diff options
Diffstat (limited to 'numpy/lib/financial.py')
| -rw-r--r-- | numpy/lib/financial.py | 8 |
1 files changed, 4 insertions, 4 deletions
diff --git a/numpy/lib/financial.py b/numpy/lib/financial.py index a3552ebc0..9c5d2753a 100644 --- a/numpy/lib/financial.py +++ b/numpy/lib/financial.py @@ -69,7 +69,7 @@ What is the future value after 10 years of saving $100 now, with an additional monthly savings of $100. Assume the interest rate is 5% (annually) compounded monthly? ->>> fv(0.05/12, 10*12, -100, -100) +>>> np.fv(0.05/12, 10*12, -100, -100) 15692.928894335748 By convention, the negative sign represents cash flow out (i.e. money not @@ -94,7 +94,7 @@ Examples What would the monthly payment need to be to pay off a $200,000 loan in 15 years at an annual interest rate of 7.5%? ->>> pmt(0.075/12, 12*15, 200000) +>>> np.pmt(0.075/12, 12*15, 200000) -1854.0247200054619 In order to pay-off (i.e. have a future-value of 0) the $200,000 obtained @@ -122,7 +122,7 @@ Examples If you only had $150 to spend as payment, how long would it take to pay-off a loan of $8,000 at 7% annual interest? ->>> nper(0.07/12, -150, 8000) +>>> np.nper(0.07/12, -150, 8000) 64.073348770661852 So, over 64 months would be required to pay off the loan. @@ -130,7 +130,7 @@ So, over 64 months would be required to pay off the loan. The same analysis could be done with several different interest rates and/or payments and/or total amounts to produce an entire table. ->>> nper(*(ogrid[0.06/12:0.071/12:0.01/12, -200:-99:100, 6000:7001:1000])) +>>> np.nper(*(np.ogrid[0.06/12:0.071/12:0.01/12, -200:-99:100, 6000:7001:1000])) array([[[ 32.58497782, 38.57048452], [ 71.51317802, 86.37179563]], |