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?

- Jack1

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- 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

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

I think, this should not be that easy as the Principal changes every month

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## More answers

- 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

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

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

- ganeshie8

That is fine, the interest is being calculated monthly, so we need to anchor to months

- 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

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

I don't think so as you only repay $38700 over the 3 years

- 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

where am i going wrong?

- campbell_st

I think the amount need to closer to $700 per week

- ganeshie8

I'm getting $3510 for monthly payment

- ganeshie8

Presuming Tim will be paying it in 4 installements each month, dividing that by 4 gives the weekly payment of $877.5

- Jack1

ok, could u help me find where im messing up my sum though plz?
https://goo.gl/cfzl0r

- ganeshie8

computation will be easier if we don't use the formula as we don't understand it much

- campbell_st

well if you make M the subject its
\[M = \frac{80000 \times r(1+r)^{36}}{(1 + r)^{36} - 1}\]

- 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

I'd suggest the repayment is about $2400 per month

- Jack1

ok, i'm following so far

- ganeshie8

what would be the Balance amount, \(B_2\) after \(2\) months ?

- 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

So it seems we're ending up with a recurrence relation
\[B_{n+1} = B_{n}(1+i) - M\]

- 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

good point! first notice, we have assumed that the payments are done at the end of each month

- 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

consequently \(B_{24}\) has to be equal to \(0\), and not \(M\)

- ganeshie8

you don't want any balance to be pending after 24 months right ?

- Jack1

agreed

- Jack1

so first month = B1 = 80334.7 - M

- ganeshie8

Lets solve M for the general case, it would be simple, trust me
80334.7 and all looks messy

- Jack1

cool, sounds good

- 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

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

Not hard to see that the bottom is a geometric series, use the partial sum formula to simplify

- Jack1

wow!... this is almost factorial looking

- Jack1

ok... whats the partial sum formula?

- ganeshie8

\[x^n+x^{n-1} + \cdots + x+1 = \dfrac{x^n-1}{x-1}\]

- Jack1

ok, so
=
\(\Large \frac {(1 + i)^{24} - 1}{(1 + i) - 1}\)
\(\Large \frac {(1 + i)^{24} - 1}{i}\) ??

- 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

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

i think i did somethin wrong, as i tried to plug that in and got 333.66 recurring as my answer?? :(

- 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

yep, found my mistake, 3510.42 = monthly repayment for 2 yr period?

- ganeshie8

Yes, for part b, simply change 24 to 36

- Jack1

yes, match!

- Jack1

awesome, thanks @ganeshie8 and @campbell_st !!

- ganeshie8

don't forget the question is asking weekly payment, not monthly

- Jack1

sáll good, can multiply it out to the total amount, then divide by 104 for 2 yrs, thanks tho

- ganeshie8

that looks like a better idea ! xD

- ganeshie8

you want to do : M*24 / 104 is it ?

- Jack1

@mathstudent55 u been typing a novel in the background this whole time... what chu workin on man?

- 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

= $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

$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

Sorry. i copied the number incorrectly.
$2398.39/month = $553.47/week

- mathstudent55

You are correct.

- Jack1

all good, thanks both!

- Jack1

and sorry it didnt turn out to be a quick question (looks like i lied in the OP) :-|

- ganeshie8

It is a quick question if you had memorized the formula

- ganeshie8

Too many complicated formulas to be memorized in financial math

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