Building a Replacement Chain

Why Build a Replacement Chain?

When mutually exclusive projects have different useful lives, comparing their NPVs over only one project cycle can be misleading. One alternative may last two years and the other four, even though the decision concerns a service or operating need expected to continue beyond either project’s first cycle. A replacement chain puts the alternatives on a common horizon by repeating each project when its useful life ends.

The method requires more than copying the initial project’s benefits. At each replacement date, the new project requires an outlay. The completed project may also have a salvage value or another terminal cash flow. The chain must include both, at the date they occur. A final project cycle that ends exactly at the comparison horizon is not replaced beyond that horizon.

Definition: A replacement chain is the sequence of cash flows formed by undertaking a project, replacing it with an identical project at the end of its useful life, and repeating this process until a chosen common horizon is reached. The chain NPV is the present value of all those cash flows, discounted to the same initial date.

A standard way to choose the comparison horizon is the least common multiple of the projects’ useful lives, measured in the same time units. For example, projects lasting two and four years can both be repeated to cover four years; projects lasting two and three years can both be repeated to cover six years. A common horizon allows the cash flows to be compared over an equal length of time, but it does not by itself establish that the projects can realistically be repeated.

Constructing the Cash-Flow Chain

Let a project have initial cost $I$, useful life $n$, and cash flow $C_t$ at the end of year $t$, for $t=1,\ldots,n$. Its initial cash flow is $C_0=-I$. If the common horizon is $T=kn$, the project is undertaken $k$ times. The first cycle begins at time 0, the next at time $n$, and each later cycle begins $n$ years after the previous one.

Suppose each completed cycle provides salvage value $S$ at its end and the next cycle costs $I$ at that same date. At an intermediate replacement date, the boundary cash flow is therefore the sum of the prior cycle’s terminal cash flow and the next cycle’s initial outlay. If operating cash flow for the final year is $B$, the boundary amount is $B+S-I$. At the end of the common horizon, there is no further replacement outlay; the terminal amount is the final cycle’s operating cash flow plus its terminal value.

Choose the horizon.
Use a horizon that is a whole number of cycles for each project, often the least common multiple of the useful lives.

Mark each cycle boundary.
At each intermediate boundary, combine the completed project’s terminal cash flow with the next project’s replacement outlay.

List all dated cash flows.
Include operating cash flows, replacement costs, salvage values, and any final-horizon terminal cash flow. Do not place a replacement beyond the horizon.

Discount and compare.
Discount every chain cash flow to time 0 at the stated effective annual rate, then compare the resulting NPVs.

The NPV of a chain with cash flows $CF_t$ and effective annual discount rate $i$ is

\operatorname{NPV}_{\text{chain}}=\sum_{t=0}^{T}\frac{CF_t}{(1+i)^t}

Here $i$ is assumed to be the appropriate rate for the project cash flows and $T$ is stated in years. With $v=1/(1+i)$, the same expression is $\sum_{t=0}^{T}CF_t v^t$. Each cash flow is discounted according to its actual calendar date in the chain, not according to its position within an isolated cycle.

Why Repeating a Cycle Gives the Chain NPV

A short derivation clarifies the timing. Let one cycle’s cash flows, including its initial outlay, be $C_0,C_1,\ldots,C_n$. The NPV of that cycle at its own start date is $\sum_{t=0}^{n}C_t v^t$. If the cycle starts at time $jn$, its cash flows are all shifted $jn$ years later, so each is discounted by an additional factor $v^{jn}$. Its present value at time 0 is therefore $v^{jn}\sum_{t=0}^{n}C_t v^t$.

For $k$ repeated cycles, adding the shifted cycle values gives

\operatorname{NPV}_{\text{chain}} =\sum_{j=0}^{k-1}v^{jn}\left(\sum_{t=0}^{n}C_t v^t\right)

This equality follows by distributing $v^{jn}$ over the cash flows in each cycle: a cash flow $C_t$ in the cycle starting at $jn$ occurs at date $jn+t$, and its discounted value is $C_t v^{jn+t}$. At a cycle boundary, the terminal cash flow of one cycle and initial cost of the next are both included at that same date. They must be added, not treated as cash flows at different dates. At the last boundary, the chain ends, so only the final cycle’s terminal cash flow is included.

