My Math Forum APR with fee payment

 Economics Economics Forum - Financial Mathematics, Econometrics, Operations Research, Mathematical Finance, Computational Finance

January 23rd, 2013, 10:56 AM   #11
Math Team

Joined: Oct 2011

Posts: 14,597
Thanks: 1038

Re: APR with fee payment

Quote:
Originally Posted by CRGreathouse
Quote:
 Originally Posted by Denis Disagree. Should be A = p(1 - 1/R^n) / (R - 1)
Depends on the assumptions. That would be correct if the first payment is immediate, mine is correct if the first payment is delayed a month (as stated in my post).
Code:
0                      1500.00
1  -510.03    15.00    1004.97
2  -510.03    10.05     504.99
3  -510.03     5.04        .00
Dat dere, she's a 3 month loan: $1,500 borrowed, 12% annual cpd monthly (R = 1.01); required monthly payment =$510.03, 1st payment 1 month after loan obtained.

Yours: pR(R^n - 1) / (R - 1) = 510.03(1.01)(1.01^3 - 1) / (1.01 - 1) = ~1561

Mine: p(1 - 1/R^n) / (R - 1) = 510.03(1 - 1/1.01^3) / (1.01 - 1) = ~1500

So me is confused...

 January 23rd, 2013, 01:07 PM #12 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: APR with fee payment To JMATH17; to illustrate reference to geometric series... Take the $1,500 loan from my previous post. It all works this way: assume no payments on loan, instead the payments are EACH put in a savings account at same rate as loan: Code: n int balance int balance int balance int balance 0 .00 1500.00 1 15.00 1515.00 .00 510.03 2 15.15 1530.15 5.10 515.13 .00 510.03 3 15.30 1545.45 5.16 520.29 5.10 515.13 .00 510.03 The savings accounts pay off the loan: 1545.45 = 520.29 + 515.13 + 510.03 1500(1 + .01)^3 = 510.03(1 + .01)^2 + 510.03(1 + .01)^1 + 510.03(1 + .01)^0 Let R = 1 + .01 ; rewrite above: AR^n = pR^(n-1) + pR^(n-2) + pR^(n-3) Rearrange, and perform following: Code: AR^n = pR^0 + pR^1 + pR^2 + ...... + pR^(n-1) AR^n * R = pR^1 + pR^2 + ...... + pR^(n-1) + pR^n ================================================================ AR^n * R - AR^n = -pR^0 + 0 + 0 + ...... + 0 + pR^n AR^n(R - 1) = p(R^n - 1) A = p(R^n - 1) / [R^n(R - 1)] Substituting back R = 1+i and simplifying: A = [1 - 1/(1+i)^n] / i Hope that helps... January 23rd, 2013, 05:18 PM #13 Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: APR with fee payment Quote:  Originally Posted by JMATH17 How do I get the APR for a loan that has a 2 year interest only period, followed by a 3 year amortising balance (interest & principal) terms are 5 year loan notional$5000 Fixed interest rate throughout the loan (6.5%) \$100 loan fee (paid immediately) 2 years interest only 3 remaining years, interest and principal My thoughts were to average the payments and use that to calculate the APR but i'm not sure if that is correct...
No change in APR. What makes you think there is?
Where this you get this problem anyway?

2 years later, owing will be 5000(1 + .065/12)^24 = 5692.14
So effectively a loan of that amount will be obtained then, at 6.5 rate...

 Tags apr, fee, payment

,

### mortgage formula in geometric series

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Denis Algebra 1 July 8th, 2016 05:22 PM Denis New Users 1 June 5th, 2013 09:45 AM newtleeg Algebra 1 March 19th, 2011 09:41 AM sivela Algebra 1 May 16th, 2010 06:48 PM Jamers328 Advanced Statistics 1 January 26th, 2009 07:40 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top