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Can people tell me how monthly payments are calculated when a mortgage has an initial rate?

What is the formula?

I have seen online calculators but not formulae.

My guess is:

We assume that the principal amount paid off each month in the inital period is as if the mortgage has no initial rate, then the payment in the initial period is adjusted for the (often lower) initial rate interest. Is this correct?

For example, let's assume I have a 25 years mortgage, which is on 3% for first 5 years, then 4% on the remaining term. How do we calculate the payment?

  • Are you asking about variable rate mortgage? – JTP - Apologise to Monica Mar 14 '16 at 23:35
  • @JoeTaxpayer I am asking: assume, I have a 25 years mortgage, which is on 3% for first 5 years, then 4% on the remaining term, how do we calculate payment? Or more generally, I have an n-year mortgage which has r% interest for first m years, followed by s% in the remaining term, how do we calculate monthly payment. (I know s% here is just SVR, but assume this is known and will never change for simplicity) – Lost1 Mar 14 '16 at 23:36
  • Is this a loan where it is known at the start what the interest rate will be at each point of time over the entire life of the loan, with a constant payment for the entire loan? Or, is it a more typical adjustable rate mortgage (ARM), where the rate for the first few years is fixed, and after that the rate can go up with a corresponding increase in monthly payment? – Ben Miller - Reinstate Monica Mar 15 '16 at 2:13
  • @BenMiller I am thinking about an adjustable mortgage and assume (for simplicity) that the adjustable rate is known after the initial period. – Lost1 Mar 15 '16 at 9:42
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In an adjustable rate mortgage (ARM), the starting interest rate is guaranteed for a certain period. After this period, the rate can go up or down.

The monthly payment on these loans is calculated as if the rate never changed over the life of the loan. However, if the rate does change, the monthly payment also changes to cover the change in interest so that the mortgage is still paid off in the same amount of time.

Using your example, let's say that you have a 25-year mortgage that is a 5-year ARM. The initial interest rate is 3%, which means that for the first 5 years, your rate is fixed at 3%. The monthly payment for those first 5 years is the same as it would be if you had a 25-year fixed rate mortgage at 3%. Here is the formula:

MathJax formula: $$P=L\frac{c(1+c)^n}{(1+c)^n-1}$$

where:

  • P = monthly payment
  • L = Loan amount
  • c = monthly interest rate. This is the annual interest rate divided by 12.
  • n = number of months in the loan (years * 12)

In our example, if the loan is $100,000, the interest rate is 3% (monthly interest rate is 0.25%, or 0.0025), and the number of months is 300 (25 years), the monthly payment will be $474.21.

Now, 5 years into a 25 year mortgage, the amortization schedule tells us that the remaining principal will be $85,505.48.

So if the rate jumps to 4% at that point in time, the monthly payment will be recalculated so that the loan is still paid off in the original 25-year time. To find the new payment, use the above formula again, but this time L=$85,505.48, c=0.04/12=0.0033333, and n=20*12=240. The new monthly payment is $518.15.

If, instead, you had a loan where the payment will be constant over the entire loan period, but the interest rate changes during the period (this is not common), there is a formula for that as well. See this StackOverflow question for the details.

  • +1 Johann Hibschman's formula for a constant payment is correct. – Chris Degnen Mar 15 '16 at 16:15
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Normally in a variable rate mortgage the payment would vary with the rate. However here is a formula for a fixed payment, (where, as the OP says, the rate adjustment is known in advance):

d = (p r1 (1 + r1)^m r2 (1 + r2)^n)/
 (-r1 + (1 + r2)^n (r1 + (-1 + (1 + r1)^m) r2))

where

d is the periodic payment
p is the loan amount
r1 is the periodic rate for the first m periods
r2 is the periodic rate for the next n periods

Here is how the formula is derived.

First, taking a simplified problem to show the workings more clearly.

Let's say a £100,000 loan repaid by 5 annual payments. The first 2 years at 3% and the following 3 years at 4%.

p  = 100,000
r1 = 0.03
m  = 2
r2 = 0.04
n  = 3

The loan amount is equal to the sum of the present value of the payments. These are the present values of the payments for each period, discounted by the interest rate(s):-

pv1 = d/(1 + r1)
pv2 = d/((1 + r1) (1 + r1))
pv3 = d/((1 + r1) (1 + r1) (1 + r2))
pv4 = d/((1 + r1) (1 + r1) (1 + r2) (1 + r2))
pv5 = d/((1 + r1) (1 + r1) (1 + r2) (1 + r2) (1 + r2))

And p = pv1 + pv2 + pv3 + pv4 + pv5

This can be expressed as a summation

enter image description here

and converted to a formula by induction:

p = ((1 + r1)^-m (1 + r2)^-n (-d r1 + 
      d (1 + r2)^n (r1 + (-1 + (1 + r1)^m) r2)))/(r1 r2)

Rearranging to give a formula for the payment:

d = (p r1 (1 + r1)^m r2 (1 + r2)^n)/
 (-r1 + (1 + r2)^n (r1 + (-1 + (1 + r1)^m) r2))

∴ d = 22078.67

Amortization table for the above result showing figures and formulas

enter image description here

Returning to the OP's example for, say, a loan of one million, with the effective rate of interest at 3% for the first 5 years and 4% for the following 20 years.

p  = 1,000,000
r1 = (1 + 0.03)^(1/12) - 1 = 0.00246627
m  = 5*12 = 60
r2 = (1 + 0.04)^(1/12) - 1 = 0.00327374
n  = (25 - 5)*12 = 240

The payment d = 5026.48

Note for the use of nominal rates

For nominal interest rates of 3% and 4% compounded monthly:

p  = 1,000,000
r1 = 0.03/12 = 0.0025
m  = 5*12 = 60
r2 = 0.04/12 = 0.00333333
n  = (25 - 5)*12 = 240

The payment d = 5057.80

  • Is possible to rearrange for r2 if we know what d is? – emehex Feb 3 '17 at 20:48
  • 1
    No, you'd have to solve for r2 numerically, e.g. like this. – Chris Degnen Feb 3 '17 at 21:21
  • Lol. so if my periods are like 3 and 33... basically no chance unless with simulation. – emehex Feb 3 '17 at 21:23
  • Chris, I am trying to extend your formula to add in a third rate of return and can quite easily extend the summation part of your formula but am struggling to convert to a usable formula. Is there any chance that you could show what this should look like please? Thanks in advance! – JWW85 Dec 17 '17 at 13:45
  • If you follow the progression of m and n to o then p = (1/(r1*r2*r3))*(1+r1)^-m*(1+r2)^-n*(1+r3)^-o*(-d*r1*r2+d*(1+r3)^o*(r1*(r2-r3)+(1+r2)^n*(r1+(-1+(1+r1)^m)*r2)*r3)) and d = -(p/(((1+r1)^-m*(1+r2)^-n*(1+r3)^-o)/r3-(1/(r1*r2*r3))(1+r1)^-m*(1+r2)^-n*(r1*(r2-r3)+(1+r2)^n*(r1+(-1+(1+r1)^m)*r2)*r3))). They should be easy to confirm with a simple check. – Chris Degnen Dec 18 '17 at 1:01

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