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# Annuities for Loans

## Calculating equal periodic payments for a loan using present values.

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Annuities for Loans

Many people buy houses they cannot afford. This causes major problems for both the banks and the people who have their homes taken. In order to make wise choices when you buy a house, it is important to know how much you can afford to pay each period and calculate a maximum loan amount.

Joanna knows she can afford to pay 12,000 a year for a house loan. Interest rates are 4.2% annually and most house loans go for 30 years. What is the maximum loan she can afford? What will she end up paying after 30 years? ### Annuities for Loans The present value can be found from the future value using the regular compound growth formula: PV(1+i)nPV=FV=FV(1+i)n\begin{align*}PV(1+i)^n &= FV\\ PV &= \frac{FV}{(1+i)^n}\end{align*} You also know the future value of an annuity: FV=R(1+i)n1i\begin{align*}FV=R \cdot \frac{(1+i)^n-1}{i}\end{align*} So by substitution, the formula for the present value of an annuity is: PV=R(1+i)n1i1(1+i)n=R(1+i)n1i(1+i)n=R1(1+i)ni\begin{align*}PV=R \cdot \frac{(1+i)^n-1}{i} \cdot \frac{1}{(1+i)^n}=R \cdot \frac{(1+i)^n-1}{i(1+i)^n}=R \cdot \frac{1-(1+i)^{-n}}{i}\end{align*} The present value of a series of equal payments R\begin{align*}R\end{align*} with interest rate i\begin{align*}i\end{align*} per period for n\begin{align*}n\end{align*} periods is: PV=R1(1+i)ni\begin{align*}PV=R \cdot \frac{1-(1+i)^{-n}}{i}\end{align*} This formula can also be used to find out other information such as how much a regular payment should be and how long it will take to pay off a loan. Take a1,000,000 house loan over 30 years with a nominal interest rate of 6% compounded monthly. You are not given the monthly payments, R\begin{align*}R\end{align*}. To find R\begin{align*}R\end{align*}, solve for R\begin{align*}R\end{align*} in the formula given above.

PV=1,000,000, R=?, i=0.005, n=360\begin{align*}PV=\ 1,000,000, \ R=?, \ i=0.005, \ n=360\end{align*} \begin{align*}PV &= R \cdot \frac{1-(1+i)^{-n}}{i}\\ 1,000,000 &= R \cdot \frac{1-(1+0.005)^{-360}}{0.005}\\ R &= \frac{1,000,000 \cdot 0.005}{1-(1+0.005)^{-360}} \approx 5995.51\end{align*} It is remarkable that in order to pay off a1,000,000 loan you will have to pay $5,995.51 a month, every month, for thirty years. After 30 years, you will have made 360 payments of$5995.51, and therefore will have paid the bank more than $2.1 million, more than twice the original loan amount. It is no wonder that people can get into trouble taking on more debt than they can afford. ### Examples #### Example 1 Earlier, you were asked about how much Joanna can afford to take out in a loan. Joanna knows she can afford to pay$12,000 a year to pay for a house loan. Interest rates are 4.2% annually and most house loans go for 30 years. What is the maximum loan she can afford? What does she end up paying after 30 years? You can use the present value formula to calculate the maximum loan:

\begin{align*}PV=12,000 \cdot \frac{1-(1+0.042)^{-30}}{0.042} \approx \202,556.98\end{align*}

For 30 years she will pay 12,000 a year. At the end of the 30 years she will have paid \begin{align*}\12,000 \cdot 30=\360,000\end{align*} total #### Example 2 How long will it take to pay off a20,000 car loan with a 6% annual interest rate compounded monthly if you pay it off in monthly installments of $500? What about if you tried to pay it off in monthly installments of$100?

\begin{align*}PV = \20,000, \ R=\500, \ i=\frac{0.06}{12}=0.005, \ n=?\end{align*}

\begin{align*}PV &= R \cdot \frac{1-(1+i)^{-n}}{i}\\ 20,000 &= 500 \cdot \frac{1-(1+0.005)^{-n}}{0.005}\\ 0.2 &= 1-(1+0.005)^{-n}\\ (1+0.005)^{-n} &= 0.8\\ n &= -\frac{\ln 0.8}{\ln 1.005} \approx 44.74 \ months\end{align*}

For the $100 case, if you try to set up an equation and solve, there will be an error. This is because the interest on$20,000 is exactly $100 and so every month the payment will go to only paying off the interest. If someone tries to pay off less than$100, then the debt will grow.

