Arithmetic Gradients with Payments at the Beginning

Why Payment Timing Matters

An arithmetic gradient changes each payment by a fixed dollar amount, but the value also depends on when each payment occurs. In an annuity due, the first payment is made at time 0 and the last of $n$ payments is made at time $n-1$. This differs by one period from an annuity-immediate, whose payments occur at times 1 through $n$. Shifting the dates carefully is a reliable way to value either an increasing or decreasing annuity due.

Assume throughout that payments are made at equal annual intervals, the effective annual interest rate is $i$, and $i>-1$. There are no payments outside the stated $n$ dates. The same timing argument works for another regular period if the rate is an effective rate for that period.

Definition: An arithmetic-increasing annuity due with first payment $P$ and increment $G$ pays $P+kG$ at time $k$, for $k=0,1,\ldots,n-1$. An arithmetic-decreasing annuity due pays $P-kG$ at those dates. Here $G\geq 0$ denotes the size of the change per payment. For a decreasing stream with nonnegative payments, require $P-(n-1)G\geq 0$.

The One-Period Shift

Compare an annuity due with an annuity-immediate that makes the same sequence of payment amounts. The immediate payments are at times 1 through $n$; moving each one period earlier produces the due payments at times 0 through $n-1$. Since each moved payment is discounted for one fewer period, the present value of the due stream is $1+i$ times the present value of that immediate stream. At a common accumulation date, each due payment earns interest for one additional period, so the accumulated value is also $1+i$ times the immediate stream's accumulated value.

\text{Due value}=(1+i)\times\text{immediate value}

For clarity, an immediate stream with first payment $P$ at time 1 and arithmetic increment $G$ has payments $P+(j-1)G$ at time $j$, for $j=1,\ldots,n$. Its present value is the sum of each payment discounted to time 0. Its accumulated value is the sum of each payment accumulated to time $n$. Applying the shift gives the following formulas.

\operatorname{PV}_{\text{due}} =(1+i)\left[ P a_{\overline{n}|} +G\sum_{j=1}^{n}(j-1)v^j \right], \qquad v=\frac{1}{1+i}

\operatorname{AV}_{n,\text{due}} =(1+i)\left[ P s_{\overline{n}|} +G\sum_{j=1}^{n}(j-1)(1+i)^{n-j} \right]

These expressions are for an increasing stream. To value a decreasing stream whose payments are $P, P-G, P-2G,\ldots$, replace $G$ with $-G$. Equivalently, subtract the gradient sum from the level-payment component. The formulas are finite sums, so they remain useful when a memorized gradient factor is unavailable.

Why the Shift Formula Holds

Let $C_j=P+(j-1)G$ be the payment at time $j$ in the immediate stream. Its present value is $\sum_{j=1}^{n}C_jv^j$. The due stream pays those same amounts at times $j-1$, so its present value is $\sum_{j=1}^{n}C_jv^{j-1}$. Since $v^{j-1}=(1+i)v^j$, factoring out $1+i$ proves the present-value shift. At time $n$, the immediate stream's accumulated value is $\sum_{j=1}^{n}C_j(1+i)^{n-j}$. The due stream accumulates each payment one additional period, giving $\sum_{j=1}^{n}C_j(1+i)^{n-j+1}$, again $1+i$ times as much. This argument relies on a constant effective rate per payment period; with rates that vary by period, a single factor $1+i$ need not apply.

Write the due payments and dates.
For an increasing stream, list $P,P+G,\ldots,P+(n-1)G$ at times $0,1,\ldots,n-1$. For a decreasing stream, use $P,P-G,\ldots,P-(n-1)G$.

Choose a value date.
For present value, discount each payment to time 0. For accumulated value, accumulate each payment to time $n$.

Use the timing shift or direct sums.
Value the same amounts as an immediate stream, then multiply by $1+i$. A direct calculation at the due dates provides an independent check.

Check the result.
Confirm the payment count, gradient direction, and value date. For a decreasing stream, also check that the last payment is nonnegative if that is required by the problem.

Present Value of an Increasing Annuity Due

The first payment of a due annuity is not discounted: it is already made at time 0. Later payments are discounted for one, two, and more periods. This makes it easy to detect a common indexing error: treating the first due payment as though it occurred at time 1.

Worked Example: Present Value with Payments at the Beginning

A fund receives four annual deposits at the beginnings of the years. The first deposit is $100, and each later deposit is $20 greater than the preceding one. If the effective annual interest rate is 25%, find the present value. Use the one-period shift, then check by valuing the due payments directly.

Method and setup. The due payments are $100, $120, $140, and $160 at times 0, 1, 2, and 3. The matching immediate stream has those amounts at times 1, 2, 3, and 4. At 25%, $v=1/1.25=0.8$.

Immediate-stream present value. Discount each payment to time 0:

$$ 100(0.8)+120(0.8)^2+140(0.8)^3+160(0.8)^4 =80+76.8+71.68+65.536 =294.016 $$

Shift to the due dates. Multiply by $1+i=1.25$:

\operatorname{PV}_{\text{due}}=294.016(1.25)=367.52

Direct check. Discount the actual due payments:

$$ 100+120(0.8)+140(0.8)^2+160(0.8)^3 =100+96+89.6+81.92 =367.52 $$

The present value is $367.52. The check confirms that the time-0 payment was left undiscounted.

Accumulated Value of an Increasing Annuity Due

For accumulated value at time $n$, the time-0 payment earns interest for $n$ periods, and the final due payment at time $n-1$ earns interest for one period. Alternatively, first accumulate the matching immediate stream to time $n$, then multiply by $1+i$.

