Debt race
Who will be the first investor to pay off their debt?
On 1st June 2008 five people A, B, C, D and E each borrow $100,000.
Each year, starting 1st June 2009, each is charged interest on the amount owed at the start of the previous year, which is added to the balance. Each year, each will pay off a lump sum of their debt, which is taken from the balance.
The rates and amounts for each borrower is as follows:
A  B  C  D  E 
3%  6%  9%  12%  15% 
$12,000  $14,000  $16,000  $18,000 
$20,000

The question is this: Which
borrower will be the first to pay off their debt?
Extension: You might like to
consider the importance of the figure $100,000 in this
calculation
Note that after the payments on the 1st June 2009 the debts will be
100,000 (1+R)  P
where R is the interest rate and P is the payment.
The interest charged in the second year will be (1+R) lots of this amount
To solve this problem we must first set up a linear difference equation and then solve the equation to find n when the debt remaining = 0.
$A_n$ =amount owed in the nth year of repayment
I = Interest rate
L= Lump sum payed annually
In general $A_n= (1+I)A_{n1} L$
$A_n  (1+I)A_{n1} =  L$
Set the right hand side = 0 and solve for the complementary function.
$z  (1+ I) = 0$
$z = 1+I$
$A_{CF} = C(1+I)^{n}$
Now solve for the particular integral:
Let $A_n = K$ then $A_{n1}$ also = K
Substituting into the general form we find:
$K (1+I)K =L \to K = \frac{L}{I}$
General solution:
$A_n = A_{CF} + A_{PI}= C(1+I)^n + \frac{L}{I}$
Now use the boundary condition that at n = 0 , $A_n = 100,000$
We find that the constant $c = (10^5  \frac{L}{I})$
Therefore $A_n = (10^5  \frac{L}{I}) (I +1)^n +\frac{L}{I}$
If we now set $A_n$ = 0 and solve for n we find:
$n =\frac{log(\frac{LI^{1}}{10^5  LI^{1}}) }{log(1+I)}$
If we now substitute the values of I and L for each person given in the intitial question we find:
Person A: n=9.73years
Person B: n= 9.60 years
Person C: n= 9.59 years
Person D: n= 9.69 years
Person E: n= 9.92 years