Annuities: Present Value

The present value of an annuity answers the opposite question to the future value: how much is a series of future payments worth today? It tells you how big a loan your payments can cover, or how much you’d need to set aside now to pay out a regular amount later.

Key ideas

What present value means

The present value $PV$ is the single amount, invested now, that would exactly fund all the future payments. Two common situations:

• A loan: the bank gives you $PV$ now, and you repay it with regular payments $R$.
• Regular withdrawals: you invest $PV$ now so that you can withdraw $R$ every period (for a scholarship, or living costs at university).

Why it’s a geometric series

Each future payment is moved back to today by dividing by $(1 + i)$ for each period, which is multiplying by $(1 + i)^{-1}$.

Three payments of $1000 at $5\%$ per year, each moved back to today.

The present value formula

Adding those values gives a geometric series, which simplifies to:

$$PV = \frac{R\big[1 - (1 + i)^{-n}\big]}{i}$$

with $R$ the payment, $i$ the rate per period, and $n$ the number of payments (ordinary simple annuity: payments at the end of each period, as often as interest is compounded).

Loan payments

To find the payment on a loan of $PV$ dollars, solve for $R$:

$$R = \frac{PV \times i}{1 - (1 + i)^{-n}}$$

The total interest paid on a loan is $nR - PV$: everything you repay, minus what you borrowed. Payments are rounded to the cent, so use the rounded payment for this.

Worked examples

Example 1: Funding regular withdrawals

How much must be deposited now so that $500 can be withdrawn at the end of every month for $2$ years, if the account pays $3\%$ per year, compounded monthly?

Solution. $R = 500$, $i = 0.0025$, $n = 24$:

$$PV = \frac{500\big[1 - (1.0025)^{-24}\big]}{0.0025} \approx 11\,632.99$$

$11 632.99 must be deposited. That’s less than the $12 000 withdrawn, because the money earns interest while it waits.

Example 2: A car loan

A $25 000 car loan is repaid with monthly payments over $5$ years at $6.6\%$ per year, compounded monthly. Find the payment and the total interest.

Solution. $i = 0.0055$, $n = 60$:

$$R = \frac{25\,000(0.0055)}{1 - (1.0055)^{-60}} \approx 490.33$$

The monthly payment is $490.33. Total repaid: $60 \times 490.33 = 29\,419.80$. Total interest: $29\,419.80 - 25\,000 = 4419.80$, so $4419.80.

Example 3: Adding it up by hand

Find the present value of $1000 paid at the end of each year for $3$ years, at $5\%$ per year compounded annually, by adding the payments. Check with the formula.

Solution. From the timeline above:

$$\frac{1000}{1.05} + \frac{1000}{1.05^2} + \frac{1000}{1.05^3} \approx 952.38 + 907.03 + 863.84 = 2723.25$$ $$PV = \frac{1000\big[1 - (1.05)^{-3}\big]}{0.05} \approx 2723.25$$

Example 4: A shorter or longer loan

A $30 000 loan is charged $5.4\%$ per year, compounded monthly. Compare the monthly payment and total interest for a $3$-year loan and a $5$-year loan.

Solution. $i = 0.0045$.

$3$ years ($n = 36$): $R = \dfrac{30\,000(0.0045)}{1 - (1.0045)^{-36}} \approx 904.52$. Total interest $= 36(904.52) - 30\,000 = 2562.72$.

$5$ years ($n = 60$): $R = \dfrac{30\,000(0.0045)}{1 - (1.0045)^{-60}} \approx 571.65$. Total interest $= 60(571.65) - 30\,000 = 4299.00$.

The longer loan has smaller payments ($571.65 vs. $904.52) but costs $1736.28 more in interest.

Common mistakes

Using the future value formula for a loan. A loan amount is received now, so it’s a present value.

Dropping the negative exponent. The formula has $(1 + i)^{-n}$. Using $(1 + i)^n$ gives a negative answer, which is a sign something’s wrong.

