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Jack1

  • one year ago

Quick Question: Compound Interest Calculation - Tim borrows 80,000 from the bank at 5.02% p.a. a) Assuming interest is calculated monthly, how much would Tim need to pay weekly to ensure the loan is paid off in 2 years? b)Assuming interest is calculated monthly, how much would Tim need to pay weekly to ensure the loan is paid off in 3 years? plz hlp?

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  1. campbell_st
    • one year ago
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    well the annual interest needs to be changed to monthly 5.02/12 = % per month the number of time period needs to be adjusted t = 2 x 12 = 24 so the future value of the loan is \[A = 80000(1 + \frac{\frac{5.02}{12}}{100})^{24}\] after calculating the future value divide it by 24 to get the monthly repayment

  2. campbell_st
    • one year ago
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    for (b) the time periods increase to 36 so calculate the future value for 36 months of interest using the same formula the divide it by 3 x 52 to get a weekly amount

  3. ganeshie8
    • one year ago
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    I think, this should not be that easy as the Principal changes every month

  4. Jack1
    • one year ago
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    that's what i was curious about... at a fixed amount, this is fine, but is there a formula that takes into account the repayments and the new Principal remaining?

  5. ganeshie8
    • one year ago
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    We may proceed the same way. To keep calculations simple, let the monthly payment be \(M\) (we can divide by 4 to get the weekly payment later)

  6. campbell_st
    • one year ago
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    don't divide monthly by 4... as this will only result in 48 repayments instread of the necessary 52 per year

  7. ganeshie8
    • one year ago
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    That is fine, the interest is being calculated monthly, so we need to anchor to months

  8. campbell_st
    • one year ago
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    well you can treat the same way as an annuity \[N = \frac{M \times (1 + r)^n -1}{r(1 + r)^n}\] N = the amount borrowed, M = montly repayments r = interest rate per compounding period and n = number of time periods

  9. Jack1
    • one year ago
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    x = ((y* ((1+(5.02/1200))^36))-1)/ (5.02/1200)*((1+(5.02/1200))^36) so is this about right? y = monthly repayments = 248.646

  10. campbell_st
    • one year ago
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    I don't think so as you only repay $38700 over the 3 years

  11. Jack1
    • one year ago
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    \(\Huge N = \frac{M \times (1 + r)^n -1}{r(1 + r)^n}\) \(\Huge 80000 = \frac{M \times (1 + \frac{5.02}{1200})^n -1}{\frac{5.02}{1200}(1 + \frac{5.02}{1200})^n}\) \(\Huge 80000 = \frac{M \times (1 + \frac{5.02}{1200})^{36} -1}{\frac{5.02}{1200}(1 + \frac{5.02}{1200})^{36}}\) ==> M = $$248.64 / month?

  12. Jack1
    • one year ago
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    where am i going wrong?

  13. campbell_st
    • one year ago
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    I think the amount need to closer to $700 per week

  14. ganeshie8
    • one year ago
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    I'm getting $3510 for monthly payment

  15. ganeshie8
    • one year ago
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    Presuming Tim will be paying it in 4 installements each month, dividing that by 4 gives the weekly payment of $877.5

  16. Jack1
    • one year ago
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    ok, could u help me find where im messing up my sum though plz? https://goo.gl/cfzl0r

  17. ganeshie8
    • one year ago
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    computation will be easier if we don't use the formula as we don't understand it much

  18. campbell_st
    • one year ago
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    well if you make M the subject its \[M = \frac{80000 \times r(1+r)^{36}}{(1 + r)^{36} - 1}\]

  19. ganeshie8
    • one year ago
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    Let \(M\) be the required monthly payment so that the loan amount \(P\) to be paid off in 24 months Let \(i\) = monthly interest rate = \(\frac{0.0502}{12}\) Then, after 1 month the balance amount, \(B_1\) is given by \[\large B_1 = P(1+i) - M\] yes ?

  20. campbell_st
    • one year ago
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    I'd suggest the repayment is about $2400 per month

  21. Jack1
    • one year ago
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    ok, i'm following so far

  22. ganeshie8
    • one year ago
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    what would be the Balance amount, \(B_2\) after \(2\) months ?

  23. ganeshie8
    • one year ago
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    at the end of 2nd month, the interest is calculated on \(B_1\) and the monthly payment is subtracted : \[\large B_2 = B_1(1+i) - M\]

  24. ganeshie8
    • one year ago
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    So it seems we're ending up with a recurrence relation \[B_{n+1} = B_{n}(1+i) - M\]

  25. ganeshie8
    • one year ago
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    You need to find \(M\) such that the loan gets paid off in 24 months, which is same as saying \(B_{24} = 0\)

  26. ganeshie8
    • one year ago
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    good point! first notice, we have assumed that the payments are done at the end of each month

  27. ganeshie8
    • one year ago
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    So the balance amount at the end of 24th month has to be 0 for the loan to be closed in 24 months

  28. ganeshie8
    • one year ago
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    consequently \(B_{24}\) has to be equal to \(0\), and not \(M\)

  29. ganeshie8
    • one year ago
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    you don't want any balance to be pending after 24 months right ?

