Question

Consider a stock priced at $30 with a standard deviation of 0.3. The risk-free rate is 0.05. There are put and call options available at exercise prices of 30 and a time to expiration of six months. The calls are priced at $2.89 and the puts cost $2.15. There are no dividends on the stock and the options are European. Assume that all transactions consist of 100 shares or one contract (100 options). Use this information to answer questions 6 through 10. Suppose the investor constructed a covered call. At expiration the stock price is $27. What is the investor's profit? a. $589 b. $289 c. $2,989 d. $2,711 e. none of the above What is the breakeven stock price at expiration for the transaction described in problem 6? a. $27.11 b. $30.00 c. $32.89 d $29.89 e. none of the above 8. If the transaction described in problem 6 is closed out when the option has three months to go and the stock price is at $36, what is the investor's profit? a. $600 b. $311 c. $889 d. $229 e. none of the above 9. What is the maximum profit from the transaction described in Question 6 if the position is held to expiration? a. $3,289 b. $289 c. infinity d. $2,711 e. none of the above 10. What is the minimum profit from the transaction described in Question 6 if the position is held to expiration? a. -$2,711 b. -$3,289 c. -$3,000 d. negative infinity e. none of the above

Answer 1

Given:

• Current stock price, So = 30
• std deviation = 0.30; risf free rate = 0.05
• Exercise price, X = 30
• Time to expiry, T = 6months = 6/12 = 0.5 years
• call option price, c = 2.89
• put option price, p =2.15
• Stock price at expiry, ST = 27

one contract = 100 options

Investors Profit:

• Covered call, profit at expiration = ST - MAX[(ST- X) , 0] + c - So
• Profit = 27 - MAX[(27-30),0] + 2.89 - 30 = 27 - 0 + 2.89 - 30 = -0.11
• For 100 options, it is -0.11 * 100 = -$11
• Answer = --$11

Break-even point = So - c = 30 - 2.89 = 27.11

Profit after 3 months, = St - MAX[ (St - X)/(1+r)^t, 0] + c - So where St = 36 and t = 3/12 = 0.25

• Profit = 36 - MAX[(36-30)/(1+0.05)^0.25 , 0] + 2.89 - 30
• Profit = 36 - MAX[5.93,0] + 2.89 - 30
• Profit = 36 - 5.93 + 2.89 - 30
• Profit = 2.96
• For 100 options, profit = 2.96 * 100 = $296

Maximum Profit = X - So + c = 30 - 30 + 2.89 = 2.89

For 100 options, 2.89 * 100 = $289

Maximum loss = So - c = 30 - 2.89 = 27.11