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Math Help - Annuity problem changing deposit after 8 years

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    Annuity problem changing deposit after 8 years

    I am totaly stuck on this problem been at it for over an hour so my search for help has guided me to this forum

    A young executive deposits $300 at the end of each month for 8 years and then increases the deposits. If the account earns 7.2%, compounded monthly, how much (to the nearest dollar) should each new deposit be in order to have a total of $400,000 after 25 years



    stuck I think i am suppose to calc the simple annuity for 8 years then minus from the total of 400,000 then calc the remander for 17 years anyone help?
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    Quote Originally Posted by joey36 View Post
    I am totaly stuck on this problem been at it for over an hour so my search for help has guided me to this forum

    A young executive deposits $300 at the end of each month for 8 years and then increases the deposits. If the account earns 7.2%, compounded monthly, how much (to the nearest dollar) should each new deposit be in order to have a total of $400,000 after 25 years



    stuck I think i am suppose to calc the simple annuity for 8 years then minus from the total of 400,000 then calc the remander for 17 years anyone help?
    After 8 years, Accumulated Value = 38,792.47. I got that here using 7.2/12 as interest per payment and 12 * 8 = 96 months:

    Annuity Immediate Accumulated Value

    You need to accumulate 400,000 -38792.47 = 361,207.53 in 12(25 - 8) = 204 months. Solving for payment on the same link, we get 907.42 monthly deposits needed to accumulate to your scenario.
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    How many times have I said it? "Basic Principles"! If all you have going for you is a formula or two, you never will make it.

    If p = Original Payment = $300.00 and
    If a = Additional Payment and
    If i = .072/12 = 0.006 and
    If c = (1+i) <== Monthly Accumulation Factor

    We have:

    [p + pc + pc^2 + ... + pc^(25*12)] + [a + ac + ac^2 + ... + ac^(17*12)] = 400000

    You should be able to create more convenient expressions for the geometric series in the square brackets.

    I get a = $374.67. You?
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    Quote Originally Posted by joey36 View Post
    I am totaly stuck on this problem been at it for over an hour so my search for help has guided me to this forum

    A young executive deposits $300 at the end of each month for 8 years and then increases the deposits. If the account earns 7.2%, compounded monthly, how much (to the nearest dollar) should each new deposit be in order to have a total of $400,000 after 25 years



    stuck I think i am suppose to calc the simple annuity for 8 years then minus from the total of 400,000 then calc the remander for 17 years anyone help?
    You're on th right track. You want to break this into two problems - first find the future value of the first 8 years of payments when they get to year 25, then add the value of an annuity from years 9 - 25 so that the sum of the two is $400K. For the first part, it's an annuity calculation for the first 8 years, followed by simple compound interest on that investment for years 19-25. You then subtract that from $400K, and the difference is how much he needs to build with an annuity between years 9 and 25.
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    I got the first answer, but i am wondering if it was not right because i was not acoutning for the 38,792.47 in interest
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    Quote Originally Posted by joey36 View Post
    I got the first answer, but i am wondering if it was not right because i was not acoutning for the 38,792.47 in interest
    You got the answer I got for the payment? 907.42? Is there something different in the back of the book.
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    Well this is an even problem and the evens are not in the back of the book
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    Quote Originally Posted by joey36 View Post
    Well this is an even problem and the evens are not in the back of the book
    Joey,

    I just redid it another way, and I get the same answer. I see something similar in Kellison's book, so I'll stick with my answer.
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    Quote Originally Posted by TKHunny View Post
    I get a = $374.67.
    Note: a + p is the value for the last 17 years. $900 is way too much. You're not accumulating interest on the $38,792.47.
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    Quote Originally Posted by TKHunny View Post
    Note: a + p is the value for the last 17 years. $900 is way too much.
    I reread the problem. If I accumulate the 38792.47 17 years, and subtract that from the 400k, I get 256467.16. 17 years @ 8%, the payment becomes 644.30.
    Last edited by mathceleb; July 8th 2008 at 11:48 AM.
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    I am leaving work right now so i have 2 hours to figure this problem out

    I have to acumulate intrest on the 38,792.47 right?

    so after i get 38,792.47 how do i acumulate the intrest then get home much i need to the next 17 years
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  12. #12
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    Closer. Still something a bit off. I couldn't find the error after a couple of checks.

    Quote Originally Posted by TKHunny View Post
    If p = Original Payment = $300.00 and
    If a = Additional Payment and
    If i = .072/12 = 0.006 and
    If c = (1+i) <== Monthly Accumulation Factor

    We have:

    [p + pc + pc^2 + ... + pc^(25*12)] + [a + ac + ac^2 + ... + ac^(17*12)] = 400000
    That appears to be off a bit. It should be...

    [p + pc + pc^{2} + ... + pc^{299}] + [a + ac + ac^{2} + ... + ac^{203}] = 400000

    Or

    p*\frac{1-c^{300}}{1-c} + a*\frac{1-c^{204}}{1-c} = 400000

    Or, since 1-c = -i = -0.006

    p*(c^{300}-1) + a*(c^{204}-1) = 2400

    Or, since p = 300 and c = 1.006

    a*(c^{204}-1) = 894.84096

    Finally, since c = 1.006, still

    a = 374.6693482

    Total Payment after eight years is a+b = 300+374.67 = 674.67

    Let's try the HP-12C

    [g][END]
    [f][CLR][FIN]
    25[g][12x]
    7.2[g][12/]
    300[CHS][PMT]
    [FV] ==> 250859.8401
    400000-[CHS] ==> 149140.1599[FV]
    17[g][12x]
    [PMT] ==> 374.6693479

    It's looking like a pretty good result.
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  13. #13
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    ...and just to emphasize the beauty and nonconfusion of the basic principles...

    In this case, p = 300 = payment for first eight years and
    b = payment for next 17 years (=p+a from the last setup)

    (p + pc + ... + pc^{95})*c^{204} + (b + bc + ... + bc^{203}) = 400000

    or

    p*\frac{1-c^{96}}{1-c}*c^{204} + b*\frac{1-c^{204}}{1-c} = 400000

    Since 1-c = -i = -0.006

    p*(c^{96}-1)*c^{204} + b*(c^{204}-1) = 2400

    Since p = 300 and c = 1.006

    b*(c^{204}-1) = 1611.345497

    More with x = 1.006

    b = 674.6693480

    Unique answers don't care how you find them.
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