# Calculating loan + interest recursively

I am trying to calculate a loan amount where the total principle depends on the payment amount.

Here's a simple example, leaving aside that a) no bank would actually do this, and b) my income is not specified:

I have $0 in the bank right now. I want to take out a loan for a $100 car, for which I will be making a regular annual payment, starting at the end of year 0 with an interest-only payment. My loan rate is 5%. In order to qualify for the loan, the bank wants to make sure that I have a balance equal to twice my annual payment amount throughout the life of the loan.

If I borrow exactly $100, my payment at the end of year 0 would be $5 ($100 x 5%). However, I need to borrow a little extra so that I have a $10 balance in the account (to satisfy the requirement that I maintain a balance of twice my annual payment). This means I actually need a loan of $110. However, a loan of $110 actually raises my payment amount to $5.50, which means I need to maintain a balance of $11 in my account.

This means that I actually need to borrow $111 to begin with, which raises my payment to $5.55, which means I actually need to borrow $111.10, etc. I have worked it out to a summation that resembles the following, although I can't be certain this is accurate: $$\sum_{n=1}^x (x-1-n)r^n$$ Where$r$is the interest rate I believe this would allow for me to iterate$x$number of times over a formula that sums the additional principle needed for each incremental amount of interest. However, I'm not sure how to carry this out to infinity. Ultimately, I need to plug this into Excel as part of a cash flow, but I wanted to think through the math first. Any ideas? - > to ensure that I have enough additional cash on hand to cover my DSCR.< What do you mean by that? Getting a bigger loan doesn't increase your income, only your payments. So the bigger loan you take, the worse off you are. Am I missing something? – fedja May 30 '13 at 17:45 @fedja The DSCR stipulates that at any one time I must have cash on hand equal to some percentage of my monthly payment (for example, 150%). So if my monthly payment is $10 on a $1,000 loan, I need $15 in the bank at all times. In this particular instance, I'm not concerned with my income (which is dealt with elsewhere) but with making sure I have $15 in the bank. My problem is that if I decide I actually need a $1,500 loan with a $12 monthly payment, but I only have $15 in the bank, how much extra do I need to borrow to have a bank balance that covers 150% of my $12 payment. – spencerrecneps May 30 '13 at 18:30 I'm still not understanding you. Let's bring the situation to its simplest terms. You have a 2 months loan. The monthly payment is 60% of the total loan amount (so the bank wants to get 20% profit; this is, by the way, always true: the monthly payment is just proportional to the loan amount if the term and the interest rate are fixed). You really need \$1000. The DSCR requirement is 100%. You have \$100 in the bank. What exactly do you intend to do? This should be a simple pen and paper exercise if you know what you are talking about. Once I see it, I think we can figure out the rest. :) – fedja May 30 '13 at 23:00 @fedja I've re-worked the question to hopefully make things more clear. I removed the references to DSCR, since I think the definition of DSCR for my purposes was causing some confusion. – spencerrecneps May 31 '13 at 1:58 Which means that you take the loan, pay the sum you need for the car, leave the rest on the account, and, most importantly, never touch the account any more but make payments from some alternative sources? If that is all correct, it makes sense and I can guide you through the relevant math. Just confirm that I'm not confused now. – fedja May 31 '13 at 2:08 ## 3 Answers Assuming that I finally understood the problem correctly, the math is as follows. Let$M$be the money you need to spend,$L$be the money borrowed,$r$the monthly interest rate,$K$the ratio of the required balance to the monthly payment. Your monthly payment (interest only) is$Lr$. You need to have the balance$KLr$(which you never touch). You have nothing now and have to spend$M$right away. The leftover from the loan is$L-M$, so we must have$L-M=KLr$, i.e.,$L=M/(1-Kr)$. Of course, this works under the assumptions I described. Other strategies may result in a different answer. - In the Wikipedia article you link to, cash on hand does not increase the DSCR. Unless the cash generates income, borrowing more lowers the DSCR by the ratio of the new payment to the old payment. If the cash can generate income greater than the old DSCR times the payment, borrowing will help the DSCR. The new income enters into the Annual Net Income term. If I inherit a bunch of money, but deposit it at zero interest, my DSCR doesn't change. Paying down the loan in a way that lowers the payment would help. - Sorry if I wasn't clear. I'm not looking to change my DSCR. See my above response to fedja for an example scenario. – spencerrecneps May 30 '13 at 18:29 If you need$x$times the loan in the bank (which is the product of the payment fraction and the coverage amount-in your example it is$0.015) then to be able to spend $p$ you need to borrow $\frac p{1-x}$. This accounts for coverage of the increased load. I don't see how interest comes into it. – Ross Millikan May 30 '13 at 20:02
I'm assuming an interest-only payment occurs in the same year that I've taken out the loan, so the interest affects my calculation because I need to make sure I borrow enough to cover my original debt needs, but also enough that I can maintain a balance in the bank equal to 150% of my payment amount. In borrowing additional money to maintain that 150% amount, I've also increased my monthly payment by some amount, which necessitates further borrowing a smaller amount of money to add to my balance in the bank to make sure I have the 150% requirement satisfied, ad infinitum. – spencerrecneps May 30 '13 at 20:51
Please state your question precisely. You don't have a DSCR, you have a covenant to maintain a certain cash balance. If you need to consider an interest only payment, say that. How many payments do you have to make and still meet the covenant? The basic logic I gave in the previous comment applies-if you have one payment to make and still meet the covenant, count that in the balance. – Ross Millikan May 30 '13 at 21:50
I'm happy to see I'm not the only one who cannot understand the question :). Let's see if spencerrecneps can show us what he means in the simple model situation I proposed. Until then, I'm at loss because I simply cannot understand how borrowing more can possibly help with anything without generating more income and the income is not part of the equation at this stage. – fedja May 30 '13 at 23:12

