In: Finance
The price of Chive Corp. stock will be either $67 or $95 at the end of the year. Call options are available with one year to expiration. T-bills currently yield 5 percent.
| a. |
Suppose the current price of the company's stock is $75. What is the value of the call option if the exercise price is $65 per share? (Do not round intermediate calculations and round your answer to 2 decimal places, e.g., 32.16.) |
| b. |
Suppose the exercise price is $90 and the current price of the company's stock is $75. What is the value of the call option now? (Do not round intermediate calculations and round your answer to 2 decimal places, e.g., 32.16.) |
a
| Upmove (U)= High price/current price=95/75=1.2667 | ||||||
| Down move (D)= Low price/current price=67/75=0.8933 | ||||||
| Risk neutral probability for up move | ||||||
| q = (e^(risk free rate*time)-D)/(U-D) | ||||||
| =(e^(0.05*1)-0.8933)/(1.2667-0.8933)=0.42305 | ||||||
| Call option payoff at high price (payoff H) | ||||||
| =Max(High price-strike price,0) | ||||||
| =Max(95-65,0) | ||||||
| =Max(30,0) | ||||||
| =30 | ||||||
| Call option payoff at low price (Payoff L) | ||||||
| =Max(Low price-strike price,0) | ||||||
| =Max(67-65,0) | ||||||
| =Max(2,0) | ||||||
| =2 | ||||||
| Price of call option = e^(-r*t)*(q*Payoff H+(1-q)*Payoff L) | ||||||
| =e^(-0.05*1)*(0.423048*30+(1-0.423048)*2) | ||||||
| =13.17 |
b
| Upmove (U)= High price/current price=95/75=1.2667 | ||||||
| Down move (D)= Low price/current price=67/75=0.8933 | ||||||
| Risk neutral probability for up move | ||||||
| q = (e^(risk free rate*time)-D)/(U-D) | ||||||
| =(e^(0.05*1)-0.8933)/(1.2667-0.8933)=0.42305 | ||||||
| Call option payoff at high price (payoff H) | ||||||
| =Max(High price-strike price,0) | ||||||
| =Max(95-90,0) | ||||||
| =Max(5,0) | ||||||
| =5 | ||||||
| Call option payoff at low price (Payoff L) | ||||||
| =Max(Low price-strike price,0) | ||||||
| =Max(67-90,0) | ||||||
| =Max(-23,0) | ||||||
| =0 | ||||||
| Price of call option = e^(-r*t)*(q*Payoff H+(1-q)*Payoff L) | ||||||
| =e^(-0.05*1)*(0.423048*5+(1-0.423048)*0) | ||||||
| =2.01 |