Present Value of Arithmetic-Decreasing Annuities

Payment Pattern and Valuation Model

An arithmetic-decreasing annuity can represent a scheduled series of repayments, withdrawals, or other cash flows that falls by the same dollar amount each period. Its present value depends both on the payment pattern and on when each payment occurs. In this tutorial, the payments are made at the end of each period, so the annuity is immediate.

Definition: An arithmetic-decreasing annuity-immediate has $n$ payments at times $1,2,\ldots,n$. The payment at time $k$ is $C-(k-1)d$, where $C$ is the first payment and $d$ is the fixed decrease per period. Unless stated otherwise, assume $n$ is a positive integer, $d\geq 0$, and interest is a constant effective rate $i\geq 0$ per payment period. The discount factor is $v=1/(1+i)$.

The payment at time 1 is $C$, not $C-d$. The last payment, at time $n$, is $C-(n-1)d$. If every payment must be nonnegative, the schedule must satisfy $C-(n-1)d\geq 0$. For $d>0$, this is equivalent to $n\leq 1+C/d$; since $n$ must be an integer, the largest allowed term is $\lfloor 1+C/d\rfloor$. If payments must be strictly positive, use $C-(n-1)d>0$ instead. These conditions concern the payment schedule, not the discounting.

Let $a_{\overline{n}|}$ denote the present-value factor for $n$ unit payments at the ends of periods 1 through $n$, and let $(Ia)_{\overline{n}|}$ denote the present-value factor for payments $1,2,\ldots,n$ at those same times. The second factor is the standard increasing arithmetic-gradient factor.

Theorem: Under the stated assumptions, the present value $P$ of payments $C,C-d,\ldots,C-(n-1)d$ is $$ P=C a_{\overline{n}|}-d\left((Ia)_{\overline{n}|}-a_{\overline{n}|}\right) =(C+d)a_{\overline{n}|}-d(Ia)_{\overline{n}|}. $$ For $i>0$, $a_{\overline{n}|}=(1-v^n)/i$. At $i=0$, the direct-sum formula is $P=nC-dn(n-1)/2$.

Why the Decreasing-Gradient Formula Holds

At time $k$, the payment can be written as $C-d(k-1)$. The level part contributes $C$ at every payment time; the decrease from that level is $d(k-1)$ at time $k$. Thus the present value is the discounted sum of the level payments minus the discounted sum of the decreases:

P=\sum_{k=1}^{n}\bigl(C-d(k-1)\bigr)v^k =C\sum_{k=1}^{n}v^k-d\sum_{k=1}^{n}(k-1)v^k.

The first sum is $a_{\overline{n}|}$. For the second, the factor $(Ia)_{\overline{n}|}$ discounts amounts $1,2,\ldots,n$, whereas the decreases have amounts $0,1,\ldots,n-1$. Subtracting the level factor shifts each of those amounts down by 1:

\sum_{k=1}^{n}(k-1)v^k =\sum_{k=1}^{n}kv^k-\sum_{k=1}^{n}v^k =(Ia)_{\overline{n}|}-a_{\overline{n}|}.

Substitution gives the stated result. This derivation also explains why simply subtracting $d(Ia)_{\overline{n}|}$ from $C a_{\overline{n}|}$ is incorrect: that would subtract $d,2d,\ldots,nd$, rather than $0,d,\ldots,(n-1)d$.

When $i>0$, the gradient factor can be evaluated from the finite sum:

(Ia)_{\overline{n}|} =\sum_{k=1}^{n}kv^k =\frac{1-(n+1)v^n+nv^{n+1}}{i^2v}.

For routine valuation, the finite-sum interpretation or the factor formula may be used. If $i=0$, expressions involving division by $i$ do not apply; sum the undiscounted payments directly, or use the zero-rate expression in the theorem.

Worked Valuations

Worked Example: Discounting a Decreasing Payment Schedule

A hypothetical grant pays $1,000 at the end of year 1, then reduces each annual payment by $100 for four more years. Find its present value at an effective annual interest rate of 5%. Confirm that the final payment is nonnegative.

Method. Write the payments using the stated end-of-year timing, then discount each one. This direct calculation makes the payment indexing visible.

There are $n=5$ payments, with $C=1000$, $d=100$, and $i=0.05$. The schedule is $1,000, $900, $800, $700, and $600. The final payment check is $1000-(5-1)100=600$, which is positive.

The discount factors are $v=1/1.05=0.952380952$, $v^2=0.907029478$, $v^3=0.863837599$, $v^4=0.822702475$, and $v^5=0.783526166$. The discounted payments are:

\begin{aligned} 1000v&=952.380952,\\ 900v^2&=816.326531,\\ 800v^3&=691.070079,\\ 700v^4&=575.891733,\\ 600v^5&=470.115700. \end{aligned}

Adding them gives $P=952.380952+816.326531+691.070079+575.891733+470.115700=3505.784995$. Rounded to the nearest cent, the present value is $3,505.78.

Check. Each discounted value is positive and less than its corresponding future payment, as expected at a positive interest rate. The five payments also decrease by exactly $100 each time.

Common mistake avoided: The first payment occurs at the end of year 1, so it is discounted once. Treating it as a time-0 payment would overstate the present value.

Worked Example: Finding the Longest Nonnegative Schedule

A hypothetical reserve is to be paid out in annual end-of-year installments. The first installment is $840, and each later installment is $120 less than the previous one. Find the largest number of installments that keeps every payment nonnegative. Then find the present value at an effective annual rate of 5%.

Method. First impose the final-payment constraint, then value the resulting schedule. The constraint must be checked before treating a zero or negative amount as an installment.

