Chapter 20, Problem 17CE

a.

To determine

Construct a payoffs table.

Explanation of Solution

On observing the given information, it is clear that the lease about for a complete part set for summer season is $560. The rental amount per day is $5. Profit for each set for the first season is $10.

If the firm leased 41 complete sets, the profit for only 41 sets rented is shown below:

$\begin{array}{c}\text{Profit for 41 sets}=\left\{\text{No}\text{. of sets rented-}\left[\left\{\begin{array}{l}\text{No}\text{. of sets leased}-\\ \text{No}\text{.of sets rented}\end{array}\right\}\times \text{rent per day}\right]\right\}\\ =41\times 10-\left(\left[41-41\right]\times 5\right)\\ =\$410\end{array}$

If the firm leased 42 complete sets, the profit for only 41 sets rented is shown below:

$\begin{array}{c}\text{Profit for 42 sets, 41 are rented}=41\times 10-\left(\left[42-41\right]\times 5\right)\\ =\$410-5\\ =\$405\end{array}$

In a similar way, the calculations are obtained.

The payoffs are computed as follows:

Event
Act	41	42	43	44	45	46
41	$410	$410	$410	$410	$410	$410
42	$405	$420	$420	$420	$420	$420
43	$400	$415	$430	$430	$430	$430
44	$395	$410	$425	$440	$440	$440
45	$390	$405	$420	$435	$450	$450
46	$385	$400	$415	$430	$445	$460

b.

To determine

Find the expected payoffs for leasing 41, 42, and 44 complete sets from the Boston firms .

Give the recommended decision.

Answer to Problem 17CE

The expected payoffs are given below:

Event	Expected payoffs
41$\left(S_1\right)$	$410
42$\left(S_2\right)$	$419.1
43$\left(S_3\right)$	$426.70
44$\left(S_4\right)$	$432.20
45$\left(S_5\right)$	$431.70
46$\left(S_6\right)$	$427.45

Explanation of Solution

The expected profit for 41 sets is shown below:

Event	Payoff value	Probability	Expected payoffs
41	$410	$0.06\left(=\frac{360}{6,000}\right)$	$\$24.6\left(=\$410\times 0.06\right)$
42	$410	$0.1\left(=\frac{600}{6,000}\right)$	$\$41\left(=\$410\times 0.1\right)$
43	$410	$0.14\left(=\frac{840}{6,000}\right)$	$\$57.4\left(=\$410\times 0.14\right)$
44	$410	$0.4\left(=\frac{2,400}{6,000}\right)$	$\$164\left(=\$410\times 0.4\right)$
45	$410	$0.25\left(=\frac{1,500}{6,000}\right)$	$\$102.5\left(=\$410\times 0.25\right)$
46	$410	$0.05\left(=\frac{300}{6,000}\right)$	$\$20.5\left(=\$410\times 0.05\right)$
Expected profit(${\displaystyle \sum \text{values}}$)			$410

The expected profit for 42 sets is shown below:

Event	Payoff value	Expected payoffs
41	$405	$\$24.3\left(=\$405\times 0.06\right)$
42	$420	$\$42\left(=\$420\times 0.1\right)$
43	$420	$\$58.8\left(=\$420\times 0.14\right)$
44	$420	$\$168\left(=\$420\times 0.4\right)$
45	$420	$\$105\left(=\$420\times 0.25\right)$
46	$420	$\$21\left(=\$420\times 0.05\right)$
Expected profit(${\displaystyle \sum \text{values}}$)		$419.10

The expected profit for 43 sets is shown below:

Event	Payoff value	Expected payoffs
41	$400	$\$24\left(=\$400\times 0.06\right)$
42	$415	$\$41.50\left(=\$415\times 0.1\right)$
43	$430	$\$60.20\left(=\$430\times 0.14\right)$
44	$430	$\$172\left(=\$430\times 0.4\right)$
45	$430	$\$107.50\left(=\$430\times 0.25\right)$
46	$430	$\$21.50\left(=\$430\times 0.05\right)$
Expected profit(${\displaystyle \sum \text{values}}$)		$426.70

Therefore, the expected profit for all the sets will be obtained in the same way.

