Moving One Payment to a New Valuation Date

Why the Valuation Date Matters

A payment has no single present value independent of context. Its value depends on the date at which it is being valued. If a payment is due at year 5, its value at year 2 is found by discounting for three years, while its value at year 8 is found by accumulating for three years. The payment date and valuation date—not simply the number of years from time 0—determine the calculation.

This tutorial considers one payment under a constant effective annual interest rate. Time is measured in whole years, and the same rate applies over every period being crossed. The earlier material on present and future value establishes that an amount grows by the accumulation factor over time and is discounted by its reciprocal when moved backward.

Definition: The payment date is the date at which the single cash flow occurs. The valuation date is the date at which its equivalent value is required. Under a constant effective annual rate $i$, moving a payment forward means accumulating it; moving it backward means discounting it.

The Single-Payment Rule Between Two Dates

Let a payment of $C$ dollars occur at time $k$, and let its value be required at time $t$. Both times are measured in years from the same origin. Assume a constant effective annual rate $i>-1$, annual compounding, and a whole-number difference between the dates. The number of periods is the distance between the dates, $|t-k|$.

V_t=C(1+i)^{t-k}

Here $V_t$ is the equivalent value at time $t$. If $t>k$, the exponent is positive and the payment is accumulated forward. If $t<k$, the exponent is negative and the payment is discounted backward. If $t=k$, the exponent is zero and $V_t=C$. This signed-exponent form handles all three cases without changing the underlying rule.

The formula assumes the rate remains constant throughout the interval and that the periods correspond to annual compounding. If the rate changes, a single factor $(1+i)^{t-k}$ is not generally appropriate. If the dates are not separated by a whole number of years, the treatment of fractional periods must be specified separately.

Worked Example: Valuing a Payment at Two Dates

A hypothetical contract pays $2,400 at the end of year 2. Assume a constant effective annual rate of 4%, compounded annually. Find the payment’s value at year 7 and at year 0. Give each answer to the nearest cent.

Method. For each requested value, compare that valuation date directly with the payment date. The payment is at year 2: year 7 is five periods later, and year 0 is two periods earlier.

Value at year 7. Since the valuation date is later, accumulate for $7-2=5$ years:

V_7=2{,}400(1.04)^5 =2{,}400(1.2166529024) =2{,}919.96696576 \approx \$2{,}919.97.

Value at year 0. Since the valuation date is earlier, discount for $2-0=2$ years:

V_0=\frac{2{,}400}{(1.04)^2} =\frac{2{,}400}{1.0816} =2{,}218.934911\ldots \approx \$2{,}218.93.

Check. Accumulating the unrounded year-0 value for two years returns the payment: $2{,}218.934911\ldots(1.0816)=2{,}400$. The year-7 value is greater than the payment, as expected when a positive rate is applied forward.

Common mistake avoided. Using seven years to calculate the value at year 7 would count from year 0, not from the payment at year 2. The relevant forward interval is $7-2=5$ years.

Why the Rule Works

The formula follows from applying the annual accumulation rule once for each period. If an amount at one year-end is $B_j$, then, at effective annual rate $i$, the amount one year later is $B_{j+1}=B_j(1+i)$. Repeated application gives the value after $n$ years:

B_{j+n} =B_j(1+i)(1+i)\cdots(1+i) =B_j(1+i)^n.

For a payment $C$ at time $k$, this proves that its value $n=t-k$ years later is $C(1+i)^{t-k}$ when $t>k$. For a date earlier than the payment, reverse the same relationship. If the payment at time $k$ is the result of accumulating a value $V_t$ forward, then $C=V_t(1+i)^{k-t}$. Solving for $V_t$ gives:

V_t=\frac{C}{(1+i)^{k-t}} =C(1+i)^{t-k} \qquad (t<k).

Thus, discounting is not a separate rule chosen by intuition: it is the amount that, when accumulated to the payment date, produces the stated payment.

Worked Example: One Payment, Dates on Both Sides

A hypothetical benefit of $5,000 is payable at the end of year 4. Use a constant effective annual rate of 6%, compounded annually, to find its equivalent value at year 1 and at year 6. Round to the nearest cent.

Method. Use the payment date, year 4, as the reference point. The value at year 1 is three years before the payment; the value at year 6 is two years after it.