Here $C=840$ and $d=120$. For $n$ installments, the final amount is $840-(n-1)120$. Requiring it to be nonnegative gives $(n-1)120\leq840$, so $n-1\leq7$ and $n\leq8$. For $n=8$, the final payment is $840-7(120)=0$. For $n=9$, it would be $840-8(120)=-120$. Thus the maximum is eight installments.

The eight scheduled amounts are $840, $720, $600, $480, $360, $240, $120, and $0. At 5%, the discount factors $v^1$ through $v^8$ are $0.952380952,\ 0.907029478,\ 0.863837599,\ 0.822702475,\ 0.783526166,\ 0.746215397,\ 0.710681330,\ 0.676839362$, respectively. Multiplying each amount by its factor gives:

\begin{aligned} 840v&=800.000000, &720v^2&=653.061224,\\ 600v^3&=518.302559, &480v^4&=394.897188,\\ 360v^5&=282.069420, &240v^6&=179.091695,\\ 120v^7&=85.281760, &0v^8&=0. \end{aligned}

The sum is $2912.703846$, so the present value rounded to the nearest cent is $2,912.70.

Check. The payment at time 8 is zero, while extending the pattern by one year would create a negative payment. The calculated value is below the undiscounted total of $3,360.

Common mistake avoided: The eight-payment schedule includes a zero final payment. If the problem required every payment to be strictly positive, eight would not be allowed; the maximum would be seven.

Worked Example: Solving for the First Payment

A hypothetical three-year payment arrangement has end-of-year payments $C$, $C-100$, and $C-200$. Its present value is $2,000 at an effective annual interest rate of 10%. Find $C$, and check the final payment.

Method. Express the present value as a coefficient of $C$ minus the present value of the fixed decreases. Then solve the resulting linear equation.

The schedule has $n=3$, $d=100$, and $v=1/1.10=0.909090909$. Its discount factors are $v^2=0.826446281$ and $v^3=0.751314801$. The level annuity factor is $a_{\overline{3}|}=v+v^2+v^3=2.486851991$. The discounted decrease amounts are $0v+1v^2+2v^3=0.826446281+1.502629602=2.329075883$.

Therefore the present-value equation is $2000=C(2.486851991)-100(2.329075883)$. The fixed decrease component has value $232.907588$, so $C=(2000+232.907588)/2.486851991=897.885\ldots$. Rounded to the nearest cent, the first payment is $897.89.

The final payment is $C-200=697.885\ldots$, or $697.89 to the nearest cent, so all three payments are positive. The equation also checks: using the unrounded value of $C$ gives a present value of $2,000.

Common mistake avoided: The first decrease is zero, not $100. The decrease amounts at the three payment times are $0, $100, and $200.

Common Indexing and Model Traps

Arithmetic-decreasing annuities are especially sensitive to the point at which the sequence is indexed. A useful check is to write the first and last payments explicitly before using a factor. In this model, the amount at time $k$ is $C-(k-1)d$, and the gradient of decreases is $0,1,\ldots,n-1$.

Common mistake: For four payments, a wrong calculation is $C a_{\overline{4}|}-d(Ia)_{\overline{4}|}$. Since $(Ia)_{\overline{4}|}$ discounts $1,2,3,4$, this subtracts decreases $d,2d,3d,4d$; it incorrectly makes the first payment $C-d$. The correct expression is $C a_{\overline{4}|}-d((Ia)_{\overline{4}|}-a_{\overline{4}|})$, which subtracts $0,d,2d,3d$. Equivalently, use $(C+d)a_{\overline{4}|}-d(Ia)_{\overline{4}|}$.

Also verify the time unit for the interest rate. If payments are annual, the rate used in $v=1/(1+i)$ must be an effective annual rate. A rate quoted for another period must first be converted to an effective rate per payment period. The formula assumes a constant rate per period and payments exactly one period apart; it does not directly apply to irregular payment dates or changing rates.

The nonnegative-payment test is separate from present-value calculation. A discount factor can value a negative cash flow, but that does not make it a permissible installment when the problem requires nonnegative payments. Check $C-(n-1)d$ against zero and interpret equality according to whether zero payments are allowed.

Key takeaway: For end-of-period payments $C,C-d,\ldots,C-(n-1)d$, subtract the present value of the decrease sequence $0,d,\ldots,(n-1)d$ from the level annuity value. Check the final amount $C-(n-1)d$ independently.

Check Your Understanding

For each problem, use the stated timing and rate assumptions. Answers are rounded to the nearest cent when a monetary present value or payment is requested.

An annual annuity-immediate pays $500 at the end of year 1, decreasing by $75 each year for six payments. At an effective annual rate of 4%, find the present value and verify the final payment. (Answer: $1,683.10; final payment $125.)
A payment schedule begins with $950 and decreases by $150 at each annual payment. What is the largest number of payments that can be made without a negative payment? (Answer: 7 payments; the seventh is $50 and the eighth would be $-$100$.)
At zero interest, an annuity-immediate pays $700 at the end of year 1 and decreases by $80 for each of the next five payments. Find its present value and final payment. (Answer: $3,000; final payment $300.)
A three-payment annuity-immediate decreases by $50 per year and has a present value of $1,000 at an effective annual rate of 5%. Find the first payment and check that the last payment is positive. (Answer: first payment $415.58; last payment $315.58.)
An annuity-immediate pays $600 at the end of year 1 and decreases by $100 annually for four payments. Find its present value at an effective annual rate of 8%, and state the final payment. (Answer: $1,522.27; final payment $300.)

What You'll Learn