I am trying to teach myself python and have no experience write code. For my first attempt I am trying to write a program that applies the snowball principle to debt reduction, but also adds in an extra set amount each payment. I can get the first debt to clear(it goes to a negative but it exits the loop). My second step wont exit the loop and I have looked at topics that dealt with nested loops but they did not help. Could someone please show me where I went wrong?
#Temp fixed number for testing use rawinput for actual program.
#name the debt
debt1 = "CC A"
#currnet balnace
balance1 = float(5000)
#APR
annualInterestRate1 = float(.1499)
#Currnet Monthly Payment
minMonthlyPayment1 = float(200)
# Exta Payment
boosterPayment = float(337)
print "The balance on ",debt1," is ",balance1
debt2 = "CC B"
balance2 = float(1000)
annualInterestRate2 = float(.1499)
minMonthlyPayment2 = float(200)
print "The balance on ",debt2," is ",balance2
debt3 = "ICCU"
balance3 = float(6000)
annualInterestRate3 = float(.0879)
minMonthlyPayment3 = float(130)
print "The balance on ",debt3," is ",balance3
debt4 = "Car"
balance4 = float(8000)
annualInterestRate4 = float(.0699)
minMonthlyPayment4 = float(200)
print "The balance on ",debt4," is ",balance4
debt5 = "Truck"
balance5 = float(15000)
annualInterestRate5 = float(.0439)
minMonthlyPayment5 = float(333)
#nubmer of payments made durning the debt reduction. Used as the index.
numPay = 0
save = 0
#For Debt1 with an APR greater then 0
intPayment1 = round(balance1*(annualInterestRate1/12),2)
while balance1 >= 0:
#payment with intrest
payment1 = minMonthlyPayment1 - intPayment1 + boosterPayment
#subtact payment from balance
balance1 -= payment1
#count Number of payments
numPay += 1
print numPay
print balance1
#For Debt2 with an APR greater then 0
#Figures monthly charge based on given APR
intPayment2 = round(balance2*(annualInterestRate2/12),2)
#Monthly payment minus intrest
standPay2 = minMonthlyPayment2 - intPayment2
while balance2 >= 0:
#payment while debt1 is being paid
#need a way to pay the payments while the other debt is being figured
backPay = numPay
while backPay >= 0:
balance2 -= standPay2
backPay += 1
#payment with intrest takes 100 away for savings
payment2 = minMonthlyPayment2 - intPayment2 + (boosterPayment-100)
#subtact payment from balance
balance2 -= payment2
#count Number of payments
numPay += 1
#keep track of how much is going to savings
save += 100
print numPay
print balance1
print save
Take a look at this loop:
while backPay >= 0:
balance2 -= standPay2
backPay += 1
Here, backPay in increased in each iteration, so the condition backPay >= 0 will always be true.
Not sure what the code is intended to do, but probably you have to do backPay -= 1 instead. However, note that since the number of iterations of the loop is known beforehand, and you are just adding a fixed number in each iteration, you could just as well replace the loop with a simple multiplication.
Related
I'm attempting to take a starting balance and increase the value by 5% each month. I then want to feed this new balance back into the equation for the next month. I've attempted to do this using a while loop but it doesn't seem to be feeding the new balance back in.
I'm using 60 months (5 years) for the equation but this can be altered
counter = 1
balance = 1000
balance_interest = balance * .05
while counter <= 60:
new_monthly_balance = (balance + balance_interest)*(counter/counter)
print(new_monthly_balance)
balance = new_monthly_balance
counter += 1
You never change balance_interest in the loop.
What do you intend to do with *(counter/counter)? This merely multiplies by 1.0, which is a no-op.
while counter <= 60:
balance *= 1.05
print(balance)
counter += 1
Better yet, since you know how many times you want to iterate, use a for:
for month in range(60):
balance *= 1.05
print(balance)
BTW, just what sort of finance has a constant 5% monthly increase???
I have an assignment where I have to prompt the user for cost of a product and amount paid, I have to output change in pennies, dimes, quarters, $1, $5, $20, $50, and $100, for example: The cost of the item is $19.99 and the client pays with a $50 bill. The change to be provided is 1 $20 bill, one $10 bill, and one penny.
