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Math Help - Annual Compounded Interest in Amortization - Please Help!

  1. #1
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    Smile Annual Compounded Interest in Amortization - Please Help!

    Hi There, this is my first post. I am taking a business mathematics course, and the present focus is on amoritization. Thus far I have not run into any major problems, but one has now come along: Compounding annually.

    Let me clarify the question, and I'll let you know what I've done so far:

    Greg and Terri have just bought a home, mortgaged for $90,000, which is being amortized over twenty years with monthly payments. If the mortage has an interest rate of 23.21 percent, compounded annually, find:

    a) The monthly payments.
    b) the amount outstanding after the 24th payment.
    c) construct a partial amortization table to show the first two payments and the last two payments of the loan.

    a) This part I think I did correctly. I'm not sure how to type the formula, but because it was compounded annually, I used "f" where f= (1+i) to the power of c - 1. My answer was a monthly payment of $1603.79.

    b) This has completely thrown me for a loop! I understand this process based off a monthly or semi-annual compound period, but I have no idea how to proceed with annual compounding. My textbook only uses monthly, quarterly, semi-annual.

    Thank you so much for your help!
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  2. #2
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    Mismatch between payment period and compounding period always presents a vexing problem. Simplest solution? Fix it.

    Is there a monthly compounding rate that is equivalent to your given annual rate?
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    reply:

    Thanks TK Hunny for your help, sorry if I seem not too bright at this: this one particular course is the one thing keeping me away from a BA in English Literature.

    I don't have any more info about compounding periods, other than what the question provided. I didn't leave out any information that was part of the question. Can you tell me how I can convert my formula to do what you suggested?
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  4. #4
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    English Literature?!

    Well, that paobably is very important information. It is then very likely that the massively complicated solutions I was contemplating simply are not applicable.

    "Compounded" is a little confusing. Fortunately, Nominal and Effective are the same for periods of exactly one year. This means we can propbably pick how we want to solve this one. The most common information is the NOMINAL interest rate. Change the nominal annual rate to monthly simply by dividing by 12. 0.2321 / 12 = .0193416666...

    Having established that, we simply need to know how to calculate the value of remaining payments.

    First, a couple of definitions:

    i = 0.19341667 = the monthly interest rate.

    1+1 = 1.19341667 = monthly accumulation factor

    v = 1/(1+i) = 0.98102535

    P is the level payment amount

    This gives: 90000 = p*(1-v^240)/i and p = $1,758.47 I really can't say how you arrived at your answer.

    In any case, the rest is easy. With the payment, we can just blast through it.

    What is the remaining balance after the last payment? 1758.47*(1-v^0)/i = 0

    What is the remaining balance after the second-to-last payment, just one remaining? 1758.47*(1-v^1)/i = 1,725.10

    What is the remaining balance after the third-to-last payment, just two remaining? 1758.47*(1-v^2)/i = 3,417.47

    240-24 = 216

    What's next?
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  5. #5
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    Quote Originally Posted by NikkiWebber View Post
    My answer was a monthly payment of $1603.79.
    Close enough.
    Accordingly, we have

    <br />
A = R\frac{{1 - \left( {1 + {\textstyle{r \over m}}} \right)^{ - \left( {tm} \right)} }}{{\left( {1 + {\textstyle{r \over m}}} \right)^{{\textstyle{m \over p}}}  - 1}}<br />
<br />
 \Leftrightarrow 90,000 = R\frac{{1 - \left( {1 + {\textstyle{{0.2321} \over 1}}} \right)^{ - \left( {20*1} \right)} }}{{\left( {1 + {\textstyle{{0.2321} \over 1}}} \right)^{{\textstyle{1 \over {12}}}}  - 1}} \Leftrightarrow R \approx {\rm{1,603}}{\rm{.76626785489}}...


    Similarly, by deriving the monthly compounding rate j that is equivalent to your given annual rate of 23.21%, we get

    <br />
\left( {1 + {\textstyle{{0.2321} \over 1}}} \right)^1  = \left( {1 + {\textstyle{j \over {12}}}} \right)^{12}  \Leftrightarrow j \approx {\rm{0}}{\rm{.210545769316394}}...<br />

    and arrive at the same result. Thus

    <br />
90,000 = R\frac{{1 - \left( {1 + {\textstyle{j \over {12}}}} \right)^{ - \left( {20*12} \right)} }}{{{\textstyle{j \over {12}}}}} \Leftrightarrow R \approx {\rm{1,603}}{\rm{.76626785489}}...<br />

    Accordingly, the amount outstanding after the 24th payment (using the prospective method) would be

    <br />
A = R\frac{{1 - \left( {1 + {\textstyle{j \over {12}}}} \right)^{ - \left[ {\left( {\frac{{20*12 - 24}}{{12}}} \right)\left( {12} \right)} \right]} }}{{{\textstyle{j \over {12}}}}} \approx {\rm{89,271}}{\rm{.4733586754}}...<br />

    Using the retrospective method, we have

    <br />
{\rm{90,000}}\left( {1 + {\textstyle{j \over {12}}}} \right)^{24}  - R\frac{{\left( {1 + {\textstyle{j \over {12}}}} \right)^{24}  - 1}}{{{\textstyle{j \over {12}}}}} \approx {\rm{89,271}}{\rm{.4733586754}}...

    Check and compare with my attached amortization schedules.
    Attached Files Attached Files
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