Checking An Equation At A Second Focal Date

Why Check an Equation at a Second Focal Date?

A value equation is built by moving cash flows to a common focal date using a stated interest model. Once an unknown amount has been solved, valuing both sides at a different date is a useful independent check. It can confirm the arithmetic and reveal whether a rate, time interval, or accumulation factor was used inconsistently.

The second-date check is not a new assumption about the transaction. It uses the same cash flows, rate, timing, and interest model as the original equation. Changing any one of these between the two calculations can create a discrepancy even when the original equation was written correctly.

Definition: A focal date is the date to which the cash flows in a value equation are moved for comparison. To check an equation at a second focal date, revalue every cash flow on both sides at another date using the same interest assumptions, then compare the resulting values.

The examples below use deterministic cash flows and an effective annual interest rate $i$, compounded annually. There is no mortality assumption or probability distribution in these examples. A cash flow at time $t$, expressed in years, is accumulated or discounted using the appropriate power of $1+i$. This tutorial focuses on checking an equation that has already been solved, rather than on methods for finding unknown rates or cash-flow dates.

Moving an Equation to a Second Date

Under compound interest at effective annual rate $i$, a dollar at time $t$ is worth $(1+i)^{s-t}$ at time $s$. If $s>t$, this factor accumulates the dollar forward; if $s<t$, the negative exponent discounts it. Every cash flow must use an exponent equal to the difference between its date and the focal date.

For example, suppose an equation balances at time $t$. Represent each cash flow on the left side as positive and each cash flow on the right side as negative. If the signed value at $t$ is zero, then moving every cash flow to $s$ multiplies that signed value by the same factor $(1+i)^{s-t}$. The result is still zero. This gives a useful justification for checking the transaction at a second date.

\sum_j C_j(1+i)^{t-t_j}=0 \quad\Longrightarrow\quad \sum_j C_j(1+i)^{s-t_j} =(1+i)^{s-t}\sum_j C_j(1+i)^{t-t_j}=0

Here $C_j$ is a signed cash flow occurring at time $t_j$: positive for one side of the transaction and negative for the other. The implication follows because $(1+i)^{s-t_j}=(1+i)^{s-t}(1+i)^{t-t_j}$. Thus, under the same compound-interest rate, equality at one focal date implies equality at every other focal date.

This result depends on using the same compound-interest model and rate for all the intervals involved. It does not authorize changing a time interval, using a different rate for one payment, or substituting a simple-interest factor for a compound-interest factor. A check that changes the assumptions is not checking the original equation.

Record the original assumptions.
Write down the effective annual rate, each payment date, and the focal date used to solve for the unknown.

Choose a second focal date.
A date at which several payments occur can reduce the number of calculations, but any date can be used if all factors are applied consistently.

Move each cash flow to that date.
Accumulate payments before the new date and discount payments after it, using the time difference in years.

Compare the two totals.
If they differ, check the payment dates, exponents, rate, and arithmetic before concluding that the transaction is not equivalent.

Worked Checks at a Second Focal Date

Worked Example: Checking a Solved Payment at the Latest Date

At an effective annual rate of $10\%$, a transaction is stated as equivalent at time 0: one schedule pays $300 at time 0 and $X$ at time 2; the other pays $400 at time 1 and $500 at time 2. Find $X$, then verify the equation at time 2. The cash flows are deterministic; no mortality assumption or probability distribution applies.

Method. First write the value equation at the original focal date, time 0. Then move every payment to time 2 using the same $10\%$ effective annual rate.

At time 0, the left schedule is worth $300+X/(1.10)^2$. The right schedule is worth $400/1.10+500/(1.10)^2$. Therefore,

300+\frac{X}{1.21} =\frac{400}{1.10}+\frac{500}{1.21}

Multiply every term by $1.21$:

363+X=440+500=940 \qquad\Longrightarrow\qquad X=577

Second-date check. At time 2, the $300 payment grows for two years, while the $400 payment grows for one year. The payments already at time 2 do not move.

\text{Left}=300(1.21)+577=363+577=940

\text{Right}=400(1.10)+500=440+500=940

Both schedules have a value of $940 at time 2, so the solved equation checks. No rounding is needed. The common mistake avoided is carrying a time-2 payment for one year, rather than two years, when valuing it at time 0.

Worked Example: Checking an Equation Whose Original Date Is Between Payments

At an effective annual rate of $20\%$, schedule A pays $300 at time 0 and $720 at time 2. Schedule B pays $600 at time 1 and $X$ at time 3. The schedules are equivalent at time 1. Find $X$ and check the equivalence at time 3. These are deterministic cash flows, with no mortality or probability model.

Method. At time 1, accumulate the time-0 payment for one year and discount the time-2 payment for one year. The payment at time 3 is discounted for two years to time 1.

300(1.20)+\frac{720}{1.20} =600+\frac{X}{(1.20)^2}

The value of schedule A at time 1 is $360+600=960$. Therefore,

960=600+\frac{X}{1.44} \qquad\Longrightarrow\qquad \frac{X}{1.44}=360 \qquad\Longrightarrow\qquad X=518.40

Second-date check. Move all payments to time 3. Schedule A’s time-0 payment grows for three years, and its time-2 payment grows for one year. Schedule B’s time-1 payment grows for two years, and its time-3 payment remains unchanged.

