Problem set 2 prob 3 JEP

#PROBLEM 3: USING BISECTION SEARCH TO MAKE THE PROGRAM FASTER (25 points possible)
#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 =28774.23

annualInterestRate = 0.2

#The program should print out one line: the lowest monthly payment that will pay off all debt in under 1 year, for example:
monthlyIntRate = annualInterestRate/12.0
low = balance/12.0
high = (balance *(1 + monthlyIntRate)**12) / 12.0
lowestPayment = (low+high)/2
finding = True

def payment(balance,annualInterestRate,lowestPayment):
    month = 1
    while month <= 12:
     
    #Monthly unpaid balance = (Previous balance) - (Minimum monthly payment)
        balance -= lowestPayment
    
	#Updated balance each month = (Monthly unpaid balance) + (Monthly interest rate x Monthly unpaid balance)
        balance += (monthlyIntRate*balance)
    
        month +=1
        
    return balance
        
while finding:
    x=payment(balance,annualInterestRate,lowestPayment)
    if  round(x) < 0:
        high = lowestPayment
        lowestPayment = (high + low)/2
    elif round(x) > 0:
        low = lowestPayment
        lowestPayment = (high + low)/2
    elif round(x) == 0:
        finding = False
        print 'Lowest Payment:',round(lowestPayment,2)

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