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

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