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Guide/Corporate finance

Annuities, perpetuities and loan amortization

Lesson 2 of 8 · 1 min read

Many financial contracts pay the same amount at regular intervals: a pension, a lease, a bond's coupons, the instalments of a mortgage. Valuing them one payment at a time works, but it is slow and hides the structure. This lesson derives closed formulas for such streams, using the discounting from "Interest, compounding and discounting", and applies them to the repayment of a loan.

Throughout the lesson payments are made once a year, the rate rrr is compounded annually, and the first payment arrives one year from today unless stated otherwise.

Ordinary annuities

An annuity is a stream of nnn equal payments CCC at the ends of years 1,2,…,n1, 2, \dots, n1,2,…,n. Its present value is the sum of the discounted payments:

PV=C1+r+C(1+r)2+⋯+C(1+r)n.\text{PV} = \frac{C}{1+r} + \frac{C}{(1+r)^2} + \cdots + \frac{C}{(1+r)^n}.PV=1+rC​+(1+r)2C​+⋯+(1+r)nC​.

This is a geometric series with first term C/(1+r)C/(1+r)C/(1+r) and ratio q=1/(1+r)q = 1/(1+r)q=1/(1+r). Using a+aq+⋯+aqn−1=a (1−qn)/(1−q)a + aq + \cdots + aq^{n-1} = a\,(1 - q^n)/(1 - q)a+aq+⋯+aqn−1=a(1−qn)/(1−q) and simplifying gives

PV=C⋅1−(1+r)−nr.\text{PV} = C \cdot \frac{1 - (1+r)^{-n}}{r}.PV=C⋅r1−(1+r)−n​.

The fraction is the annuity factor: the present value of 1 a year for nnn years. Compounding the present value forward by nnn years gives the future value at the date of the last payment:

FV=PV (1+r)n=C⋅(1+r)n−1r.\text{FV} = \text{PV}\,(1+r)^n = C \cdot \frac{(1+r)^n - 1}{r}.FV=PV(1+r)n=C⋅r(1+r)n−1​.

Example: 1,000 a year for 10 years at r=6%r = 6\%r=6%. The annuity factor is (1−1.06−10)/0.06≈7.3601(1 - 1.06^{-10})/0.06 \approx 7.3601(1−1.06−10)/0.06≈7.3601, so PV≈7,360.09\text{PV} \approx 7{,}360.09PV≈7,360.09 and FV≈13,180.79\text{FV} \approx 13{,}180.79FV≈13,180.79. The ten payments add up to 10,000, but they are worth only 7,360.09 today, because most of them arrive years from now.

Annuities due

If each payment is made at the start of the year instead of the end, as with rent, the stream is an annuity due. Every payment arrives one year earlier, so every discounted term is multiplied by (1+r)(1+r)(1+r):

PVdue=(1+r)⋅C⋅1−(1+r)−nr.\text{PV}_{\text{due}} = (1+r) \cdot C \cdot \frac{1 - (1+r)^{-n}}{r}.PVdue​=(1+r)⋅C⋅r1−(1+r)−n​.

In the example, 1.06×7,360.09≈7,801.691.06 \times 7{,}360.09 \approx 7{,}801.691.06×7,360.09≈7,801.69.

Perpetuities

A perpetuity pays CCC every year forever. Letting n→∞n \to \inftyn→∞ in the annuity formula, (1+r)−n→0(1+r)^{-n} \to 0(1+r)−n→0, so

PV=Cr.\text{PV} = \frac{C}{r}.PV=rC​.

A growing perpetuity pays CCC after one year and grows at the constant rate ggg after that: CCC, C(1+g)C(1+g)C(1+g), C(1+g)2C(1+g)^2C(1+g)2, and so on. The series is again geometric, with ratio (1+g)/(1+r)(1+g)/(1+r)(1+g)/(1+r), and it converges only when g<rg < rg<r:

PV=Cr−g,g<r.\text{PV} = \frac{C}{r - g}, \qquad g < r.PV=r−gC​,g<r.

At r=6%r = 6\%r=6%, a perpetuity of 1,000 is worth 1,000/0.06≈16,666.671{,}000/0.06 \approx 16{,}666.671,000/0.06≈16,666.67. If the payment grows by 2% a year, it is worth 1,000/0.04=25,0001{,}000/0.04 = 25{,}0001,000/0.04=25,000. The growing perpetuity is the formula behind the terminal value in company valuation.

Annuity factor (1−(1+r)−n)/r(1 - (1+r)^{-n})/r(1−(1+r)−n)/r, perpetuity C/rC/rC/r, growing perpetuity C/(r−g)C/(r - g)C/(r−g) for g<rg < rg<r. Each is a geometric sum of discounted payments.

The payment on a loan

A level-payment loan of amount LLL over nnn years is an annuity seen from the lender's side: the lender pays LLL today and receives nnn equal payments CCC. The loan is fairly priced when the present value of the payments equals LLL, so

C=L1−(1+r)−nr=L r1−(1+r)−n.C = \frac{L}{\dfrac{1 - (1+r)^{-n}}{r}} = \frac{L\,r}{1 - (1+r)^{-n}}.C=r1−(1+r)−n​L​=1−(1+r)−nLr​.

