The Equation of Value

Money at different times can't be compared directly — you have to move every cash flow to one common point first. Pick a comparison date, accumulate or discount every inflow and outflow to that date, and set the two sides equal — that balance is the equation of value, the master tool behind almost every time-value problem.

By the end you'll be able to set up an equation of value at any comparison date, and use it to solve for an unknown accumulated value, present value, rate \(i\), or time.

Predict: if you slide the comparison date later, does the balance point (the solution rate) change? Move it and check.

You lend 1000 at time 0 (green, an inflow to whoever is owed) and are repaid 600 at year 2 and 554.02 at year 4 (both in warm orange, outflows from the lender's side of the equation). Drag the comparison-date marker along the timeline and the rate i slider below it — every cash flow is valued at the marker's date T using one formula, \((1+i)^{(T-t)}\), which accumulates forward when T is after t and discounts back when T is before it. Watch the readouts below balance to $0 exactly at \(i = 5\%\) — and stay balanced there no matter where you drag the marker.

Drag the diamond marker on the timeline (or focus it and use the arrow keys) to move the comparison date T.

Cash-flow timeline — drag the marker to change the comparison date T
inflow (loan) outflow (repayment) comparison date T

Inflow value at T: $1045.31    Outflow value at T: $1105.73

Net (in − out): −$60.42

Every time-value problem is really the same problem: move all the money to one moment and check that what came in equals what went out. Once you fix that moment, the rest is just accumulation and discounting.

Formal

At comparison date T, a cash flow C at time t is worth \(C\cdot(1+i)^{T-t}\) — accumulated forward if \(T > t\), discounted back if \(T < t\); the exponent's sign flips automatically. The equation of value sets (value of inflows at T) equal to (value of outflows at T) and solves for whichever quantity is unknown: an accumulated value, a present value, the rate \(i\), or the time \(t\). Under compound interest, moving the comparison date from \(T_1\) to \(T_2\) multiplies every term — both sides — by the same \((1+i)^{T_2-T_1}\), so that factor cancels and the solution never depends on which date you chose.

Applied

This is why loan payments, bond cash flows, and pension contributions can all be compared on one timeline even though they land on different dates: value every flow at a single comparison date and add. Since the date is free to choose, pick whichever one makes the algebra simplest — often time 0, or the date of the payment you don't know yet.

Worked example

You borrow 1000 today and repay 600 at the end of year 2 and X at the end of year 4, at \(i = 5\%\). Valuing everything at time 0 (\(v = 1/1.05\)): \(1000 = 600v^2 + Xv^4 = 600(0.907029) + X(0.822702)\), so \(X = (1000 - 544.22)/0.822702 \approx\) 554.02. Check by valuing the same flows at time 4 instead: \(1000(1.05)^4 = 600(1.05)^2 + X \Rightarrow 1215.51 = 661.5 + X\), giving \(X \approx\) 554.01 — the same answer (rounding aside), exactly as the timeline demo above shows staying balanced wherever you drag the marker.

Your turn

Same idea, different numbers: you lend 2000 today and are repaid 1000 at the end of year 3 and Z at the end of year 6, at \(i = 4\%\). Value everything at time 0: \(2000 = 1000v^3 + Zv^6\). With \(v = 1/1.04 = 0.961538\), \(v^3 = 0.888996\) and \(v^6 = 0.790315\). \(1000v^3 = 888.996\), so \(Zv^6 = 2000 - 888.996 = 1111.004\). Now finish it: \(Z = 1111.004 / 0.790315 = \) ____

Reveal the answer

\(Z = 1111.004 / 0.790315 \approx\) 1405.77. Checking at time 6 instead: \(2000(1.04)^6 = 1000(1.04)^3 + Z \Rightarrow 2530.64 = 1124.86 + Z\), giving \(Z \approx 1405.77\) — matching (rounding aside), no matter which date you value it at.

More info — why the comparison date cancels out

Suppose an equation of value balances at date \(T_1\): \(\sum \text{inflows} \cdot (1+i)^{T_1-t} = \sum \text{outflows} \cdot (1+i)^{T_1-t}\). Move to a new date \(T_2\) and every single term on both sides gets multiplied by the same extra factor \((1+i)^{T_2-T_1}\) — inflows and outflows alike, since it doesn't depend on which cash flow it's attached to. Multiplying both sides of an equality by the same nonzero number preserves the equality, so the balance still holds at \(T_2\) with the exact same \(i\). That's the algebra behind the marker in the timeline above never un-balancing as you drag it. See the Marcel Finan guide in Dive deeper below for more worked equation-of-value problems.

Check your understanding

Question 1 of 3

You borrow 1500 today and agree to repay 800 at the end of year 2 and W at the end of year 5, at an effective annual rate \(i = 7\%\). Find W.

Question 2 of 3

Under compound interest, if you move the comparison date of an equation of value later, what happens to the interest rate i that balances it?

Question 3 of 3

In the equation of value \(1000 = 600v^2 + Xv^4\), what does the factor \(v^2\) represent?

Recap

  • An equation of value sets the value of all inflows at a comparison date T equal to the value of all outflows at T, moving each cash flow by \((1+i)^{T-t}\).
  • Under compound interest, the choice of T never changes the solution — shifting T multiplies every term by the same \((1+i)^{\Delta}\), which cancels.
  • The same equation solves for whichever piece is unknown: an accumulated value, a present value, the rate \(i\), or the time \(t\).
  • Pick T to make the algebra easiest — often time 0, or the date of the unknown payment.

Dive deeper

Sources

  • The Equation of Value