Equivalent Values at Different Dates

Changing the Date of an Equivalent Value

A finite level annuity is a sequence of equal payments made at regular intervals for a fixed number of periods. Its present value and accumulated value express the same scheduled payments as single amounts at different dates. Once either value is known, compound interest lets us move that single equivalent amount to another date.

The key is to keep two ideas separate: when the payments occur and when the equivalent value is stated. Changing the date of an equivalent value does not move, add, or remove any payment. This distinction is useful whenever a problem gives an annuity value at one date but asks for an equivalent amount at another.

Definition: Two amounts at different dates are equivalent under a specified interest assumption if accumulating or discounting one amount to the date of the other gives the same value. With an effective rate $i$ per period, a value moved forward $k$ periods is multiplied by $(1+i)^k$; a value moved backward $k$ periods is divided by $(1+i)^k$.

Move a Value by Counting Periods Between Dates

Use the effective interest rate for the period that matches the timeline. For example, with an effective annual rate, count years; with an effective monthly rate, count months. If a value is stated at time $t$ and is to be stated at time $s$, the date-shift rule is:

V_s = V_t(1+i)^{s-t}

Here, $V_t$ and $V_s$ are single equivalent amounts at times $t$ and $s$, measured in the same period units. If $s>t$, the exponent is positive and the amount is accumulated. If $s<t$, the exponent is negative, which is equivalent to discounting. The rule assumes the same effective rate applies over every period between the two dates.

For an annuity-immediate of $n$ payments of $R$, the payments occur at times $1,2,\ldots,n$. Its present value at time 0 and accumulated value at time $n$ are:

PV_0 = R a_{\overline{n}|} \qquad\text{and}\qquad AV_n = R s_{\overline{n}|}

The present value is located before the first payment, at time 0; the accumulated value is located at the last payment date, at time $n$. Either amount can be shifted to another date by counting the number of periods between its stated date and the requested date. For an annuity-due, the payments instead occur at times $0,1,\ldots,n-1$, so its accumulated value is at time $n-1$. Always identify the actual date attached to the value before shifting it.

Worked Example: Move an Annuity-Immediate Value Past the Last Payment

A fund receives three annual payments of $200 at the ends of years 1, 2, and 3. The effective annual interest rate is 10%. Find the equivalent single amount at the end of year 5. Retain full precision in intermediate calculations and round the final answer to cents.

Method. Find the present value at time 0, then accumulate that single equivalent amount five years. As a check, accumulate the scheduled payments to time 3 and move that accumulated value forward two years.

The discount factor is $v=1/1.10=0.909090909\ldots$. The present value is:

$$ PV_0 =200(v+v^2+v^3) =200(0.909090909+0.826446281+0.751314801) =497.370398 $$

Time 5 is five periods after time 0. Accumulate the present value by $1.10^5=1.61051$:

V_5 = 497.370398\ldots(1.61051)=801.02

Check. At time 3, the first payment has earned interest for two years, the second for one year, and the third for none:

$$ AV_3=200(1.10^2+1.10+1) =200(1.21+1.10+1) =662 $$

Moving this amount from time 3 to time 5 requires two periods: $662(1.10)^2=662(1.21)=801.02$, which agrees. The common mistake is to multiply $PV_0$ by $(1.10)^3$, which only moves the value to time 3, not time 5.

Why the Date-Shift Rule Works

Compound interest gives a single amount $V_t$ the value $V_t(1+i)$ one period later. Applying the same rule again gives $V_t(1+i)^2$ two periods later. Repeating the process $k$ times gives the forward rule. Reversing those steps divides by $1+i$ each time, giving the backward rule.

The rule also explains why shifting an annuity’s equivalent value preserves the original payment schedule. At time 0, the present value of an annuity-immediate equals the sum of its payments discounted to that date. Accumulating that sum to time $n$ gives:

R a_{\overline{n}|}(1+i)^n =R\left(\sum_{j=1}^{n}v^j\right)(1+i)^n =R\sum_{j=1}^{n}(1+i)^{n-j} =R s_{\overline{n}|}

Each payment has simply earned interest from its scheduled date to time $n$. No payment has been rescheduled. Moving either equivalent amount farther forward or backward applies the same compound-interest rule again.

Moving an Annuity-Due Value

The date shift is not specific to annuities-immediate. For an annuity-due, first locate the value on the timeline. With $n$ payments of $R$ at times $0$ through $n-1$, the accumulated value at time $n-1$ is found by accumulating each payment to that date. From there, move the single accumulated value to the requested date. The number of periods in the shift depends on the dates, not on the number of payments.