The expected profit all the events is shown below:

Event	Expected payoffs
41$\left(S_1\right)$	$410
42$\left(S_2\right)$	$419.1
43$\left(S_3\right)$	$426.70
44$\left(S_4\right)$	$432.20
45$\left(S_5\right)$	$431.70
46$\left(S_6\right)$	$427.45

c.

To determine

Identify the most profitable alternative based on the expected daily profit.

Explanation of Solution

The expected profit for the event $S_4$ is greater when compared to other events.

Thus, the event $S_4$ would be recommended.

Hence, order 44 sets because the expected profit of $432.20 is the highest.

d.

To determine

Obtain the expected opportunity loss for 41, 42, and 43.

Answer to Problem 17CE

The expected opportunity loss table is obtained as follows:

	Act
	41$\left(S_1\right)$	42$\left(S_2\right)$	43$\left(S_3\right)$	44$\left(S_4\right)$	9$\left(S_5\right)$	45$\left(S_6\right)$
Expected opp. loss	$28.3	$19.2	$11.6	$6.10	$6.60	$10.85

Explanation of Solution

Opportunity loss:

$\text{Opportunity loss=}\left\{\begin{array}{l}\text{Difference between the best payoffs and }\\ \text{remaining payoffs}\text{.}\end{array}\right\}$

From the table in Part (a) if the Supply $A_1$, then the best payoff value is $410 when the state of nature $S_1$.

Then, the opportunity loss is obtained by taking the difference between $410 and $410 That is, $\$410-\$410=0$.

Thus, the opportunity loss for $A_1$, given a state of nature is $S_1$, is $0.

In the same way, the remaining values in the table are obtained. If the Supply $A_2$, then the value is $15 when the state of nature is $S_1$.

Then, the opportunity loss is obtained by taking the difference between $410 and $405, that is, $\$410-\$405=\$5$.

Thus, the opportunity loss for $A_2$, given a state of nature is $S_1$, is $5.

The opportunity loss table is obtained as follows:

Supply	Opportunity loss
	41$\left(S_1\right)$	42$\left(S_2\right)$	43$\left(S_3\right)$	44$\left(S_4\right)$	45$\left(S_5\right)$	46$\left(S_6\right)$
41$\left(A_1\right)$	$0	$10	$20	$30	$40	$50
42$\left(A_2\right)$	$5	$0	$10	$20	$30	$40
43$\left(A_3\right)$	$10	$5	$0	$10	$20	$30
44$\left(A_4\right)$	$15	$10	$5	$0	$10	$20
45$\left(A_5\right)$	$20	$15	$10	$5	$0	$10
46$\left(A_6\right)$	$25	$20	$15	$10	$5	$0

Then, the expected opportunity loss for 41 sets is calculated as follows:

$\begin{array}{c}\text{Expected}\text{opportunity}\text{loss}\text{for}\text{41 sets}=\left\{\begin{array}{l}P\left(S_1\right)R\left(A_1,S_1\right)+P\left(S_2\right)R\left(A_1,S_2\right)+\\ P\left(S_3\right)R\left(A_1,S_3\right)+P\left(S_4\right)R\left(A_1,S_4\right)+\\ P\left(S_5\right)R\left(A_1,S_5\right)+P\left(S_6\right)R\left(A_1,S_6\right)\end{array}\right\}\\ =\left\{\begin{array}{l}0\left(0.06\right)+10\left(0.1\right)+20\left(0.14\right)+\\ 30\left(0.4\right)+40\left(0.25\right)+50\left(0.05\right)\end{array}\right\}\\ =28.30\end{array}$

Thus, the expected opportunity loss for 41 sets is $28.3.

The expected opportunity loss for 42 sets is calculated as follows:

$\begin{array}{c}\text{Expected}\text{opportunity}\text{loss}\text{for}\text{42 sets}=\left\{\begin{array}{l}P\left(S_1\right)R\left(A_2,S_1\right)+P\left(S_2\right)R\left(A_2,S_2\right)+\\ P\left(S_3\right)R\left(A_2,S_3\right)+P\left(S_4\right)R\left(A_2,S_4\right)+\\ P\left(S_5\right)R\left(A_2,S_5\right)+P\left(S_6\right)R\left(A_2,S_6\right)\end{array}\right\}\\ =\left\{\begin{array}{l}5\left(0.06\right)+0\left(0.1\right)+10\left(0.14\right)+\\ 20\left(0.4\right)+30\left(0.25\right)+40\left(0.05\right)\end{array}\right\}\\ =19.2\end{array}$

Thus, the expected opportunity loss for 42 sets is $19.2.

