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Annuities investment question

  1. Jun 6, 2005 #1
    i cant get these questions.

    how much is each payment for an investment of $32000... an interest rate of 12% per annum, compounded semi-annually, with payments every 6 months, starting in half a year.

    - i tried it but i cant seem to get the right answer... i think it might be the n value... i put one, but then it doesnt say how many years

    i used the formula and i used...

    A lottery to raise funds for a hospital is adverstuising a $240,000 prize. The winner will receieve $1000 every month for 20 years, starting a year from now. If the interest rate is 8.9% per annum, compounded annually, how much must be invested now to have the money to pay this prize?

    the answer i got was like around 1600 short of the real answer in the back of the textbook... but im not sure why.

    i used
    i= 0.089/12
  2. jcsd
  3. Jun 6, 2005 #2
    Your first question is unclear to me. Usually investments have a return on money, so maybe your payment is 0.12/2*32000($1920) every six months.
    In your 2nd question, interest is compounded annually ( paid once at the end of the year ) but the lottery winner is paid monthly so the equivalent interest rate should be (1.089^(1/12))=0.007130287 and not 0.089/12. You seem to be using a calculator so make sure your payment is set for the beginning of the period and not the end.
  4. Jun 6, 2005 #3


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    So in such circumstances where the balance is not static throughout the compounding period should we assume that the interest is calulated on the average balance over the compounding period or on the final balance or perhaps something else. I hate ambiguous problems.
  5. Jun 6, 2005 #4
    Sorry, I didn't read your second question carefully.
    Amount needed now = sum(1000*(1.089^(k/12)), k=0,1,2...239)/1.089. Since one year elapses before the first payment takes place, the sum is divided by 1.089 for the interest earned during the deferral period of one year. If Amt(k)= amount of money right after payment k then Amt(k)*(1.089^(1/12))-1000=Amt(k+1). Interest is calculated on the outstanding amount right after a payment is paid.
  6. Jun 6, 2005 #5
    The exponent should be negative in the sum.
    Amount needed now = sum(1000*(1.089^(-k/12)), k=0,1,2...239)/1.089.
    Amount needed now = 106,131.39
    Using a financial calculator: Pmt@Begin, I%=100*(1.089^(1/12)-1), Pmt=1000/1.089, N=240, calculate PV.
  7. Jun 7, 2005 #6
    i copied the questions right out of my book...im not using a financial calculator just a scientific one... i use the formula that they give me...

    i use the formula...


    and then for interest its the i, its the interest over the compounding periods per year...

    i just substitute numbers in... im not sure what you talking about :confused: cuz im not using a financial or graphing calculator. i asked my dad and he calculated it fine in his financial calculator, but im supposed to be using a scientific one... so im still kinda at sea.
  8. Jun 7, 2005 #7
    Try the following on your scientific calculator: Pv=(R/(1.089))*[1-(1+i)^-n]/i*(1+i) where
    Hopefully this will bring you back to land.
  9. Jun 9, 2005 #8


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    Well in that case I get a slightly differnt answer than anyone else here. I get $108,146.81. Here is the working.

    Let P bet the amount invested and r = (1+I/100) = 1.089

    The amount after each year is as follows.
    0 : P
    1 : rP
    2 : (rP - 12000)r = Pr^2 - 12000r
    3 : (Pr^2 - 12000r -12000)r = Pr^3 - 12000(r^2 + r)
    ... see the pattern ...

    20 : Pr^20 - 12000(r^19 + r^18 + ... r^1)

    Therefore Pr^19 = 12000(r^18 + r^17 + ... 1)
    = 12000 (r^19-1) / (r-1)

    P = 12000 ( 1 - r^(-19) ) / (r-1)
    = $108,146.81
  10. Jun 9, 2005 #9


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    Or if you want to save a bit of money then make arrange to have the final payment for each year deducted just after the interest is added (instead of just before). This cuts the amount down to $107,410.27

    In this case the working is as follows.

    0 : P
    1 : rP
    2 : (rP-11000)r - 1000 = Pr^2 - 11000r - 1000
    3 : (Pr^2 - 11000r -12000)r - 1000 = Pr^3 - 11000r^2 - 12000r - 1000
    4 : (Pr^3 - 11000r^2 - 12000r - 12000)r - 1000 = Pr^4 - 11000r^3 - 12000(r^2+r) - 1000

    20 : Pr^20 - 11000r^19 - 12000(r^18 + r^17 + ... r) - 1000

    So Pr^19 = 11000r^18 + 12000 (r^18 - 1)/(r-1) + 1000/r

    P = 11000/r + 1000/r^20 + 12000/r (1 - r^(-18)) / (r-1)
    = 107,410.27
  11. Jun 9, 2005 #10
    im still a bit confused about what you did... the answer is actually $111,943.89 in the back of my textbook for #2 and the financial calculator confirms this...

    when i calculated it using my formula, i got $9193.97
  12. Jun 9, 2005 #11
    For the first question, what amount of time is this being paid off? Or is that what you mean when you say "but then it doesnt say how many years"?

    I only glanced over LittleWolf's equation, but it looked correct. You should be able to solve the problem using his or her way.
  13. Jun 10, 2005 #12
    I backed into your book's answer of $111,943.89. This answer requires 8.9% interest compounded MONTHLY where the first $1000 monthly payment is expected in one MONTH. Please check if the problem you posted matches the problem in your book. This is the calc for your book 111943.89=100*(1-(1+.089/12)^(-240))/(.089/12).
  14. Jun 10, 2005 #13


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    Confirmed :

    r=1 + 0.089/12

    0 : P
    1 : Pr - 1000
    2 : Pr^2 - 1000r -1000
    3 : Pr^3 - 1000(r^2 + r + 1)

    240 : Pr^240 - 1000(r^219 + r^218 + ... 1)

    P = 1000 (1 - r^(-240))/(r-1)
    = $111943.89
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