C. Discrete Functions · Financial Applications: Interest, Present Value and AnnuitiesThe Present Value of an Ordinary Simple Annuity
C3.7 C3.5About 70 minutes
Learning goals- I can explain what the present value of an annuity means, for savings that will be withdrawn and for loans.
- I can show that the present value of an annuity is the sum of a geometric series.
- I can use PV = R[1 − (1 + i)⁻ⁿ] ÷ i to find the present value.
- I can find the total interest paid on a loan and compare it with the amount borrowed.
Words to know
present value of an annuity (PV)
The single amount that, invested today at rate i per period, would provide all the future payments exactly. For a loan, it is the amount borrowed.
loan
Money borrowed now (the present value) and paid back with equal payments that include interest.
amortize
To pay off a loan with a series of equal payments, each covering that period's interest and part of the principal.
total interest on a loan
Total of all payments minus the amount borrowed: nR − PV.
Money that will come out, one payment at a time
Suppose you want to receive $500 at the end of every month for 3 years while you study, starting one month from now. How much must be in the bank today? Or think about a car loan: you agree to pay $425 a month for 5 years. How much money are you really borrowing today?
Both questions ask for the present value of an annuity: the value today of a series of future payments.
Building it from present values
From Lesson 4, the present value of one payment R made n periods from now is R(1 + i)⁻ⁿ. Do that for every payment and add:
PV = R(1 + i)⁻¹ + R(1 + i)⁻² + … + R(1 + i)⁻ⁿ
The formula
The series above is geometric with first term a = R(1 + i)⁻¹, common ratio r = (1 + i)⁻¹ and n terms. Summing it (see the Challenge questions) gives
PV = R[1 − (1 + i)⁻ⁿ] ÷ i
A second way to see it: the present value is just the future value of the annuity moved back n periods:
PV = FV(1 + i)⁻ⁿ = R[(1 + i)ⁿ − 1] ÷ i × (1 + i)⁻ⁿ = R[1 − (1 + i)⁻ⁿ] ÷ i.
The same conditions apply: payments at the end of each period, and the payment period equals the compounding period.
Worked exampleExample 1: Funding monthly withdrawals
How much must be invested today, at 4.2% compounded monthly, to provide $500 at the end of every month for 3 years?
- R = 500, i = 0.042 ÷ 12 = 0.0035, n = 36.
- PV = 500[1 − (1.0035)⁻³⁶] ÷ 0.0035.
- Calculator: 1.0035 ^ (−36) ≈ 0.881809; 1 − that ≈ 0.118191; ÷ 0.0035 ≈ 33.7689; × 500.
- PV ≈ $16 884.46.
- Check: the total withdrawn is 36 × 500 = $18 000. The account pays out more than you put in, because the money still in the account keeps earning interest: about $1 115.54.
Worked exampleExample 2: How much is the car loan?
Priya can afford car payments of $425 at the end of every month for 5 years. The interest rate is 6.9% compounded monthly. (a) What is the most she can borrow? (b) How much interest will she pay in total? (c) Compare the interest with the amount borrowed.
- R = 425, i = 0.069 ÷ 12 = 0.00575, n = 60.
- (a) PV = 425[1 − (1.00575)⁻⁶⁰] ÷ 0.00575 ≈ $21 514.57.
- (b) Total paid: 60 × 425 = $25 500. Interest = 25 500 − 21 514.57 ≈ $3 985.43.
- (c) The interest is about 18.5% of the amount borrowed. For every dollar borrowed she pays back about $1.19.
She can borrow about $21 514.57 and will pay about $3 985.43 in interest.
Each $1 000 payment is worth less today the further away it is. The present value of the whole annuity is the sum of all ten bars: about $7 360.09.Worked exampleExample 3: Lump sum or payments?
A scholarship offers either $50 000 now or $6 000 at the end of each year for 10 years. Money can be invested at 5% compounded annually. Which is worth more?
- Compare the values today.
- PV of the payments = 6000[1 − (1.05)⁻¹⁰] ÷ 0.05 ≈ $46 330.41.
- $50 000 now is worth more than $46 330.41, so the lump sum is the better choice at 5%.
- (The payments total $60 000, but that number ignores the time value of money.)
Take the $50 000 now (at 5%).
PV or FV?
Ask: "Where on the timeline is the single amount I want?"
- At the end, after the payments (a savings goal): use FV = R[(1 + i)ⁿ − 1] ÷ i.
- At the start, before the payments (a loan, or money set aside for withdrawals): use PV = R[1 − (1 + i)⁻ⁿ] ÷ i.
Just for funA loan is a time machine for money: it brings future dollars into the present. Like every time machine in the movies, it has a cost. Here the cost is called interest.
Hands-on activityCar shopping with math
You need: a calculator and, with a grown-up, a car dealer or car-buying website (or a newspaper ad).
- Find a used car you like and note its price.
- Suppose the loan rate is 7.2% compounded monthly. Use the present value formula to find the most you could borrow if you can pay $300, $400 or $500 a month for 4 years (48 payments).
- Which payment would let you buy your car (ignoring taxes and fees)?
- For that payment, find the total interest and compare it with the amount borrowed.
What to look for: with i = 0.006 and n = 48, the present values are about $12 479.65 ($300 a month), $16 639.53 ($400) and $20 799.41 ($500). For $400 a month, the total paid is $19 200, so the interest is about $2 560.47, roughly 15% of the amount borrowed. Real loans also involve taxes, fees and credit checks, which this model ignores.
