Part 2 TKM0844 11E IM Ch13
Chapter 13
Risk Analysis and Project Evaluation
13-1.
Crusik Distribution Company thinks that there are two possible outcomes for its new facial care product: Either it will be very successful, or customers will not appreciate its “unique appeal.” The two outcomes are equally likely, but the successful outcome obviously comes with higher revenues. We can picture the situation like this:
50%
40%
30%
20%
10%
0%
$1,000,000
$5,000,000
Thus Crusik’s revenues will be either $1M or $5M. The expected value summarizes these two outcomes by weighting each by its probability of occurring: expected revenues = (revenues if highly successful) ∗ (probability of success) +
(revenues if less successful) ∗ (probability of success)
= ($5M) ∗ (50%) + ($1M) ∗ (50%)
= $3M.
In this case, the expected outcome is exactly between the two possible outcomes; this is the same result we would have gotten if we had simply found the arithmetic average. However, this only occurred because the two outcomes were equally likely. (For example, we will not get this result in Problem 13-2, where the three outcomes do not each have a 1/3 chance of happening.)
We can visualize the calculations using the spreadsheet below:
outcome extremely successful not as accepted
A
B
C = A*B
probability
50%
50%
annual sales $5,000,000
$1,000,000
expected value =
prob*sales
$2,500,000
$500,000
$3,000,000
©2011 Pearson Education, Inc. Publishing as Prentice Hall
Solutions to End of Chapter Problems—Chapter 13
13-2.
355
Peterson Trucking estimates that Armour Transport’s cash flow next year will be −$50,000,
$150,000, or $250,000, with the following probabilities:
50%
40%
30%
20%
10%
0%
($50,000)
$150,000
$250,000
A. To find the expected value, we will weigh each of the possible outcomes by its associated probability, then add the products. We show this calculation using the spreadsheet below:
state of economy recession normal expansion A
B
C = A*B
probability
30%
50%
20%
cash flow
Risk Analysis and Project Evaluation
13-1.
Crusik Distribution Company thinks that there are two possible outcomes for its new facial care product: Either it will be very successful, or customers will not appreciate its “unique appeal.” The two outcomes are equally likely, but the successful outcome obviously comes with higher revenues. We can picture the situation like this:
50%
40%
30%
20%
10%
0%
$1,000,000
$5,000,000
Thus Crusik’s revenues will be either $1M or $5M. The expected value summarizes these two outcomes by weighting each by its probability of occurring: expected revenues = (revenues if highly successful) ∗ (probability of success) +
(revenues if less successful) ∗ (probability of success)
= ($5M) ∗ (50%) + ($1M) ∗ (50%)
= $3M.
In this case, the expected outcome is exactly between the two possible outcomes; this is the same result we would have gotten if we had simply found the arithmetic average. However, this only occurred because the two outcomes were equally likely. (For example, we will not get this result in Problem 13-2, where the three outcomes do not each have a 1/3 chance of happening.)
We can visualize the calculations using the spreadsheet below:
outcome extremely successful not as accepted
A
B
C = A*B
probability
50%
50%
annual sales $5,000,000
$1,000,000
expected value =
prob*sales
$2,500,000
$500,000
$3,000,000
©2011 Pearson Education, Inc. Publishing as Prentice Hall
Solutions to End of Chapter Problems—Chapter 13
13-2.
355
Peterson Trucking estimates that Armour Transport’s cash flow next year will be −$50,000,
$150,000, or $250,000, with the following probabilities:
50%
40%
30%
20%
10%
0%
($50,000)
$150,000
$250,000
A. To find the expected value, we will weigh each of the possible outcomes by its associated probability, then add the products. We show this calculation using the spreadsheet below:
state of economy recession normal expansion A
B
C = A*B
probability
30%
50%
20%
cash flow