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Math Help - PLease help!!

  1. #1
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    PLease help!!

    Please help me solve this problem

    Parents have two children aged 10 and 5 years now. The parents set up a savings account at an interest rate of 11,5% per hatf-year that will provide each child with and equal lump sum of Rx on their 18th birthday. If the parents deposit consecutive, equal, annual amounts of R11500 into the savings account starting one year from now and ending when the last child turns eighteen, then the value of Rx (to the nearest rand) is:

    thanks in advance
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  2. #2
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    Re: PLease help!!

    In your class notes or your text, what formulas do you have for present and future value? What formula do you have (if any) for an annuity?
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  3. #3
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    Re: PLease help!!

    I've got a future value annuity that's fv= payment((1+I)^n -1)/I and pv=payment(1-(1+I)^-n)/I
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    Re: PLease help!!

    OK. First there is something of an ambiguity in how the problem is given. What is the relationship of birthdays to when the deposits are made? What is the relationship between the birthdays of the two children? Are there fractional years involved? Is there a payment made on the 18th birthday? Another ambiguity is exactly what value is being asked for. The intuitive answer is how much is given to each child, is that correct?

    Probably it is best to give the exact and complete problem so we are not guessing.

    How many deposits will be made, assuming that the younger child turned 5 today, the first deposit will occur 1 year from today, and the last will occur 13 years from today? What will the future value of those deposits be 13 years from today.
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    Re: PLease help!!

    That is the while question.
    The possible answers are(rounded off):
    199844
    230841
    189951
    213445
    200233

    I'd be great full if you could just give me a basic idea on how to solve it, how do I make it such that both children get the same about?
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  6. #6
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    Re: PLease help!!

    I just spent an hour writing you an answer, which has disappeared. I'll rewrite it, but I now have other things I must do. It may take a while.
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  7. #7
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    Re: PLease help!!

    Ah man, thank you so much, I don't mind waiting another 3 hours,
    Could you please send it by then ?
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  8. #8
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    Re: PLease help!!

    This is a nightmare to answer because the problem is grossly under-specified. It is not even clear what amount you are supposed to be finding. So I cannot be sure that I am giving you good guidance. I do get an answer that is close to one that is provided to you. The difference probably comes from conventions on rounding and truncation, but I can't be sure.

    I am going to assume that the two children have the same birthday and that today is that birthday. This assumption avoids having to deal with fractional years. I am also assuming that the desired answer is the total amount that is divided evenly between the two children.

    Let's get some preliminaries out of the way.

    The relevant interest rate is $\left(1 + \dfrac{11.5}{2 * 100}\right)^2 - 1 \approx 0.118,306 = 11.8306\%.$ Now your problem may use a different approximation.

    Now this is a question about future values so we will be using those formulas.

    How many deposits are there.

    Well the first occurs next year, when the children turn 6 and 11 respectively. I am assuming that a final deposit is made on the younger child's 18th birthday.

    So that gives us 13 deposits at ages 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, and 18, but we cannot use our formula for those 13 deposits because there is a withdrawal when the older child turns 18. There will be an equal withdrawal when the younger children turns 18.

    $Let\ w = amount\ of\ withdrawal \implies 2w = amount\ given\ to\ children.$

    The older child will turn 18 in 8 years. What will be in the account after the eighth deposit is made.

    The answer is: $11,500 * \dfrac{\left(1 + 0.118,306\right)^8 - 1}{0.118,306} \approx 140,574.99.$ That part is just the formula.

    How much is left in the account when the older child receives w. Obviously, $140,574.99 - w.$

    Now 5 more deposits will be made at the start of years 9, 10, 11, 12, and 13. How much will that accumulate to over the five years from year 8 to year 13. Again this is just the formula.

    $11,500 * \dfrac{\left(1 + 0.118,306\right)^5 - 1}{0.118,306} \approx 72,812.23.$

    But this may not be all that will be in the account. Some money may have been left in the account back when the withdrawal was made for the older child. If so, that money will still be there plus accumulated interest. So we have to use the other future value formula.

