Question

Problem 1:

a) An annuity pays into an account 100 at end of year 2, 200 at end of year 3, ..., up to 900 at end of year 10. Interest rate is 7% per year. At end of year 12, how much is in the account

b) An annuity pays 800 at end of year 1, 900 at end of year 2,..., 2000 at end of year 13. What is the present value of the annuity? Use i = .07.

c) An annuity pays 100 per year at end of first 6 years, 100(1.2) per year at end of second 6 years, 100(1.2)2 per year at end of third 6 years, ..., 100(1.2)8 at end of ninth group of 6 years. Find the PV. Use i = .03 as annual interest rate.

Answer 1

a.

FW of account = 100*(F/G,7%,10)*(F/P,7%,2)

= 100*54.520685*1.144900

= 6242.07

b.

PW of account = 800*(P/A,7%,13) + 100*(P/G,7%13)

= 800*8.357651 + 100*42.330185

= 10919.14

c.

FW at end of first 6 yr period = 100*(F/A,3%,6) = 100*6.468410 = 646.84

Effective interest rate for 6 yr period = (1+0.03)^6 - 1

= (1.03)^6 - 1

= 1.194052297 - 1

= 0.194052297

Present worth of geometric series = A*[1-(1+g)^n/(1+i)^n]/(i-g)

PV of given series = 646.84*[1-(1+0.2)^9/(1+0.194052297)^9]/(0.194052297-0.2)

= 646.84*[1-(1.2)^9/(1.194052297)^9]/(-0.005947703)

= 646.84 * 7.68929444

= 4973.74