}\right) - 3\left(\dfrac{13}{37}\right) + 0 = 0\) or \(V - 3\left(\dfrac{7}{37}\right) - 7\left(\dfrac{13}{37}\right) + 0 = 0\) | M1 | 3.1a |
| \(V = \dfrac{112}{37} \Rightarrow \dfrac{112}{37} - 2\left(\dfrac{7}{37}\right) - (k + 4)\left(\dfrac{17}{37}\right) - \left(\dfrac{13}{37}\right) + 0 = 0\) | dM1 | 3.4 |
| \(k = 1\) | A1 | 2.2a |
| (4) | ||
| (17 marks) |
Notes
B1: \(p_3\) correctly stated
M1: Attempts to calculate \(V\) using either equation from (d) not involving \(k\), or attempts to eliminate \(V\) from two equations, one involving \(k\)
dM1: Dependent on previous M mark – either uses equation in \(k\) and their \(V\) to calculate \(k\), or eliminates \(V\) from two equations to calculate \(k\)
A1: CAO \(k = 1\)