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 interestonly 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:
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  
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. 
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:
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:
 

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