At year 1, discount three periods:

V_1=\frac{5{,}000}{(1.06)^3} =\frac{5{,}000}{1.191016} =4{,}198.0964\ldots \approx \$4{,}198.10.

At year 6, accumulate two periods:

V_6=5{,}000(1.06)^2 =5{,}000(1.1236) =\$5{,}618.00.

Check. The year-1 amount is below $5,000 because it is valued before the payment date. Accumulating its unrounded value for three years gives $4{,}198.0964\ldots(1.191016)=5{,}000$. The year-6 amount is above $5,000 because it is valued two years later.

Common mistake avoided. The distance from year 0 to year 6 is not the period count for the year-6 value. The payment occurs at year 4, so only the two periods from year 4 to year 6 are relevant.

Read the Timeline Before Choosing the Exponent

A useful way to prevent a wrong exponent is to write the two dates explicitly before calculating. The exponent is the valuation date minus the payment date. Its sign then records the direction of movement: positive for a later valuation date and negative for an earlier one.

Mark the payment date.
Record the time $k$ at which the stated amount $C$ occurs.

Mark the valuation date.
Record the requested time $t$; do not assume it is time 0.

Subtract the dates in the stated order.
Calculate $t-k$. The absolute value is the number of periods, and the sign indicates whether the value is moved forward or backward.

Apply the rate factor and check direction.
Use $C(1+i)^{t-k}$. At a positive rate, a forward value should exceed $C$, and a backward value should be less than $C$.

The direction check is useful but not sufficient: an incorrect period count can still produce a value on the expected side of the payment. Always verify the dates as well.

Worked Example: Correcting the Period Count

A hypothetical payment of $3,000 is due at the end of year 9. Find its value at year 6 using a constant effective annual rate of 5%, compounded annually. Round to the nearest cent.

Method. The requested date is three years before the payment, since $9-6=3$. Discount the payment for those three periods:

V_6=\frac{3{,}000}{(1.05)^{9-6}} =\frac{3{,}000}{1.157625} =2{,}591.512795\ldots \approx \$2{,}591.51.

Check. Accumulating the unrounded value for three years returns the payment: $2{,}591.512795\ldots(1.157625)=3{,}000$, subject only to displayed precision.

Common mistake avoided. Discounting for nine years would value the payment at year 0, not year 6. The incorrect setup is $3{,}000/(1.05)^9$; the correct number of periods is the distance from year 6 to year 9, which is three. The valuation date determines the calculation.

Common Exam Traps

Counting from the origin instead of the payment date. If a payment occurs at year 4 and is valued at year 7, the period count is $7-4=3$, not 7.
Discounting when the requested date is later. A later valuation date requires accumulation. At a positive rate, that value should be larger than the payment.
Accumulating when the requested date is earlier. An earlier valuation date requires discounting. At a positive rate, that value should be smaller than the payment.
Using the right number of periods in the wrong direction. For dates 6 and 9, three periods separate them, but valuing at year 6 from a payment at year 9 requires discounting.
Rounding too early. Keep the unrounded factor and value through the calculation, then round the final dollar amount to cents.

Key takeaway: For one payment $C$ at time $k$, its value at time $t$ under a constant effective annual rate is $C(1+i)^{t-k}$. The difference $t-k$ identifies both the number of periods and the direction of movement.

Check Your Understanding

For each problem, assume annual compounding at the stated constant effective annual rate. Round monetary answers to the nearest cent.

A hypothetical $1,200 payment occurs at year 3. Find its value at year 8 using a 3% effective annual rate. (Answer: $1,391.13)
A hypothetical $8,400 payment is due at year 7. Find its value at year 4 using a 4% effective annual rate. (Answer: $7,467.57)
A hypothetical $2,500 payment occurs at year 2. Find its value at year 5 using a 2.5% effective annual rate. (Answer: $2,692.23)
A hypothetical $10,000 payment is due at year 8. Find its value at year 3 using a 5% effective annual rate. (Answer: $7,835.26)
A hypothetical $6,000 payment occurs at year 4. Find its values at year 0 and year 7 using a 3% effective annual rate. State the period count and direction for each value. (Answer: At year 0, discount 4 years to $5,330.92; at year 7, accumulate 3 years to $6,556.36.)

What You'll Learn