There are only two possible outcomes, so the probability of failure is \(P(f) = 1 - P(s) = 3/5\). Thus, \[E = w_sp_s + w_fp_f = \frac{2}{5} \times 55000 + \frac{3}{5} \times (-1750) = 20950\]
The expected values of the two policies are based on the temperature and the profits: \[\begin{aligned} E(cola) &= 0.3 \times 1500 + 0.7 \times 5000 = 3950\\ E(coffee) &= 0.3 \times 4000 + 0.7 \times 1000 = 1900 \end{aligned}\] The expected value of selling cola is higher; therefore the firm should purchase cola. This makes intuitive sense, as cola has a higher profit under the more likely weather scenario. This is further illustrated in the decistion tree below.
Basedon the on the decision tree presented below, the resort should be operated. This decision has an expected value of $116,000.
If oil is found, the net profit is $6M - $1M = $5M. If no oil is found, a -$1M net profit (i.e. a $1M loss) is realized. If a geologist is hired to perform testing, each of these profit numbers is lowered by $0.1M (resulting in a $4.9M profit and $1.1M loss). This information is reflected in teh decision tree below:
As indicated in the decision tree, the oil company should hire a geologist to perform testing. It is indicated that they should drill – this is because the expected value (profit) is greater than 0.
The expected values are given by \(\sum w_ip_i\) for the table: \[\begin{aligned} E(A) &= 0.35 \times 1100 + 0.3 \times 900 + 0.25 \times 400 + 0.1 \times 300 = 785\\ E(B) &= 0.35 \times 850 + 0.3 \times 1500 + 0.25 \times 1000 + 0.1 \times 500 = 1047.50\\ E(A) &= 0.35 \times 700 + 0.3 \times 1200 + 0.25 \times 500 + 0.1 \times 900 = 820 \end{aligned}\]
The highest expected value is for alternative B, so that should be chosen if the criteria is maximized expected value.
The regret table is composed by selecting each entry from the column maximum to get the regret under each state of nature. The expected regret is then calculated by \(\sum r_ip_i\):
# enter payoffs and probabilities
X <- matrix(
c(1100, 900, 400, 300,
850, 1500, 1000, 500,
700, 1200, 500, 900),
nrow = 3, byrow = TRUE)
p <- c(0.35, 0.3, 0.25, 0.1)
# get regret matrix
reg <- apply(X, 2, function(x) {max(x) - x})
# calculate expected regrets
reg <- cbind(reg, apply(reg, 1, function(x) {sum(x * p)}))
| 1 | 2 | 3 | 4 | Expected Regret | |
|---|---|---|---|---|---|
| A | 0 | 600 | 600 | 600 | 390 |
| B | 250 | 0 | 0 | 400 | 127.5 |
| C | 400 | 300 | 500 | 0 | 355 |
The lowest expected regret occurs for alternative B, so this should also be selected if the criteria is minimized expected regret.