Description
You have just purchased a stereo system that cost $1000 on the following credit plan:
no down payment, an interest rate of 18% per year (and hence 1.5% per month), and
monthly payments of $50. The monthly payment of $50 is used to pay the interest, and
whatever is left is used to pay part of the remaining debt. Hence, the first month you
pay 1.5% of $1000 in interest. That is $15 in interest. The remaining $35 is deducted
from your debt, which leaves you with a debt of $965.00. The next month you pay
interest of 1.5% of $965.00, which is $14.48. Hence, you can deduct $35.52 (which is
$50–$14.48) from the amount you owe.
Write a program that will tell you how many months it will take you to pay off this loan
in particular and any loan in general. Your program also needs to calculate the total
amount of interest paid over the life of any loan. Use a loop to calculate the amount of
interest and the size of the debt after each month. Your program must output the
monthly amount of interest paid and remaining debt. Use a variable to count the
number of loop iterations and hence the number of months until the debt is zero. You
may want to use other variables as well. The last payment may be less than $50 if the
debt is small, but do not forget the interest. If you owe $50, then your monthly payment
of $50 will not pay off your debt, although it will come close. One month’s interest on
$50 is only 75 cents.
Here is a sample dialog (where the user input is depicted as Bold, but you do not need
to display user input in bold.):
Loan Amount: 1000
Interest Rate (% per year): 18
Monthly Payments: 50
2
Your program’s output should match the style of the sample output (including a left
lined-up style).
In what follows, you find a second case used to test the correctness of your program.
******************************************************
Amortization Table
******************************************************
Month Balance Payment Rate Interest Principal
0 $1000.00 N/A N/A N/A N/A
1 $965.00 $50.00 1.5 $15.00 $35.00
2 $929.48 $50.00 1.5 $14.47 $35.53
3 $893.42 $50.00 1.5 $13.94 $36.06
4 $856.82 $50.00 1.5 $13.40 $36.60
5 $819.67 $50.00 1.5 $12.85 $37.15
6 $781.97 $50.00 1.5 $12.30 $37.70
7 $743.70 $50.00 1.5 $11.73 $38.27
8 $704.85 $50.00 1.5 $11.16 $38.84
9 $665.42 $50.00 1.5 $10.57 $39.43
10 $625.40 $50.00 1.5 $9.98 $40.02
11 $584.79 $50.00 1.5 $9.38 $40.62
12 $543.56 $50.00 1.5 $8.77 $41.23
13 $501.71 $50.00 1.5 $8.15 $41.85
14 $459.24 $50.00 1.5 $7.53 $42.47
15 $416.13 $50.00 1.5 $6.89 $43.11
16 $372.37 $50.00 1.5 $6.24 $43.76
17 $327.95 $50.00 1.5 $5.59 $44.41
18 $282.87 $50.00 1.5 $4.92 $45.08
19 $237.11 $50.00 1.5 $4.24 $45.76
20 $190.67 $50.00 1.5 $3.56 $46.44
21 $143.53 $50.00 1.5 $2.86 $47.14
22 $95.68 $50.00 1.5 $2.15 $47.85
23 $47.12 $50.00 1.5 $1.44 $48.56
24 $0.00 $47.83 1.5 $0.71 $47.12
******************************************************
It takes 24 months to pay off the loan.
Total interest paid is: $197.83
Loan Amount: 2000
Interest Rate (% per year): 12
Monthly Payments: 80
3
Special Cases:
1. Please think about how to deal with the last payment, which is smaller than
regular payment. For example, in the above table, the regular payments are
$80.00 whereas the last payment is only 72.98.
2. Your program needs to ensure that regular payments are larger than any
monthly interest. For example, in the above amortization table, your program
must test if the regular monthly payment (i.e., $80.00) is larger than the
monthly interest (e.g., $20.00 in the first month).
3. If you do not address the above issue, your program may not terminate in some
special cases. See the example below:
