CHAPTER 18
OPTION VALUATION AND STRATEGIES
Teaching Guides for Questions and Problems in the Text
QUESTIONS
18-1. The Black/Scholes option valuation model specifies variables that affect the value of an
option. As these variables change, so does the value of an option.
a. As risk increases, the value of a call option increases. This is intuitive; increased variability
b. An increase in interest rates also increases the value of a call option. This is not intuitive
since throughout the text, an increase in interest rates always decreased the value of the
c. As the option approaches expiration, the value of the option declines. This is also intuitive,
18-2. Put-call parity links stocks, bonds, and put and call options. A change in the value of one
of the components causes the prices of the other securities to change. Through the process of
18-3. One of the inputs in the Black/Scholes model is the variability of the underlying stock’s
18-4. The hedge ratio determines the number of options necessary to hedge (offset) price
movements in a stock. Since the investor wants to transfer the gain on a stock from one year to
another, that individual wants a long position in an option that offsets the price movement in the
short position in the stock. The hedge ratio determines the number of call options this
18-5. The individual who writes a straddle (sells a put and a call with the same expiration date
and same strike price) is anticipating that the price of the stock will not fluctuate. If the price of
the stock rises, that may produce a loss on the sale of the call. If the price of the stock falls, that
may produce a loss on the put. Thus, the straddle produces a profit for the writer if the price of
the stock does not change in either direction or experiences only minor price fluctuations.
18-6. Spreads consist of taking opposite long and short positions in options with different
strike prices. To establish a bear spread, the investor buys the call option with the higher strike
18-7. One strategy that reduces the risk of loss when a stock is purchased is to buy a put on the
stock. This is the protective put strategy. The same concept applies to selling short. If an
18-8. Options are often used with other securities to help manage risk. If an investor has a long
position in a security, the opposite or short position in an option reduces the risk exposure
The collar is used to protect (lock in) an existing gain in the price of a stock. The investor sells
a call with a higher strike price and buys a put with a lower stock price. If the price of the stock
The hedge ratio determines the number of options necessary to hedge and offset price
The protective call is the opposite of the protective put. If the investor sells a stock short,
buying the call reduces the risk exposure associated with the short position in the stock. If the
18-9. A straddle requires buying or writing a put and a call with the same strike price. If the
investor thought that the price of the stock would remain stable, the investor sells (writes) each
options. If the price of the stock remains stable, the two options will expire and the write
profits.
18-10. Black/Scholes indicates that the value of a call option is related to the difference
between the price of the underlying stock and the present value of the strike price (i.e., the price
of the stock minus the present value of the strike price). An increase in interest rates will
decrease the present value of the strike price and increase the value of a call.
The value of a put option is related to the difference between the present value of the strike
PROBLEMS
18-1. The Black/Scholes option valuation says that the value of a call is the price of the stock
times a function (d1) minus the present value of the strike price times a function (d2). In this
initial problem d1 and d2 are equal to 1, and the value of the call option is
If the student uses e instead of 1.1 to determine the value of the call, the answer is
18-2. This problem covers the variables that affect an option’s value according to the
Black/Scholes option valuation model. (These answers were derived using a computer
program. Several are readiy available through the internet.)
a. In this part the price of the stock is varied:
Price of Standard Time Interest Value of
the stock deviation (days) rate the call
b. In this part time is varied:
Price of Standard Time Interest Value of
the stock deviation (days) rate the call
c. In this part the interest rates varied:
Price of Standard Time Interest Value of
the stock deviation (days) rate the call
d. In this part the standard deviation is varied:
Price of Standard Time Interest Value of
the stock deviation (days) rate the call
(Rounding may produce marginally different answers.)
e. The generalizations are
1. As the price of the stock rises, the value of the call
2. As the time the expiration diminishes, the value of
3. As the interest rate increases, the value of the call
4. As the time the standard deviation declines, the value
18-3. a. The investor owns 600 shares of stock and seeks to use the hedge ratio to offset any
price movement in the stock. The number of call options necessary to hedge is
b. If the investor buys one call option on 100 shares, that option will hedge 60 shares (100
shares X 0.6) sold short.
18-4. To execute the arbitrage:
buy the call -3.00
Stock Profit Profit Profit Profit Net
price call put stock bond profit
$45 ($3) ($3.50) $5 $4.50 $3
Since the position generates a profit if the price of the stock rises or falls, it illustrates a riskless
arbitrage opportunity.
