9.7: Remaining Loan Balance

With loans, it is often desirable to determine what the remaining loan balance will be after some number of years. For example, if you purchase a home and plan to sell it in five years, you might want to know how much of the loan balance you will have paid off and how much you have to pay from the sale.

To determine the remaining loan balance after some number of years, we first need to know the loan payments, if we don’t already know them. Remember that only a portion of your loan payments go towards the loan balance; a portion is going to go towards interest. For example, if your payments were $1,000 a month, after a year you will not have paid off $12,000 of the loan balance.

To determine the remaining loan balance, we can think “how much loan will these loan payments be able to pay off in the remaining time on the loan?”

Example 13

If a mortgage at a 6% interest rate has payments of $1,000 a month, how much will the loan balance be 10 years from the end the loan?

Solution

To determine this, we are looking for the amount of the loan that can be paid off by $1,000 a month payments in 10 years. In other words, we’re looking for $P_0$ when

$\begin{array}{ll} d = \$1,000 & \text{the monthly loan payment} \\ r = 0.06 & 6\% \text{ annual rate} \\ k = 12 & \text{since we’re doing monthly payments, we’ll compound monthly} \\ N = 10 & \text{since we’re making monthly payments for 10 more years} \end{array}$

$$\begin{align*} P_{0} & =\dfrac{1000\left(1-\left(1+\dfrac{0.06}{12}\right)^{-10(12)}\right)}{\left(\dfrac{0.06}{12}\right)} \\[4pt] &=\dfrac{1000\left(1-(1.005)^{-120}\right)}{(0.005)} \\[4pt] &=\dfrac{1000\left(1-(1.005)^{-120}\right)}{(0.005)} \\[4pt] &=\dfrac{1000(1-0.5496)}{(0.005)} \\[4pt] &=\$ 90,073.45 \end{align*} \nonumber$$

The loan balance with 10 years remaining on the loan will be $\$ 90,073.45$

Often times answering remaining balance questions requires two steps:

1. Calculating the monthly payments on the loan
2. Calculating the remaining loan balance based on the remaining time on the loan

Example 14

A couple purchases a home with a $\$ 180,000$ mortgage at $4 \%$ for 30 years with monthly payments. What will the remaining balance on their mortgage be after 5 years?

Solution

First we will calculate their monthly payments.

We’re looking for $d$.

$\begin{array}{ll} r = 0.04 & 4\% \text{ annual rate} \\ k = 12 & \text{since they’re paying monthly} \\ N = 30 & \text{30 years} \\ P_0 = \$180,000 & \text{the starting loan amount} \end{array}$

We set up the equation and solve for $d$.

$$\begin{align*} 180,000 &=\dfrac{d\left(1-\left(1+\dfrac{0.04}{12}\right)^{-30(12)}\right)}{\left(\dfrac{0.04}{12}\right)} \\[4pt] 180,000 &=\dfrac{d\left(1-(1.00333)^{-360}\right)}{(0.00333)} \\[4pt] 180,000 &=d(209.562) \\[4pt] d &=\dfrac{180,000}{209.562} \\[4pt] &=\$ 858.93 \end{align*} \nonumber$$

Now that we know the monthly payments, we can determine the remaining balance. We want the remaining balance after 5 years, when 25 years will be remaining on the loan, so we calculate the loan balance that will be paid off with the monthly payments over those 25 years.

$\begin{array}{ll} d = \$858.93 & \text{the monthly loan payment we calculated above} \\ r = 0.04 & 4\% \text{ annual rate} \\ k = 12 & \text{since they’re paying monthly} \\ N = 25 & \text{since they’d be making monthly payments for 25 more years} \end{array}$

$$\begin{align*} P_{0} &=\dfrac{858.93\left(1-\left(1+\dfrac{0.04}{12}\right)^{-25(12)}\right)}{\left(\dfrac{0.04}{12}\right)} \\[4pt] &=\dfrac{858.93\left(1-(1.00333)^{-300}\right)}{(0.00333)} \\[4pt] &=\dfrac{858.93(1-0.369)}{(0.00333)} \\[4pt] & \approx \$ 162,758 \end{align*} \nonumber$$

The loan balance after 5 years, with 25 years remaining on the loan, will be $\$ 162,758$

Over that 5 years, the couple has paid off $\$ 180,000-\$ 162,758=\$ 17,242$ of the loan balance. They have paid a total of $\$ 858.93$ a month for 5 years ( 60 months), for a total of $\$ 51,535.80$, so $\$ 51,535.80-\$ 17,242=\$ 34,292.80$ of what they have paid so far has been interest.