Question

1. The Moon company currently sells for 100 $. The annual stock price volatility is %10 and risk- free interest rate %8, the price of a call on a company’s stock with strike price 200 $ and time period 2 months. Find the stock option price with Black and Scholes Model. If option market value 320 $ what is the option strategy?

2. The Moon company currently (t1) sells for 100 $ . The annual stock price volatility is %10 and risk- free interest rate %8, the price of a call on a company’s stock with strike price 120 $ and time period 2 months. Changes in parameters in period t2 are as shown in the table.

t2

Price 110

volatility %8

period 3 months

Risk free interest rate %5

a) What is the option price changes if delta value 0.05?

b) What is the Gamma value of the stock?

c) What is the Vega value of the stock?

d) What is the option price changes if theta value 0.04?

e) What is the Rho value of stock?

3. A loan of 20000 at a rate of 5% is paid off in one years, compute the revised payment and create payment table.

Answer 1

1. find the formula:

$c=S0\times N(d1)-K\times N(d2)\times DF$

$d1=\frac{ln(\frac{S0}{K})+(r+\frac{\sigma^2}{2})\times t}{ \sigma\times t^{0.5}}$

$d2=d1-\sigma\times t^{0.5}$

$DF=e^{-r\times t}$

S0=100

K=200

$\sigma=0.1$

r=0.08, 0.0067 years

t=2, 0.167 years

$DF=e^{-0.0067\times0.167}=0.999$

$d1=\frac{ln(\frac{100}{200})+(0.0067+\frac{0.1^2}{2})\times 0.167}{ 0.1\times0.167^{0.5}}=-0.6455$

$d2=-0.6455-0.1\times 0.167^{0.5}=0.6863$

$c=100\times N(-0.6455)-200\times N(-0.6863)\times0.999=-23.27$

K=320

$DF=e^{-0.0067\times0.167}=0.999$

$d1=\frac{ln(\frac{100}{320})+(0.0067+\frac{0.1^2}{2})\times 0.167}{ 0.1\times0.167^{0.5}}=-1.1155$

$d2=-1.1155-0.1\times 0.167^{0.5}=-1.1563$

$c=100\times N(-1.1155)-320\times N(-1.1563)\times0.999=-26.33$

2.

similar to the first task

2 month:

$DF=e^{-0.004167\times0.167}=0.999$

$d1=\frac{ln(\frac{100}{120})+(0.0067+\frac{0.1^2}{2})\times 0.167}{ 0.1\times0.167^{0.5}}=-0.1347$

$d2=-0.1347-0.1\times 0.167^{0.5}=-0.1755$

$c=100\times N(-0.1347)-120\times N(-0.1755)\times0.999=-6.94$

3 month:

$DF=e^{-0.0067\times0.25}=0.999$

$d1=\frac{ln(\frac{110}{120})+(0.004167+\frac{0.08^2}{2})\times 0.25}{ 0.08\times0.25^{0.5}}=-0.0409$

$d2=-0.0409-0.1\times 0.25^{0.5}=-0.0809$

$c=110\times N(-0.0409)-120\times N(-0.0809)\times0.999=-2.87$

a)

d2=0.05

$d2=d1-\sigma\times t^{0.5}$

$d2+ \sigma\times t^{0.5}=d1$

$0.05+0.1\times0.167^{0.5}=0.0908$

$c=100\times N(0.0908)-120\times N(0.05)\times0.999=-8.71$

b)

$G=\frac{2,71828182845904^{-n}}{(S\times \omicron)(\times2\pi\times t)^{0.5}}$

$n=\frac{d1^2}{2}+d\times t$

G=0.9808

c)

$V=\frac{1}{100}\times S0\times 2,71828182845904^{-qt}\times\frac{1}{(2\pi)^{0.5}}\times2,71828182845904^{\frac{-d1^2}{2}}$

V=0.1589

d)$T=\frac{1}{T}(-(\frac{S0\sigma2,71828182845904^{-qt}}{2t^0.5}\times\frac{1}{(2\pi)^{0.5}}-rX2,71828182845904^{-rt}N(d2)+qS02,71828182845904^{-qt}N(d1)$

T=0.0205

e)$Rho=K\times t\times2,71828182845904^{-r\times t}\times N(d2)$

Rho=8.60

3.the amount of the monthly payment will be calculated according to the formula:

$x=S(P+\frac{P}{(1+P)^N-1})=20 000(0.004167+\frac{0.004167}{(1+0.004167)^12-1})=1712.15$

we will make a payment schedule

In the first case, the repayment of the annuity monthly loan. In the second case, if we repay the entire loan amount in the first month, then no further payments need be made.