A 65 year old man intends to use his retirement funds to purchase an annuity from a life insurance company.?

July 17, 2010 - 1:01 pm 1 Comment

given the amount of money the man has available to invest, the insurance company is able to offer two alternatives. the first option is to receive $2785 each month for as long as he lives; the second option is to receieve $3500 each month, but for only 20 years (payments will be made to his estate if he should die before that time) the relevant interest rate is 6 percent per year. how long must the man live so that the first option is a better deal?

can someone plz help me answer this question for my finance assignment?

Let:
p be the amount of each payment,
r be the fractional interest rate per year,
n be the number of years,
k be the number of payments and compounding periods per year,
s be the sum invested.

The present value of payments is the sum of a geometric series with first term p and common ratio (1 + r / k)^(-1):
s = p sum(i = 0 to nk – 1) (1 + r / k)^(-ki)
s= p(1 – (1 + r / k)^(-nk)) / (r / k)
s = (pk / r)(1 – (1 + r / k)^(-nk)) …(1)

Solving for n:
sr / kp = 1 – (1 + r / k)^(-nk)
1 – sr / kp = (1 + r / k)^(-nk)
log(1 – sr / kp) = – nk log(1 + r / k)
n = – log(1 – sr / kp) / k log(1 + r / k) …(2)

For the 20 year option, (1) gives:
s = (3500*200)(1 – 1.005^(-240))
s = $488,532.70

For the lifetime option, (2) gives:
n = – log(1 – 488,532.70 * 0.06 / (12 * 2785)) / 12 log(1.005)
n = 35.02yr.

One Response to “A 65 year old man intends to use his retirement funds to purchase an annuity from a life insurance company.?”

  1. Rackbrane Says:

    Let:
    p be the amount of each payment,
    r be the fractional interest rate per year,
    n be the number of years,
    k be the number of payments and compounding periods per year,
    s be the sum invested.

    The present value of payments is the sum of a geometric series with first term p and common ratio (1 + r / k)^(-1):
    s = p sum(i = 0 to nk – 1) (1 + r / k)^(-ki)
    s= p(1 – (1 + r / k)^(-nk)) / (r / k)
    s = (pk / r)(1 – (1 + r / k)^(-nk)) …(1)

    Solving for n:
    sr / kp = 1 – (1 + r / k)^(-nk)
    1 – sr / kp = (1 + r / k)^(-nk)
    log(1 – sr / kp) = – nk log(1 + r / k)
    n = – log(1 – sr / kp) / k log(1 + r / k) …(2)

    For the 20 year option, (1) gives:
    s = (3500*200)(1 – 1.005^(-240))
    s = $488,532.70

    For the lifetime option, (2) gives:
    n = – log(1 – 488,532.70 * 0.06 / (12 * 2785)) / 12 log(1.005)
    n = 35.02yr.
    References :

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