Question

Prepare an amortization schedule for the first three months on a loan of $87000 at 6% for 20 years

Answer 1

First, we need to calculate the value of the monthly payments. We can use the general ordinary anuity formula:

$PV=\text{PMT}\cdot\lbrack\frac{1-(1+i)^{-n}}{i}\rbrack$

Where PV=$87000

n=20

i=0.06

Replace the given values and solve for PMT:

$\begin{gathered} 87000=\text{PMT}\cdot\lbrack\frac{1-(1+0.06)^{-20}}{0.06}\rbrack \\ 87000=\text{PMT}\cdot\lbrack\frac{1-0.3118}{0.06}\rbrack \\ 87000=\text{PMT}\cdot\lbrack\frac{0.6882}{0.06}\rbrack \\ 87000=\text{PMT}\cdot11.4699 \\ \text{PMT}=\frac{87000}{11.4699} \\ \text{PMT}=7585.06 \end{gathered}$

Now that you have the payment, you can construct the table of the amortization schedule with the know information:

To calculate the missing information, start by calculating the interest component of the payment (interest paid) by multiplying the periodic interest by the remaining principal.

The monthly interest is the yearly interest divided by 12, then:

$MI=\frac{0.06}{12}=0.005$

And the interest paid is then:

$\begin{gathered} IP=MI\cdot\text{ Remaining principal} \\ IP=0.005\cdot87000\text{ (for the first payment)} \\ IP=435 \end{gathered}$

Now, calculate the principal paid by subtracting the interest paid from the payment amount:

$\text{ Principal paid=7585.06-435}=7150.06\text{ (first payment)}$

Then, by putting the values on the amortization schedule:

The remaining principal is 87000-7150.06 (principal paid).

Now for the second payment, calculate the interest paid with the new remaining principal:

$\begin{gathered} IP=0.005\cdot79849,94\text{ (for the second payment)} \\ IP=399,25 \end{gathered}$

And the principal paid is:

$\text{ Principal paid=7585.06-399.25}=7185.81\text{ (second payment)}$

The remaining principal is:

$RP=79849.94-7185.81=72664.13$

Thus:

And for the third month, you apply the same calculations and the amortization schedule is: