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Lifecycle costs, cash flows and comparing alternatives

The cheaper item at purchase can cost more to operate. The reverse can also happen when the service period is short or expected savings do not occur. Engineering economics makes the comparison’s service, cash amounts, timing and assumptions explicit before ranking options.

Use design and measurement for comparisons and evidence; systems engineering develops configuration and interfaces. The supplied discount rates are arithmetic inputs, not current official rates or investment guidance. Professional codes and standards are identified by source and scope; actual legal and contractual duties require the applicable context.

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After reading, you can

Compare equal-service alternatives using dated lifecycle cash flows and a consistent valuation basis.

01

Compare the same required service

State the function, performance, availability and study period that each alternative must provide. Include relevant acquisition, installation, operation, maintenance, replacement and retirement costs, with any justified residual value. A lower-cost option that provides less of the required service is a different comparison, not automatically a saving.

Identify the perspective of the account. A payment that is a cost to one organisation can be income to another; public effects can remain outside a purchaser’s budget. Keep cash costs, physical resource use and nonmonetised consequences distinguishable. An attractive total cannot establish affordability in each budget period or satisfaction of nonfinancial requirements.

02

Use present value consistently

For a supplied constant discount rate r, a cost C occurring t periods after the base date has present value C/(1+r)^t. Sum corresponding dated terms for each alternative. Use a consistent base date and timing convention. The discount rate is a model input with a decision context, not a measured physical constant.

Our original equal-service exercise supplies three years, no residual or other cost differences, and a real rate of 5%. A costs 1000 initially plus 200 at each year end; B costs 1300 plus 80 at each year end. All amounts are in base-date purchasing-power cost units. The three discount factors sum to about 2.723248. A’s present cost is 1544.65; B’s is 1517.86.

Cash amounts need dates before comparisonOriginal three-year equal-service alternatives. A costs 1000 at time 0 and 200 at each year end 1, 2, 3. B costs 1300 initially and 80 at each year end. Supplied real discount rate 5%, base-date cost units, no residual or other differences.Original equal-service case · costs in constant purchasing-power unitst=0t=1t=2t=3AB10002002002001300808080At supplied 5%: PV(A)≈1544.65; PV(B)≈1517.86
Costs are positive amounts. A: 1000 at 0 and 200 at each end-year 1–3; B: 1300 at 0 and 80 at each end-year 1–3. With supplied real 5% rate and constant-price amounts, annuity factor Σ1/(1.05)^t≈2.723248. PV costs 1544.65 and 1517.86; B is lower by 26.79 under the stated exclusions.
03

Interpret the result within its boundary

Constant-price cash amounts require a compatible real rate; nominal amounts that include general inflation require a compatible nominal rate. Do not inflate the amounts and then discount them using a real rate as though the bases matched. Also distinguish general inflation from a particular input’s relative price change.

The constructed B alternative is lower by about 26.79 present cost units under the given assumptions. That result is not a current product price, prescribed discount rate or recommendation for an actual investment. Examine useful life, performance evidence, omitted costs and uncertainty before extending a cost ranking to a project decision. A simple payback calculation would ask a narrower question and omit some of this timing account.

Reconstruct the cost comparison

A’s cost is 1000+200/1.05+200/1.05²+200/1.05³≈1544.65. B’s is 1300+80/1.05+80/1.05²+80/1.05³≈1517.86. The initial premium 300 is offset by the present value of 120 annual savings, about 326.79. Their difference is 26.79.

At a separately supplied real rate of 10%, the factor sum is about 2.486852. The savings then have present value 298.42, below the same 300 premium by about 1.58. The model ranking reverses. This illustrates why the timing and valuation assumptions belong beside the result rather than remaining hidden in a spreadsheet.

Try the reasoning

  1. Why must service be comparable before ranking lifecycle costs?

    Show the answer and reasoning

    A lower total can result from providing a different function, performance or duration rather than using resources more efficiently for the same service.

  2. Can constant-price amounts be combined with an inflation-inclusive rate without adjustment?

    Show the answer and reasoning

    No. The price basis and discount basis must be compatible.

  3. Does B’s lower cost in the 5% exercise establish an actual purchase decision?

    Show the answer and reasoning

    No. The supplied model excludes other requirements, costs and uncertainty and describes no actual product.

What to carry forward

Compare a defined service over a common period, date every cash flow and state the basis that turns those flows into a conditional cost ranking.

References

Teaching resources and research supporting this lesson’s explanations and examples.

  1. Life Cycle Costing Manual, HB135e2025, §§3.1–3.3 and 8Joshua Kneifel and David Webb; NISTReference guide · PDFnvlpubs.nist.govRead source
  2. Take Five: Life Cycle Costs for AcquisitionU.S. Department of Energy, FEMPReference guideenergy.govRead source