Worked Example: A Two-Year Project Repeated Against a Four-Year Project

Alternative G costs $100,000, produces net cash flow of $60,000 at the end of each year, lasts two years, and has no salvage value. Alternative H costs $170,000, produces $50,000 at the end of each year, lasts four years, and has no salvage value. Either project can be repeated identically. The effective annual discount rate is $10\%$. Compare the alternatives over the four-year common horizon. Assume all cash flows occur at year-end, and neither project has an additional terminal cash flow.

Method. The useful lives are two and four years, so the common horizon is four years. G is undertaken at time 0 and replaced at time 2. H is undertaken once and is not replaced at time 4. Let $v=1/1.10=0.909090909$.

G’s dated cash flows. At time 0 the outlay is $-$100,000$. At year 1 the cash flow is $+$60,000$. At year 2, the first project’s final operating cash flow of $60,000 is combined with the replacement cost of $100,000, giving $-$40,000$. Years 3 and 4 each produce $60,000. Thus the chain is $(-100{,}000,\ 60{,}000,\ -40{,}000,\ 60{,}000,\ 60{,}000)$.

Discounting each amount gives

\begin{aligned} \operatorname{NPV}_G &=-100{,}000+60{,}000v-40{,}000v^2+60{,}000v^3+60{,}000v^4\\ &=-100{,}000+54{,}545.45-33{,}057.85+45{,}078.89+40{,}980.81\\ &\approx \$7{,}547.30 \end{aligned}

H’s dated cash flows. H has one initial outlay and four annual cash flows of $50,000. Its chain NPV is

\begin{aligned} \operatorname{NPV}_H &=-170{,}000+50{,}000(v+v^2+v^3+v^4)\\ &=-170{,}000+50{,}000(3.169865446)\\ &\approx -\$11{,}506.73 \end{aligned}

The four-year chain has the higher NPV for G: about $7,547, compared with about $-$11,507$ for H. The calculation includes the replacement cost for G at year 2 but no replacement at the common horizon. A frequent error is to omit G’s year-2 replacement cost; that would overstate its NPV by $100{,}000v^2$, or about $82,645.

Worked Example: Salvage Value and Replacement Cost at a Boundary

A hypothetical maintenance system R costs $120 thousand and lasts three years. It produces $52 thousand at each year-end and can be sold for $15 thousand at the end of each three-year cycle. Replacing it costs $130 thousand. System S costs $90 thousand, lasts two years, produces $54 thousand at each year-end, and has a $8 thousand salvage value at each cycle end. A replacement for S costs $95 thousand. Both systems can be repeated identically. The effective annual discount rate is $8\%$, and the common horizon is six years. Compare chain NPVs in thousands of dollars.

Method and timing. The lives are three and two years, so both chains extend to year 6. For R, the year-3 cash flow is $52+15-130=-63$: the operating cash flow and salvage proceeds are offset by the replacement outlay. The same replacement occurs at year 3, leaving a final cycle to end at year 6. For S, the combined cash flow at each intermediate two-year boundary is $54+8-95=-33$. Those replacement boundaries are years 2 and 4. At year 6, S receives its final operating cash flow and salvage value, but no replacement outlay.

The discount factors for years 1 through 6 are $0.925925926,\ 0.857338820,\ 0.793832241,\ 0.735029853,\ 0.680583197,\ 0.630169627$. The resulting dated cash flows are:

Year	R cash flow ($000)	S cash flow ($000)
0	-120	-90
1	52	54
2	52	-33
3	-63	54
4	52	-33
5	52	54
6	67	62

For R, the present-value contributions in thousands are $-120$, $52(0.925925926)=48.148148$, $52(0.857338820)=44.581620$, $-63(0.793832241)=-50.011432$, $52(0.735029853)=38.221552$, $52(0.680583197)=35.390326$, and $67(0.630169627)=42.221367$. Their sum is approximately $38.552 thousand.

For S, the contributions in thousands are $-90$, $54(0.925925926)=50.000000$, $-33(0.857338820)=-28.292181$, $54(0.793832241)=42.866941$, $-33(0.735029853)=-24.255985$, $54(0.680583197)=36.751493$, and $62(0.630169627)=39.070517$. Their sum is approximately $26.141 thousand.

R’s chain NPV exceeds S’s by about $38.552-26.141=12.411$ thousand dollars. The comparison depends on combining salvage and replacement cash flows at the same date; recording the salvage but leaving out the replacement outlay would distort the result.

