5) Solve the following annuity problems. a) An investor deposits $1000 to an acc

Question

5) Solve the following annuity problems. a) An investor deposits $1000 to an account every December 31 starting in 2005. If the account earns 7.5% from 2005-10 and 6% thereafter, find the accumulated value at December 31, 2017. b) An insurer expects payments of $500m made at the end of each year for the next 25 years. Find the present value assuming interest of 5% for the next 10 years and 3% after 5 years. c) Payments of $300 are made at the end of every year, starting in one year and continuing forever. Find the present value of the payments using 6% interest for the first 10 years and 4.5% thereafter. d) Payments are due at the end of each year for 15 years. The payments start at $50 and increase by 5% every year. Find the present value at the start of the first year, using 6% interest for the first 5 years and 4% thereafter.

Explanation / Answer

1

Value At the end of 2010=1000*1.075^5+1000*1.075^4+1000*1.075^3+1000*1.075^2+1000*1.075^1+1000=7244.02

Value at the end of 2017=7244.02*1.075^7+1000*1.06^6+1000*1.06^5+1000*1.06^4+1000*1.06^3+1000*1.06^2+1000*1.06+1000

=$20412.02

2

Question is incorrect: interest of 5% for the next 10 years and 3% after 5 years..

3.

PV=300/1.06+300/1.06^2+...........300/1.06^10+300/(1.06^10*1.045)+300/(1.06^10*1.045^2)

=300/1.06*(1-1/1.06^10)/(1-1/1.06)+300/(1.06^10*1.045)/(1-1/1.045)

=300/0.06*(1-1/1.06^10)+300/(1.06^10*1.045)/(1-1/1.045)

=$5930.658

4

PV=50/1.06+50*1.05/1.06^2+50*1.05^2/1.06^3+50*1.05^3/1.06^4+50*1.05^4/1.06^5+50*1.05^5/(1.06^5*1.04)+50*1.05^6/(1.06^5*1.04^2)+50*1.05^7/(1.06^5*1.04^3)+..........

=$710.3131

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