Rearrange summation into simpler terms:

$$\sum_{n=1}^x (x-1-n)r^n$$

$$\sum_{n=1}^x (x - 1)r^n - \sum_{n=1}^x nr^n$$

$$(x - 1)\sum_{n=1}^xr^n - \sum_{n=1}^x nr^n$$

Carry out to infinity, as you ask:

$$\lim_{x\to\infty}\left[(x - 1)\sum_{n=1}^xr^n - \sum_{n=1}^x nr^n\right]$$

Now we look here for the closed forms of these finite sum series.

$$\lim_{x\to\infty}\left[(x - 1)\left(\frac{1 - r^{x+1}}{1 - r} - 1\right) - r\frac{1-(x+1)r^x+xr^{x+1}}{(1-r)^2}\right]$$

But unfortunately, it looks like this won't converge to a limit. Since $0 \leq r < 1$, the $r^x$ and $r^{x+1}$ factors vanish and we are left with:

$$\lim_{x\to\infty}\left[(x - 1)\left(\frac{1}{1 - r} - 1\right) - r\frac{1}{(1-r)^2}\right]$$

This is just $Ax + B$ which has no limit as $x$ grows large.

But maybe the above ideas hint you in the direction of finding a closed form for your calculation so that you can avoid iterating or recursing.

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Hmmm. The limit ought to approach 0. I wonder if my stated summation is not correct. Conceptually, each iteration of calculating the needed principle amount to maintain the DSCR should be increasingly small. Likewise, the additional interest paid on that additional principle will be slightly less than the previous iteration, and will increase my need for cash on hand slightly less each time. Eventually, the additional principle, interest, and resulting change to cash on hand becomes infinitesimal. – spencerrecneps May 30 '13 at 19:49
@spencerrecneps Since $x$ represents the number of iterations, which we intend to grow large, maybe it should not be factored into the formula in that way. Likewise, did I screw up somewhere ... – Kaz May 30 '13 at 20:23