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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?
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?

This Question is Closed

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1well 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
 one year ago
Best ResponseYou've already chosen the best response.1for (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
 one year ago
Best ResponseYou've already chosen the best response.3I think, this should not be that easy as the Principal changes every month

Jack1
 one year ago
Best ResponseYou've already chosen the best response.2that'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
 one year ago
Best ResponseYou've already chosen the best response.3We 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
 one year ago
Best ResponseYou've already chosen the best response.1don't divide monthly by 4... as this will only result in 48 repayments instread of the necessary 52 per year

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3That is fine, the interest is being calculated monthly, so we need to anchor to months

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1well 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
 one year ago
Best ResponseYou've already chosen the best response.2x = ((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
 one year ago
Best ResponseYou've already chosen the best response.1I don't think so as you only repay $38700 over the 3 years

Jack1
 one year ago
Best ResponseYou've already chosen the best response.2\(\Huge N = \frac{M \times (1 + r)^n 1}{r(1 + r)^n}\) \(\Huge 80000 = \frac{M \times (1 + \frac{5.02}{1200})^n 1}{\frac{5.02}{1200}(1 + \frac{5.02}{1200})^n}\) \(\Huge 80000 = \frac{M \times (1 + \frac{5.02}{1200})^{36} 1}{\frac{5.02}{1200}(1 + \frac{5.02}{1200})^{36}}\) ==> M = $$248.64 / month?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1I think the amount need to closer to $700 per week

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3I'm getting $3510 for monthly payment

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Presuming Tim will be paying it in 4 installements each month, dividing that by 4 gives the weekly payment of $877.5

Jack1
 one year ago
Best ResponseYou've already chosen the best response.2ok, could u help me find where im messing up my sum though plz? https://goo.gl/cfzl0r

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3computation will be easier if we don't use the formula as we don't understand it much

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.1well if you make M the subject its \[M = \frac{80000 \times r(1+r)^{36}}{(1 + r)^{36}  1}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Let \(M\) 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
 one year ago
Best ResponseYou've already chosen the best response.1I'd suggest the repayment is about $2400 per month

Jack1
 one year ago
Best ResponseYou've already chosen the best response.2ok, i'm following so far

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3what would be the Balance amount, \(B_2\) after \(2\) months ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3at 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
 one year ago
Best ResponseYou've already chosen the best response.3So it seems we're ending up with a recurrence relation \[B_{n+1} = B_{n}(1+i)  M\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3You need to find \(M\) such that the loan gets paid off in 24 months, which is same as saying \(B_{24} = 0\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3good point! first notice, we have assumed that the payments are done at the end of each month

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3So the balance amount at the end of 24th month has to be 0 for the loan to be closed in 24 months

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3consequently \(B_{24}\) has to be equal to \(0\), and not \(M\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3you don't want any balance to be pending after 24 months right ?

Jack1
 one year ago
Best ResponseYou've already chosen the best response.2so first month = B1 = 80334.7  M

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Lets solve M for the general case, it would be simple, trust me 80334.7 and all looks messy

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Let \(n+1=24\), 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
 one year ago
Best ResponseYou've already chosen the best response.3Maybe 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
 one year ago
Best ResponseYou've already chosen the best response.3Not hard to see that the bottom is a geometric series, use the partial sum formula to simplify

Jack1
 one year ago
Best ResponseYou've already chosen the best response.2wow!... this is almost factorial looking

Jack1
 one year ago
Best ResponseYou've already chosen the best response.2ok... whats the partial sum formula?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\[x^n+x^{n1} + \cdots + x+1 = \dfrac{x^n1}{x1}\]

Jack1
 one year ago
Best ResponseYou've already chosen the best response.2ok, so = \(\Large \frac {(1 + i)^{24}  1}{(1 + i)  1}\) \(\Large \frac {(1 + i)^{24}  1}{i}\) ??

Jack1
 one year ago
Best ResponseYou've already chosen the best response.2\(\large M = \dfrac{B_0(1+i)^{24}}{(1+i)^{24}+(1+i)^{23} + \cdots + (1+i)+1}\) \(\Large M = \frac{i \times B_0(1+i)^{24}}{(1+i)^{24} + 1} \) ??

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3looks good ! plugging that in and rearranging we end up with a nice looking formula for monthly payment : \[M = B_0*i*\dfrac{(1+i)^n}{(1+i)^n1}\] where \(B_0\) = loan amount (starting principal) \(i\) = periodic interest rate \(n\) = total number of periods

Jack1
 one year ago
Best ResponseYou've already chosen the best response.2i think i did somethin wrong, as i tried to plug that in and got 333.66 recurring as my answer?? :(

Jack1
 one year ago
Best ResponseYou've already chosen the best response.2yep, found my mistake, 3510.42 = monthly repayment for 2 yr period?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Yes, for part b, simply change 24 to 36

Jack1
 one year ago
Best ResponseYou've already chosen the best response.2awesome, thanks @ganeshie8 and @campbell_st !!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3don't forget the question is asking weekly payment, not monthly

Jack1
 one year ago
Best ResponseYou've already chosen the best response.2sáll good, can multiply it out to the total amount, then divide by 104 for 2 yrs, thanks tho

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3that looks like a better idea ! xD

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3you want to do : M*24 / 104 is it ?

Jack1
 one year ago
Best ResponseYou've already chosen the best response.2@mathstudent55 u been typing a novel in the background this whole time... what chu workin on man?

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.0\(\large A = \dfrac{Pr(1 + r)^n}{(1 + r)^n  1} \) \(\large A = \dfrac{80000\times \frac{0.0502}{12}(1 + \frac{0.0502}{12})^{24}}{(1 + \frac{0.0502}{12})^{24}  1} \) \(\large A = $3510.43/month \) \(\large A = $810.10/week\)  \(\large A = \dfrac{Pr(1 + r)^n}{(1 + r)^n  1} \) \(\large A = \dfrac{80000\times \frac{0.0502}{12}(1 + \frac{0.0502}{12})^{36}}{(1 + \frac{0.0502}{12})^{36}  1} \) \(\large A = $2398.39/month \) \(\large A = $532.98/week\)

Jack1
 one year ago
Best ResponseYou've already chosen the best response.2= $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
 one year ago
Best ResponseYou've already chosen the best response.3$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%5E361%29%29*36%2F%2852*3%29

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.0Sorry. i copied the number incorrectly. $2398.39/month = $553.47/week

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.0You are correct.

Jack1
 one year ago
Best ResponseYou've already chosen the best response.2and sorry it didnt turn out to be a quick question (looks like i lied in the OP) :

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3It is a quick question if you had memorized the formula

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Too many complicated formulas to be memorized in financial math
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