Comparison of alternatives and replacement analysis

Why it matters

Most engineering decisions are choices between alternatives: a carbon-steel or a stainless exchanger, a cheap pump with high power bills or a dearer efficient one, keeping an old compressor or replacing it. The right answer depends on first cost, running cost, life and the time value of money taken together. Comparing them on a common basis — and ignoring money already spent — is the core skill of this topic.

Key ideas

Mutually exclusive alternatives do the same job, so only one is chosen. If revenues are the same for all, compare costs only; the alternative with the lowest equivalent cost wins.

Common bases of comparison.

• Present worth of costs — discount all costs to time zero. Valid directly only when lives are equal (or over a common multiple of lives).
• Equivalent uniform annual cost (EUAC) — convert first cost and salvage to an annual amount with the capital-recovery factor and add the annual operating cost. Because it is per year, it compares alternatives of unequal life directly, assuming each would be replaced by an identical unit.
• Capitalised cost — the present worth of providing the service for ever; equivalent to EUAC divided by i.
• Incremental return — when a dearer alternative saves operating cost, the extra investment is justified only if the return on the increment (annual saving after tax ÷ extra investment) is at least the minimum acceptable rate. Always compare each alternative with the next cheaper one that is itself acceptable, not with the cheapest.

Replacement analysis. The existing asset is the defender, the candidate new one the challenger.

• Sunk costs are irrelevant. What was paid for the defender, and its book value, do not affect the decision (except through tax effects). The defender's capital cost is its present market (resale) value — the money given up by keeping it.
• Use the defender's remaining life and future operating costs, and the challenger's full life.
• Compare EUACs: replace if the challenger's EUAC is lower.
• Reasons for replacement: physical deterioration (rising maintenance, falling efficiency), obsolescence (a better technology), inadequacy (capacity too small), and changed regulations.

Economic life is the number of years of service that minimises an asset's EUAC. Capital-recovery cost per year falls with life, while operating and maintenance costs rise with age; the sum has a minimum. Economic life is usually shorter than physical life.

Non-monetary factors — safety, reliability, flexibility, environmental compliance, spares standardisation — can override a small cost difference; a sensitivity check shows how close the decision is.

Formulas

Capital-recovery factor: (A/P, i, n) = i·(1 + i)ⁿ / [(1 + i)ⁿ − 1] EUAC with salvage: EUAC = (P − S)·(A/P, i, n) + S·i + C_op Present worth of costs (equal lives): PW = P + C_op·(P/A, i, n) − S/(1 + i)ⁿ, with (P/A, i, n) = 1/(A/P, i, n) Capitalised cost: K = EUAC / i Incremental rate of return (no time value): r_inc = (annual saving after tax) / (extra investment) Straight-line approximation (interest ignored): annual cost = (P − S)/n + C_op

Symbols: P first cost or, for a defender, current market value (₹); S salvage at end of life (₹); n life (years); i interest rate (decimal); C_op annual operating cost (₹/yr); EUAC in ₹/yr.

Worked examples

Example 1 (standard). Two pumps for the same duty, i = 10%. Pump X: ₹6 lakh, life 5 years, no salvage, operating cost ₹1.5 lakh/yr. Pump Y: ₹10 lakh, life 10 years, salvage ₹1 lakh, operating cost ₹1.0 lakh/yr. Which is cheaper?

1. (A/P, 10%, 5) = 0.1 × 1.1⁵/(1.1⁵ − 1) = 0.26380; (A/P, 10%, 10) = 0.16275.
2. X: EUAC = 6 × 0.26380 + 1.5 = 1.583 + 1.5 = ₹3.083 lakh/yr.
3. Y: EUAC = (10 − 1) × 0.16275 + 1 × 0.10 + 1.0 = 1.465 + 0.10 + 1.0 = ₹2.565 lakh/yr.
4. Pump Y is cheaper by about ₹0.52 lakh per year. Unequal lives are handled automatically by the annual basis.

Example 2 (GATE level). An old compressor has a book value of ₹7 lakh, could be sold today for ₹4 lakh, will last 4 more years with no salvage, and costs ₹6 lakh/yr to run. A new one costs ₹15 lakh, lasts 8 years, has salvage ₹3 lakh and costs ₹3 lakh/yr to run. i = 12%. Should it be replaced?

1. The book value ₹7 lakh is sunk; the defender's capital cost is its market value ₹4 lakh.
2. (A/P, 12%, 4) = 0.32923; (A/P, 12%, 8) = 0.20130.
3. Defender: EUAC = 4 × 0.32923 + 6 = 1.317 + 6 = ₹7.317 lakh/yr.
4. Challenger: EUAC = (15 − 3) × 0.20130 + 3 × 0.12 + 3 = 2.416 + 0.360 + 3 = ₹5.776 lakh/yr.
5. Replace: the challenger costs about ₹1.54 lakh/yr less. Using the ₹7 lakh book value would wrongly inflate the defender's cost.

Common mistakes

• Comparing present worths of alternatives with different lives without a common study period.
• Counting the defender's original price or book value (sunk cost) instead of its market value.
• Forgetting the S·i term (or recovering salvage) in EUAC.
• Choosing the alternative with the highest total ROI instead of checking the incremental return on extra investment.
• Ignoring interest altogether in long-life comparisons, which favours capital-intensive options.

For GATE CH

Expect EUAC or present-worth comparisons of two alternatives (often with unequal lives), capitalised-cost comparisons, a replacement decision with a sunk cost to ignore, or the simple straight-line annual cost when interest is not given. Practise the capital-recovery factor and reading which costs are relevant.

Quick check

1. Why can EUAC compare a 5-year and a 10-year option directly?
2. Equipment ₹5 lakh, salvage ₹0.5 lakh, life 10 years, operating ₹0.3 lakh/yr; interest ignored. Annual cost?
3. In replacement analysis, what is the capital cost of the defender?
4. An extra ₹4 lakh investment saves ₹1 lakh/yr after tax. Incremental return?

Answers: 1. It assumes each is replaced by an identical unit, so a cost per year is comparable. 2. 4.5/10 + 0.3 = ₹0.75 lakh/yr. 3. Its present market (resale) value. 4. 25% per year.