Example: a loan of 200,000 at 6% over 10 years. The annuity factor is 7.3601, so the annual payment is 200,000/7.3601≈27,173.59200{,}000 / 7.3601 \approx 27{,}173.59200,000/7.3601≈27,173.59. Over the life of the loan the borrower pays 10×27,173.59≈271,735.9210 \times 27{,}173.59 \approx 271{,}735.9210×27,173.59≈271,735.92, of which 71,735.92 is interest.

The amortization schedule

Each payment is split in two. The interest part is the rate times the balance still owed at the start of the year. The rest of the payment is principal and reduces the balance:

It=r Bt−1,Pt=C−It,Bt=Bt−1−Pt.I_t = r\,B_{t-1}, \qquad P_t = C - I_t, \qquad B_t = B_{t-1} - P_t.It​=rBt−1​,Pt​=C−It​,Bt​=Bt−1​−Pt​.

For the loan above, with B0=200,000B_0 = 200{,}000B0​=200,000, r=6%r = 6\%r=6% and C=27,173.59C = 27{,}173.59C=27,173.59, the first two rows of the schedule are:

Worked example

The first two years of the amortization schedule

  1. Interest in year 1: the rate times the opening balance.

    I1=r B0\displaystyle I_1 = r\,B_0I1​=rB0​=12,000
  2. The rest of the payment repays principal.

    P1=C−I1\displaystyle P_1 = C - I_1P1​=C−I1​=15,173.59
  3. Balance still owed at the end of year 1.

    B1=B0−P1\displaystyle B_1 = B_0 - P_1B1​=B0​−P1​=184,826.41
  4. Interest in year 2 is smaller, because the balance is smaller.

    I2=r B1\displaystyle I_2 = r\,B_1I2​=rB1​=11,089.58
  5. So more of the same payment goes to principal.

    P2=C−I2\displaystyle P_2 = C - I_2P2​=C−I2​=16,084.01
  6. Balance at the end of year 2.

    B2=B1−P2\displaystyle B_2 = B_1 - P_2B2​=B1​−P2​=168,742.4

Solving for the term

The same equation can be turned around. If the borrower can afford a payment CCC, the number of years needed to repay LLL follows from L=C (1−(1+r)−n)/rL = C\,(1 - (1+r)^{-n})/rL=C(1−(1+r)−n)/r by taking logarithms:

n=−ln⁡(1−L rC)ln⁡(1+r).n = -\frac{\ln\left(1 - \dfrac{L\,r}{C}\right)}{\ln(1+r)}.n=−ln(1+r)ln(1−CLr​)​.

Paying 20,000 a year on the 200,000 loan at 6% gives n=−ln⁡(1−0.6)/ln⁡1.06≈15.73n = -\ln(1 - 0.6)/\ln 1.06 \approx 15.73n=−ln(1−0.6)/ln1.06≈15.73 years, so in practice 16 payments, the last one smaller. The formula also shows a limit: if C≤L r=12,000C \le L\,r = 12{,}000C≤Lr=12,000, the payment does not even cover the first year's interest and the loan is never repaid. Solving for the rate instead has no closed form; it is found numerically, exactly like the internal rate of return in the next lesson.

How the payments change

The payment stays the same, but its make-up shifts. Early on the balance is large, so most of the payment is interest; as the balance falls, the interest part shrinks and the principal part grows. By construction the balance reaches exactly zero with the last payment.

The figure draws the whole schedule. Each bar is one year's payment, split into interest (bottom) and principal (top); the dashed line is the balance left at the end of each year, on the right-hand axis. Click a year, or move to it with Tab and press Enter, to see that row of the schedule worked out below the chart. Start with year 1 at the defaults (10 years, 6%), then switch the term to 30 years: the payment falls to about 14,530, but 82.6% of the first payment is now interest, and total interest rises to more than the loan itself.

Amortization schedule: interest and principal year by yearOpen in Time Value of Money →
Term
Rate

Click a year (or Tab to it and press Enter) to see that row of the schedule below.

  • Interest
  • Principal
  • Remaining balance (end of year)
Year 1
Opening balance
200,000.00
Payment
27,173.59
Interest
Interest = 6% × 200,000.00 = 12,000.00
Principal
Principal = 27,173.59 − 12,000.00 = 15,173.59
Closing balance
Closing balance = 200,000.00 − 15,173.59 = 184,826.41
Payment
27,173.59
Total interest paid
71,735.92
Interest share of the first payment
44.2%

Summary

  • An annuity of CCC for nnn years is worth C (1−(1+r)−n)/rC\,(1 - (1+r)^{-n})/rC(1−(1+r)−n)/r today; an annuity due is worth (1+r)(1+r)(1+r) times as much.
  • A perpetuity is worth C/rC/rC/r, and a growing perpetuity C/(r−g)C/(r - g)C/(r−g) as long as g<rg < rg<r.
  • A level loan payment is the loan amount divided by the annuity factor.
  • Each payment is interest on the opening balance plus principal; early payments are mostly interest, and longer terms shift the mix further towards interest.

The next lesson, "Net present value and internal rate of return", values projects whose cash flows are not level, by discounting each one separately with the same tools.

Try itTry regular contributions and compound growth in the Time Value of Money lab →
2 questions on this lesson at the end of the series →
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Time Value of Money