You'll notice that in Problem 2, your monthly payment had to be a multiple of $10. Why did we make it that way? You can try running your code locally so that the payment can be any dollar and cent amount (in other words, the monthly payment is a multiple of $0.01). Does your code still work? It should, but you may notice that your code runs more slowly, especially in cases with very large balances and interest rates. (Note: when your code is running on our servers, there are limits on the amount of computing time each submission is allowed, so your observations from running this experiment on the grading system might be limited to an error message complaining about too much time taken.)
Well then, how can we calculate a more accurate fixed monthly payment than we did in Problem 2 without running into the problem of slow code? We can make this program run faster using a technique introduced in lecture - bisection search!
The following variables contain values as described below:
balance - the outstanding balance on the credit card
annualInterestRate - annual interest rate as a decimal
To recap the problem: we are searching for the smallest monthly payment such that we can pay off the entire balance within a year. What is a reasonable lower bound for this payment value? $0 is the obvious anwer, but you can do better than that. If there was no interest, the debt can be paid off by monthly payments of one-twelfth of the original balance, so we must pay at least this much every month. One-twelfth of the original balance is a good lower bound.
What is a good upper bound? Imagine that instead of paying monthly, we paid off the entire balance at the end of the year. What we ultimately pay must be greater than what we would've paid in monthly installments, because the interest was compounded on the balance we didn't pay off each month. So a good upper bound for the monthly payment would be one-twelfth of the balance, after having its interest compounded monthly for an entire year.
In short:
Monthly interest rate = (Annual interest rate) / 12.0
Monthly payment lower bound = Balance / 12
Monthly payment upper bound = (Balance x (1 + Monthly interest rate)12) / 12.0
Write a program that uses these bounds and bisection search (for more info check out the Wikipedia page on bisection search) to find the smallest monthly payment to the cent (no more multiples of $10) such that we can pay off the debt within a year. Try it out with large inputs, and notice how fast it is (try the same large inputs in your solution to Problem 2 to compare!). Produce the same return value as you did in Problem 2.
Note that if you do not use bisection search, your code will not run - your code only has 30 seconds to run on our servers.
Test Cases to Test Your Code With. Be sure to test these on your own machine - and that you get the same output! - before running your code on this webpage!
Click to See Problem 3 Test Cases
The code you paste into the following box should not specify the values for the variables balance or annualInterestRate - our test code will define those values before testing your submission.
To solve this problem, we can use the bisection search algorithm. Here's a step-by-step guide on how to use it:
1. Calculate the monthly interest rate by dividing the annual interest rate by 12.
2. Calculate the lower bound for the monthly payment by dividing the balance by 12. This represents the minimum monthly payment if there were no interest.
3. Calculate the upper bound for the monthly payment by multiplying the balance by (1 + monthly interest rate) raised to the power of 12, and then dividing the result by 12. This represents the maximum monthly payment if the entire balance was paid off at the end of the year.
4. Set the initial values for the lower and upper bounds as the lower bound and upper bound calculated in steps 2 and 3, respectively.
5. While the difference between the upper and lower bounds is greater than a small tolerance (e.g., $0.01), do the following steps:
a. Calculate the midpoint between the current lower and upper bounds.
b. Set the balance variable to the initial balance value.
c. For each month in a range of 12, do the following:
i. Subtract the current monthly payment from the balance.
ii. Add the interest on the remaining balance.
d. Check if the balance after 12 months is positive. If it is, update the lower bound of the monthly payment to the midpoint value; otherwise, update the upper bound to the midpoint value.
6. Once the while loop terminates, the midpoint value will be the smallest monthly payment that pays off the debt within a year.
Here is an example implementation in Python:
```python
monthly_interest_rate = annualInterestRate / 12.0
lower_bound = balance / 12
upper_bound = (balance * (1 + monthly_interest_rate)**12) / 12.0
while upper_bound - lower_bound > 0.01:
monthly_payment = (lower_bound + upper_bound) / 2
temp_balance = balance
for _ in range(12):
temp_balance -= monthly_payment
temp_balance += temp_balance * monthly_interest_rate
if temp_balance > 0:
lower_bound = monthly_payment
else:
upper_bound = monthly_payment
smallest_payment = round((upper_bound + lower_bound) / 2, 2)
print("Lowest Payment:", smallest_payment)
```
In this implementation, the program starts with the calculated lower and upper bounds for the monthly payment. It continuously updates the lower or upper bound based on whether the balance after 12 months is positive or negative. Finally, it calculates the midpoint of the bounds as the smallest monthly payment and rounds it to the nearest cent.