I am confused how to get an output like that though, any help would be greatly appreciated, heres what I have so far
cost = float(input('Cost: '))
amount_paid = float(input('Amount paid: '))
penny = 0.01
dime = 0.10
quarter = 0.25
dollar_1 = 1.00
dollar_5 = 5.00
dollar_10 = 10.00
dollar_20 = 20.00
dollar_50 = 50.00
dollar_100 = 100.00
change = cost - amount_paid
if amount_paid < cost:
print('Error')
I dont know what to do next
A common misstep here is to use floats. You should instead convert everything to the smallest whole unit (a cent) and use integer math. Floating point math is...fuzzy.
currencies = {"penny": 1,
"nickel": 5,
"dime": 10,
"quarter": 25,
"dollar": 1_00,
"five": 5_00,
"ten": 10_00,
"twenty": 20_00,
"fifty": 50_00,
"hundred": 100_00}
# never seen that numeric notation before? It's safe to embed underscores
# in numerical literals! It's often used for large numbers in place of
# commas, but it makes sense here in place of a period.
Then you should only need to define a dictionary for the result, and use divmod to find how many of the denomination can fit in the amount left due.
change_due = {}
for denomination, amt in reversed(currencies.items()):
if amt < amt_due:
d, m = divmod(amt_due, amt)
change_due[denomination] = d
amt_due = m
welcome to stackoverflow! I wrote the code for you and this is how it works. Basically it sees each currency and uses integer division // to see how many integers can fit in. It then subtracts that amount from the remaining change and the process continues. Please ask if you don't understand something, or if you think there is an error.
Code:
cost = float(input('Cost: '))
amount_paid = float(input('Amount paid: '))
penny = 0.01
dime = 0.10
quarter = 0.25
dollar_1 = 1.00
dollar_5 = 5.00
dollar_10 = 10.00
dollar_20 = 20.00
dollar_50 = 50.00
dollar_100 = 100.00
changeTypes = {dollar_100:0,dollar_50:0,dollar_20:0,dollar_10:0,dollar_5:0,dollar_1:0,quarter:0,dime:0,penny:0}
change = amount_paid-cost
if amount_paid < cost:
print('Error: InsufficientFunds')
for changeType in changeTypes:
numAmount = max(0,change//changeType)
change-=numAmount*changeType
changeTypes[changeType] = int(numAmount)
print(changeTypes)
P.S you should make this a function, it shouldn't be too hard.
You could do this good with dictionaries, but without using still there are many ways to go about this, options are endless, heres one idea
def get_bills(change, value):
if change//value > 0:
bills = change//value
change -= bills * value
return bills, change
else:
return 0, change
cost = float(input('Cost: '))
paid = float(input('Amount Paid: '))
while paid < cost:
paid = float(input('Amount Paid: '))
change = paid - cost
hundreds, change, = get_bills(change, 100)
fifties, change, = get_bills(change, 50)
twenties, change = get_bills(change, 20)
tens, change = get_bills(change, 10)
fives, change = get_bills(change, 5)
ones, change = get_bills(change, 1)
quarters, change = get_bills(change, .25)
dimes, change = get_bills(change, .1)
nickels, change = get_bills(change, .05)
pennies = round(change * 100)
print(f"Hundreds: {hundreds}, Fifties: {fifties}, Twenties: {twenties}," +
f" Tens: {tens}, Fives: {fives}, Ones: {ones}, Quarters: {quarters}," +
f" Dimes: {dimes}, Nickels: {nickels}, Pennies: " +
f"{pennies}")
bill = float(input())
paid = float(input())
Available = {100.0:0,50.0:0,20.0:0,10.0:0,5.0:0,1.0:0,0.25:0,0.10:0,0.01:0}
due = paid-bill
for change in sorted(Available,reverse = True):
amt= max(0,due//change)
due-=amt*change
Available[change] = int(amt)
print(Available)
I know this is a late response but maybe it can help someone.
Below is a code for doing exactly what you want. The program iterates through the notes available from largest to smallest and calculates how many times the current note may be used to deduct from the remaining change to be given.
Finally returning a list containing the notes used to reach the sum required.
# Available notes for change
notes = [500, 200, 100, 50, 20, 10]
def change_notes(change, notes):
notes_out = []
for note in notes:
print(f"current note is {note}")
sleep(1)
while change > 0 and note <= change:
if change - note >= 0:
change -= note
notes_out.append(note)
print(f"change is {change}")
sleep(1)
if change == 0:
break
return notes_out
I've written a solution for an EdX course homework problem using iteration.
