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6 Pages

### Chapter 23 Solutions

Course: FIN FIN/554, Summer 2006
School: Phoenix
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Word Count: 1686

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24: Chapter Options and Corporate Finance: Extensions and Applications 24.1 a. The inputs to the Black-Scholes model are the current price of the underlying asset (S), the strike price of the option (K), the time to expiration of the option in fractions of a year (t), the variance of the underlying asset (s2), and the continuously-compounded risk-free interest rate (r). Mr. Levin has been granted 20,000 European...

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24: Chapter Options and Corporate Finance: Extensions and Applications 24.1 a. The inputs to the Black-Scholes model are the current price of the underlying asset (S), the strike price of the option (K), the time to expiration of the option in fractions of a year (t), the variance of the underlying asset (s2), and the continuously-compounded risk-free interest rate (r). Mr. Levin has been granted 20,000 European call options on Mountainbrooks stock with 4 years until expiration. Since these options were granted at-the-money, the strike price of each option is equal to the current value of one share, or \$50. Therefore, the Black-Scholes inputs are: S = \$50 K = \$50 t=4 s2 = 0.25 r = 0.06 After identifying the inputs, solve for d1 and d2: d1 = [ln(S/K) + (r + s2)(t) ] / (s2t)1/2 = [ln(50/50) + {0.06 + (0.25)}(4) ] / (0.25*4)1/2 = 0.7400 d2 = d1 - (s2t)1/2 = 0.7400 - (0.25*4)1/2 = -0.2600 Find N(d1) and N(d2), the area under the normal curve from negative infinity to d1 and negative infinity to d2, respectively. N(d1) = N(0.7400) = 0.7704 N(d2) = N(-0.2600) = 0.3974 According to the Black-Scholes formula, the price of a European call option (C) on a nondividend paying common stock is: C = SN(d1) Ke-rtN(d2) = (50)(0.7704) (50)e-(0.06)(4) (0.3974) = \$ 22.8897 The Black-Scholes Price of one call option is \$ 22.8897. Since Mr. Levin was granted 20,000 options, the current value of his options package is \$457,794 (= 20,000 * \$ 22.8897). b. Because Mr. Levin is risk-neutral, you should recommend the alternative with the highest net present value. Since the expected value of the stock option package is worth more than \$450,000, Mr. Levin would prefer to be compensated with the options rather than with the immediate bonus. Answers to End-of-Chapter Problems B-367 c. If Mr. Levin is risk-averse, he may or may not prefer the stock option package to the immediate bonus. Even though the stock option package has a higher net present value, he may not prefer it because it is undiversified. The fact that he cannot sell his options prematurely makes it much more risky than the immediate bonus. Therefore, we cannot say which alternative he would prefer. Kimberleighs total compensation package consists of an annual salary of \$500,000 for 3 years in addition to 10,000 at-the-money stock options. First, 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, discounted at 10%. PV(Annual Salary Payments) = \$500,000 A30.10 = \$1,243,426 24.2 The present value of Kimberleighs three annual salary payments is \$1,243,426. Next, use the Black-Scholes model to determine the value of the stock options. Kimberleigh was granted 10,000 call options on Blubells stock. Blubells current stock price is \$30 per share. Since these options were granted at-the-money, the strike price of each option is also \$30. Therefore, the Black-Scholes inputs are: S = \$30 K = \$30 t=3 s2 = 0.1225 r = 0.05 After identifying the inputs, solve for d1 and d2: d1 = [ln(S/K) + (r + s2)(t) ] / (s2t)1/2 = [ln(30/30) + {0.05 + (0.1225)}(3) ] / (0.1225*3)1/2 = 0.5505 d2 = d1 - (s2t)1/2 = 0.5505 - (0.1225*3)1/2 = -0.0557 Find N(d1) and N(d2), the area under the normal curve from negative infinity to d1 and negative infinity to d2, respectively. N(d1) = N(0.5505) = 0.7090 N(d2) = N(-0.0557) = 0.4778 According to the Black-Scholes formula, the price of a European call option (C) on a nondividend paying common stock is: C = SN(d1) Ke-rtN(d2) = (30)(0.7090) (30)e-(0.05)(3) (0.4778) = \$8.93 The Black-Scholes Price of the call option is \$8.93 Answers to End-of-Chapter Problems B-368 Since Kimberleigh was granted 10,000 of these options, his stock option package is worth \$89,300 (= 10,000 * \$8.93). Kimberleighs total compensation package is equal to the sum of the values of his salary payments and the stock option package. Therefore, the total value of Kimberleighs compensation package at the date the contract is signed is \$1,332,726 (= \$1,243,426 + \$89,300). 