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Grade 11 · MCR3U · Functions

MCR3U: Annuities: Future and Present Value

An annuity is a series of equal payments made at equal intervals: monthly savings deposits, car loan payments, a pension. Each payment earns compound interest for a different length of time, so adding them up is a geometric series, and that series gives the two annuity formulas.

How It Works

Ordinary annuities. In MCR3U each payment is made at the end of its period, and the interest compounds as often as the payments are made. Use RR for the regular payment, ii for the interest rate per period and nn for the number of payments.

Future value. This is what all the payments are worth at the end. The last payment earns no interest, the one before it earns one period, and the first earns n−1n - 1 periods. So FV=R+R(1+i)+⋯+R(1+i)n−1FV = R + R(1 + i) + \dots + R(1 + i)^{n - 1}, a geometric series with first term RR and ratio 1+i1 + i. Its sum is FV=R((1+i)n−1)iFV = \frac{R\left((1 + i)^n - 1\right)}{i}.

Present value. This is the single amount today that would fund all the payments, like the amount of a loan or the size of a fund you will draw from. Bring each payment back to today: PV=R(1+i)−1+R(1+i)−2+⋯+R(1+i)−nPV = R(1 + i)^{-1} + R(1 + i)^{-2} + \dots + R(1 + i)^{-n}, which sums to PV=R(1−(1+i)−n)iPV = \frac{R\left(1 - (1 + i)^{-n}\right)}{i}.

Which formula. Saving up toward a goal is future value. Borrowing now and paying it back, or withdrawing from a fund until it runs out, is present value.

Finding the payment. Put the known FVFV or PVPV into its formula and solve for RR. For a loan, the total interest is everything you pay, nRnR, minus the amount borrowed.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

  1. Easy

    200200 dollars is deposited at the end of every month for 33 years, at 6%6\% a year compounded monthly. Find the future value and the interest earned.
    Show solution
    1. R=200R = 200, i=0.0612=0.005i = \frac{0.06}{12} = 0.005 and n=36n = 36.
    2. FV=200((1.005)36−1)0.005≈7867.22FV = \frac{200\left((1.005)^{36} - 1\right)}{0.005} \approx 7867.22.
    3. The deposits total 36(200)=720036(200) = 7200, so the interest is 7867.22−7200=667.227867.22 - 7200 = 667.22.

    Answer: About 7867.227867.22 dollars, with 667.22667.22 dollars of interest.

  2. Easy

    Find the present value of 500500 dollars paid at the end of each year for 1010 years, at 4%4\% a year compounded annually.
    Show solution
    1. R=500R = 500, i=0.04i = 0.04 and n=10n = 10.
    2. PV=500(1−(1.04)−10)0.04≈4055.45PV = \frac{500\left(1 - (1.04)^{-10}\right)}{0.04} \approx 4055.45.

    Answer: About 4055.454055.45 dollars.

  3. Medium

    Sam wants 1500015000 dollars in 55 years. What must he deposit at the end of each quarter, at 4.8%4.8\% a year compounded quarterly?
    Show solution
    1. Saving toward a goal: future value, with i=0.0484=0.012i = \frac{0.048}{4} = 0.012 and n=20n = 20.
    2. 15000=R((1.012)20−1)0.01215000 = \frac{R\left((1.012)^{20} - 1\right)}{0.012}, so R=15000(0.012)(1.012)20−1≈668.07R = \frac{15000(0.012)}{(1.012)^{20} - 1} \approx 668.07.

    Answer: About 668.07668.07 dollars a quarter.

  4. Medium

    A 2200022000 dollar car loan is repaid monthly over 55 years at 3.6%3.6\% a year compounded monthly. Find the payment and the total interest.
    Show solution
    1. A loan: present value, with i=0.03612=0.003i = \frac{0.036}{12} = 0.003 and n=60n = 60.
    2. R=22000(0.003)1−(1.003)−60≈401.20R = \frac{22000(0.003)}{1 - (1.003)^{-60}} \approx 401.20.
    3. Total paid: 60(401.20)=24072.0060(401.20) = 24072.00, so the interest is 24072.00−22000=2072.0024072.00 - 22000 = 2072.00.

    Answer: About 401.20401.20 dollars a month, with about 2072.002072.00 dollars of interest.

  5. Test level

    Mia wants to withdraw 20002000 dollars at the end of every month for 2020 years after she retires, from a fund earning 4.2%4.2\% a year compounded monthly. How much does the fund need when she retires? If she builds it with equal monthly deposits over 3030 years at the same rate, what is each deposit?
    Show solution
    1. The withdrawals are a present value at retirement: i=0.04212=0.0035i = \frac{0.042}{12} = 0.0035 and n=240n = 240.
    2. PV=2000(1−(1.0035)−240)0.0035≈324374.79PV = \frac{2000\left(1 - (1.0035)^{-240}\right)}{0.0035} \approx 324374.79.
    3. That amount is the future value of her deposits, with n=360n = 360: R=324374.79(0.0035)(1.0035)360−1≈450.94R = \frac{324374.79(0.0035)}{(1.0035)^{360} - 1} \approx 450.94.

    Answer: About 324374.79324374.79 dollars at retirement, built with deposits of about 450.94450.94 dollars a month.

Learn It on Martian Lab

Martian Lab teaches this in the Annuities lesson of its MCR3U Discrete Functions unit. Watch it, then take the Discrete Functions practice tests to see if it stuck.

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Free Discrete Functions practice test, with solutions

More Discrete Functions topics: Arithmetic Sequences and Series · Geometric Sequences and Series · Pascal's Triangle and the Binomial Theorem · Simple and Compound Interest

More MCR3U topics: Inverse Functions · Transformations of Functions · Sine Law and the Ambiguous Case · Domain, Range and Function Notation · Simplifying Rational Expressions · Operations with Rational Expressions · Max/Min and Vertex Form · Linear-Quadratic Systems · Quadratic Word Problems · Simplifying Radicals · Exponent Laws and Rational Exponents · Graphing Exponential Functions · Exponential Growth and Decay · Special Angles and the CAST Rule · Cosine Law and 3D Problems · Transformations of Sine and Cosine · Sinusoidal Word Problems

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