#### Example 3

It saves money to pay off debt faster in order to save money on interest. As shown earlier, interest can more than double the cost of a 30 year mortgage. This example shows how much money can be saved by paying off more than the minimum.

Suppose a 300,000 loan has 6% interest convertible monthly with monthly payments over 30 years. What are the monthly payments? How much time and money would be saved if the monthly payments were larger by a fraction of \begin{align*}\frac{13}{12}\end{align*}? This is like making 13 payments a year instead of just 12. First you will calculate the monthly payments if 12 payments a year are made. \begin{align*}PV &= R \cdot \frac{1-(1+i)^{-n}}{i}\\ 300,000 &= R \cdot \frac{1-(1+0.005)^{-360}}{0.005}\\ R &= \1,798.65\end{align*} After 30 years, you will have paid647,514.57, more than twice the original loan amount.

If instead the monthly payment was \begin{align*}\frac{13}{12} \cdot 1798.65=1948.54\end{align*}, you would pay off the loan faster. In order to find out how much faster, you will make your unknown.

\begin{align*}PV &= R \cdot \frac{1-(1+i)^{-n}}{i}\\ 300,000 &= 1948.54 \cdot \frac{1-(1+0.005)^{-n}}{0.005}\\ 0.7698 &= 1-(1+0.005)^{-n}\\ (1+0.005)^{-n} &= 0.23019\\ n &= -\frac{\ln 0.23019}{\ln 1.005} \approx 294.5 \ months\end{align*}

294.5 months is about 24.5 years. Paying fractionally more each month saved more than 5 years of payments.

\begin{align*}294.5 \ months \cdot \1,948.54=\573,847.99\end{align*}

The loan ends up costing $573,847.99, which saves you more than$73,000 over the total cost if you had paid over 30 years.

Mackenzie obtains a 15 year student loan for 160,000 with 6.8% interest. What will her yearly payments be? \begin{align*}PV=\160,000, \ R=?, \ n=15, \ i=0.068\end{align*} \begin{align*}160,000 &= R \cdot \frac{1-(1+0.068)^{-15}}{0.068}\\ R & \approx \17,345.88\end{align*} #### Example 5 How long will it take Francisco to pay off a16,000 credit card bill with 19.9% APR if he pays 800 per month? Note: APR in this case means nominal rate convertible monthly. \begin{align*}PV=\16,000, \ R=\600, \ n=?, \ i=\frac{0.199}{12}\end{align*} \begin{align*}16,000 &= 600 \cdot \frac{1-\left(1+\frac{0.199}{12}\right)^{-n}}{\frac{0.199}{12}}\\ n &= 24.50 \ months\end{align*} ### Review For problems 1-10, find the missing value in each row using the present value for annuities formula.  Problem Number \begin{align*}PV\end{align*} \begin{align*}R\end{align*} \begin{align*}n\end{align*} (years) \begin{align*}i\end{align*} (annual) Periods per year 1.4,000 7 1.5% 1 2. $15,575 5 5% 4 3.$4,500 $300 3% 12 4.$1,000 12 2% 1 5. $16,670 10 10% 4 6.$400 4 2% 12 7. $315,000$1,800 5% 12 8. $500 30 8% 12 9.$1,000 40 6% 4 10. $10,000 6 7% 12 11. Charese obtains a 15 year student loan for$200,000 with 6.8% interest. What will her yearly payments be?

12. How long will it take Tyler to pay off a $5,000 credit card bill with 21.9% APR if he pays$300 per month? Note: APR in this case means nominal rate convertible monthly.

13. What will the monthly payments be on a credit card debt of $5,000 with 24.99% APR if it is paid off over 3 years? 14. What is the monthly payment of a$300,000 house loan over 30 years with a nominal interest rate of 2% convertible monthly?

15. What is the monthly payment of a \$270,000 house loan over 30 years with a nominal interest rate of 3% convertible monthly?

To see the Review answers, open this PDF file and look for section 13.7.

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### Vocabulary Language: English

annuity

An annuity is a series of equal payments that occur periodically.

future value

In the context of earning interest, future value stands for the amount in the account at some future time $t$.

loan

A loan is borrowed money. Loans are commonly repaid with interest.

mortgage

A loan is money borrowed that has to be paid back with interest. A mortgage is a loan specifically for a house.

present value

In the context of earning interest, present value stands for the amount in the account at time 0.