Corporate_Finance_9th_edition_Solutions_Manual_FINAL0

# 07 probabilityrise2667 1 probabilityrise30

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Unformatted text preview: a synthetic call option with identical payoffs to the call option described above. In order to do this, the company will need to buy gold and borrow at the riskfree rate. The amount of gold to buy is based on the delta of the option, where delta is defined as: Delta = (Swing of option) / (Swing of price of gold) Since the call option will be worth \$100 if the price of gold rises and \$0 if it falls, the swing of the call option is \$100 (= \$100 – 0). Since the price of gold will either be \$975 or \$740 at the time of the option’s expiration, the swing of the price of gold is \$235 (= \$975 – 740). Given this information the delta of the call option is: Delta = (Swing of option) / (Swing of price of gold) Delta = (\$100 / \$235) Delta = 0.43 Therefore, the first step in creating a synthetic call option is to buy 0.43 of an ounce of gold. Since gold currently sells for \$815 per ounce, the company will pay \$346.81 (= 0.43 × \$815) to purchase 0.43 of an ounce of gold. In order to determine the amount that should be borrowed, compare the payoff of the actual call option to the payoff of delta shares at expiration: Call Option If the price of gold rises to \$975: If the price of gold falls to \$740: Delta Shares If the price of gold rises to \$975: If the price of gold falls to \$740: Payoff = \$100 Payoff = \$0 Payoff = (0.43)(\$975) = \$414.89 Payoff = (0.43)(\$740) = \$314.89 The payoff of this synthetic call position should be identical to the payoff of an actual call option. However, buying 0.43 of a share leaves us exactly \$314.89 above the payoff at expiration, whether the price of gold rises or falls. In order to decrease the company’s payoff at expiration by \$314.89, it should borrow the present value of \$314.89 now. In three months, the company must pay \$314.89, which will decrease its payoffs so that they exactly match those of an actual call option. So, the amount to borrow today is: Amount to borrow today = \$314.89 / 1.06501/4 454 Amount to borrow today = \$309.97 d. Since the company pays \$346.81 in order to purchase gold and borrows \$309.97, the total cost of the synthetic call option is \$36.83 (= \$346.81 – 309.97). This is exactly the same price for an actual call option. Since an actual call option and a synthetic call option provide identical payoff structures, the company should not expect to pay more for one than for the other. 29. To construct the collar, the investor must purchase the stock, sell a call option with a high strike price, and buy a put option with a low strike price. So, to find the cost of the collar, we need to find the price of the call option and the price of the put option. We can use Black-Scholes to find the price of the call option, which will be: Price of call option with \$110 strike price: d1 = [ln(\$85/\$110) + (.07 + .502/2) × (6/12)] / (.50 × d2 = –.4535 – (.50 × N(d1) = .3251 N(d2) = .2098 Putting these values into the Black-Scholes model, we find the call price is: C = \$85(.3251) – (\$110–.07(6/12))(.2098) = \$5.35 Now we can use Black-Scholes and put-call parity to find the price of the put option with a strike price of \$65. Doing so, we find: Price of put option with \$65 strike price: d1 = [ln(\$85/\$65) + (.07 + .502/2) × (6/12)] / (.50 × d2 = 1.0345 – (.50 × N(d1) = .8496 N(d2) = .7521 Putting these values into the Black-Scholes model, we find the call price is: C = \$85(.8496) – (\$65e–.07(6/12))(.7521) = \$25.01 Rearranging the put-call parity equation, we get: P = C – S + Xe–Rt P = \$25.01 – 85 + 65e–.07(6/12) P = \$2.77 (6 / 12) ) = 1.0345 (6 / 12) ) = –.4535 6 / 12 ) = –0.8070 6 / 12 ) = 0.6810 455 So, the investor will buy the stock, sell the call option, and buy the put option, so the total cost is: Total cost of collar = \$85 – 5.35 + 2.77 Total cost of collar = \$82.43 Challenge 30. a. Using the equation for the PV of a continuously compounded lump sum, we get: PV = \$40,000 × e–.05(2) = \$36,193.50 b. Using Black-Scholes model to value the equity, we get: d1 = [ln(\$19,000/\$40,000) + (.05 + .602/2) × 2] / (.60 × d2 = –.3352 – (.60 × N(d1) = .3687 N(d2) = .1183 Putting these values into Black-Scholes: E = \$19,000(.3687) – (\$40,000e–.05(2))(.1183) = \$2,725.75 And using put-call parity, the price of the put option is: Put = \$40,000e–.05(2) + 2,725.75 – 19,000 = \$19,919.25 c. The value of a risky bond is the value of a risk-free bond minus the value of a put option on the firm’s equity, so: Value of risky bond = \$36,193.50 – 19,919.25 = \$16,274.25 Using the equation for the PV of a continuously compounded lump sum to find the return on debt, we get: \$16,274.25 = \$40,000e–R(2) .40686 = e–R2 RD = –(1/2)ln(.40686) = .4496 or 44.96% d. The value of the debt with five years to maturity at the risk-free rate is: PV = \$40,000 × e–.05(5) = \$31,152.03 2 ) = –1.1837 2 ) = –.3352 456 Using Black-Scholes model to value the equity, we get: d1 = [ln(\$19,000/\$40,000) + (.05 + .602/2) × 5] / (.60 × d2 = .3023 – (.60 × N(d1) = .6188 N(d2) = .1493 Putting these values into Black-Sc...
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## This note was uploaded on 07/10/2010 for the course FIN 6301 taught by Professor Eshmalwi during the Spring '10 term at University of Texas-Tyler.

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