To recap our scenario:
We have a loan with an initial balance $b, that has an annual interest rate of r%. We are making a monthly payment of $p every month. In Part 1, we were finding out how long to pay off the balance.
In Part 2, we asking the same question, but with an added component: this time assume that there is a monthly charge of $c to the balance. The charge stops when the balance is paid.
Variables:
b = a_0 = initial balance
p = monthly payment
r = annual rate
c = monthly additional charge
a_n = balance after n payments
Let the recursion formula be:
a_(n+1) = (a_n + c) + (a_n + c) × r/1200 - p
with the initial condition a_0 = b.
Let θ = r/1200, then:
a_(n+1) = (a_n + c) × (1 + θ) - p
Let's get a_1, a_2, and a_3 in terms of a_0.
a_1 = (a_0 + c) × (1 + θ) - p
a_2 = (a_1 + c) × (1 + θ) - p
a_2 = ((a_0 + c) × (1 + θ) - p + c) × (1 + θ) - p
a_2 = (a_0 + c) × (1 + θ)^2 - p × (1 + θ) + c × (1 + θ) - p
a_3 = (a_2 + c) × (1 + θ) - p
a_3 = a_2 × (1 + θ) + c × (1 + θ) - p
a_3 = ((a_0 + c) × (1 + θ)^2 - p × (1 + θ) + c × (1 + θ) - p) × (1 + θ) + c × (1 + θ) - p
a_3 = (a_0 + c) × (1 + θ)^3 - p × (1 + θ)^2 + c × (1 + θ)^2 - p × (1 + θ) + c × (1 + θ) - p
Noticing a pattern...
a_n = (a_0 + c) × (1 + θ)^n - p × Σ((1 + θ)^k, k=0, n-1) + c × Σ((1 + θ)^k, k=1, n-1)
a_n = (a_0 + c) × (1 + θ)^n - p × ((1 + θ)^n - 1)/θ + c × ((1 + θ)^n - (1 + θ))/θ
With a_0 = b:
a_n = (b + c) × (1 + θ)^n - p × ((1 + θ)^n - 1)/θ + c × ((1 + θ)^n - (1 + θ))/θ
Example 1:
Sandra has a credit card with a 15% APR. The initial balance is $1,500.00. Sandra uses a premium credit card that charges $20.00 each month there is a balance.
b = 1500.00
c = 20.00
r = 15%
p = 500.00
Then θ = 15/1200 = 1/80
The resulting sequence is:
a_0 = 1500.00
a_1 = 1039.00
a_2 = 572.23
a_3 = 99.64
In 3 payments , Sandra has knocked the balance to below the payment amount of $1,500.00. In month 4, the amount will be ($99.64 + $20.00) × (1 + 15/1200) = $121.14, and her debt will be over.
Example 2:
Greg has taken a payday loan, for the amount of $1,893.64. The interest rate is 14.99% and the loan incurs a $14.99 holding fee for each month the balance exists. Greg makes a $350.00 payment every month.
b = 1893.64
c = 14.99
r = 15.99%
p = 350.00
Then θ = 15.99/1200
The sequence generated is:
a_0 = 1893.64
a_1 = 1584.06
a_2 = 1270.36
a_3 = 952.48
a_4 = 630.36
a_5 = 303.95
At month 6, the final balance will be ($303.95 + $14.99) × (1 + 14.99/1200) = $322.92.
We can find out when the balance is zero by solving for n.
When a_n = 0:
n = ln((c × (1 + θ)/θ - p/θ) / (b + c - p/θ +c/θ)) × (ln (1 + θ))^(-1)
Then the integer part of n, int(n), is the nth payment when the balance is less than the payment amount.
Recalling our examples:
Example 1:
b = 1500.00
c = 20.00
r = 15%
p = 500.00
Then θ = 15/1200 = 1/80, and n ≈ 3.21
This means the balance is below the payment after then 3rd payment.
Example 2:
b = 1893.64
c = 14.99
r = 15.99%
p = 350.00
Then θ = 15.99/1200 and n ≈ 5.91
The balance becomes below the payment after the 5th payment.
Have a great day everyone!
Eddie
This blog is property of Edward Shore. 2013
A blog is that is all about mathematics and calculators, two of my passions in life.