24.3 When solving a question dealing with real options, begin by identifying the option-like features of the situation. First, since Webber 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 in downtown Vancouver would be worth \$10 million today. This amount can be viewed as the current price of the underlying asset (S). Third, it will cost Webber \$10.5 million to construct such an office building. This amount can be viewed as the strike price of a call option (K), since it is the amount that the firm must pay in order to exercise its right to erect an office building. Finally, since the firms right to build on the land lasts only 1 year, the time to expiration (t) of the real option is one year. The Webber Company can use a Two-State model to value its option to build on the land. Value of an Office Building (in millions) Today 1 Year 12.5 10 8 ? 0 = max(0, 8-10.5) Webber's Real Call Option with a Strike of \$10.5 (in millions) Today 1 Year 2 = max(0, 12.5-10.5) If demand increases and the value of the building rises, the return on the value of the building over the period is 25% [= (12.5/10) 1]. If demand decreases and the value of the building falls, the return on the value of the building over the period is 20% [= (8/10) 1]. Use the following expression to determine the risk-neutral probability of a rise in the value of the building: Risk-Free Rate = (ProbabilityRise)(ReturnRise) + (ProbabilityFall)(ReturnFall) = (ProbabilityRise)(ReturnRise) (1 + - ProbabilityRise)(ReturnFall) 0.025 = (ProbabilityRise)(0.25) + (1 ProbabilityRise)(-0.20) ProbabilityRise = 0.50 ProbabilityFall = 1 - ProbabilityRise = 1 0.50 = 0.50 Answers to End-of-Chapter Problems B-369 The risk-neutral probability of a rise in the value of the building is 50%, and the risk-neutral probability of a fall in the value of the building is 50%. Using these risk-neutral probabilities, determine the expected payoff of Webbers real option at expiration. Expected Payoff at Expiration = (.50)(\$2,000,000) + (.50)(\$0) = \$1,000,000 Since this payoff will occur 1 year from now, it must be discounted at the risk-free rate of 2.5% in order to find its present value: PV(Expected Payoff at Expiration) = (\$1,000,000 / 1.025) = \$975,610 A call option with a strike price of \$10.5 million and 1 year until expiration is worth \$975,610. Therefore, the right to build on office building in downtown Vancouver over the next year is worth \$975,610 today. Since \$750,000 is less than the value of the real option to build, Webber should not accept the offer from his competitor. Instead, Webber should retain the right to erect an office building on the land. 24.4 When solving a question dealing with real options, begin by identifying the option-like features of the situation. First, since Jet Black 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 60,000 barrels of oil and the current price of oil is \$25 per barrel, the current price of the underlying asset (S) to be used in the Black-Scholes model is \$1,500,000 (= 60,000 barrels * \$25 per barrel). Third, since Jet Black will not drill unless the price of oil in one year will compensate its excavation costs, \$1.75 million can be viewed as the real options strike price (K). Finally, since the winner of the auction has the right to drill for oil in one year, the real option can be viewed as having a time to expiration (t) of one year. Therefore, the inputs to the Black-Scholes formula are: S = \$1,500,000 K = \$1,750,000 t=1 After identifying the inputs, solve for d1 and d2: d1 = [ln(S/K) + (r + s2)(t) ] / (s2t)1/2 = [ln(1,500,000/1,750,000) + {0.10 + (0.36)}(1) ] / (0.36*1)1/2 = 0.2907 d2 = d1 - (s2t)1/2 =0.2907- (0.36*1)1/2 = -0.3903 Find N(d1) and N(d2), the area under the normal curve from negative infinity to d1 and negative infinity to d2, respectively. Answers to End-of-Chapter Problems B-370 s2 = 0.36 r = 0.10 N(d1) = N(0.2907) = 0.6144 N(d2) = N(-0.3903) = 0.3482 According to the Black-Scholes formula, the price of a European call option (C) on a nondividend paying common stock is: C = SN(d1) Ke-rtN(d2) = (1,500,000)(0.6144) (1,750,000)e-(0.10)(1) (0.3482) = \$370,237 The Black-Scholes Price of the call option is \$370,237 Therefore, the maximum bid that Jet Black should be willing to make at the auction is \$370,237. 