Jack1
  • Jack1
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?
Mathematics
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SOLVED
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chestercat
  • chestercat
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campbell_st
  • campbell_st
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
campbell_st
  • campbell_st
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
ganeshie8
  • ganeshie8
I think, this should not be that easy as the Principal changes every month

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Jack1
  • Jack1
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?
ganeshie8
  • ganeshie8
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)
campbell_st
  • campbell_st
don't divide monthly by 4... as this will only result in 48 repayments instread of the necessary 52 per year
ganeshie8
  • ganeshie8
That is fine, the interest is being calculated monthly, so we need to anchor to months
campbell_st
  • campbell_st
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
Jack1
  • Jack1
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
campbell_st
  • campbell_st
I don't think so as you only repay $38700 over the 3 years
Jack1
  • Jack1
\(\Huge N = \frac{M \times (1 + r)^n -1}{r(1 + r)^n}\) \(\Huge 80000 = \frac{M \times (1 + \frac{5.02}{1200})^n -1}{\frac{5.02}{1200}(1 + \frac{5.02}{1200})^n}\) \(\Huge 80000 = \frac{M \times (1 + \frac{5.02}{1200})^{36} -1}{\frac{5.02}{1200}(1 + \frac{5.02}{1200})^{36}}\) ==> M = $$248.64 / month?
Jack1
  • Jack1
where am i going wrong?
campbell_st
  • campbell_st
I think the amount need to closer to $700 per week
ganeshie8
  • ganeshie8
I'm getting $3510 for monthly payment
ganeshie8
  • ganeshie8
Presuming Tim will be paying it in 4 installements each month, dividing that by 4 gives the weekly payment of $877.5
Jack1
  • Jack1
ok, could u help me find where im messing up my sum though plz? https://goo.gl/cfzl0r
ganeshie8
  • ganeshie8
computation will be easier if we don't use the formula as we don't understand it much
campbell_st
  • campbell_st
well if you make M the subject its \[M = \frac{80000 \times r(1+r)^{36}}{(1 + r)^{36} - 1}\]
ganeshie8
  • ganeshie8
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 ?
campbell_st
  • campbell_st
I'd suggest the repayment is about $2400 per month
Jack1
  • Jack1
ok, i'm following so far
ganeshie8
  • ganeshie8
what would be the Balance amount, \(B_2\) after \(2\) months ?
ganeshie8
  • ganeshie8
at the end of 2nd month, the interest is calculated on \(B_1\) and the monthly payment is subtracted : \[\large B_2 = B_1(1+i) - M\]
ganeshie8
  • ganeshie8
So it seems we're ending up with a recurrence relation \[B_{n+1} = B_{n}(1+i) - M\]
ganeshie8
  • ganeshie8
You need to find \(M\) such that the loan gets paid off in 24 months, which is same as saying \(B_{24} = 0\)
ganeshie8
  • ganeshie8
good point! first notice, we have assumed that the payments are done at the end of each month
ganeshie8
  • ganeshie8
So the balance amount at the end of 24th month has to be 0 for the loan to be closed in 24 months
ganeshie8
  • ganeshie8
consequently \(B_{24}\) has to be equal to \(0\), and not \(M\)
ganeshie8
  • ganeshie8
you don't want any balance to be pending after 24 months right ?
Jack1
  • Jack1
agreed
Jack1
  • Jack1
so first month = B1 = 80334.7 - M
ganeshie8
  • ganeshie8
Lets solve M for the general case, it would be simple, trust me 80334.7 and all looks messy
Jack1
  • Jack1
cool, sounds good
ganeshie8
  • ganeshie8
Let \(n+1=24\), and \(B_{23+1} = 0\) \[B_{n+1} = B_{n}(1+i) - M\] becomes \[B_{23+1} =B_{23}(1+i)-M = 0 \implies M = B_{23}(1+i)\]
ganeshie8
  • ganeshie8
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 \[M = \dfrac{B_0(1+i)^{24}}{(1+i)^{24}+(1+i)^{23} + \cdots + (1+i)+1}\]
ganeshie8
  • ganeshie8
Not hard to see that the bottom is a geometric series, use the partial sum formula to simplify
Jack1
  • Jack1
wow!... this is almost factorial looking
Jack1
  • Jack1
ok... whats the partial sum formula?
