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November 11th, 2015, 07:46 AM   #21
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Joined: May 2008

Posts: 301
Thanks: 81

Beer soaked opinion follows.
Quote:
 Originally Posted by Denis Code: MONTH PAYMENT BALANCE 84 750.00 B ..... 96 750.00 B - 400 (400 net to principal) B = balance owing after 84th payment (end 7th year). p = payment = 750.00 i = 1.09^(1/12) - 1 : u (1 + i)^12 Bu - p(u - 1)/i = B - 400 B = 99,616.379123.....? B - 400 = 99,216.379123... You can do this portion yourself: Calculate n associated with B - 400 : n = 428 (rounded) Total months = 428 + 96 = 524 Calculate PV of 524 payments: 101,645.380261..... SO: the loan is for 101,645, over 524 months (43 years 8 months). Works out fine: checked it...
Extremely close but no cigar.
I can't seem to find my precious notes and I forgot exactly how I arrived at the unknown loan amount; after 2 hours of contemplation, finally got an idea how I may have done it.
Be back in 2 or 3 days. Got to go on a road trip.
Maybe Sir Dexter will have better luck as soon as he stops filibusterng from his book of whatever.

November 11th, 2015, 08:39 AM   #22
Math Team

Joined: Oct 2011

Posts: 14,597
Thanks: 1038

Quote:
 Originally Posted by jonah Extremely close but no cigar.
Of course I'm not "right on"; I told you I rounded
the final payment date from 427.0909... to 428.00

The original loan contract must call for 523 payments
of 750 and a final payment of 750 * .0909... = 68.18

However, borrower needs to make that final payment on
the 2.727th day of the final month (using 30 days!), so such
should be part of the agreement, and also part of the
agreement should be:
.....else 68.18 plus interest for 30 - 2.727 = 27.273 days
if paid on last day of month.

2.727th day means at 5:26:53 PM of 3rd day, so if receipt
of payment is not dated as such, then the whole thing is
out of whack and not precise...make sure your precise
answer keeps that in mind, else I'll sue you!!

Last edited by Denis; November 11th, 2015 at 08:48 AM.

 November 13th, 2015, 01:28 PM #23 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 To wrap this up: amount of loan = 101,629.56025... number of months: 523.09097... owing end of 523rd month: 67.96135... final payment amount: 750 * .09097 = 68.22751... (depends on day and time payment made!)
November 20th, 2015, 12:57 PM   #24
Senior Member

Joined: May 2008

Posts: 301
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Beer soaked computations follow.
Quote:
 Originally Posted by jonah A loan at an annual effective rate of 9% or 9% compounded annually or $\displaystyle {j_1} = 9\%$ is being repaid by monthly payments of \$750 each. The total principal repaid in the 12 monthly installments of the 8th year is \$400. What is the total interest paid in the 12 installments of the 10th year?
The total interest paid in the 12 installments of the 10th year is
$\displaystyle 750(12) - 400(1.09)^{10-8} = 8,524.76$
This answer (less intuitively) makes use of the fact that the entries in the principal repaid column (especially in a no rounding scenario) are in the ratio of 1 + i.
A question that dawned on me at the time was whether it's possible to solve for the amount of the loan and loan term from the given information.
I realized that solving for the loan amount is just a less direct way (but very intuitive way) of solving the total interest paid in the 12 installments of the 10th year. Thus, with
$\displaystyle i=(1.09)^{1/12}-1$
and A = loan amount, we get
$\displaystyle [A(1.09)^{7}-750\frac{{(1 + i)^{7(12)} - 1}}{i}]-[A(1.09)^{8}-750\frac{{(1 + i)^{8(12)} - 1}}{i}] = 400$
where $\displaystyle A = 101,629.560257857...$
In this way,
$\displaystyle [A(1.09)^{9}-750\frac{{(1 + i)^{9(12)} - 1}}{i}]-[A(1.09)^{10}-750\frac{{(1 + i)^{10(12)} - 1}}{i}] = 8,524.76$
It follows then that from
$\displaystyle A=750\frac{{1 - (1 + i)^{-n}}}{i}$
we get n = 523.090970999108...
Quote:
 Originally Posted by Denis To wrap this up: amount of loan = 101,629.56025... number of months: 523.09097... owing end of 523rd month: 67.96135... final payment amount: 750 * .09097 = 68.22751... (depends on day and time payment made!)
Management hereby gives notice that Sir Denis should avoid rounding equivalent monthly rates in future computations. Thus,
$\displaystyle [A(1+i)^{523}-750\frac{{(1 + i)^{523} - 1}}{i}](1+i)=68.4511700645338$
(This according to excel.)
Management thanks you profusely for your comments and computations, Sir D.T. Sober. Rest assured
that such will be duly noted in your Personnel (yes, 2 n's) File, and seriously taken in
consideration at your forthcoming Annual Performance Review (not to be confused with
the APR associated with financial jargon).
Management also wish to apologize to Sir D.T. Sober for the late reply.
Management got involved in some kind of team building thingamajig that didn't allow for beer consumption.

November 20th, 2015, 04:52 PM   #25
Math Team

Joined: Oct 2011

Posts: 14,597
Thanks: 1038

Quote:
 Originally Posted by jonah The total interest paid in the 12 installments of the 10th year is $\displaystyle 750(12) - 400(1.09)^{10-8} = 8,524.76$ This answer (less intuitively) makes use of the fact that the entries in the principal repaid column (especially in a no rounding scenario) are in the ratio of 1 + i.
Forgot to mention that part: I also get exactly that.
But that's simply using the balances owing at end of:
month#108: 98,780.37912...
month#120: 98,305.13912...
Both easily obtainable by future value formulas.

750 * 12 - 98780.367912 + 98305.139.12 = 8524.76

All other amount you show I agree with to the penny;
so why don't I get promoted to the "decision makers group"
to get you transferred to the mail room!!

Anyhoo...perhaps I should go back to drinking:
I used to think I knew everything back then
But...booze gave me a loud mouth, plus
disconnected it from my brain

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