CoCalc Public FilesProjects / S21 / mort_Proj4 / 2021-03-11-project 4 Han Zhang .ipynbOpen with one click!

In [1]:

x=1+0.052/12 c=-208000*(x**360)*(1-x)/(1-x**360) c

1

-1142.150632443497

for original 30 years with rate 5.2% , monthly payment is 1142.1506

2

In [2]:

1142.1506*360-208000

3

203174.21599999996

In [ ]:

for 30 years , the owner should pay total interests is 203174.216

4

In [3]:

p=208000*(x**120)-1142.15*(1-x**120)/(1-x) p

5

170202.578411025

now we count the refinance option , after the first ten years original loan, there are 170202.578 left

6

In [4]:

x=1+0.039/12 C=(12000-(170202.578)*(x**360))*(1-x)/(1-x**360) C

7

-785.1912588058652

for additional 30 years , the owner should pay $785.19 per month

8

In [5]:

1142.1506*120+785.191*30*12-208000+12000

9

223726.832

then 40 years for the interests , the owner should pay 223726.832. compare with 30 years ($203174.216), I THOUGHT THE REFINANCE OPTION IS WORTHWHILE BECAUSE IT IS NOT A BIG DIFFERENCE IN THE INTERESTS BUT EXTRA 10 YEARS COULD BE CONTINUED AND EVERYMONTH PAY LESS THEN THE 30 YEARS . SO I WILL CHOOSE REFINANCE OPTION.

10

The $12,000 is not part of the principal sum for the refinance option, but that is besides the point. You gave the $12,000 the wrong sign, making your monthly payment smaller than expected. Be sure to include an introduction to the problem at the top of your document. What are you trying to solve? 9/10

11

In [1]:

x=1+0.167/12 c=-3500*x**12*(1-x)/(1-x**12) c

12

-318.71854813021145

In [2]:

3500*x**12/12

13

344.2817028530176

In [4]:

3500*0.167/12

14

48.708333333333336

In [5]:

x=1+0.082/12 c=20000*x**48*(1-x)/(1-x**48) c

15

490.1382323808376

In [6]:

x=1+0.042/12 c=250000*x**360*(1-x)/(1-x**360) c

16

1222.5429342838258

In [7]:

1222.54*360

17

440114.39999999997

In [8]:

440114.4-250000

18

190114.40000000002

In [ ]:

19