Example

You can afford $200 per month as a car payment. If you can get an auto loan at 3% interest for 60 months (5 years), how expensive of a car can you afford? In other words, what amount loan can you pay off with $200 per month?

Answer: In this example,

d = $200	the monthly loan payment
r = 0.03	3% annual rate
k = 12	since we’re doing monthly payments, we’ll compound monthly
N = 5	since we’re making monthly payments for 5 years

We’re looking for P₀, the starting amount of the loan. [latex-display]\begin{align}&{{P}_{0}}=\frac{200\left(1-{{\left(1+\frac{0.03}{12}\right)}^{-5(12)}}\right)}{\left(\frac{0.03}{12}\right)}\\&{{P}_{0}}=\frac{200\left(1-{{\left(1.0025\right)}^{-60}}\right)}{\left(0.0025\right)}\\&{{P}_{0}}=\frac{200\left(1-0.861\right)}{\left(0.0025\right)}=\$11,120 \\\end{align}[/latex-display] You can afford a $11,120 loan. You will pay a total of $12,000 ($200 per month for 60 months) to the loan company. The difference between the amount you pay and the amount of the loan is the interest paid. In this case, you’re paying $12,000-$11,120 = $880 interest total.

Details of this example are examined in this video. https://youtu.be/5NiNcdYytvY  

Example

You want to take out a $140,000 mortgage (home loan). The interest rate on the loan is 6%, and the loan is for 30 years. How much will your monthly payments be?

Answer: In this example, we’re looking for d.

r = 0.06	6% annual rate
k = 12	since we’re paying monthly
N = 30	30 years
P0 = $140,000	the starting loan amount

In this case, we’re going to have to set up the equation, and solve for d. [latex-display]\begin{align}&140,000=\frac{d\left(1-{{\left(1+\frac{0.06}{12}\right)}^{-30(12)}}\right)}{\left(\frac{0.06}{12}\right)}\\&140,000=\frac{d\left(1-{{\left(1.005\right)}^{-360}}\right)}{\left(0.005\right)}\\&140,000=d(166.792)\\&d=\frac{140,000}{166.792}=\$839.37 \\\end{align}[/latex-display] You will make payments of $839.37 per month for 30 years. You’re paying a total of $302,173.20 to the loan company: $839.37 per month for 360 months. You are paying a total of $302,173.20 - $140,000 = $162,173.20 in interest over the life of the loan.

View more about this example here. https://youtu.be/BYCECTyUc68

Example

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?

Answer: 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 P₀ when

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

[latex-display]\begin{align}&{{P}_{0}}=\frac{1000\left(1-{{\left(1+\frac{0.06}{12}\right)}^{-10(12)}}\right)}{\left(\frac{0.06}{12}\right)}\\&{{P}_{0}}=\frac{1000\left(1-{{\left(1.005\right)}^{-120}}\right)}{\left(0.005\right)}\\&{{P}_{0}}=\frac{1000\left(1-0.5496\right)}{\left(0.005\right)}=\$90,073.45 \\\end{align}[/latex-display] The loan balance with 10 years remaining on the loan will be $90,073.45.

This example is explained in the following video: https://youtu.be/fXLzeyCfAwE

Example

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?

Answer: First we will calculate their monthly payments. We’re looking for d.

r = 0.04	4% annual rate
k = 12	since they’re paying monthly
N = 30	30 years
P0 = $180,000	the starting loan amount

We set up the equation and solve for d. [latex-display]\begin{align}&180,000=\frac{d\left(1-{{\left(1+\frac{0.04}{12}\right)}^{-30(12)}}\right)}{\left(\frac{0.04}{12}\right)}\\&180,000=\frac{d\left(1-{{\left(1.00333\right)}^{-360}}\right)}{\left(0.00333\right)}\\&180,000=d(209.562)\\&d=\frac{180,000}{209.562}=\$858.93 \\\end{align}[/latex-display]   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.

d = $858.93	the monthly loan payment we calculated above
r = 0.04	4% annual rate
k = 12	since they’re doing monthly payments
N = 25	since they’d be making monthly payments for 25 more years

[latex-display]\begin{align}&{{P}_{0}}=\frac{858.93\left(1-{{\left(1+\frac{0.04}{12}\right)}^{-25(12)}}\right)}{\left(\frac{0.04}{12}\right)}\\&{{P}_{0}}=\frac{858.93\left(1-{{\left(1.00333\right)}^{-300}}\right)}{\left(0.00333\right)}\\&{{P}_{0}}=\frac{858.93\left(1-0.369\right)}{\left(0.00333\right)}=\$155,793.91 \\\end{align}[/latex-display] The loan balance after 5 years, with 25 years remaining on the loan, will be $155,793.91. Over that 5 years, the couple has paid off $180,000 - $155,793.91 = $24,206.09 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 - $24,206.09 = $27,329.71 of what they have paid so far has been interest.

More explanation of this example is available here: https://youtu.be/-J1Ak2LLyRo