Worked Example: Do Not Replace Beyond the Common Horizon

A hypothetical two-year equipment project costs $80 thousand, produces $50 thousand at each year-end, and has a salvage value of $10 thousand at the end of each cycle. A replacement costs $85 thousand. A competing three-year project costs $130 thousand, produces $56 thousand at each year-end, and has a salvage value of $20 thousand at each cycle end. Its replacement cost is $140 thousand. Both are repeatable. The effective annual discount rate is $10\%$. Compare them over the six-year common horizon, assuming no other terminal cash flows.

Method. The common horizon is six years. For the two-year project, the intermediate boundary cash flow is $50+10-85=-25$ thousand at years 2 and 4. The year-6 terminal cash flow is $50+10=60$ thousand, because no replacement is made after the horizon. Its dated chain is $(-80,50,-25,50,-25,50,60)$.

For the three-year project, the intermediate boundary cash flow at year 3 is $56+20-140=-64$ thousand. At year 6 the final cash flow is $56+20=76$ thousand, with no further replacement. Its dated chain is $(-130,56,56,-64,56,56,76)$. The discount factors at years 1 through 6 are $0.909090909,\ 0.826446281,\ 0.751314801,\ 0.683013455,\ 0.620921323,\ 0.564473930$.

The two-year project’s present-value contributions, in thousands, are $-80$, $50v=45.454545$, $-25v^2=-20.661157$, $50v^3=37.565740$, $-25v^4=-17.075336$, $50v^5=31.046066$, and $60v^6=33.868436$. Therefore its chain NPV is about $30.198 thousand.

The three-year project’s contributions are $-130$, $56v=50.909091$, $56v^2=46.280992$, $-64v^3=-48.084147$, $56v^4=38.248753$, $56v^5=34.771594$, and $76v^6=42.900019$. Its chain NPV is about $35.026 thousand. The three-year project has the higher chain NPV by approximately $4.828 thousand. The final-year amounts include salvage but exclude a replacement cost after year 6.

Assumptions and Common Traps

A replacement-chain comparison is informative only if its repeatability assumptions are defensible. It assumes that a replacement project can be obtained when needed and that later versions have the same relevant cash-flow pattern as the first. In practice, costs, operating benefits, technology, capacity, and availability may change. The comparison also assumes the same appropriate discount rate applies to the cash flows being compared and that both alternatives meet the same underlying need over the common horizon.

Omitting a replacement outlay. At an intermediate boundary, include the cost of the next project at that date.
Counting the boundary twice or at the wrong date. Add the old cycle’s terminal cash flow and the new cycle’s initial outlay at the same boundary date.
Replacing at the horizon. If the comparison ends when the current cycle ends, do not include an outlay for a project that would operate only after the horizon.
Forgetting terminal cash flows. Include salvage or other relevant terminal cash flows at the end of each cycle, including the final one, when the problem specifies them.
Assuming repeatability without support. If a project cannot be repeated under comparable conditions, its chain may not represent a feasible alternative.

Key takeaway: A replacement chain compares alternatives over a shared horizon by repeating each project and discounting the resulting dated cash flows. The crucial bookkeeping rule is to include both the completed cycle’s terminal amount and the next cycle’s outlay at each intermediate boundary, but no replacement beyond the horizon. This method is aligned with the SOA VEE Accounting and Finance syllabus.

Check Your Understanding

For each problem, construct the dated chain before comparing or calculating its NPV. Monetary amounts are in dollars unless stated otherwise.

A two-year project costs $40,000, pays $24,000 at each year-end, and has no salvage value. It can be repeated identically to cover four years. At an effective annual rate of $5\%$, calculate its chain NPV. (Answer: approximately $8,822.)
A three-year project costs $100 thousand, pays $40 thousand at each year-end, and has a $5 thousand salvage value at each cycle end. It costs $110 thousand to replace and is repeated to cover six years. At an effective annual rate of $10\%$, calculate its chain NPV. (Answer: approximately $-$1.855$ thousand.)
A two-year project costs $70,000, pays $42,000 at each year-end, and has no salvage value. It can be repeated to cover four years. At an effective annual rate of $9\%$, calculate its chain NPV. (Answer: approximately $7,151.)
Project A costs $60,000, lasts two years, pays $38,000 at each year-end, and has no salvage value. Project B costs $105,000, lasts four years, pays $31,000 at each year-end, and has no salvage value. Both are repeatable; compare them over four years at an effective annual rate of $10\%$. Which has the higher chain NPV? (Answer: A, approximately $10,868; B is approximately $-$6,734$.)

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