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April 15

Determining effective interest rate?

I'm looking to find out how to determine the "real" interest rate for a loan, if the repayment includes money besides interest. For example, if I have a $100,000 loan with a 6% interest for 240 months, but there is a $80 per-month cost regardless of balance for the first 60 months, what is the "real" interest rate for the first 5 years? This isn't a homework question. It's based on a real-life problem, PMI. I can use an amortization calculator to determine payments based on the interest and loan term, but I don't know how to get an interest rate out when there is a fixed cost involved. RudolfRed (talk) 03:34, 15 April 2013 (UTC)[reply]

You need to find the discount rate at which the present value of all future cash flows (interest payments, insurance premiums and principal repayment at the end of the loan) equals the current value of the loan, which is $100,000. Since your additional insurnace premiums only apply for the first 5 years of the loan, this probably isn't something that a vanilla amortization calculator can handle, unless it allows you to vary the amounts of payments during the term of the loan. A quick "back of an envelope" calculation suggests that the effective interest rate in your case is approximately 6.35%. Gandalf61 (talk) 10:15, 15 April 2013 (UTC)[reply]

(ec)

Mathematically this is quite simple. You have

0=\sum _{i}\;C_{i}\;(1+p)^{-t_{i}}

where the

C_{i}

are the cash flows, changing sign on whether you receive or pay money. The

t_{i}

is the time for the i-th cash flow, as difference in years from the first cash flow. Now you only have to solve the equation for p, which is the interest rate (math-style, so p=0.05 would be 5 percent). Solving the equation is mostly done by approximation, or, in standard cases, by specialized formula.

In real live this gets more complicated. Think of what to do with leap years and if you are allowed (or forced) to assume every month has 30 days. See Present_value. 95.112.217.255 (talk) 10:28, 15 April 2013 (UTC)[reply]

Suppose I lended you $100 two years ago, last year you payed me $230, and now I pay you $132 and we are even. The effective interest rate may be 10% because 100 × 1.1² − 230 × 1.1 + 132 = 121 − 253 + 132 = 0. The effective interest rate may also be 20% because 100 × 1.2² − 230 × 1.2 + 132 = 144 − 276 + 132 = 0. Bo Jacoby (talk) 13:00, 15 April 2013 (UTC).[reply]

Is that relevant here? I don't think multiple solutions can happen when the unique payment in one direction (e.g., a typical unidirectional loan) is at the beginning. --Tardis (talk) 14:02, 15 April 2013 (UTC)[reply]

Really? :Suppose I lended you $100 two years ago, last year you payed me $10, and now you pay me $72 and we are even. The effective interest rate may be −10% because 100 × 0.9² − 10 × 0.9 − 72 = 81 − 9 − 72 = 0. The effective interest rate may also be −180% because 100 × (−0.8)² − 10 × (−0.8) + 72 = 64 + 8 − 72 = 0. Bo Jacoby (talk) 21:50, 15 April 2013 (UTC).[reply]

When we consider that interests should act exponentially, instead of (1-p) it really should be

e^{\lambda \;t_{i}}

, so the second result from your second example is excluded. But your first example still gives me trouble. I can't find anything wrong neither in the formula nor in your example. What stuns me most is the fact that only a total of 2$ effectively change owner. I guess I can choose from the interest rates, depending if I have to pay income tax or if I can use it for tax deduction. 77.3.188.115 (talk) 10:28, 16 April 2013 (UTC)[reply]

Bo's first example is, in effect, a sequence of two one-year loans in opposite directions. There is a loan from A to B of $100, repaid at the end of year 1 with interest of $x, then a loan from B to A of $130 - x, repaid at the end of year 2 with interest of $2 + x. The combining of the repayment of the first loan with the start of the second loan hides the value of x. The annual interest rates for the two loans are respectively x/100 and (2 + x)/(130 - x). Bo is simply pointing out that there are two values of x which make these two interest rates equal i.e. x = $10 and x = $20. Which is interesting, but not very relevant to the OP's question. Gandalf61 (talk) 11:05, 16 April 2013 (UTC)[reply]

If p is the effective interest rate, and c_i is the cash flow at year no i = 1,2,...,n, then x = 1+p satisfies Σ_i c_i xⁿ⁻ⁱ = 0. It is the rule, rather than the exception, that equations of degree n−1 have n−1 solutions. See Fundamental theorem of algebra. Bo Jacoby (talk) 13:54, 16 April 2013 (UTC).[reply]

True in general, but beware of spherical cows. When finding the effective interest rate for a loan with repayments c_i we are solving Σ_i c_i x⁻ⁱ = P where P is the loan principal, the c_i are all positive and we are only interested in solutions for positive x. Since Σ_i c_i x⁻ⁱ is a monotonically decreasing function of x which tends to infinity as x tends to 0 and tends to 0 as x tends to infinity, there will be exactly one feasible solution for a given positive value of P. Other solutions only appear if you remove the constraint that all c_i are positive (as in your first example) or if you allow negative x (as in your second example). Gandalf61 (talk) 14:26, 16 April 2013 (UTC)[reply]

(outdent) What is "the loan principal"? If you have some contractor building "a new town somewhere in the west", his bank account will frequently change sign, being negative when he ordered building materials or paid monthly wages, or positive when he receives intermediate payments. I this case, the polynomial in question really 'is' very general. 77.3.188.115 (talk) 17:51, 16 April 2013 (UTC)[reply]

The loan principal or principal sum is the original amount of the loan. Yes, there can be examples where there are mixed cash flows in both directions, but this does not occur in the case of a simple loan analysis. Your example has gone further afield, into the more complex area of project finance. Gandalf61 (talk) 08:19, 17 April 2013 (UTC)[reply]

I did not assume that mixed cash flow does not occur, nor that the interest factor is positive. Gandalf makes these extra assumption in order to avoid extra solutions. (cf. spherical cow). Bo Jacoby (talk) 08:23, 18 April 2013 (UTC).[reply]

Here's how I'd do it:

1) Calculate the total amount (T) you will repay for the loan. The bank should provide you with a payment schedule. If they don't outright give you the total, then you will have to total up all the payments. If they don't include the $80 fee for each of the first 60 months, you can add that in, too ($80 × 60 = $4800).

2) Use this formula to determine the effective rate (R_e): $100000(1 + (R_e/100))²⁰ = T. There's a way to solve this with logs, or you can just do trial and error for different values of R_e until you close in on the answer.

3) Note that this does not include the effect of inflation. Since the fees are in the first few years, and inflation is low now, this effect should be fairly minimal. That is, the value of the $4800 is pretty close to $4800 today.

And, by the way, I agree with your approach. Banks toss all sorts of fees into loans and don't count those as interest, making it near impossible to compare loans. Also note that, if comparing different loans with the same periods, you can just compare the total of all fees and interest for each. Back-end fees (those due at the end of the loan) would really need to be adjusted downward for inflation, but banks generally charge their fees up front, in which case the effect of inflation is minimal. One tricky bit is the effect of late payments. You'd need to realistically calculate the number of late payments you might make, and figure the late fees, and any effect on the interest rate, into the total. StuRat (talk) 13:16, 15 April 2013 (UTC)[reply]

StuRat - your formula $100000(1 + (R_e/100))²⁰ = T assumes a single bullet repayment of amount T at the end of the loan period. This won't work when there are regular monthly repayments throughout the term of the loan - it will underestimate the actual interest rate. Gandalf61 (talk) 14:10, 15 April 2013 (UTC)[reply]

Thanks for all the replies RudolfRed (talk) 03:20, 16 April 2013 (UTC)[reply]