she wants to begin her retirement with monthly payouts of \$1,500. the retirement account will last 50 years. What lump sum deposit must she make into her payout annuity at the beginning of her retirement? part b) mary will use an ordinary annuity with an annual interest rate of 12% at Gringotts to come up with that deposit amount. She will make monthly deposits for the next 30 years. find her monthly payments.

When she retires, Mary will want to receive \$1500 a month for 50 years, that’s 50×12=600 months. The total amount she would receive over 50 years=600×1500=\$900,000. But she gains interest on the lump sum deposit. The interest decreases with each payout. We need to work out the lump sum D (to be funded by her annuity). The monthly interest rate is 10.25/12=0.8542% approx=0.008542. Let’s assume that she gets interest immediately on the lump sum, so on her first payout the lump sum will grow to 0.1025D/12+D or D(1+0.1025/12). The payout reduces this to D(1+0.1025/12)-1500. This then accrues interest:

(D(1+0.1025/12)-1500)(1+0.1025/12) and then the next payout occurs:

(D(1+0.1025/12)-1500)(1+0.1025/12)-1500. Expanding this we get:

D(1+0.1025/12)²-1500-1500(1+0.1025/12)=

D(1+0.1025/12)²-1500(1+(1+0.1025/12)).

Over 50 years (600 months) we get:

D(1+0.1025/12)⁶⁰⁰-1500(1+(1+0.1025/12)+(1+0.1025/12)​²+...+(1+0.1025/12)​⁵⁹⁹).

This expression must equal zero because after 50 years there is no payout. The expression can be simplified:

D(1+0.1025/12)⁶⁰⁰-1500((1+0.1025/12)⁶⁰⁰-1)/(0.1025/12).

Therefore D=(1500((1+0.1025/12)⁶⁰⁰-1)/(0.1025/12​))÷((1+0.1025/12)⁶⁰⁰).

Let G, the growth factor,=(1+0.1025/12)⁶⁰⁰=164.55 approx. So D=18000(G-1)/(0.1025G).

From this, D=\$174,542.57 and will ensure Mary will receive \$1500 a month for 50 years of retirement, that is, will cover \$900,000 in payouts.

Her retirement is funded by whatever she earns in her annuity. The annual interest rate is 12% so the monthly interest rate is 1%, or 0.01. If she makes a monthly payment of m for 30 years (360 months) she will accrue m×1.01³⁶⁰=35.95m approx with the first payment, m×1.01³⁵⁹=35.59m approx with the second payment, and so on, until the last payment which accrues 1.01m. Add these increments together and we get a series:

m(1.01+1.01²+...+1.01³⁶⁰)=1.01m(1+1.01+...+1.01³⁵⁹). This can be expressed as 1.01m(1.01³⁶⁰-1)/0.01=101m(1.01³⁶⁰-1).

This will be her lump sum deposit D when she retires. This amount must cover her pension payouts, so:

1.01m(1.01³⁶⁰-1)/0.01=D; 3529.91m=D; so m=D/3529.91.

Using the calculated value of D from the first part, m=\$49.45.

by Top Rated User (610k points)