The code takes a credit card balance and annual interest rate to calculate the minimum monthly payment required to pay off the balance (plus any interest) in 12 months.
Here is my iterative code:
def minimum_payment_iter(ann_interest, balance):
month_int = ann_interest/12
remaining = balance
payment = 10
months = 1
while remaining > 0:
months = 1
payment += 10
remaining = balance
while months < 13:
remaining -= payment
remaining += remaining*month_int
months += 1
return payment
I've taken a stab at doing the recursive version, but I've exceeded the maximum recursion depth:
def minimum_payment_recur(ann_interest, balance, payment = 10):
month_int = ann_interest/12
remaining = balance
month = 1
if remaining <= 0:
return payment
else:
remaining -= payment
remaining += remaining*month_int
month += 1
return minimum_payment_recur(ann_interest, balance, payment + 10)
I just did this one. You need to add a counter in the recursion, which is what you would do in a loop.
So, something like:
def function(a,b,counter=12):
"""
Docstring goes here
"""
if counter == 1:
return a+b
else:
return function(a,b,counter-1)
I find it tough to get real help in these online courses on edX as well because the courses are so strict on what you can post, and I never got an offer to speak with an individual outside of the "Discussion". I'd pay a premium if I could get 1 on 1 help from edX courses, honestly.
Anyway, hope that helps you.
I'm struggling with a problem that calculates the minimum fixed monthly payment needed in order pay off a credit card balance within 12 months. By a fixed monthly payment, I mean a single number which does not change each month, but instead is a constant amount, multiple of 10 and the same for all months, that will be paid each month. ( it is possible for the balance to become negative using this payment scheme, which is OK)
So as input I have
original_balance = 3329
annualInterestRate = 0.2
From this I'm calculating the followings :
after_12_months_interest = original_balance
monthlyInterestRate = round(annualInterestRate/12.0,2)
monthly_payment = 10
total_paid = 0
for i in range(0,12):
after_12_months_interest = after_12_months_interest + (annualInterestRate/12.0)*after_12_months_interest
while total_paid < after_12_months_interest:
new_balance = 0
unpaid_balance = original_balance - monthly_payment
total_paid = 0
for i in range(0, 13):
total_paid = total_paid + monthly_payment
if total_paid < after_12_months_interest:
monthly_payment = monthly_payment + 10
print "Lowest Payment: ", monthly_payment
My problem I have is that I end up having a monthly_payment just a little more than I should have it. In this case the return for the monthly_payment is 320 instead of 310. For all use cases I've tried the monthly_payment it's slightly more than it should be.
Anyone can give me a hint or an idea on what I'm doing wrong please. Thank you
Mandatory one-liner
from itertools import count
print(next(payment for payment in count(0, 10)
if sum(payment*(1+monthly)**i for i in range(12)) > original_balance*(1+annual)))
What this does:
next takes the first element of an iterator.
count tries values from 0 to infinity (only each time next is called, and only until a value is returned)
sum(payment*(1+monthly)**i for i in range(12)) That's the combined value of the payments. Each payment is worth itself plus all the saved interests (the earlier you repay, the less interest you'll owe later)
original_balance*(1+annual) is indeed the total value if nothing is repayed.
Alternative
print(next(payment for payment in count(0, 10)
if reduce(lambda x,_:(x - payment)*(1+monthly), range(12), original_balance) <= 0))
This one computes the combined remainder of the debt by reduceing the original_balance 12 times.
How to calculate the monthly fee on a loan?
Given is:
a: an amount to loan.
b: the loan period (number of months).
c: the interest rate p.a. (interests is calculated and added every month, 1/12 of the interest is added. So if the interest is on 12%, 1% interest is added every month).
d: the amount of money owed after the end of the period.
This problem is a bit different than the usual since, the goal is not to have the loan payed after the loan period has ended, but to still owe an amount that is given. I have been able to find an algorithm so solve the problem if I wanted to pay the entire amount, but it will of course not work for this problem where the goal is to end up owing a given amount rather than not owing anything.
I managed to make a solution to this problem by starting with an guess and then keep on improving that guess until it was close enough. I wondered however, if there is a better way to simply calculate this, rather than just guessing.
