R
Rose95
Member
My solution to the problem was:
d1 = [ln(34.55/40) + (2.5%+0.5*0.1^2)*3]/0.1*sqrt(3) = -0.32604 ≈ -0.33
d2 = d1 – 0.1 * sqrt(3) = -0.4992 ≈ -0.50
Φ(d2) = Φ(-0.50) = 1-Φ(0.50) = 1- 0.69146 = 0.30854
Φ(d1) = Φ(-0.33)=1- Φ(0.33) = 1- 0.62930 = 0.3707
c_t = 34.55 * Φ(d1) - 40 * exp (-0.025 * 3) * Φ(d2)
= 1.36
However, the memo gives a slightly higher answer of 1.3998 and different Φ values . Not sure how we are differing. Please help.
d1 = [ln(34.55/40) + (2.5%+0.5*0.1^2)*3]/0.1*sqrt(3) = -0.32604 ≈ -0.33
d2 = d1 – 0.1 * sqrt(3) = -0.4992 ≈ -0.50
Φ(d2) = Φ(-0.50) = 1-Φ(0.50) = 1- 0.69146 = 0.30854
Φ(d1) = Φ(-0.33)=1- Φ(0.33) = 1- 0.62930 = 0.3707
c_t = 34.55 * Φ(d1) - 40 * exp (-0.025 * 3) * Φ(d2)
= 1.36
However, the memo gives a slightly higher answer of 1.3998 and different Φ values . Not sure how we are differing. Please help.