The expected opportunity loss for 43 sets is calculated as follows:

$\begin{array}{c}\text{Expected}\text{opportunity}\text{loss}\text{for}\text{43 sets}=\left\{\begin{array}{l}P\left(S_1\right)R\left(A_3,S_1\right)+P\left(S_2\right)R\left(A_3,S_2\right)+\\ P\left(S_3\right)R\left(A_3,S_3\right)+P\left(S_4\right)R\left(A_3,S_4\right)+\\ P\left(S_5\right)R\left(A_3,S_5\right)+P\left(S_6\right)R\left(A_3,S_6\right)\end{array}\right\}\\ =\left\{\begin{array}{l}10\left(0.06\right)+5\left(0.1\right)+0\left(0.14\right)+\\ 10\left(0.4\right)+20\left(0.25\right)+30\left(0.05\right)\end{array}\right\}\\ =11.6\end{array}$

Thus, the expected opportunity loss for 42 sets is $11.6.

In the same way, the remaining values are obtained. The expected opportunity losses are shown below:

	Act
	41$\left(S_1\right)$	42$\left(S_2\right)$	43$\left(S_3\right)$	44$\left(S_4\right)$	9$\left(S_5\right)$	45$\left(S_6\right)$
Expected opp. loss	$28.3	$19.2	$11.6	$6.10	$6.60	$10.85

e.

To determine

Obtain the most profitable course of action to take based on the expected opportunity loss table.

State whether the decision made in Part (c) is agreed in this context.

Explanation of Solution

On observing the table in Part (d), the opportunity loss when ordering 44 sets is minimum. Therefore, ordering 44 sets is most profitable.

Both the results stated in Part (c) and Part (e) conclude with the same results. Therefore, the result obtained in Part (c) is agreed with result in this context.

f.

To determine

Obtain the expected value of perfect information.

Answer to Problem 17CE

The expected value of perfect information is $6.10.

Explanation of Solution

It is given that $P\left(S_1\right)=0.06$, $P\left(S_2\right)=0.1$, $P\left(S_3\right)=0.14$, $P\left(S_4\right)=0.4$, $P\left(S_5\right)=0.25$, and $P\left(S_6\right)=0.05$.

The above calculations show that the EMV for the ordering 44 sets is the maximum at $432.20.

In order to maximize the profit, it is appropriate to order 44.

The expected value of perfect information can be calculated using the following formula:

$\begin{array}{l}\text{Expected value of perfect information}\\ =\text{Expected value under conditions of certainty}\\ -\text{Expected value under conditions of uncertainty}\text{.}\end{array}$

The expected monetary values calculated above provide the expected values under conditions of uncertainty.

Now, under conditions of certainty, the maximum rent $\left(S_1\right)$ is $410 when 41 sets are ordered; the maximum rent $\left(S_2\right)$ is $420 when 42 sets are ordered; the maximum rent $\left(S_3\right)$ is $430 when 43 sets are ordered; the maximum rent $\left(S_4\right)$ is $440 when 44 sets are ordered; the maximum rent $\left(S_5\right)$ is $450 when 45 sets are ordered; and the maximum rent $\left(S_6\right)$ is $460 when 45 sets are ordered. The expected value of uncertainty is $432.20.

Using the probabilities, the expected value under conditions of certainty is calculated below:

$\begin{array}{c}\text{Expected value under conditions of certainty}=\left\{\begin{array}{l}410\left(0.06\right)+420\left(0.1\right)+430\left(0.14\right)+\\ 440\left(0.4\right)+450\left(0.25\right)+460\left(0.05\right)\end{array}\right\}\\ =\mathrm{438.3.}\end{array}$

Hence, the expected value of perfect information is calculated below:

$\begin{array}{c}\text{Expected value of perfect information}=438.3-432.20\\ =\$\mathrm{6.10.}\end{array}$

Thus, the expected value of perfect information is $6.10.

On observing the evidence stated in the previous parts, the maximum amount to be paid for the perfect information is $6.10.