Edit: Here's how I'm doing it now.
def find_payment(start, end, months, interest):
difference = start
guess = int(start / months * interest)
while True:
total = start
for month in range(1, months + 1):
ascribe = total * interest / 12
total = total + ascribe - guess
difference = total - end
# See if the guess was good enough.
if abs(difference) > start * 0.001:
if difference < 0:
if abs(difference) < guess:
print "payment is %s" % guess
return evolution(start, guess, interest, months)
else:
mod = int(abs(difference) / start * guess)
if mod == 0:
mod = 1
guess -= mod
else:
mod = int(difference / start * guess)
if mod == 0:
mod = 1
guess += mod
else:
print "payment is %s" % guess
return evolution(start, guess, interest, months)
evolution is just a function that displays how the loan would look like payment for payment and interest for interest, summing up total amount of interest paid etc.
An example would be if I wanted to find out the monthly payments for a loan starting with $100k and ending at $50k with an interest of 8% and a duration of 70 months, calling
>>> find_payment(100000, 50000, 70, 0.08)
payment is 1363
In the above case I would end up owing 49935, and I went through the loop 5 times. The amount of times needed to go through the loop depends on how close I want to get to the amount and it varies a bit.
This is a basically a mortgage repayment calculation.
Assuming that start is greater than end, and that interest is between 0 and 1 (i.e. 0.1 for 10% interest)
First consider the part of the payment you want to pay off.
Principal = start - end
The monthly payment is given by:
pay_a = (interest / 12) / (1 - (1+interest/12) ^ (-months))) * Principal
You then need to consider the extra interest. Which is just equal to the remaining principal times the monthly interest
pay_b = interest / 12 * end
So the total payment is
payment = (interest / 12) * (1 / (1 - (1+interest/12) ^ (-months))) * Principal + end)
On the example you gave of
Start: 100000
End: 50000
Months: 70
Interest: 8%
pay_a = 896.20
pay_b = 333.33
Payment = 1229.54
When I tested these values in Excel, after 70 payments the remaing loan was 50,000. This is assuming you pay the interest on the notional before the payment is made each month.
Perhaps the easiest way to think about this is to split the loan in two parts, one part which is to be repaid in full and another part where you don't pay off anything. You have already computed the monthly fee for the first part.
You can keep paying the interest of every month; then, you will alway owe the same amont.
Owe_1 = a
Int_2 = Owe_1*(InterestRate/12)
Pay_2 = Int_2
Owe_2 = Owe_1 + Int_2 - Pay_2 # ==> Owe_1 + Int_2 - Int_2 = Owe_1
Int_3 = Owe_2*(InterestRate/12)
Pay_3 = Int_3
Owe_3 = Owe_2 + Int_3 - Pay_3 # ==> Owe_2 + Int_3 - Int_3 = Owe_2 = Owe_1
python code to calculate emi
class EMI_CALCULATOR(object):
# Data attributes
# Helps to calculate EMI
Loan_amount = None # assigning none values
Month_Payment = None # assigning none values
Interest_rate = None #assigning none values
Payment_period = None #assigning none values
def get_loan_amount(self):
#get the value of loan amount
self.Loan_amount = input("Enter The Loan amount(in rupees) :")
pass
def get_interest_rate(self):
# get the value of interest rate
self.Interest_rate = input("Enter The Interest rate(in percentage(%)) : ")
pass
def get_payment_period(self):
# get the payment period"
self.Payment_period = input("Enter The Payment period (in month): ")
pass
def calc_interest_rate(self):
# To calculate the interest rate"
self.get_interest_rate()
if self.Interest_rate > 1:
self.Interest_rate = (self.Interest_rate /100.0)
else:
print "You have not entered The interest rate correctly ,please try again "
pass
def calc_emi(self):
# To calculate the EMI"
try:
self.get_loan_amount() #input loan amount
self.get_payment_period() #input payment period
self.calc_interest_rate() #input interest rate and calculate the interest rate
except NameError:
print "You have not entered Loan amount (OR) payment period (OR) interest rate correctly,Please enter and try again. "
try:
self.Month_Payment = (self.Loan_amount*pow((self.Interest_rate/12)+1,
(self.Payment_period))*self.Interest_rate/12)/(pow(self.Interest_rate/12+1,
(self.Payment_period)) - 1)
except ZeroDivisionError:
print "ERROR!! ZERO DIVISION ERROR , Please enter The Interest rate correctly and Try again."