Worked Example: Move an Annuity-Due Accumulated Value Forward

A hypothetical savings plan receives four annual deposits of $500 at times 0, 1, 2, and 3. The effective annual interest rate is 5%. Find the equivalent amount at time 6. Retain full precision in intermediate calculations and round the final answer to cents.

Method. First accumulate the deposits to the last deposit date, time 3. Then accumulate that value three periods to time 6.

At time 3, the deposits made at times 0, 1, 2, and 3 have earned interest for 3, 2, 1, and 0 years, respectively:

AV_3 =500(1.05^3+1.05^2+1.05+1) =500(1.157625+1.1025+1.05+1) =2{,}155.0625

There are three periods from time 3 to time 6, so:

V_6=2{,}155.0625(1.05)^3 =2{,}155.0625(1.157625) =2{,}494.7542265625

The equivalent amount is $2,494.75. As a direct check, accumulate each deposit to time 6: the required exponents are 6, 5, 4, and 3. Thus $500(1.05^6+1.05^5+1.05^4+1.05^3)=2{,}494.7542265625$. The check confirms that the shift preserves all four original deposit dates.

Moving a Value Backward

If the requested date is earlier than the date of a known equivalent amount, discount rather than accumulate. Count the entire gap between dates, even if the requested date is before the annuity’s first payment. The result is a single equivalent amount at that earlier date; it does not imply that the actual annuity payments have moved.

Worked Example: Discount an Accumulated Value to an Earlier Date

An annuity-immediate consists of two annual payments of $100 at times 1 and 2. The effective annual interest rate is 10%. Find the equivalent single amount at time $-1$, one year before time 0. Retain full precision in intermediate calculations and round the final answer to cents.

Method. Accumulate the payments to time 2, then discount that amount three periods, from time 2 back to time $-1$.

At time 2, the payment at time 1 has earned one year of interest; the payment at time 2 has earned none:

$$ AV_2=100(1.10)+100=210 $$

The date gap from time $-1$ to time 2 is three years. Therefore:

V_{-1}=\frac{210}{(1.10)^3} =\frac{210}{1.331} =157.776108\ldots

The equivalent amount at time $-1$ is $157.78. To check, discount the payments individually from times 1 and 2 to time $-1$: the gaps are two and three years. Then $100/(1.10)^2+100/(1.10)^3=82.644628\ldots+75.131480\ldots=157.776108\ldots$. The common mistake is to discount only two years because time $-1$ is one year before time 0; the relevant gap starts at the accumulated value’s date, time 2.

Common Timing Traps

A date-shift calculation is usually short, but a misidentified starting date or period count can change the answer substantially. Before calculating, write down the date of the known value and the date requested. Then count the periods between those two dates.

Using the payment count as the shift. The exponent is the gap between valuation dates, not automatically the number of payments. A value at time 3 moved to time 6 shifts three periods, whether the annuity has four payments or forty.
Confusing the annuity endpoint dates. An annuity-immediate with $n$ payments ends at time $n$; an annuity-due with $n$ payments ends at time $n-1$. Identify the payment timing before locating its accumulated value.
Changing the payment schedule. Moving an equivalent single amount does not move the actual payments. Keep the original payment dates when checking by individual cash flows.
Using the wrong rate period. The rate and the counted periods must match. If the timeline is monthly, use the effective rate per month rather than applying an annual rate once per month.
Rounding too early. Keep full precision through the shift and round only the final money amount to cents unless the problem specifies another convention.

Key takeaway: To state an annuity’s equivalent value at a new date, start from a known value date, count the periods to the requested date, and apply compound accumulation or discounting. The payment schedule remains unchanged.

Check Your Understanding

For each problem, identify the date of the known value before applying the date shift. Round final dollar amounts to cents.

An annuity-immediate pays $250 at the end of each year for three years. The effective annual rate is 4%. Find its equivalent value at time 5 by accumulating the value at time 3. (Answer: $844.08)
An annuity-due pays $120 at times 0, 1, and 2. The effective annual rate is 5%. Find its equivalent value at time 5 by first accumulating to the last payment date. (Answer: $437.93)
An annuity-immediate pays $400 at times 1 and 2. The effective annual rate is 6%. Find its equivalent value at time $-1$ by discounting from time 2. (Answer: $691.85)
An annuity-immediate pays $90 at times 1, 2, 3, and 4. The effective annual rate is 10%. Find its equivalent value at time 6 by accumulating the value at time 4. (Answer: $505.40)
A payment stream has an accumulated value of $1,000 at time 4. The effective annual rate is 8%. Find the equivalent amount at time 1. (Answer: $793.83)

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