Corporate_Finance_9th_edition_Solutions_Manual_FINAL0

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Unformatted text preview: ure 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 = 45,000 steel rods(\$24 – 16) Amount received = \$360,000 The amount received can be viewed as the put option’s strike price (K). Third, since the project requires Sardano to purchase 400 tons of steel and the current price of steel is \$630 per ton, the current price of the underlying asset (S) to be used in the Black-Scholes formula is: “Stock” price = 400 tons(\$630 per ton) “Stock” price = \$252,000 Finally, since Sardano must decide whether to purchase the steel or not 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/K) + (R + σ 2/2)(t) ] / (σ 2t)1/2 d1 = [ln(\$252,000/\$360,000) + (.045 + .452/2) × (6/12)] / (.45 × d2 = –0.8911 – (.45 × 6 / 12 ) = –0.8911 6 / 12 ) = –1.2093 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) = N(–0.8911) = 0.1864 N(d2) = N(–1.2093) = 0.1133 Now we can find the value of call option, which will be: C = SN(d1) – Ke-–RtN(d2) C = \$252,000(0.1864) – (\$360,000e–.045(6/12))(0.1133) C = \$7,110.89 Now we can use put-call parity to find the price of the put option, which is: C = P + S – Ke–Rt \$7,110.89 = P + \$252,000 – \$360,000e–.045(6/12) P = \$107,101.33 This is the most the company should be willing to pay for the lease. 470 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 = \$8,200,000 – 7,000,000 Value of put with low demand = \$1,200,000 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 = (\$13,400,000 – 11,600,000) / \$11,600,000 Percentage increase with high demand = .1552 or 15.52% And the percentage decrease in the value of the technology with low demand is: Percentage decrease with high demand = (\$7,000,000 – 11,600,000) / \$11,600,000 Percentage decrease with high demand = –.3966 or –39.66% Now we can find the risk-neutral probability of a rise in the value of the technology as: Risk-free rate = (ProbabilityRise)(ReturnRise) + (ProbabilityFall)(ReturnFall) Risk-free rate = (ProbabilityRise)(ReturnRise) + (1 – ProbabilityRise)(ReturnFall) 0.06 = (ProbabilityRise)(0.1552) + (1 – ProbabilityRise)(–.3966) ProbabilityRise = 0.8275 So, a probability of a fall is: ProbabilityFall = 1 – ProbabilityRise ProbabilityFall = 1 – 0.8275 ProbabilityFall = 0.1725 Using these risk-neutral probabilities, we can determine the expected payoff of the real option at expiration. With high demand, the option is worthless since the technology will not be sold, and the value of the technology with low demand is the \$1.2 million we calculated previously. So, the value of the option to abandon is: Value of option to abandon = [(.8275)(0) + (.1725)(\$1,200,000)] / (1 + .06) Value of option to abandon = \$195,283.02 471 8. Using the binomial mode, we will find the value of u and d, which are: u=e σ/ n u = e.70/ 12 u = 1.2239 d=1/u d = 1 / 1.2239 d = 0.8170 This implies the percentage increase if the stock price increases will be 22 percent, and the percentage decrease if the stock price falls will be 18 percent. The monthly interest rate is: Monthly interest rate = 0.05/12 Monthly interest rate = 0.0042 Next, we need to find the risk neutral probability of a price increase or decrease, which will be: 0.0042 = 0.22(Probability of rise) + –0.18(1 – Probability of rise) Probability of rise = 0.4599 And the probability of a price decrease is: Probability of decrease = 1 – 0.4599 Probability of decrease = 0.5401 The following figure shows the stock price and put price for each possible move over the next two months: Stock price (D) Put price Stock price (B) Put price Stock price(A) Put price \$ \$ 58.00 11.05 Stock price (C) Put price \$ \$ 47.39 17.34 Stock price (F) Put price \$ \$ 70.99 3.77 Stock price (E) Put price \$ \$ 86.89 0 \$ \$ 58.00 7.00 \$ \$ 38.72 26.28 472 The stock price at node (A) is the current stock price. The stock price at node (B) is from an up move, which means: Stock price (B) = \$58(1.2239) Stock price (B) = \$70.99 And the stock price at node (D) is two up moves, or: Stock price (D) = \$58(1.2239)(1.2239) Stock price (D) =...
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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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