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CHAPTER 23
OPTIONS AND CORPORATE FINANCE:
EXTENSIONS AND APPLICATIONS
Answers to Concepts Review and Critical Thinking Questions
1. One of the purposes to giving stock options to CEOs (instead of cash) is to tie the performance of the
3. Virtually all projects have embedded options, which are ignored in NPV calculations and likely leads
to undervaluation.
4. As the volatility increases, the value of an option increases. As the volatility of coal and oil increases,
6. The company has an option to abandon the mine temporarily, which is an American put. If the option
7. Your colleague is correct, but the fact that increased volatility increases the value of an option is an
important part of option valuation. All else the same, a call option on a venture that has higher volatility
8. Real option analysis is not a technique that can be applied in isolation. The value of the asset in real
9. Insurance is a put option. Consider your homeowner’s insurance. If your house were to burn down,
10. In a market with competitors, you must realize that the competitors have real options as well. The
decisions made by these competitors may often change the payoffs for your company’s options. For
example, the first entrant into a market can often be rewarded with a larger market share because the
Solutions to Questions and Problems
NOTE: All end-of-chapter problems were solved using a spreadsheet. Many problems require multiple
steps. Due to space and readability constraints, when these intermediate steps are included in this solutions
manual, rounding may appear to have occurred. However, the final answer for each problem is found
without rounding during any step in the problem.
1. a. The inputs to the Black-Scholes model are the current price of the underlying asset (S), the strike
price of the option (E), the time to expiration of the option in fractions of a year (t), the variance
(2) of the underlying asset, and the continuously-compounded risk-free interest rate (R). Since
these options were granted at-the-money, the strike price of each option is equal to the current
value of one share, or $45. We can use Black-Scholes to solve for the option price. Doing so, we
find:
d1 = [ln(S/E) + (R + 2/2)(t) ]/(2t)1/2
d1 = [ln($45/$45) + (.06 + .582/2) (5)]/(.58
5
) = .8798
Now we can find the value of each option, which will be:
C = SN(d1) – Ee–RtN(d2)
C = $45(.8105) – ($45e–.06(5))(.3383)
2. The total compensation package consists of an annual salary in addition to 25,000 at-the-money stock
options. First, we will find the present value of the salary payments. Since the payments occur at the
end of the year, the payments can be valued as a three-year annuity, which will be:
Next, we can use the Black-Scholes model to determine the value of the stock options. Doing so, we
find:
Find N(d1) and N(d2), the area under the normal curve from negative infinity to d1 and negative
infinity to d2, respectively. Doing so:
N(d1) = .7313
Now we can find the value of each option, which will be:
C = SN(d1) – Ee–RtN(d2)
3. Since the contract is to sell up to 5 million gallons, it is a call option, so we need to value the contract
accordingly. Using the binomial mode, we will find the value of u and d, which are:
u = 1.3364
d = 1/u
d = 1/1.3364
d = .7483
This implies the percentage increase if gasoline increases will be 33.64 percent, and the percentage
decrease if prices fall will be –25.17 percent. So, the price in three months with an up or down move
will be:
The option is worthless if the price decreases. If the price increases, the value of the option per gallon
is:
Value with price increase = $2.35 – 1.95
Value with price increase = $.40
4. When solving a question dealing with real options, begin by identifying the option-like features of the
situation. First, since the company will exercise its option to build if the value of an office building
rises, the right to build the office building is similar to a call option. Second, an office building would
be worth $58.7 million today. This amount can be viewed as the current price of the underlying asset
$70.4
$5.4
=Max($70.4 – 65, 0)
$58.7
?
Risk-free rate = (ProbabilityRise)(ReturnRise) + (ProbabilityFall)(ReturnFall)
Risk-free rate = (ProbabilityRise)(ReturnRise) + (1 – ProbabilityRise)(ReturnFall)
.048 = (ProbabilityRise)(.1993) + (1 – ProbabilityRise)(–.0596)
ProbabilityRise = .4156
5. When solving a question dealing with real options, begin by identifying the option-like features of the
situation. First, since the company will only choose to drill and excavate if the price of oil rises, the
right to drill on the land can be viewed as a call option. Second, since the land contains 495,000 barrels
of oil and the current price of oil is $60 per barrel, the current price of the underlying asset (S) to be
N(d1) = .0831
N(d2) = .0297
Now we can find the value of the call option, which will be:
6. When solving a question dealing with real options, begin by identifying the option-like features of the
situation. First, since the company will only choose to manufacture the steel rods if the price of steel
falls, the lease, which gives the firm the ability to manufacture steel, can be viewed as a put option.