18-5. Like the previous problem, this problem illustrates put-call parity. If the sum of the prices
of the stock and the put is not equal to the sum of the prices of the call and the bond, an
opportunity for arbi-trage exists. This problem works through various combinations of stock
prices, option prices, and bond prices. It also illustrates that the arbitrage profit is independent
of what happens to the price of the stock.
a. The call is undervalued:
cash inflow
buy the call -10
Stock Profit Profit Profit Profit Net
price call put stock bond profit
$90 ($10) ($5) $15 $10 $10
b. The put is undervalued:
cash inflow
sell the call 20
buy the put -3
Stock Profit Profit Profit Profit Net
price call put stock bond profit
$90 $20 $7 ($15) ($10) $2
c. The bond is overvalued:
cash inflow
sell the call 20
buy the put -5
Stock Profit Profit Profit Profit Net
price call put stock bond profit
$90 $20 $5 ($15) ($5) $5
d. The bond is undervalued:
cash inflow
buy the call -20
sell the put 5
e. The stock is overvalued:
cash inflow
buy the call -20
net cash 7
Stock Profit Profit Profit Profit Net
price call put stock bond profit
$90 ($20) ($5) $22 $10 $7
f. The stock is undervalued:
cash inflow
sell the call 20
buy the put -5
Stock Profit Profit Profit Profit Net
price call put stock bond profit
$90 $20 $5 ($11) ($10) $4
The opportunity for arbitrage ceases when the put-call parity indicates that there is no
opportunity for profit. The implication is that if one of the four markets is in disequilibrium, the
demand and supply for each security will be affected which causes their prices to change.
18-6. The previous problems illustrate arbitrage when the markets are in disequilibrium. This
problem illustrates that a long position in the stock and the put generates the same profits as a
comparable position in the call and the bond. To verify there is no opportunity for arbitrage,
consider the following:
buy the call -5.38
To demonstrate that buying the stock and the put generates the same profits as buying the call
and the bond, construct the following profit/loss profile:
Stock Profit Profit Net Profit Profit Net
price call bond stock put
Both buying the call and the bond generates the same profit as buying the stock and the put. If
the markets are in equilibrium, the two positions are equivalent.
18-7. To verify that a short on the stock is mimicked by a short in the call and the bond (i.e.,
borrowing and paying interest) plus a long in the put, set up the same type of table used in the
previous problem.
Stock Net on Interest + Profit + Profit = Net
price stock paid short call long put
$110 ($10) ($6) ($1) ($3) ($10)
105 (5) (6) 4 (3) (5)
100 0 (6) 9 (3) 0
The two sides are equal at all prices of the stock, so the short on the stock is mimicked by the
short in the bond plus the short in the call plus the long in the put.
18-8. In this problem, the investor incorrectly executes the arbitrage and always sustains losses.
buy the stock -50.00
buy the put -4.00
sell the bond (borrow) 47.67
sell the call 4.38
net cash -2.00
Stock Profit Profit Profit Profit Net
price stock put call bond profit
$40 ($10) $1.00 $4.38 ($2.38) ($2)
Since the position always generates a loss if the price of the stock rises or falls, it illustrates
executing the wrong arbitrage.
18-9. This problem is another arbitrage problem and illustrates why the call with the lower
strike price must sell for more than the call with the higher strike price. The investor would buy
the call at $25 for $2, sell the call at $30 for $4, and have a net cash inflow of $2. This position
generates a profit independently of what happens to the price of the stock.
Price Intrinsic value: Profit:
of the Call Call Buying the Selling the Total
stock at $25 at $30 $25 call $30 call profit
for $2 for $4
$10 $0 $0 ($2) $4 $2
15 0 0 (2) 4 2
Given the prices of the two calls, the investor cannot lose. As all investors seek to execute this
arbitrage opportunity, the prices of the calls must change. The price of the call at $25 rises
while the price of the call at $30 falls. These price movements will continue until the price of
18-10. This problem reverses problem #9 and applies the concept to puts. The put with the
higher strike price must command a higher price or an opportunity for arbitrage exists. The
investor would buy the put at $60 for $3, sell the put at $55 for $6, and have a net cash inflow
of $3. This position generates a profit independently of what happens to the price of the stock.
Price Intrinsic value: Profit:
of the Put at Put at Buying the Selling the Total
stock $55 $60 $60 put $55 put profit
for $3 for $6
$50 $5 $10 $7 $ 1 $8
55 0 5 2 6 8
56 0 4 1 6 7
18-11. a. This problem is another illustration of a riskless arbitrage. The investor should:
buy the call with six months to expiration ($4.50 cash outflow)
b. Price Intrinsic value: Profit: Net
of the put at $100 put sold put bought
stock for $7 for $4.50
$85 $15 $0 ($8) $10.50 $2.50
90 10 0 (3) 5.50 2.50
c. The six month option that the investor purchased has three months to expiration. It should
continue to sell for more than its intrinsic value (i.e., it continues to command a time premium).
The net gain determined above assumes both options are selling for their intrinsic value.
18-12. This problem reverses the protective put strategy. It applies that concept to limiting
losses from the short sale of a stock.
a. The maximum possible profit is $26 ($29 – minus the $3 cost of the call) which occurs if
the price of the stock were to decline to $0.