Question

E 4. Assume a 10% interest rate and find the following: a. Value of C if it is at t=0 0 - 1 - b. Value of F if it is at t=4 Jo 25 /DO .-2-3-4 c. Value of A if it is the uniform series from t=1 to t=4 6 - 1-2-3-4

Answer 1

Answer a

The cash flow is a gradient of $25

Hence PV of gradient = G*((1+i)^n-I*n-1)/((1+i)^n*i^2)

=25*((1+10%)^4-10%*4-1)/((1+10%)^4*10%^2)

=25*(1.1^4-0.4-1)/(1.1^4*0.01)

=25*(1.4641-1.4)/(1.4614*0.01)

=25*(0.0641)/0.014641

=$109.45

Answer b

We can break the cash flow into an annuity of $25 for 4 years and a gradient of $25 (same as cashflow in question a)

Hence PV of annuity =A*(1-(1+r)^-n)/r

=25*(1-(1+10%)^-4)/10%

=25*(1-1.1^-4)/0.1

=25*(1-0.683)/0.1

=25*(0.317/0.1)

=$79.25

Hence total PV = 79.25+109.45 = $188.7

FV at t=4 = 188.7*(1+10%)^4 =188.7*1.1^4=188.7*1.4641=$276.27

Answer c

We can break the cash flow into an annuity of $40 for 3 years and a gradient of $40 of 3 years

Hence PV of annuity =A*(1-(1+r)^-n)/r

=40*(1-(1+10%)^-3)/10%

=40*(1-1.1^-3)/0.1

=40*(1-0.7513)/0.1

=40*(0.2487/0.1)

=$99.47

PV of gradient = G*((1+i)^n-I*n-1)/((1+i)^n*i^2)

=40*((1+10%)^4-10%*3-1)/((1+10%)^3*10%^2)

=40*(1.1^3-0.3-1)/(1.1^3*0.01)

=40*(1.331-1.3)/(1.331*0.01)

=40*(0.031)/0.01331

=$93.16

Total PV = 99.47+93.16 = $192.63

Hence annuity for 4 years will be given by PV of annuity =A*(1-(1+r)^-n)/r

or, 192.63=A*(1-(1+10%)^-4)/10%

or, 192.63=A*(1-1.1^-4)/0.1

or, 192.63=A*(1-0.683)/0.1

or, 192.63=A*0.317/0.1

or, A = 192.63*0.1/0.317

or, A = $60.77