My Math Forum Problem with circular logic for loan payments...

 Linear Algebra Linear Algebra Math Forum

 August 16th, 2017, 05:35 PM #1 Newbie   Joined: Sep 2016 From: Florida Posts: 17 Thanks: 0 Problem with circular logic for loan payments... I think this is in the right section. Here is my problem: I am trying to figure out how much of a loan I need to take out. The project costs \$200,000 and the loan is 6% over 30 years. It is interest-only for the first 10 years, and I will pay the whole principal back after 4 months. So I will have to make 4 interest payments and that's it. The problem is that I need to take out more money to account for the loan payments that I will make. The interest payments come to \$1000/month, so that is \$4000 for 4 months. But this in turn means that I need to take out \$204,000 to cover those payments. This comes to \$1,020/month or \$4,080 over 4 months. But now this means that I need to take out \$204,080. This is \$1,020.40/month or \$4,081.60 over 4 months. The problem keeps repeating itself. Is there a way to solve this mathematically? Last edited by skipjack; August 16th, 2017 at 07:54 PM. August 16th, 2017, 05:46 PM #2 Senior Member Joined: May 2016 From: USA Posts: 1,047 Thanks: 430 Quote:  Originally Posted by farmerjohn1324 I think this is in the right section. Here is my problem: I am trying to figure out how much of a loan I need to take out. The project costs 200,000 dollars and the loan is 6% over 30 years. It is interest-only for the first 10 years, and I will pay the whole principal back after 4 months. So I will have to make 4 interest payments and that's it. The problem is that I need to take out more money to account for the loan payments that I will make. The interest payments come to 1000 dollars per month, so that is 4000 dollars for 4 months. But this in turn means that I need to take out 204,000 dollars to cover those payments. This comes to 1,020 dollars per month or 4,080 dollars over 4 months. But now this means that I need to take out 204,080 dollars. This is 1,020.40 dollars per month or$4,081.60 dollars over 4 months. The problem keeps repeating itself. Is there a way to solve this mathematically?
Dollar signs have a special meaning on this site. DON'T USE THEM.

Now that I know what the problem is, I can think about it.

 August 16th, 2017, 06:45 PM #3 Senior Member   Joined: Aug 2017 From: United Kingdom Posts: 202 Thanks: 60 Math Focus: Algebraic Number Theory, Arithmetic Geometry You would want your loan to be 204082 dollars (to the nearest dollar).
August 16th, 2017, 07:19 PM   #4
Newbie

Joined: Sep 2016
From: Florida

Posts: 17
Thanks: 0

Quote:
 Originally Posted by cjem You would want your loan to be 204082 dollars (to the nearest dollar).
Okay but do you have a formula for this so I can find the answer for different loan amounts?

 August 16th, 2017, 08:27 PM #5 Global Moderator   Joined: Dec 2006 Posts: 19,162 Thanks: 1638 If you borrow \$204,081.64, you have \$200,000 for your project and \$4,081.64 left over, which will cover four monthly interest payments of \$1,020.41. If the interest will total an nth part of the project cost, divide that cost by n - 1 (and round up to the next cent) to find how much extra you need to borrow. You might be able to round down if the monthly interest payments are also rounded down. Thanks from JeffM1
 August 16th, 2017, 08:55 PM #6 Senior Member   Joined: May 2016 From: USA Posts: 1,047 Thanks: 430 It's late. I may be wrong because I am very sleepy, but I think this is a limit problem. As the amount borrowed goes up, the amount of interest goes up. But it goes up by a declining amount. It may just be the limit of an arithmetic series.
 August 16th, 2017, 11:26 PM #7 Global Moderator   Joined: Dec 2006 Posts: 19,162 Thanks: 1638 It's the limit of a geometric series.
 August 17th, 2017, 05:25 AM #8 Senior Member   Joined: May 2016 From: USA Posts: 1,047 Thanks: 430 Now that I am awake. Let i = monthly interest rate. Let m = months until payoff. Let n = needed amount. Let b = borrowed amount. $b = n + bim \implies n = b - bim = b(1 - im) \implies$ $b = \dfrac{n}{1 - im}.$ In your case, n = 200000; i = 0.06 / 12 = 0.005, and n = 4. $b = \dfrac{200000}{1 - 4 * 0.005} = \dfrac{200000}{1 - 0.02} = \dfrac{200000}{0.98} \approx 204081.63.$ Virtually the same answer as skipjack. Let's see how it works. Monthly interest due = $0.005 * 204081.63 \approx 1020.41.$ Total interest = $4 * 1020.41 = 4081.64.$ Sorry about last night. Brain fogged by fatigue. I should have done nothing til morning.
August 18th, 2017, 09:57 AM   #9
Newbie

Joined: Sep 2016
From: Florida

Posts: 17
Thanks: 0

Quote:
 Originally Posted by skipjack If you borrow \$204,081.64, you have \$200,000 for your project and \$4,081.64 left over, which will cover four monthly interest payments of \$1,020.41. If the interest will total an nth part of the project cost, divide that cost by n - 1 (and round up to the next cent) to find how much extra you need to borrow. You might be able to round down if the monthly interest payments are also rounded down.
What is the value of n in my case?

Can you write that in a formula?

August 18th, 2017, 09:58 AM   #10
Newbie

Joined: Sep 2016
From: Florida

Posts: 17
Thanks: 0

Quote:
 Originally Posted by JeffM1 Now that I am awake. Let i = monthly interest rate. Let m = months until payoff. Let n = needed amount. Let b = borrowed amount. $b = n + bim \implies n = b - bim = b(1 - im) \implies$ $b = \dfrac{n}{1 - im}.$ In your case, n = 200000; i = 0.06 / 12 = 0.005, and n = 4. $b = \dfrac{200000}{1 - 4 * 0.005} = \dfrac{200000}{1 - 0.02} = \dfrac{200000}{0.98} \approx 204081.63.$ Virtually the same answer as skipjack. Let's see how it works. Monthly interest due = $0.005 * 204081.63 \approx 1020.41.$ Total interest = $4 * 1020.41 = 4081.64.$ Sorry about last night. Brain fogged by fatigue. I should have done nothing til morning.
Thanks.

 Tags circular, loan, logic, payments, problem

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post farmerjohn1324 Algebra 3 September 21st, 2016 11:55 AM farmerjohn1324 Linear Algebra 1 September 9th, 2016 07:12 AM Russell368 Economics 3 July 7th, 2011 09:50 PM mikjel Algebra 1 January 10th, 2010 06:19 AM symmetry Algebra 3 January 30th, 2007 01:04 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top