#### Subject 13. The Black model

To price European options on futures we can use the Black model.

c = e -r(c) T [f0(T) N(d1) - X N(d2)]

p = e -r(c) T {X [1 - N(d2)] - f0(T) [1 - N(d1)]}

where

d1 = [ln(f0(T)/X) + (σ2/2)T] / (σ T1/2)
d2 = d1 - σ T1/2
f0(T) = the futures price

The volatility, σ, is the volatility of the continuously compounded change in the futures price.

Note:

• This model can be obtained directly from the Black-Scholes-Merton formula.
• As with the Black-Scholes-Merton formula, this model applies to European options only. We can also use the model for American options on forwards as they are never exercised early.

Example

A forward contract is priced at 65. European options on the forward contract have an exercise price of 70 and expire in 180 days. The continuously compounded risk-free rate is 5.25%, and volatility is 0.17. Calculate the prices of a call option and a put option on the forward contract.

First calculate T: T = 180/365 = 0.4931.

Then calculate the values of d1 and d2.
d1 = [ln(65/70) + (0.17)2/2 x 0.4931] / (0.17 x 0.49311/2) = -0.5612. d2 = -0.5612 - 0.17 x 0.49311/2 = -0.6805.

Using the normal distribution table,
N(d1) = N(-0.5612) = 1 - N(0.5612) = 1 - 0.7123 = 0.2877.
N(d2) = N(-0.6805) = 1 - N(0.6805) = 1 - 0.7517 = 0.2483.

c = e-0.0525 x 0.4931(65 x 0.2877 - 70 x 0.2483) = 1.2858.
p = e-0.0525 x 0.4931[70 x (1 - 0.2483) - 65 x (1 - 0.2877)] = 6.1580.

Calculate the value of a European interest rate option using the Black model.

An interest rate call option based on a 90-day underlying rate has an exercise rate of 5.5% and expires in 90 days. The forward rate is 5.25%, and the volatility is 0.08. The continuously compounded risk-free rate is 4%. Let's calculate the price of the interest rate call option using the Black model.

The time to maturity is T = 90/365 = 0.2466.
f0(T) = 0.0525.
X = 0.055.
d1 = [ln(f0(T)/X) + (σ2/2)T] / (σ T1/2) = [ln(0.0525/0.055) + 0.082/2 x 0.2466] / (0.08 x 0.2466)1/2 = -1.1512.
d2 = d1 - σ T1/2 = -1.1512 - 0.08 x 0.2466 1/2 = -1.1910.
N(d1) = N(-1.1512) = 1 - N(1.1512) = 0.1251.
N(d2) = N(-1.1910) = 1 - N(1.1910) = 0.117.

c = e-0.04 x 0.2466 (0.0525 x 0.1251 - 0.055 x 0.117) = 0.000131.

We need to make two adjustments:

• The answer is given under assumption that the option payoff occurs at the option expiration. However, this interest rate option expires in 90 days and pays off 90 days that that. Therefore, we need to use the forward rate to discount the result back from day 180 to day 90: 0.000131 x e -0.0525 x (90/365) = 0.00013.

• As the underlying rate and exercise rate are expressed as annual rates, the answer is an annual rate. However, interest rate option prices are often quoted as periodic rates. We need to convert the result to periodic rate based on a 90-day rate and using the customary 360-day year: 0.00013 x (90/360) = 0.0000325.

#### Practice Question 1

A forward contract is priced at 129. A European option on the forward contract has an exercise price of 135 and expires in 49 days. The continuously compounded risk-free rate is 3.75%, and volatility is 0.25. Calculate the prices of a call option and a put option on the forward contract.
T = 49/365 = 0.1342.
Then calculate the values of d1 and d2.
d1 = [ln(129/135) + (0.25)2/2 x 0.1342] / (0.25 x 0.13421/2) = -0.4505.
d2 = -0.4505 - 0.25 x 0.13421/2 = -0.5421.

Using the normal distribution table,
N(d1) = N(-0.4505) = 1 - N(0.4505) = 1 - 0.6736 = 0.3264.
N(d2) = N(-0.5421) = 1 - N(0.5421) = 1 - 0.7054 = 0.2946.

c = e-0.0375 x 0.1342 (129 x 0.3264 - 135 x 0.2946) = 2.3229.
p = e-0.0375 x 0.1342 [135 x (1 - 0.2946) - 129 x (1 - 0.3264)] = 8.2927.

#### Practice Question 2

An interest rate put option based on a 90-day underlying rate has an exercise rate of 5.5% and expires in 150 days. The forward rate is 5.25%, and the volatility is 0.08. The continuously compounded risk-free rate is 4%. Calculate the price of the interest rate put option using the Black model. The notional principal is \$10 million.
Correct Answer: The time to maturity is T = 150/365 = 0.4110.
f0(T) = 0.0525.
X = 0.055.
d1 = [ln(f0(T)/X) + (σ2/2)/T] / (σ T1/2) = [ln(0.0525/0.055) + 0.082/2 x 0.4110] / (0.08 x 0.41101/2 = -0.8815.
d2 = d1 - σ T1/2 = -0.8815 - 0.08 x 0.41101/2 = -0.9327.
N(d1) = N(-0.8815) = 1 - N(0.8815) = 0.1894.
N(d2) = N(-0.9327) = 1 - N(0.9327) = 0.1762.

p = e-0.04 x 0.4110 [0.055 x (1 - 0.1762) - 0.0525 x (1 - 0.0.1894)] = 0.002708.