Second, since the firm will receive a fixed amount of money if it chooses to manufacture the rods:
Amount received = 175,000 steel rods($34 – 21)
Amount received = $2,275,000
The amount received can be viewed as the put option’s strike price (E). Third, since the project requires
the company to purchase 500 tons of steel and the current price of steel is $3,700 per ton, the current
price of the underlying asset (S) to be used in the Black-Scholes formula is:
Finally, since the company must decide whether or not to purchase the steel in six months, the firm’s
real option to manufacture steel rods can be viewed as having a time to expiration (t) of six months.
In order to calculate the value of this real put option, we can use the Black-Scholes model to determine
the value of an otherwise identical call option then infer the value of the put using put-call parity.
Using the Black-Scholes model to determine the value of the option, we find:
d1 = [ln(S/E) + (R + 2/2)(t) ]/(2t)1/2
Find N(d1) and N(d2), the area under the normal curve from negative infinity to d1 and negative
infinity to d2, respectively. Doing so:
N(d2) = .2060
Now we can find the value of the call option, which will be:
C = SN(d1) – Ee–RtN(d2)
7. In one year, the company will abandon the technology if the demand is low since the value of
abandonment is higher than the value of continuing operations. Since the company is selling the
technology in this case, the option is a put option. The value of the put option in one year if demand
is low will be:
Value of put with low demand = $11,800,000 – 9,400,000
Value of put with low demand = $2,400,000
Of course, if demand is high, the company will not sell the technology, so the put will expire worthless.
We can value the put with the binomial model. In one year, the percentage gain on the project if the
demand is high will be:
Percentage increase with high demand = ($17,800,000 – 14,300,000)/$14,300,000
Percentage increase with high demand = .2448, or 24.48%
8. Using the binomial model, we will find the value of u and d, which are:
u = 1.2239
d = 1/u
d = 1/1.2239
d = .8170
The following figure shows the stock price and put price for each possible move over the next two
months:
Stock price (D)
$122.84
Put price
$0
Stock price (B)
$100.36
Put price
$4.30
And the stock price at node (D) is two up moves, or:
Stock price (D) = $82(1.2239)(1.2239)
Stock price (D) = $122.84
The stock price at node (C) is from a down move, or:
Stock price (C) = $82(.8170)
put option at these nodes is the maximum of the strike price minus the stock price, or zero. So:
Put value (D) = Max($90 – 122.84, $0)
Put value (D) = $0
Put value (E) = Max($90 – 82, $0)
Put value (E) = $8
Put value (C) = [.4599($8) + .5401($35.26)]/1.0042
Put value (C) = $22.63
Using the put values at nodes (B) and (C), we can now find the value of the put today, which is:
9. Since the exercise style is now American, the option can be exercised prior to expiration. At node (B),
we would not want to exercise the put option since it would be out of the money at that stock price.
However, if the stock price falls next month, the value of the put option if exercised is:
Value if exercised = $90 – 67.00
10. Using the binomial model, we will find the value of u and d, which are:
d = 1/u
d = 1/1.2363
d = .8089
This implies the percentage increase if the stock price increases will be 23.63 percent, and the
month steps:
Value (D)
$90,595,230
Call price
$27,595,230
Value pre-payment
$74,178,667
Value post-payment (B)
$73,278,667
Value (G)
$38,527,093
Call price
$0
First, we need to find the building value at every step along the binomial tree. The building value at
node (A) is the current building value. The building value at node (B) is from an up move, which
means:
To find the building value at node (E), we multiply the after-payment building value at node (B) by
the down move, or:
Building value (E) = $73,278,667(.8089)
Building value (E) = $59,272,028
The building value at node (C) is from a down move, which means the building value will be:
Finally, the building value at node (G) is from a down move from node (C), so the building value is:
Building value (G) = $47,631,474(.8089)
Building value (G) = $38,527,093
Note that because of the accrued rent payment in six months, the binomial tree does not recombine
during the next step. This occurs whenever a fixed payment is made during a binomial tree. For
Call value (F) = Max($58,887,320 – 63,000,000, $0)
Call value (F) = $0
Call value (G) = Max($38,527,093 – 63,000,000, $0)
Call value (G) = $0
The value of the call at node (B) is the present value of the expected value. We find the expected value
the call option early. The reason is that the potential gain is unlimited. In contrast, the potential gain
on a put option is limited by the strike price, so it may be valuable to exercise an American put option
early if it is deep in the money.
We can value the call at node (C), which will be the present value of the expected value of the call at
nodes (F) and (G) since those are the only two possible building values after node (C). Since neither