ganeshie8
  • ganeshie8
\[x^n+x^{n-1} + \cdots + x+1 = \dfrac{x^n-1}{x-1}\]
Jack1
  • Jack1
ok, so = \(\Large \frac {(1 + i)^{24} - 1}{(1 + i) - 1}\) \(\Large \frac {(1 + i)^{24} - 1}{i}\) ??
Jack1
  • Jack1
\(\large M = \dfrac{B_0(1+i)^{24}}{(1+i)^{24}+(1+i)^{23} + \cdots + (1+i)+1}\) \(\Large M = \frac{i \times B_0(1+i)^{24}}{(1+i)^{24} + 1} \) ??
ganeshie8
  • ganeshie8
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
Jack1
  • Jack1
i think i did somethin wrong, as i tried to plug that in and got 333.66 recurring as my answer?? :(
ganeshie8
  • ganeshie8
http://www.wolframalpha.com/input/?i=80000*0.0502%2F12*%281%2B0.0502%2F12%29%5E24%2F%28%281%2B0.0502%2F12%29%5E24-1%29
Jack1
  • Jack1
yep, found my mistake, 3510.42 = monthly repayment for 2 yr period?
ganeshie8
  • ganeshie8
Yes, for part b, simply change 24 to 36
Jack1
  • Jack1
yes, match!
Jack1
  • Jack1
awesome, thanks @ganeshie8 and @campbell_st !!
ganeshie8
  • ganeshie8
don't forget the question is asking weekly payment, not monthly
Jack1
  • Jack1
sáll good, can multiply it out to the total amount, then divide by 104 for 2 yrs, thanks tho
ganeshie8
  • ganeshie8
that looks like a better idea ! xD
ganeshie8
  • ganeshie8
you want to do : M*24 / 104 is it ?
Jack1
  • Jack1
@mathstudent55 u been typing a novel in the background this whole time... what chu workin on man?
mathstudent55
  • mathstudent55
\(\large A = \dfrac{Pr(1 + r)^n}{(1 + r)^n - 1} \) \(\large A = \dfrac{80000\times \frac{0.0502}{12}(1 + \frac{0.0502}{12})^{24}}{(1 + \frac{0.0502}{12})^{24} - 1} \) \(\large A = $3510.43/month \) \(\large A = $810.10/week\) -------------------------------------- \(\large A = \dfrac{Pr(1 + r)^n}{(1 + r)^n - 1} \) \(\large A = \dfrac{80000\times \frac{0.0502}{12}(1 + \frac{0.0502}{12})^{36}}{(1 + \frac{0.0502}{12})^{36} - 1} \) \(\large A = $2398.39/month \) \(\large A = $532.98/week\)
Jack1
  • Jack1
= $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?
ganeshie8
  • ganeshie8
$553.47 looks correct http://www.wolframalpha.com/input/?i=%2880000*0.0502%2F12*%281%2B0.0502%2F12%29%5E36%2F%28%281%2B0.0502%2F12%29%5E36-1%29%29*36%2F%2852*3%29
mathstudent55
  • mathstudent55
Sorry. i copied the number incorrectly. $2398.39/month = $553.47/week
mathstudent55
  • mathstudent55
You are correct.
Jack1
  • Jack1
all good, thanks both!
Jack1
  • Jack1
and sorry it didnt turn out to be a quick question (looks like i lied in the OP) :-|
ganeshie8
  • ganeshie8
It is a quick question if you had memorized the formula
ganeshie8
  • ganeshie8
Too many complicated formulas to be memorized in financial math

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