else:
print "Monthly Payment is : %r"%self.Month_Payment
pass
if __name__ == '__main__':# main method
Init = EMI_CALCULATOR() # creating instances
Init.calc_emi() #to calculate EMI
for more info visit : https://emilgeorgejames.wordpress.com/2015/07/29/python-emi-equated-monthly-installment-calculator/
This rather a detailed way but will give the whole payment as well
# Mortgage Loan that gives the balance and total payment per year
# Function that gives the monthly payment
def f1 (principle,annual_interest_rate,duration):
r = annual_interest_rate/1200
n = duration*12
a=principle*r*((1+r)**n)
b= (((1+r)**n)- 1)
if r > 0 :
MonthlyPayment = (a/b)
else :
MonthlyPayment = principle/n
return MonthlyPayment
# Function that gives the balance
def f2 (principle,annual_interest_rate,duration,number_of_payments):
r = annual_interest_rate/1200
n = duration*12
a= ((1+r)**n)
b= ((1+r)**number_of_payments)
c= (((1+r)**n)-1)
if r > 0 :
RemainingLoanBalance = principle*((a-b)/c)
else :
RemainingLoanBalance = principle*(1-(number_of_payments/n))
return RemainingLoanBalance
# Entering the required values
principle=float(input("Enter loan amount: "))
annual_interest_rate=float(input("Enter annual interest rate (percent): "))
duration=int(input("Enter loan duration in years: "))
# Output that returns all useful data needed
print ("LOAN AMOUNT:",principle,"INTEREST RATE (PERCENT):",annual_interest_rate)
print ("DURATION (YEARS):",duration,"MONTHLY PAYMENT:",int(f1(principle,annual_interest_rate,duration)))
k=duration+1
BALANCE=principle
total=0
for i in range (1,k):
TOTALPAYMENT= f1(BALANCE,annual_interest_rate,k-i)*12
total+= TOTALPAYMENT
BALANCE= f2(principle,annual_interest_rate,duration,12*i)
print("YEAR:",i,"BALANCE:",int(BALANCE),"TOTAL PAYMENT",int(total))
How about this?
def EMI_calc(principle, rate, time, frequency):
return (principle / ((1-((1+(rate/frequency))**(-1*(time*frequency))))/(rate/frequency)))
print("""
----- Welcome to EMI programe for Python -----
""")
print("\n You have chosen to know the EMI for Loan.\n")
input('\nTo Continue Press ENTER --- to ABORT Press ctrl+c > \n')
print("\nPlease Enter amount of Loan to be taken: >\n")
principle = int(input())
print("\nEnter rate of interst (%): >\n")
rate = float(input())/100
print("\nEnter Term (Years): >\n")
time = float(input())
print("\nPlease enter the frequency of installments) : >\n")
frequency = int(input())
EMI = round(EMI_calc(principle, rate, time, frequency),0)
print("""
---------------------------------------------------------------------
""")
print(f"""
The EMI for Loan of Rs.{principle};
at interest rate of {rate*100} % for {time} years;
would be: Rs.""", EMI)
print("""
---------------------------------------------------------------------
""")
Here is a code snippet using numpy functions. This shows you the payment, principal, interest, instalment and total_amount each month. Run it and see the output. You can also check the syntax for Excel "IPMT()" and "PPMT()" functions for more explanation of the arguments.
https://docs.scipy.org/doc/numpy-1.13.0/reference/generated/numpy.pmt.html#numpy.pmt
import math
import numpy as np
rate = 0.08
start_amount = 100000.0
end_amount = 50000.0
diff_amount = start_amount - end_amount
# nr_years = 4
payment_frequency = int (12)
nr_months = 70 # = nr_years * payment_frequency
per_np = np.arange (nr_months) + 1 # +1 because index starts with 1 here
pay_b = rate / payment_frequency * end_amount
ipmt_np = np.ipmt (rate / payment_frequency, per_np, nr_months, diff_amount) - pay_b
ppmt_np = np.ppmt (rate / payment_frequency, per_np, nr_months, diff_amount)
for payment in per_np:
idx = payment - 1
principal = math.fabs (ppmt_np [idx])
start_amount = start_amount - principal
interest = math.fabs (ipmt_np [idx])
instalment = principal + interest
print payment, "\t", principal, "\t", interest, "\t\t", instalment, "\t\t", start_amount
print np.sum (ipmt_np)