    Amount left over plus accumulated interest = $(140,574.99 - w) * (1 + 0.118,306)^5 = 245,873.28 - 1.749,054w.$

    So the total in the account to give to the younger child is $72,812.23 + 245,873.28 - 1.749,054w = 318,685.51 - 1.749,054w.$

    Wonderful, but what is w? Well the amount in the account when the younger child turns 18 is to be w. Oh goody: we have an equation.

    $w = 318,685.51 - 1.749,054w \implies 2.749,054w = 318,685.51 \implies w = 115,925.52 \implies 2w = 231,851.04.$

    That looks pretty close to one of the answers you are given to choose among. When I work this out using excel, it comes out within 0.03. See what happens if you use 11.83% instead of 11.8306%.
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    Re: PLease help!!

    Thanks a million, you a genius
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  10. #10
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    Re: PLease help!!

    Quote Originally Posted by Zdesignzi9250 View Post
    That is the while question.
    The possible answers are(rounded off):
    199844
    230841
    189951
    213445
    200233

    I'd be great full if you could just give me a basic idea on how to solve it, how do I make it such that both children get the same about?
    None of your possible answers came close to mine. Consider the following expression:
    11,500* [1-(1+w)^(-13)]/w = x (1+w)^(-13) + x (1+w)^(-8)

    where w=(1+.115/2)^2-1=.11830625
    Solve for x.

    I'm somewhat surprised that the indefatigable math knight-errant Sir Wilmer/Denis didn't come to your aid sooner. For the record, I just finished my 6th beer bottle; so I wouldn't bet on me solution too much. Not too worry though, I'm sure Sir Wilmer/Denis will come along shortly to challenge my answer and guide you along, and lecture me for posting drunk again. Cheers.
    Last edited by jonah; May 9th 2014 at 11:33 PM.
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    Re: PLease help!!

    Quote Originally Posted by jonah View Post
    None of your possible answers came close to mine. Consider the following expression:
    11,500* [1-(1+w)^(-13)]/w = x (1+w)^(-13) + x (1+w)^(-8)

    where w=(1+.115/2)^2-1=.11830625
    Solve for x.

    I'm somewhat surprised that the indefatigable math knight-errant Sir Wilmer/Denis didn't come to your aid sooner. For the record, I just finished my 6th beer bottle; so I wouldn't bet on me solution too much. Not too worry though, I'm sure Sir Wilmer/Denis will come along shortly to challenge my answer and guide you along, and lecture me for posting drunk again. Cheers.
    I used to have a friend who posted as denis at a different site. Not seen him here unfortunately. It appears to me, however, that you are treating this as a present value problem when it is a future value problem. Not saying that there is not a more elegant way to solve the problem than what I suggested, but this is not it.
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  12. #12
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    Re: PLease help!!

    Quote Originally Posted by JeffM View Post
    I used to have a friend who posted as denis at a different site. Not seen him here unfortunately. It appears to me, however, that you are treating this as a present value problem when it is a future value problem. Not saying that there is not a more elegant way to solve the problem than what I suggested, but this is not it.
    Oh ye math knight-errant of little faith! Trust me Sir JeffM, our good friend Sir Wilmer/Denis will come sooner or later; unless of course he's too busy jousting with other math knight-errants at the moment.
    Anyways, I take it that you're under the impression that my take on this situation is rather hazy; that my present value approach is not the way to do it.
    The beauty of equations of value involving compound interest problems is that the answer will not vary with the choice of location of the focal or comparison date. I simply chose the present as comparison/focal date out of convenience (I didn't look at your analysis at the time as it seemed too lengthy for my heavily intoxicated brain at the time and because it was pure punishment using a smartphone what with the letters being too small and all). Your suggested solution, on the other hand, used the end of 15 years as the comparison date (with a lot of analysis and intermediate rounding involved).
    Equivalently, you could have just as easily summarized the whole problem with the end of 15 years again as comparison date with the following equation of value:
    11,500 * [ (1+w)^13-1]/w = x ( 1 + w ) ^ 5 + x
    where w=(1+.115/2)^2-1=.11830625

    Without any rounding of the effective rate w, you'd see for yourself that x=115,925.620233917...
    (vs. your w=115,925.52, not much difference)
    Btw, if I seem redundant, it's because I just had some strong brandy. Also, how come you specified twice the value of x (w in your case) as the final answer when the OP's problem was just asking for the value of x, thus my comment "None of your possible answers came close to mine." to his/her list of possible answers.
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  13. #13
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    Re: PLease help!!

