Question

question: Write a program that uses these bounds and bisection search (for more info check out the Wikipedia page here) 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.

Output Phython: Enter the outstanding balance on your credit card: 320000 Enter the annual credit card interest rate as a decimal: .2 RESULT Monthly payment to pay off debt in 1 year: 29643.05 Number of months needed: 12 Balance: -0.1

Answer 1

# In python 3

balance = float(input("Enter the outstanding balance on your credit card: ")) # Asking the user to input outstanding balance
  
annualInterestRate = float(input("Enter the annual interest rate as a decimal: ")) # Asking the user to input annual interest rate
newbalance = balance # Storing the balance in newbalance to use in calculations

monthlyInterestRate = annualInterestRate/12.0 # Calculating the monthly interest rate

lowerbound = newbalance/12.0 # Lower bound of the monthlyPayment is balance/12.0

upperbound = (newbalance *(1+monthlyInterestRate)**12)/12.0 # Upper bound of the monthlyPayment is calculated using the compound interest formula
#The minimumMonthlyPay lies somewhere in between the lower bounbd and the upper bound. We need to search for it

minimumMonthlyPay = (lowerbound + upperbound)/2.0 # We start searching for the minimumMonthlyPay using the average of lower and upper bounds just like in binary search

i = 1 # Inintializing the looping variable

while(i <= 12):
monthlyUnpaidBal = newbalance - minimumMonthlyPay # Unpaid balance for each month
newbalance = monthlyUnpaidBal + monthlyInterestRate * monthlyUnpaidBal #New balance because of the interest rate
i += 1
  
if(i == 13 and round(newbalance,2) > 0.00): #Checking whether the minimum monthly pay is greater than the average of lower and upperbounds
# if so we search between average and upper bound
i = 1
lowerbound = minimumMonthlyPay

minimumMonthlyPay = (lowerbound + upperbound)/2.0
newbalance = balance
  
elif(i == 13 and round(newbalance,2) <0.00 ): # Checking if the minimum monthlypay is lesser than average of lower and upper bounds
#if so we search between lowerbound and average
i = 1
upperbound = minimumMonthlyPay
minimumMonthlyPay = (lowerbound + upperbound)/2.0
newbalance = balance   
  
  
#Printing out the minimumMonthlyPay   
print ("Monthly Payment to pay off debt in 1 year:", round(minimumMonthlyPay,2))