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Give step-by-step solution with explanation and final answer:4. An investor is interested in purchasing a European call option on Altec, Inc., a non-dividend- paying common stock. The option has an exercise price of $15 and three months until expiration. Altec’s stock is currently trading at $14 per share and the annual standard deviation of its continuously compounded returns is 0.28. The risk-free rate of interest is 5% per year. Compute (i) the value of the call option, using the Black Scholes option pricing model; and (ii) the value of the put option basing on your answer in part (i) and the put-call parity theorem.

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Give step-by-step solution with explanation and final answer:Uploaded Image4. An investor is interested in purchasing a European call option on Altec, Inc., a non-dividend- paying common stock. The option has an exercise price of $15 and three months until expiration. Altec’s stock is currently trading at $14 per share and the annual standard deviation of its continuously compounded returns is 0.28. The risk-free rate of interest is 5% per year. Compute (i) the value of the call option, using the Black Scholes option pricing model; and (ii) the value of the put option basing on your answer in part (i) and the put-call parity theorem.

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Asked by: Veera manikanta
Created at: 2025-08-16 00:57:12

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Let's solve the problem step by step: ### **Given Data** - \( S = \$14 \) (Current stock price) - \( K = \$15 \) (Exercise/Strike price) - \( T = 0.25 \) years (3 months is 0.25 years) - \( r = 0.05 \) (risk-free rate, per year) - \( \sigma = 0.28 \) (annual standard deviation) - No dividends. --- ### **(i) Value of the Call Option (Black-Scholes Formula)** The Black-Scholes formula for a European call option is: \[ C = S N(d_1) - K e^{-rT} N(d_2) \] Where: \[ d_1 = \frac{\ln(S/K) + (r + \frac{1}{2}\sigma^2)T}{\sigma\sqrt{T}} \] \[ d_2 = d_1 - \sigma\sqrt{T} \] --- #### **Step 1: Calculate \(d_1\) and \(d_2\)** First, calculate each term: - \(\ln(S/K) = \ln(14/15) = \ln(0.9333) \approx -0.069\) - \((r + \frac{1}{2}\sigma^2) = 0.05 + 0.5 \times (0.28^2) = 0.05 + 0.5 \times 0.0784 = 0.05 + 0.0392 = 0.0892\) - \((r + \frac{1}{2}\sigma^2)T = 0.0892 \times 0.25 = 0.0223\) - \(\sigma\sqrt{T} = 0.28 \times \sqrt{0.25} = 0.28 \times 0.5 = 0.14\) Now, \[ d_1 = \frac{-0.069 + 0.0223}{0.14} = \frac{-0.0467}{0.14} \approx -0.334 \] \[ d_2 = -0.334 - 0.14 = -0.474 \] --- #### **Step 2: Find \(N(d_1)\) and \(N(d_2)\)** From the standard normal table: - \(N(-0.334) \approx 0.369\) - \(N(-0.474) \approx 0.317\) --- #### **Step 3: Plug into Black-Scholes Formula** \[ C = 14 \times 0.369 - 15 \times e^{-0.05 \times 0.25} \times 0.317 \] First, calculate \(e^{-0.05 \times 0.25} = e^{-0.0125} \approx 0.9876\) \[ C = 14 \times 0.369 - 15 \times 0.9876 \times 0.317 \] \[ = 5.166 - (15 \times 0.9876 \times 0.317) \] \[ = 5.166 - (14.814 \times 0.317) \] \[ = 5.166 - 4.695 \] \[ = 0.471 \] **So, the value of the call option is approximately \( \boxed{0.47} \).** --- ### **(ii) Value of the Put Option (Put-Call Parity)** Put-Call Parity: \[ P = C + Ke^{-rT} - S \] Plug in the values: - \(C = 0.47\) - \(K = 15\) - \(e^{-rT} = 0.9876\) - \(S = 14\) \[ P = 0.47 + 15 \times 0.9876 - 14 \] \[ = 0.47 + 14.814 - 14 \] \[ = 0.47 + 0.814 \] \[ = 1.284 \] **So, the value of the put option is approximately \( \boxed{1.28} \).** --- ## **Final Answers** - **Value of Call Option:** \( \boxed{0.47} \) - **Value of Put Option:** \( \boxed{1.28} \)

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