Consider the following decision table with rewards in $ millions. $\begin{array}{r}\text { State of Nature }\\\begin{array}{|c|c|c|c|}\hline \text { Decision Alternatives } & S_{1} & S_{2} & S_{3} \\\hline d_{1} & -1 & 2 & 8 \\\hline d_{2} & -3 & 7 & 5 \\\hline d_{3} & -0.5 & 0.75 & 1 \\\hline d_{4} & 0 & 0 & 0 \\\hline d_{5} & -1 & -1 & -1\\\hline\end{array}\end{array}$
Using the Hurwicz criterion with alpha = 0.1, the appropriate choice would be ________.

Question

Consider the following decision table with rewards in $ millions.

\begin{array}{r}\text { State of Nature }\\\begin{array}{|c|c|c|c|}\hline \text { Decision Alternatives } & S_{1} & S_{2} & S_{3} \\\hline d_{1} & -1 & 2 & 8 \\\hline d_{2} & -3 & 7 & 5 \\\hline d_{3} & -0.5 & 0.75 & 1 \\\hline d_{4} & 0 & 0 & 0 \\\hline d_{5} & -1 & -1 & -1\\\hline\end{array}\end{array}

Using the Hurwicz criterion with alpha = 0.1, the appropriate choice would be ________.

Quizplus · Accepted Answer

The Hurwicz criterion is a decision-making rule that combines the optimism and pessimism of the decision-maker by considering both the best and worst outcomes of each decision alternative. It uses a coefficient of optimism, alpha ($\alpha$), to weigh the best outcome and (1-$\alpha$) to weigh the worst outcome for each alternative. Given $\alpha = 0.1$, we calculate the Hurwicz value for each decision alternative as follows:- For $d_1$: $0.1 \times 8 + 0.9 \times (-1) = 0.8 - 0.9 = -0.1$- For $d_2$: $0.1 \times 7 + 0.9 \times (-3) = 0.7 - 2.7 = -2$- For $d_3$: $0.1 \times 1 + 0.9 \times (-0.5) = 0.1 - 0.45 = -0.35$- For $d_4$: $0.1 \times 0 + 0.9 \times 0 = 0$- For $d_5$: $0.1 \times (-1) + 0.9 \times (-1) = -1$The decision alternative with the highest Hurwicz value is $d_4$ with a value of 0, making it the appropriate choice according to the Hurwicz criterion with $\alpha = 0.1$.

Consider the Following Decision Table with Rewards in $ Millions