\text{A}=300(1.20)^3+720(1.20) =518.40+864=1{,}382.40

\text{B}=600(1.20)^2+518.40 =864+518.40=1{,}382.40

The two values agree at $1,382.40. The amount $X$ was already stated in cents, and no further rounding is needed. The common mistake avoided is discounting the time-2 payment for two years to time 1; it is only one year away.

What a Mismatch Can Tell You

If the original equation balances but the second-date values do not, the two calculations cannot both represent the same transaction under the stated assumptions. Do not immediately alter the solved amount to force the second check to balance. Instead, identify which factor or payment date was applied differently. A useful diagnostic is to compare each individual cash flow’s time interval with its exponent.

The following example deliberately shows an incorrect factor. It illustrates how a second-date check exposes the error and how to correct it.

Worked Example: Finding an Inconsistent Discount Factor

At an effective annual rate of $10\%$, schedule A pays $1,000 at time 0 and $500 at time 2. Schedule B pays $400 at time 1 and $X$ at time 2. A calculation reports $X=$1,200$, after discounting the time-2 payments by $1.10$ to time 0. Check the reported amount at time 2 and find the correct $X$. These are deterministic cash flows; there is no mortality assumption or probability distribution.

Method. Revalue the reported schedules at time 2 using the stated effective annual rate. Then write the original time-0 equation with the correct two-year discount factor for each time-2 payment.

At time 2, schedule A is worth $1{,}000(1.10)^2+500=1{,}210+500=1{,}710$. The reported schedule B is worth $400(1.10)+1{,}200=440+1{,}200=1{,}640$. The difference is $1{,}710-1{,}640=$70$, so the reported amount does not check.

The time-2 payments are two years from time 0, so their correct discount factor is $(1.10)^2=1.21$, not $1.10$. The correct time-0 equation is

1{,}000+\frac{500}{1.21} =\frac{400}{1.10}+\frac{X}{1.21}

Multiplying by $1.21$ gives

1{,}210+500=440+X \qquad\Longrightarrow\qquad X=1{,}270

Check at time 2: schedule A is still worth $1,710, and schedule B is worth $400(1.10)+1{,}270=440+1{,}270=$1{,}710$. The corrected amount is $1,270.00, with no rounding beyond cents. The common mistake avoided is treating a two-year interval as one year. The $70 discrepancy in the incorrect check is a warning that the reported value equation used inconsistent factors.

Using the Check Carefully

A second-date check is most informative when the calculations are organized so that each factor is visible. For every payment, note its date, the new focal date, and the number of years between them. Then apply $(1+i)^n$ to accumulate forward or $(1+i)^{-n}$ to discount backward, where $n$ is the number of years separating the payment and focal date.

Keep the same rate. If the original equation uses a $6\%$ effective annual rate, the check must use that same rate.
Keep the same timing. A payment at time 2 is two years from time 0, even if another payment occurs at time 1.
Keep the focal date consistent within each comparison. Both sides must be valued at time 2, for example, rather than one side at time 2 and the other at time 1.
Retain precision until the comparison is complete. Rounding intermediate discounted values can make equal totals appear slightly different.

A successful check supports the original calculation, but it does not repair a misstated cash-flow schedule. If the same incorrect payment dates or rate assumptions are used at both focal dates, the calculation can remain internally consistent while describing the wrong transaction. Confirm that the data in the value equation match the stated arrangement.

Key takeaway: Under compound interest at a fixed effective annual rate, equivalent schedules remain equivalent when valued at a second focal date. Recompute both sides using the same rate and the correct time interval for every cash flow; a discrepancy points to inconsistent assumptions, factors, timing, or arithmetic.

Check Your Understanding

In each problem, use the stated effective annual rate and verify the transaction at the requested second focal date. Cash flows are deterministic; no mortality or probability distribution is involved.

At $5\%$, schedule A pays $200 at time 0 and $315 at time 2. Schedule B pays $250 at time 1 and $X$ at time 2. The schedules are equivalent at time 0. Find $X$ and check at time 2. (Answer: $X=$273.00$; both values at time 2 are $535.50.)
At $8\%$, schedule A pays $500 at time 0 and $648 at time 2. Schedule B pays $400 at time 1 and $X$ at time 3. They are equivalent at time 1. Find $X$ and check at time 3. (Answer: $X=$863.14$; both values at time 3 are $1,329.70.)
At $10\%$, schedule A pays $500 at time 0 and $600 at time 2. Schedule B pays $200 at time 1 and $X$ at time 2. They are equivalent at time 0. Find $X$, then verify at time 2. (Answer: $X=$985.00$; both values at time 2 are $1,205.00.)
At $7\%$, schedule A pays $1,000 at time 0 and $300 at time 2. Schedule B pays $400 at time 1 and $X$ at time 2. A report gives $X=$900$. Check the report at time 2 and calculate the corrected amount that makes the schedules equivalent. (Answer: the reported values differ by $116.90; the corrected amount is $X=$1,016.90$.)

What You'll Learn