Worked Example: Accumulating an Increasing Due Stream

A savings plan makes three deposits at the beginnings of three consecutive years. The first deposit is $200, and each subsequent deposit is $50 greater. At an effective annual rate of 10%, find the accumulated value at time 3.

Method and setup. The due payments are $200, $250, and $300 at times 0, 1, and 2. The matching immediate stream pays the same amounts at times 1, 2, and 3.

Accumulate the immediate stream to time 3. The three payments earn interest for three, two, and zero years, respectively:

$$ 200(1.1)^2+250(1.1)+300 =242+275+300 =817 $$

Apply the shift. The due stream earns one additional year of interest in total:

\operatorname{AV}_{3,\text{due}}=817(1.1)=898.70

Direct check. At time 3, the actual due payments have accumulated for three, two, and one years:

$$ 200(1.1)^3+250(1.1)^2+300(1.1) =266.20+302.50+330.00 =898.70 $$

The accumulated value is $898.70. The most recent due payment earns one year of interest; it does not earn zero years merely because it is the last payment.

Decreasing Payments and an Unknown Gradient

A decreasing annuity due uses the same timing formulas, with a negative gradient. For accumulated value, the level-payment component can be calculated first, then the accumulated effect of the decreases is subtracted. This decomposition is especially useful when the first payment and target value are known but the decrease $G$ is unknown.

Worked Example: Solving for the Decrease

A four-payment annuity due pays $900 at time 0 and decreases by a fixed amount $G$ at each subsequent payment. At an effective annual rate of 10%, its accumulated value at time 4 is $3,536.94. Find $G$, and verify the resulting payments and accumulated value.

Method and setup. The payments are $900,\ 900-G,\ 900-2G,\ 900-3G$ at times 0, 1, 2, and 3. Accumulate each payment to time 4. The interest accumulation factors are $1.1^4=1.4641$, $1.1^3=1.331$, $1.1^2=1.21$, and $1.1$.

Separate the level amount and the decreases. If every payment were $900, the accumulated value would be:

$$ 900(1.4641+1.331+1.21+1.1)=900(5.1051)=4594.59 $$

The decreases have sizes $0,G,2G,3G$. Their accumulated coefficient is:

$$ 0(1.4641)+1(1.331)+2(1.21)+3(1.1) =1.331+2.42+3.3 =7.051 $$

Thus the accumulated value is $4594.59-7.051G$. Set this equal to the stated target and solve:

$$ 4594.59-7.051G=3536.94 $$

7.051G=1057.65, \qquad G=\frac{1057.65}{7.051}=150

Verify. The payments are $900, $750, $600, and $450, all nonnegative. Their accumulated value is:

$$ 900(1.4641)+750(1.331)+600(1.21)+450(1.1) =1317.69+998.25+726+495 =3536.94 $$

Therefore the decrease is $150 per payment. The check confirms both the gradient and the stated target.

Common Timing Traps

A frequent mistake is to recognize the payment amounts correctly but value them at the wrong dates. For example, discounting the first payment of an annuity due by one period changes the cash-flow schedule. In the first worked example, discounting all four payments as though they occurred at times 1 through 4 gives $294.016; that is the immediate-stream value, not the due value of $367.52. The correction is to multiply the immediate value by $1+i=1.25$, or to discount the actual due payments at times 0 through 3.

For accumulated values, the analogous error is to omit the extra period of interest on every payment after shifting from immediate to due dates. In the second worked example, $817 is the immediate stream's value at time 3. It is not the due stream's value: the due value is $817(1.1)=898.70$. Directly checking how long each actual payment earns interest is an effective way to catch the error.

Count the payments, not the intervals. There are $n$ payments from time 0 through time $n-1$, and the matching immediate stream runs from time 1 through time $n$.
Keep the gradient index aligned with the first payment. The first payment is $P$, so its gradient component is zero; the next is $G$, not $2G$.
Preserve the sign. In a decreasing stream, the gradient contribution is subtracted. Check the final payment against the stated assumptions.
Use the appropriate period rate. The one-period factor is $1+i$ only when $i$ is effective for the payment interval.

Key takeaway: An arithmetic annuity due can be valued by taking the same arithmetic payment sequence as an immediate annuity and shifting every payment one period earlier. Under a constant effective rate per period, multiply either its present value or its accumulated value by $1+i$. For a decreasing stream, use a negative gradient and verify the payment amounts.

Check Your Understanding

For each problem, identify the payment dates before calculating. Answers are rounded to the nearest cent where needed.

An annuity due pays $400, $450, and $500 at an effective annual rate of 5%. Find its present value. (Answer: $1,282.09)
A four-payment annuity due pays $120, $150, $180, and $210. At an effective annual rate of 10%, find its accumulated value at time 4. (Answer: $824.14)
A three-payment annuity due pays $500, $450, and $400 at an effective annual rate of 10%. Find its present value. (Answer: $1,239.67)
A three-payment annuity due pays $700 at time 0 and decreases by $G$ at each later payment. At 10% effective annually, its accumulated value at time 3 is $2,207.70. Find $G$, and state the three payments. (Answer: $G=$$100; payments are $700, $600, and $500)
An immediate annuity pays $100, $125, and $150 at times 1, 2, and 3. A due annuity pays the same amounts at times 0, 1, and 2. At 25% effective annually, find each present value and their difference. (Answer: immediate $236.80; due $296.00; difference $59.20)

What You'll Learn