Using the annual rate or years. $i$ is per period and $n$ is the number of payments.

Forgetting what “total interest” means. It’s total repaid minus the amount borrowed: $nR - PV$.

Practice

1. (Warm-up) Does each situation need a future value or a present value?

• (a) How much will monthly deposits grow to in $5$ years?
• (b) How much can you borrow if you can afford $300 a month?
• (c) How much is needed now to fund $1000 a year for $4$ years of university?

Solution

(a) Future value.

(b) Present value (the loan is received now).

(c) Present value.

2. (Warm-up) Find the present value of $200 paid at the end of every quarter for $3$ years at $4\%$ per year, compounded quarterly.

Solution

$i = 0.01$, $n = 12$:

$$PV = \frac{200\big[1 - (1.01)^{-12}\big]}{0.01} \approx 2251.02$$

The present value is $2251.02.

3. (Warm-up) A $20 000 loan is repaid with $48$ monthly payments of $450. How much interest is paid?

Solution

$48 \times 450 - 20\,000 = 21\,600 - 20\,000 = 1600$, so $1600.

4. (Core) A $1800 laptop is bought with a loan repaid in $12$ monthly payments at $9\%$ per year, compounded monthly. Find the payment.

Solution

$i = 0.0075$, $n = 12$:

$$R = \frac{1800(0.0075)}{1 - (1.0075)^{-12}} \approx 157.41$$

The payment is $157.41.

5. (Core) A scholarship will pay $2000 at the end of each year for $4$ years. How much must be invested now at $3.5\%$ per year, compounded annually?

Solution

$$PV = \frac{2000\big[1 - (1.035)^{-4}\big]}{0.035} \approx 7346.16$$

$7346.16 must be invested.

6. (Core) You can afford $350 a month for $4$ years. What’s the largest loan you can take at $7.2\%$ per year, compounded monthly?

Solution

$i = 0.006$, $n = 48$:

$$PV = \frac{350\big[1 - (1.006)^{-48}\big]}{0.006} \approx 14\,559.59$$

You can borrow up to $14 559.59.

7. (Core) For the loan in Example 2, explain why the total interest isn’t just $6.6\% \times 25\,000 \times 5$.

Solution

$0.066 \times 25\,000 \times 5 = 8250$ would be simple interest on the full $25 000 for all $5$ years. But each payment repays part of the loan, so the balance owing keeps shrinking, and interest is only charged on what’s still owed. That’s why the actual interest, $4419.80, is much less.

8. (Challenge) Show that the sum of present values $R(1 + i)^{-1} + R(1 + i)^{-2} + \dots + R(1 + i)^{-n}$ gives the present value formula.

Solution

This is a geometric series with $a = R(1 + i)^{-1}$, $r = (1 + i)^{-1}$, and $n$ terms. Using $S_n = \dfrac{a(1 - r^n)}{1 - r}$:

$$PV = \frac{R(1 + i)^{-1}\big[1 - (1 + i)^{-n}\big]}{1 - (1 + i)^{-1}}$$

Multiply the top and bottom by $(1 + i)$:

$$PV = \frac{R\big[1 - (1 + i)^{-n}\big]}{(1 + i) - 1} = \frac{R\big[1 - (1 + i)^{-n}\big]}{i}$$

9. (Challenge) For Example 1, find the future value of the same payments ($500 a month for $24$ months at $0.25\%$ per month), then divide it by $(1.0025)^{24}$. What do you notice, and why?

Solution

$$FV = \frac{500\big[(1.0025)^{24} - 1\big]}{0.0025} \approx 12\,351.41$$$$\frac{12\,351.41}{(1.0025)^{24}} \approx 11\,632.99$$

It’s the same as the present value. The present value is just the future value moved back $n$ periods: $PV = FV(1 + i)^{-n}$. Both describe the same payments, measured at different times.