Two useful functions for calculating lifetime mortgage payments

mortgage.py

## Author: Yotam Gingold <yotam@yotamgingold.com>
## License: Public domain, CC0: https://creativecommons.org/share-your-work/public-domain/cc0/
## URL: https://gist.github.com/yig/6eb1c3170e6e2e05df810fca8e47f3ca

"""
You can calculate your monthly payment for a loan for X years with interest rate Y%. For example:

    > monthly_payment_for_termination_in_X_years( years = 30, principal = 300_000, interest_rate_per_year =  3.75/100 )
    $1389.35

At the end of your loan, how much will you have paid?

    > $1389.35/month * 30 years * 12 months/year
    = $500,166

But you could also pay more per month. Let's say you are willing to spend a fixed budget of $2182 on your mortgage each year. Then we can ask how long it would take to pay off the loan:

    > years_to_payoff_loan_given_payment_amount( principal = 300_000, interest_rate_per_year = 3.75/100, payment_amount = 2182 )
    15.00 years

So by paying $2182/month, you've turned your 30-year loan into a 15-year loan with a higher interest rate. How much will you pay in total over the lifetime of the loan?

    > $2182/month * 15 years * 12 months/year
    = $392,760

You saved $500,166 - $392,760 = $107,406.

Let's refinance. We'll refinance for a 15 year loan with a better interest rate: 3.65%. Your required monthly payments would be:

    > monthly_payment_for_termination_in_X_years( years = 15, principal = 300_000, interest_rate_per_year =  3.65/100 )
    $2166.81

At the end of this 15-year loan, you will have paid:

    > $2166.81/month * 15 years * 12 months/year
    $390,025.80

Refinancing saved you $2,734.20 over a 15-year period.

But you were willing to pay $2182/month. How long would it take you to pay off the refinanced loan at the lower interest rate?

    > years_to_payoff_loan_given_payment_amount( principal = 300_000, interest_rate_per_year = 3.65/100, payment_amount = 2182 )
    14.86 years

If you pay that, over the lifetime of the loan you will have paid:

    > $2182/month * 14.86 years * 12 months/year
    $389,094.24

So if your budget allows you to pay $2182/month, you will save $3,665.76 by refinancing over a 15-year period.

If your budget is fixed, the only thing that matters is the interest rate. In this case, it saved you $21/month.
"""

from __future__ import print_function, division

def monthly_payment_for_termination_in_X_years( years, principal, interest_rate_per_year ):
    '''
    Given:
        years: The number of years to pay off the loan in (e.g. 30 for a 30 year loan)
        principal: The loan amount
        interest_rate_per_year: The amount of interest paid per year (e.g. 3.875/100 for a loan whose interest rate is 3.875)
    Returns:
        Monthly payment so that the loan is paid off after `months` months.
    '''
    ## Formula from: https://medium.com/towards-artificial-intelligence/mortgage-calculator-python-code-94d976d25a27
    months = years*12
    interest_rate_per_month = interest_rate_per_year/12
    R = 1 + interest_rate_per_month
    return principal * R**months * ( 1 - R ) / ( 1 - R**months )

def years_to_payoff_loan_given_payment_amount( principal, interest_rate_per_year, payment_amount, pay_until_balance = 0 ):
    '''
    Given:
        principal: The loan amount
        interest_rate_per_year: The amount of interest paid per year (e.g. 3.875/100 for a loan whose interest rate is 3.875)
        payment_amount: The amount paid per month
    Returns:
        The number of years the loan will be paid off in (e.g. 30 for a 30 year loan paid at the rate returned by monthly_payment_for_termination_in_X_years())
    '''
    ## Formula from: https://medium.com/towards-artificial-intelligence/mortgage-calculator-python-code-94d976d25a27
    interest_rate_per_month = interest_rate_per_year/12
    R = 1 + interest_rate_per_month
    P0 = principal
    X = payment_amount
    B = pay_until_balance
    
    '''
    P_m = P0 R^m - X ( 1 - R^m ) / ( 1 - R )
    where
    P0 is the initial loan amount `principal`
    P_m is the remaining balance after m months
    X is the monthly payment amount
    
    We want to solve for m given X, R, P0.
    In other words, find m such that P_m = 0.
      P_m = 0 = P0 R^m - X ( 1 - R^m ) / ( 1 - R )
    <=>
      0 = P0 R^m - X / ( 1 - R ) + X R^m / ( 1 - R )
    <=>
      X / (1-R) = R^m ( P0 + X / ( 1 - R ) )
    <=>
      X / ( (1-R) * ( P0 + X/(1-R) ) ) = R^m
    <=>
      log( X / ( (1-R) * ( P0 + X/(1-R) ) ) ) = m log R
    <=>
      m = log( X / ( (1-R) * ( P0 + X/(1-R) ) ) ) / log R
    '''
    from math import log
    # months = log( X / ( (1-R) * ( P0 + X/(1-R) ) ) ) / log(R)
    
    '''
    The derivation is similar if we want to find m such that P_m = B.
        P_m = B = P0 R^m - X ( 1 - R^m ) / ( 1 - R )
    <=>
        B = P0 R^m - X / ( 1 - R ) + X R^m / ( 1 - R )
    <=>
        B + X / (1-R) = R^m ( P0 + X / ( 1 - R ) )
    <=>
        ( B + X/(1-R) ) / ( P0 + X/(1-R) ) = R^m
    <=>
        log( B + X/(1-R) ) / ( P0 + X/(1-R) ) = m log R
    <=>
        m = log( B + X/(1-R) ) / ( P0 + X/(1-R) ) / log R
    '''
    
    months = log( ( B + X/(1-R) ) / ( P0 + X/(1-R) ) ) / log(R)
    return months/12