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CH22

Fundamentals of Corporate Finance + Standard & Poor's Educational Version of Market Insight

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CHAPTER 22 Real Options Answers to Practice Questions 1. a. A five-year American call option on oil. The initial exercise price is \$32 a barrel, but the exercise price rises by 5 percent per year. b. An American put option to abandon the restaurant at an exercise price of \$5 million. The restaurant’s current value is (\$700,000/r). The annual standard deviation of the changes in the value of the restaurant as a going concern is 15 percent. c. A put option, as in (b), except that the exercise price should be interpreted as \$5 million in real estate value plus the present value of the future fixed costs avoided by closing down the restaurant. Thus, the exercise price is: \$5,000,000 + (\$300,000/0.10) = \$8,000,000. Note: The underlying asset is now PV(revenue – variable cost), with annual standard deviation of 10.5 percent. d. A complex option that allows the company to abandon temporarily (an American put) and (if the put is exercised) to subsequently restart (an American call). e. An in-the-money American option to choose between two assets; that is, the developer can defer exercise and then determine whether it is more profitable to build a hotel or an apartment building. By waiting, however, the developer loses the cash flows from immediate development. f. A call option that allows Air France to fix the delivery date and price. 2. A commitment to invest in the Mark II would have a negative NPV. The option to invest has a positive NPV. The value of the option more than offsets the negative NPV of the Mark I. 3. a. P = 467 EX = 800 σ = 0.35 t = 3.0 r f = 0.10 .7194 ) 3.0 (0.35 .1132 t σ d d .1132 /2 ) 3.0 (0.35 ) 3.0 )]/(0.35 00/1.10 log[467/(8 /2 t σ t X)]/σ log[P/PV(E d 1 2 3 1 0 0 0 - = × - - = - = - = × + × = + = N(d 1 ) = N(-0.1132) = 0.4549 N(d 2 ) = N(-0.7194) = 0.2359 Call value = [N(d 1 ) × P] – [N(d 2 ) × PV(EX)] 203

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= [0.4549 × 467] – [0.2359 × (800/1.10 3 )] = \$70.65 b. P = 500 EX = 900 σ = 0.35 t = 3.0 r f = 0.10 0.8010 ) 3.0 (0.35 0.1948 t σ d d 0.1948 /2 ) 3.0 (0.35 ) 3.0 )]/(0.35 00/1.10 log[500/(9 /2 t σ t X)]/σ log[P/PV(E d 1 2 3 1 - = × - - = - = - = × + × = + = N(d 1 ) = N(-0.1948) = 0.4228 N(d 2 ) = N(-0.8010) = 0.2116 Call value = [N(d 1 ) × P] – [N(d 2 ) × PV(EX)] = [0.4228 × 500] – [0.2116 × (900/1.10 3 )] = \$68.32 c. P = 467 EX = 900 σ = 0.20 t = 3.0 r f = 0.10 1.2417 ) 3.0 (0.20 0.8953 t σ d d 0.8953 /2 ) 3.0 (0.20 ) 3.0 )]/(0.20 00/1.10 log[467/(9 /2 t σ t X)]/σ log[P/PV(E d 1 2 3 1 - = × - - = - = - = × + × = + = N(d 1 ) = N(-0.8953) = 0.1853 N(d 2 ) = N(-1.2417) = 0.1072 Call value = [N(d 1 ) × P] – [N(d 2 ) × PV(EX)] = [0.1853 × 467] – [0.1072 × (900/1.10 3 )] = \$14.05 4. P = 1.7 EX = 2 σ = 0.15 t = 1.0 r f = 0.12 .4029 ) 1.0 (0.15 0.2529 t σ d d 0.2529 /2 ) 1.0 (0.15 ) 1.0 )]/(0.15 /1.12 log[1.7/(2 /2 t σ t X)]/σ log[P/PV(E d 1 2 1 1 0 - = × - - = - = - = × + × = + = N(d 1 ) = N(-0.2529) = 0.4002 N(d 2 ) = N(-0.4029) = 0.3435 Call value = [N(d 1 ) × P] – [N(d 2 ) × PV(EX)] = [0.4002 × 1.7] – [0.3435
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CH22 - CHAPTER 22 Real Options Answers to Practice...

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