Results 1 to 6 of 6

Math Help - Interest rate

  1. #1
    Junior Member
    Joined
    Mar 2010
    Posts
    60

    Interest rate

    Hello

    A credit K (100'000) is to be repayed within t (5) years, including interest rate p (5.25%), by an constant annual sum b.

    My idea was the following: Generally each year K/t $ must be payed back. Additionnal to that, the interest rate on the credit must be payed. In the first year the interested rate must be payed on 100'000$, the second year on 80'000$ (...) and in the end (fift year) on 20'000$

    As the ammount payed each month must be constant, i thought that the whole interest rate, that must be payed, can be found out and afterwards distributed to the annual payments. So the whole interest rate to be payed equals p(100'000 + 80'000 + 60'000 (...) + 20'000). The part within the brakets is the sum of the arithmetic series K - (t-1)(K/t). The formula for this sum equals t(K+K/t)/2.

    Thus the whole interest rate to be payed is p(t(K+K/t)/2). This must be "distributed" to each annual payment, which is the reason why the whole formula must be divided by t.
    To sum up: the formula for b is
    K/t + (p(t(K+K/t)/2))/t which is the same as K((t+1)p+2)/2t

    Unfortunately i get the wrong results. Anyone an idea why??
    thank you

    Schdero
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,035
    Thanks
    49
    It might just be me, but I think this is much easier in reverse - say you pay b every year for 5 years into a venture that loses r % per year of any investment. (Make sure you're sure whether to use simple or compound, of course.) Initially you pay b, and a year later your investment is worth (100-r)/100 b, plus another b that you then invest. After another year you'll have [(100-r)/100]^2 b (or (100-2r)/100 b if simple) left from the initial layout, (100-r)/100 b from the second, and another b. Etc. Five years later, what's the expression in terms of b, and what's it equal to?

    r isn't 5.25, though, and percentages are muddling, anyway - better to say that a year after investing b you have 1/1.0525 of b left. (Plus the new investment of b.)

    After 2 years, (1/1.0525)^2 b + 1/1.0525 b + b. ... or, 1/1.105 b + 1/1.0525 b + b if simple interest.

    Etc.

    Just in case it helps, here's a picture of your leaky investment...



    d = 1/1.0525. The last column is of course the final value. (Or the initial loan, to reverse back again.)

    PS: I hope you see why d = 1/1.0525 (... because its the reverse of 105.25%). Substitute this number for d in the last column and set the whole thing equal to 100 000 and you will indeed get the same result as with the formula below.
    Last edited by tom@ballooncalculus; December 26th 2010 at 02:56 PM. Reason: PS
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,861
    Thanks
    742
    Hello, Schdero!

    This is an Amortization problem.
    Are you expected to derive the formula? . . . ack!


    A loan of $100,000 at an interest rate of 5.25%
    is to be repaid in five equal annual payments.

    Find this annual payment.

    Amortization formula: . A \;=\;P\,\dfrac{i(1+i)^n}{(1+i)^n-1}

    . . where: . \begin{Bmatrix}A &=& \text{periodic payment} \\ P &=& \text{principal borrowed} \\ i &=& \text{periodic interest rate} \\ n &=& \text{number of periods} \end{Bmatrix}


    We are given: . \begin{Bmatrix}P \:=\:100,\!000 \\ i \:=\: 0.0525 \\ n \:=\:5 \end{Bmatrix}

    Therefore: . A \;=\;100,\!000\,\dfrac{(0.0525)(1.0525)^5}{(1,0525  )^5-1} \;=\;23,\!257.33168


    You will make an annual payment of $23,257.33.

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Mar 2010
    Posts
    60
    Thank you!
    But you know, what ails me is the logic behind it:
    For sure every year 20'000 are to be repayed. But this doesn't include the interest rate:
    In the first year you need to pay the interest on the whole 100'000 as nothing is repayed yet. In the second year, 20'000 are already payed back, which is why you need to pay the interest rate on the remaining 80'000 only. This goes on like that, until you pay back the last 20'000 plus the interest rate on them. So the total interest you pay is 5.25% on 100'000 + 80'000 + 60'000 + 40'000 + 20'000 = 0.0525*300'000 which is 15'750. This money must be distributed equally to the five payments of 20'000 which are annually made. Thus the annual payment equals 23'150 (20'000 + 15'750/5).
    This is close to the correct result but not quite right. Can anybody help to see the mistake in the logic??
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Oct 2008
    Posts
    1,035
    Thanks
    49
    Quote Originally Posted by Schdero View Post
    For sure every year 20'000 are to be repayed.
    Nope!

    Quote Originally Posted by Schdero View Post
    In the second year, 20'000 are already payed back
    Nope!

    Quote Originally Posted by Schdero View Post
    But you know, what ails me is the logic behind it:

    Can anybody help to see the mistake in the logic??
    Each payment b is a fifth of the total re-payment, not a fifth of the total loan.

    But of course that's bad news, because the total repayment isn't something you're given - it's pretty much the whole problem.

    Reversing everything as I suggested is one way (not the only, but it worked for me, not knowing the formula!) to see how the payments do have to relate to the total loan. Then you have a simple equation to solve.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,115
    Thanks
    68
    Say you pay your calculated $23,150 at end of 1st year: owing becomes 100,000 - 23,150 = 76,850;
    but also owing is the 1st year's interest of 100000*.0525 = 5250: so really owing is 76,850 + 5,250 = 82,100.

    So interest for year#2 is calculated on 82,100 (not on 80,000 ).
    Hence Soroban's 23,257.33 instead of your 23,150.
    The extra 107.33 each year takes care of also paying off the portion of interest that "compounds".

    Your payment would be correct IF the lender agreed to reduce the rate to ~5.0822...%:
    Code:
    YEAR  PAYMENT  INTEREST    BALANCE
     0                       100000.00
     1  -23150.00   5082.20   81932.20
     2  -23150.00   4163.98   62946.18
     3  -23150.00   3199.06   42995.24***
     4  -23150.00   2185.11   22030.35
     5  -23150.00   1119.65        .00
    Example: *** 62946.18 * .050822 = 3199.06

    But don't lose any sleep over it!
    You'll suddenly "see" it later
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Interest Rate Help
    Posted in the Business Math Forum
    Replies: 5
    Last Post: June 11th 2011, 10:47 AM
  2. Calculate interest rate based on interest in parcels
    Posted in the Business Math Forum
    Replies: 1
    Last Post: February 17th 2011, 10:38 PM
  3. Interest Rate
    Posted in the Calculus Forum
    Replies: 14
    Last Post: November 17th 2010, 01:56 PM
  4. Interest Rate
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: January 26th 2010, 07:37 PM
  5. interest rate
    Posted in the Business Math Forum
    Replies: 2
    Last Post: June 9th 2007, 07:37 AM

Search Tags


/mathhelpforum @mathhelpforum