A pure python implementation of HP-12C financial calculator.
pip install hpy12cFirst import hpy12c module and use any financial function provided by the library:
import hpy12c as hp
hp.pmt(0.075 / 12, 12 * 15, 200_000) # ==> -1854.0247200054619What is the future value after 10 years of saving $100 now, with an additional monthly savings of $100 with the interest rate at 5% (annually) compounded monthly?
hp.fv(0.05 / 12, 10 * 12, -100, -100) # ==> 15692.928894335748By convention, the negative sign represents cash flow out (i.e. money not available today). Thus, saving $100 a month at 5% annual interest leads to $15,692.93 available to spend in 10 years.
Suppose one invests 100 units and then makes the following withdrawals at regular (fixed) intervals: 39, 59, 55, 20. Assuming the ending value is 0, one's 100 unit investment yields 173 units; however, due to the combination of compounding and the periodic withdrawals, the "average" rate of return is neither simply 0.73/4 nor (1.73)^0.25-1.
hp.irr([-100, 39, 59, 55, 20]) # ==> 0.28095So, the internal rate of return is 28.09%
What is the interest amount of a payment in the 8th period (i.e., 8th month), having a $5,000 loan to be paid in 2 years at an annual interest rate of 7.5%?
hp.ipmt(0.075 / 12, 8, 12 * 2, 5_000.00) # ==> -22.612926783996798So, in the 8th payment, $22.61 are the interest amount.
What is the principal amount of a payment in the 8th period (i.e., 8th month), having a $5,000 loan to be paid in 2 years at an annual interest rate of 7.5%?
hp.ppmt(0.075 / 12, 8, 12 * 2, 5_000.00) # ==> -202.38503647412284So, in the 8th payment, $202.39 are the principal amount.
If you only had $150/month to pay towards the loan, how long would it take to pay-off a loan of $8,000 at 7% annual interest?
hp.nper(0.07 / 12, -150, 8000) # ==> 64.07334877066185So, over 64 months would be required to pay off the loan.
Calculates the Net Present Value of an investment
hp.npv(0.281, [-100, 39, 59, 55, 29]) # ==> 2.6025181340459755What is the monthly payment needed to pay off a $200,000 loan in 15 years at an annual interest rate of 7.5%?
hp.pmt(0.075 / 12, 12 * 15, 200_000) # ==> -1854.0247200054619In order to pay-off (i.e., have a future-value of 0) the $200,000 obtained
today, a monthly payment of $1,854.02 would be required. Note that this
example illustrates usage of fv (future value) having a default value of 0.
What is the present value (e.g., the initial investment) of an investment that needs to total $20,000.00 after 10 years of saving $100 every month? Assume the interest rate is 5% (annually) compounded monthly.
hp.pv(0.05 / 12, 12 * 10, -100, 20_000) # ==> -2715.0857731569663By convention, the negative sign represents cash flow out (i.e., money not available today). Thus, to end up with $20,000.00 in 10 years saving $100 a month at 5% annual interest, an initial deposit of $2715,09 should be made.
Suppose you take a loan of $50,000.00 to pay in 3 years with a monthly payment of $2,500.00. What is the rate applied to this loan?
hp.rate(12 * 3, 2_500, -50_000) # ==> 0.036006853458478955So, the rate applied is 3.60%.
Most of the algorithms here was heavily inspired by the NumPy project. The main difference, is that HPy12C does not need any dependency, as it was written in Pure Python. There are some special cases that influenced the development of this library:
- Having a C (or Rust) binding dependency in some environments is not feasable;
- Having the full blown NumPy library could be too much (e.g AWS Lambda);
- Cases when you just want to do some simple financial calculation;
hpy12c is released under the MIT License.
A special thanks goes to the python NumPy project and Exonio, which was the source for most of the functions.