  30. Jack1
    • one year ago
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    agreed

  31. Jack1
    • one year ago
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    so first month = B1 = 80334.7 - M

  32. ganeshie8
    • one year ago
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    Lets solve M for the general case, it would be simple, trust me 80334.7 and all looks messy

  33. Jack1
    • one year ago
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    cool, sounds good

  34. ganeshie8
    • one year ago
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    Let \(n+1=24\), and \(B_{23+1} = 0\) \[B_{n+1} = B_{n}(1+i) - M\] becomes \[B_{23+1} =B_{23}(1+i)-M = 0 \implies M = B_{23}(1+i)\]

  35. ganeshie8
    • one year ago
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    Maybe next substitute \(B_{23} = B_{22}(1+i) -M\) and get \[M = (B_{22}(1+i)-M)(1+i) = B_{22}(1+i)^2 - M(1+i)\] recursively substituting we end up with \[M = (B_{22}(1+i)-M)(1+i) = B_{0}(1+i)^{24} - M(1+i)^{23} - M(1+i)^{22} - \cdots -M(1+i)\] isolating \(M\) we get \[M = \dfrac{B_0(1+i)^{24}}{(1+i)^{24}+(1+i)^{23} + \cdots + (1+i)+1}\]

  36. ganeshie8
    • one year ago
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    Not hard to see that the bottom is a geometric series, use the partial sum formula to simplify

  37. Jack1
    • one year ago
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    wow!... this is almost factorial looking

  38. Jack1
    • one year ago
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    ok... whats the partial sum formula?

  39. ganeshie8
    • one year ago
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    \[x^n+x^{n-1} + \cdots + x+1 = \dfrac{x^n-1}{x-1}\]

  40. Jack1
    • one year ago
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    ok, so = \(\Large \frac {(1 + i)^{24} - 1}{(1 + i) - 1}\) \(\Large \frac {(1 + i)^{24} - 1}{i}\) ??

  41. Jack1
    • one year ago
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    \(\large M = \dfrac{B_0(1+i)^{24}}{(1+i)^{24}+(1+i)^{23} + \cdots + (1+i)+1}\) \(\Large M = \frac{i \times B_0(1+i)^{24}}{(1+i)^{24} + 1} \) ??

  42. ganeshie8
    • one year ago
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    looks good ! plugging that in and rearranging we end up with a nice looking formula for monthly payment : \[M = B_0*i*\dfrac{(1+i)^n}{(1+i)^n-1}\] where \(B_0\) = loan amount (starting principal) \(i\) = periodic interest rate \(n\) = total number of periods

  43. Jack1
    • one year ago
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    i think i did somethin wrong, as i tried to plug that in and got 333.66 recurring as my answer?? :(

  44. Jack1
    • one year ago
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    yep, found my mistake, 3510.42 = monthly repayment for 2 yr period?

  45. ganeshie8
    • one year ago
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    Yes, for part b, simply change 24 to 36

  46. Jack1
    • one year ago
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    yes, match!

  47. Jack1
    • one year ago
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    awesome, thanks @ganeshie8 and @campbell_st !!

  48. ganeshie8
    • one year ago
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    don't forget the question is asking weekly payment, not monthly

  49. Jack1
    • one year ago
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    sáll good, can multiply it out to the total amount, then divide by 104 for 2 yrs, thanks tho

  50. ganeshie8
    • one year ago
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    that looks like a better idea ! xD

  51. ganeshie8
    • one year ago
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    you want to do : M*24 / 104 is it ?

  52. Jack1
    • one year ago
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    @mathstudent55 u been typing a novel in the background this whole time... what chu workin on man?

  53. mathstudent55
    • one year ago
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    \(\large A = \dfrac{Pr(1 + r)^n}{(1 + r)^n - 1} \) \(\large A = \dfrac{80000\times \frac{0.0502}{12}(1 + \frac{0.0502}{12})^{24}}{(1 + \frac{0.0502}{12})^{24} - 1} \) \(\large A = $3510.43/month \) \(\large A = $810.10/week\) -------------------------------------- \(\large A = \dfrac{Pr(1 + r)^n}{(1 + r)^n - 1} \) \(\large A = \dfrac{80000\times \frac{0.0502}{12}(1 + \frac{0.0502}{12})^{36}}{(1 + \frac{0.0502}{12})^{36} - 1} \) \(\large A = $2398.39/month \) \(\large A = $532.98/week\)

  54. Jack1
    • one year ago
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    = $810.10 per week for a 2 yr loan, and = $553.47 per week for a 3 yr loan... i think? wait how did u get 532 per week?

  55. mathstudent55
    • one year ago
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    Sorry. i copied the number incorrectly. $2398.39/month = $553.47/week

  56. mathstudent55
    • one year ago
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    You are correct.

  57. Jack1
    • one year ago
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    all good, thanks both!

  58. Jack1
    • one year ago
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    and sorry it didnt turn out to be a quick question (looks like i lied in the OP) :-|

  59. ganeshie8
    • one year ago
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    It is a quick question if you had memorized the formula

  60. ganeshie8
    • one year ago
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    Too many complicated formulas to be memorized in financial math

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