12.5 Practice Problems
Problem 12.5.1. In Example 12.1.1, for which values of the Hurwicz coefficient \(\alpha \) is the large plant preferred to the medium one?
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Solution. The Hurwicz values are \(320\alpha - 120\) (large), \(150\alpha - 30\) (medium) and \(40\alpha + 20\) (small). Large beats medium when \(320\alpha - 120 > 150\alpha - 30\), that is \(\alpha > \frac 9{17} = 0.529\). Over the whole range: small beats medium when \(\alpha < \frac 5{11} = 0.455\) and beats large when \(\alpha < \frac 12\), so the small plant is chosen for \(\alpha < 0.455\), the medium for \(0.455 < \alpha < 0.529\), and the large for \(\alpha > 0.529\), which includes the \(\alpha = 0.6\) of Software 12.4.1.
Problem 12.5.2. Verify that in Example 12.1.1 the expected opportunity loss of the medium plant equals the EVPI.
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Solution. The regrets of the medium plant are \(200 - 120 = 80\), \(80 - 80 = 0\) and \(20 - (-30) = 50\); their expectation is \(0.3 \times 80 + 0.5 \times 0 + 0.2 \times 50 = 34\), the EVPI. The best action always has expected regret equal to the EVPI.
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Solution. Row minima \(1, 3\), maximin \(3\); column maxima \(5, 4\), minimax \(4\): no saddle point. With \(a - b - c + d = 5 - 1 - 3 + 4 = 5\): \(p_1 = \frac {4 - 3}5 = \frac 15\), \(q_1 = \frac {4 - 1}5 = \frac 35\), value \(\frac {20 - 3}5 = 3.4\).
Problem 12.5.4. Use dominance to reduce and solve \(\begin {pmatrix} 3 & 2 & 4 \\ 1 & 4 & 2 \\ 2 & 1 & 3 \end {pmatrix}\).
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Solution. Row \(3\) is dominated by row \(1\) (\(3 \geq 2\), \(2 \geq 1\), \(4 \geq 3\)) and is deleted. In the remaining rows column \(3\), \((4, 2)\), is dominated by column \(1\), \((3, 1)\), which the column player prefers, and is deleted. The game \(\begin {pmatrix} 3 & 2 \\ 1 & 4 \end {pmatrix}\) has no saddle point (maximin \(2\), minimax \(3\)), and \(p_1 = \frac {4 - 1}4 = \frac 34\), \(q_1 = \frac {4 - 2}4 = \frac 12\), value \(\frac {12 - 2}4 = 2.5\).
Problem 12.5.5. A decision maker with utility \(u(x) = \sqrt x\) can take Rs. 100 for certain or a gamble paying Rs. 250 or nothing with equal probability. Which does she choose, and what is the gamble’s certainty equivalent?
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Solution. The gamble’s expected value is \(125\), but its expected utility is \(\frac 12\sqrt {250} = 7.91 < \sqrt {100} = 10\), so she takes the sure Rs. 100. The certainty equivalent is \(7.91^2 = 62.5\): she would exchange the gamble for any sure amount above Rs. 62.5. The difference \(125 - 62.5\) is her risk premium.
Problem 12.5.6. Show that a pure strategy that is used with positive probability in an optimal mixed strategy earns exactly the value of the game against the opponent’s optimal strategy.
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Solution. Let \(p\), \(q\) be optimal and \(v\) the value. Each row earns \((Aq)_i \leq v\), by optimality of \(q\). Then \(v = p^{\mathsf T}Aq = \sum _ip_i(Aq)_i \leq \sum _ip_iv = v\), so equality holds term by term: \(p_i > 0\) forces \((Aq)_i = v\). This is complementary slackness for the pair of LPs in the proof of Theorem 12.3.1.
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