    Jonah

    First of all, I tend to be erring rather than errant.

    Right you are. I solved your equation and got an answer that differed from mine and assumed that there was something wrong with your answer. I think I probably made a mistake in entering the equation into my calculator. Of course, I may also have not noticed that you and I were answering different questions.

    Anyway, it was dumb of me not to notice that you were equating the present value of two equal but non-simultaneous future values with the present value of the annuity used to create those future values. Very elegant solution. I wish I had thought of it.

    Now as to why we got different answers. The question was very badly specified, were the birthdays on the same date or not, how far away from the present were the birthdays, etc. Worse, it appeared to be a multiple choice problem. Consequently I could not work backwards from the answer to what the premises of the problem actually were. Very frustrating. I eventually concluded that the question was asking for the total amount to be paid out at the two different birthdays. That did give me an answer close to one of the ones that were proposed. You made a different guess as to what the actual question proposed was. I also had to make a guess as to how the problem was dealing with fractional amounts: truncation, rounding, frequency, etc. Who knows what the author of the problem intended? I don't particularly like multiple choice questions.

    Now what is most interesting to me is why I didn't hit on your elegant solution. (Well one possibility is that I am stupid, but I tend to reject that answer without many grounds for doing so.) I think the main reason is that I tended to present solutions to such problems in the form of a spreadsheet because people understood the spreadsheet and feared the formulas as magic. (I used the formulas myself to make sure I had not erred in constructing the spreadsheet.) Anyway that got me focused on the accrual aspect of the residual in the account after the first withdrawal.

    Anyway, in case any student comes along and goes over this thread, it may be worthwhile to show that there are different methods to use the four basic formulas to arrive at the same answer. The formulas will almost always give an approximate answer because the actual answer will depend on the specific details of how interest computations are rounded or truncated.

    My way, which is future and process oriented, sees that the account will be accumulating on an annuity basis for 8 years, then will have a withdrawal that leaves a residual, which will earn interest for five years, to which will be added an annuity for 5 years, leading to a final and equal withdrawal.

    That gives the following equation for w, the amount of one withdrawal, and uses both future value formulas.

    $\left(11,500 * \dfrac{1.11830625^8-1}{0.11830625} - w\right)(1.11830625)^5 + \left(11,500 * \dfrac{(1.11830625^5-1)}{0.11830625}\right) = w \implies$

    $2.7490562w \approx 11,500(12.2239232 * 1.7490562 + 6.3315) \implies w \approx \dfrac{11,500 * (21.38033 + 6.3315}{2.7490562} \approx 115,925.62.$

    This differs from my answer above because of a more exact approximation of the relevant interest rate. Furthermore, if the question is asking about the total amount disbursed, then 2w is the correct answer. That is approximately 231,851, which is close to one of the proposed answers, namely 230,840.

    Jonah's way, which is present oriented and elegant, sees that, over time, the value of the deposits and withdrawals must equal, which means that their present values are equal. So he equates the present value of the deposits, using the present value of an annuity formula) to the sum of the present values of the two withdrawals.

    That gives the following equation for w, the amount of one withdrawal, and uses both present value formulas.

    $11,500 * \dfrac{1 - 1.11830625^{-13}}{0.11830625} = w\left(1.11830625^{-8} + 1.11830625^{-13}\right) \implies$

    $0.64253w \approx 74,485.7623 \implies w \approx 115,925.73.$

    The difference in the answers is negligible and is the result of rounding and truncation errors.
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  14. #14
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    Re: PLease help!!

    Quote Originally Posted by JeffM View Post
    Jonah's way, which is present oriented and elegant...
    It is indeed present oriented but a week of no alcohol made me realize that I must have been more drunk (or enchanted, as Don Quixote would claim) than I thought at the time to type "I simply chose the present as comparison/focal date out of convenience ...". The end of 15 years is definitely the better comparison date as it involved less computations and no neqative exponents. Anyways, just finished me 5th beer and me beer goggles just appeared. Cheers.
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