Your turn
1Find the present value of $1 000 at the end of each year for 3 years at 5% compounded annually.
1000[1 − (1.05)⁻³] ÷ 0.05 ≈ $2 723.25.
2How much must be invested today at 4.2% compounded monthly to provide $500 at the end of each month for 3 years?
500[1 − (1.0035)⁻³⁶] ÷ 0.0035 ≈ $16 884.46.
3A loan is repaid with $300 at the end of every month for 2 years at 4.8% compounded monthly. How much was borrowed?
i = 0.004, n = 24: 300[1 − (1.004)⁻²⁴] ÷ 0.004 ≈ $6 852.15.
4How much interest is paid on the loan in the last question?
24 × 300 − 6 852.15 ≈ $347.85.
5Find the present value of $2 000 at the end of every half-year for 5 years at 5% compounded semi-annually.
i = 0.025, n = 10: ≈ $17 504.13.
6Find the present value of $150 at the end of every month for 1 year at 6% compounded monthly.
i = 0.005, n = 12: ≈ $1 742.84.
7Which is worth more today at 5% compounded annually: $50 000 now or $6 000 at the end of each year for 10 years?
PV of the payments ≈ $46 330.41, so $50 000 now is worth more.
More practice: warm-up
1For a loan, what does the present value of the payments represent?
2Write the present value of $100 paid in 2 years at 10% compounded annually.
3Find the present value of $100 at the end of each year for 2 years at 10% compounded annually.
100(1.1)⁻¹ + 100(1.1)⁻² ≈ 90.91 + 82.64 = $173.55.
4A loan has 48 payments of $250. What is the total paid?
5Is the present value of an annuity more or less than the total of its payments (when i > 0)?
Less, because each future payment is discounted.
More practice: on your own
1Priya pays $425 a month for 5 years at 6.9% compounded monthly. How much did she borrow, and how much interest does she pay?
PV ≈ $21 514.57; interest ≈ 25 500 − 21 514.57 = $3 985.43.
2A business will receive $12 000 at the end of each year for 4 years. What is that worth today at 4% compounded annually?
3$350 at the end of every month for 4 years at 5.4% compounded monthly. Find the present value.
i = 0.0045, n = 48: ≈ $15 079.10.
4Show that you get the same answer for Example 1 by finding the future value of the annuity and then moving it back 36 months.
FV = 500[(1.0035)³⁶ − 1] ÷ 0.0035 ≈ $19 147.52. Then 19 147.52(1.0035)⁻³⁶ ≈ $16 884.46 ✓.
5A $15 000 loan is repaid with 60 payments of $250 at 7.2% compounded monthly. Is that enough to pay it off? Explain.
PV of the payments = 250[1 − (1.006)⁻⁶⁰] ÷ 0.006 ≈ $12 565.53, which is less than $15 000, so no. 60 × $250 = $15 000 would only repay the loan if there were no interest.
Challenge
1Use the geometric series formula with a = R(1 + i)⁻¹ and r = (1 + i)⁻¹ to derive PV = R[1 − (1 + i)⁻ⁿ] ÷ i.
Sₙ = a(1 − rⁿ) ÷ (1 − r) = R(1 + i)⁻¹[1 − (1 + i)⁻ⁿ] ÷ [1 − (1 + i)⁻¹]. Multiply top and bottom by (1 + i): the bottom becomes (1 + i) − 1 = i and the top becomes R[1 − (1 + i)⁻ⁿ]. So PV = R[1 − (1 + i)⁻ⁿ] ÷ i.
2A used car is bought with a $5 000 down payment plus $400 at the end of every month for 3 years at 4.8% compounded monthly. What is the equivalent cash price of the car?
PV of payments = 400[1 − (1.004)⁻³⁶] ÷ 0.004 ≈ $13 386.35. Cash price ≈ 5000 + 13 386.35 = $18 386.35.
3Prize A: $2 500 at the end of every quarter for 5 years. Prize B: $1 500 at the end of every month for 2 years followed by nothing. At 7% compounded quarterly for A and 3.6% compounded monthly for B, which prize is worth more today? (Assume you can invest at each prize's rate.)
A: 2500[1 − (1.0175)⁻²⁰] ÷ 0.0175 ≈ $41 882.20. B: 1500[1 − (1.003)⁻²⁴] ÷ 0.003 ≈ $34 684.40. Prize A is worth more today.
Quiz
1. The present value of a loan's payments equals:
2. Which formula gives the present value of an ordinary simple annuity?
3. A loan is repaid by 36 payments of $300. The present value of the payments is $10 000. The total interest is:
4. In the present value series R(1 + i)⁻¹ + R(1 + i)⁻² + …, the common ratio is:
5. You need money in the bank today to make 24 monthly withdrawals of $200. Which do you calculate?
6. At 5% compounded annually, the present value of $6 000 a year for 10 years is about $46 330.41. Compared with $50 000 now, the payments are:
Big idea: Expectation C3.7 asks students to solve problems involving the present value of an ordinary simple annuity, including real-world loans, and C3.5 connects annuities to geometric series. The curriculum's own example is to calculate the total interest paid over the life of a loan and compare it with the original principal.
How to help: If your family has a car loan or other instalment plan, you might look at the monthly payment and the number of payments together (no need to share private details) and ask, "How much is that in total? How much of it is interest?"
Common difficulties:
- Choosing FV when PV is needed. A timeline settles it: is the single amount at the start or the end?
- Calculator errors with the negative exponent: 1.0035 ^ (−36).
- Thinking the total of the payments is what the annuity is worth today.