24.5 When solving a question dealing with real options, begin by identifying the option-like features of the situation. First, since Sardano 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, \$1,000,000 (= 5,000 steel rods * {\$300-\$100}) can be viewed as the put options strike price (K). Third, since the project requires Sardano to purchase 400 tons of steel and the current price of steel is \$3,000 per ton, the current price of the underlying asset (S) to be used in the Black-Scholes formula is \$1,200,000 (= 400 tons * \$3,000 per ton). Finally, since Sardano must decide whether to purchase the steel or not in six months, the firms 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, 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. Therefore, the inputs to the Black-Scholes formula are: S= \$1,200,000 K =\$1,000,000 t = 0.50 After identifying the inputs, solve for d1 and d2: d1 = [ln(S/K) + (r + s2)(t) ] / (s2t)1/2 = [ln(1,200,000/1,000,000) + {0.03 + (0.3025)}(0.50) ] / (0.3025*0.50)1/2 = 0.7018 d2 = d1 - (s2t)1/2 =0.7018- (0.3025*0.50)1/2 = 0.3129 Find N(d1) and N(d2), the area under the normal curve from negative infinity to d1 and negative infinity to d2, respectively. N(d1) = N(0.7018) = 0.7586 Answers to End-of-Chapter Problems B-371 s2 = 0.3025 r = 0.03 N(d2) = N(0.3129) = 0.6228 According to the Black-Scholes formula, the price of a European call option (C) on a nondividend paying common stock is: C = SN(d1) Ke-rtN(d2) = (1,200,000)(0.7586) (1,000,000)e-(0.03)(0.50) (0.6228) = \$296,792 The Black-Scholes Price of the call option is \$296,792. According to Put-Call Parity: where C P S PV(K) C = P + S PV(K) = the cost of a call option = the cost of a put option = the price of the underlying asset = the present value of the strike price In this problem: C = \$296,792 S = \$1,200,000 PV(K) = [\$1,000,000 / e(0.03*0.50) ] = \$985,112 Rearranging the Put-Call Parity equation: P = C S + PV(K) = \$296,792 - \$1,200,000 + \$985,112 = \$81,904 According to Put-Call Parity, the Black-Scholes price of a put option with the characteristics described above is \$81,904. Therefore, the maximum amount that Sardano and Sons should be willing to pay for the lease of the warehouse is \$81,904. Answers to End-of-Chapter Problems B-372
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Lecture November 20, 2008 10 true false, 5 to 10 short answer, 1 to 2 essays (no graph, more essay questions) Most on readings Tuesday before = review sessionChanging media News media forms change over time Changing carrying capacity
Cornell - DSOC - 207
Lecture November 30 Globalization, poverty and inequality a. Globalization, poverty and inequality a. Globalization: the popular view b. Thomas Friedman. The Lexus and the Olive tree. Globalization is the new international system that replaces the Co
Cornell - DSOC - 207
Lecture October 9th, 2008 Dsoc 2070: Crime, crime trends and crime statistics a. What ever happened to crime a. What ever happened to crime? As a nation, we're not talking a lot about it these days. Law enforcement and criminal policy has largely bee
Cornell - DSOC - 207
Lecture October 14: Sociological theories of crime a. Problems with official poverty measure a. Out of touch with standards of living and consumption patterns: i. Childcare (working women with children under 6 increased from 15% to 58%) ii.transporta
Cornell - DSOC - 207
Lecture: DSoc 207 October 23, 2008 Sociological theories of crime All have policy education within them! Social learning theory (Sutherland) A learned behavior as opposed to being determined by birth Geographic factor Key conflict: an acquired
Cornell - DSOC - 207
Lecture October 30: Era of mass incarceration and felon disenfranchisement The Prison Population Past three decades number of inmates has increased by more than 500% 2 million plus inmates, 2 million on probation and 775 thousand on parole There