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Sunk cost, ESL, replacement strategy from AEC table

Short answer

Sunk costs are past irrecoverable costs ignored in replacement decisions. Economic life minimizes AEC. With given AECs, defender ESL is 2 years (AEC 5,200), challenger ESL is 4 years (5,600). Best strategy: keep defender for 2 years, then replace with challengers used for 3 years each, total PW Rs 29,541 (AEC 5,537).

Explain about the sunk cost, economic life and reasons for replacement of an asset. The annual equivalent cost of defender and challenger are given in the table below. What is the best replacement strategy? Use MARR = 10%. The planning horizon of the project is 8 years.

End of year (n)123456
(AEC)D540052005500570062006600
(AEC)C770062005700560056805900

Sunk cost

A sunk cost is a cost that has already been incurred and cannot be recovered whatever is decided now, for example the book value of an old machine above its present market value. It must be ignored in replacement analysis; only the present market value of the defender and future cash flows matter. Treating the sunk cost as part of the defender's investment would wrongly favour keeping the old asset.

Economic life

Economic life (economic service life) is the period of service that minimises the asset's equivalent annual cost (AEC). Before it, capital recovery cost is falling faster than operating cost rises; after it, rising operating and maintenance cost dominates.

Reasons for replacement

  • Physical impairment: wear and tear raises maintenance cost and reduces output or quality.
  • Obsolescence: a new machine gives the same service at lower cost or better quality.
  • Inadequacy: capacity is no longer enough for the demand.
  • Requirement changes: changes in product, law or safety standards.
  • Financial reasons: lease or buy decisions, tax effects, or the risk of high repair bills.

Strategy from the AEC table (MARR = 10%, 8-year horizon)

From the table: defender minimum AEC = 5,200 at n = 2 (ESL = 2 years); challenger minimum AEC = 5,600 at n = 4 (ESL = 4 years). Because the defender's minimum is lower than the challenger's, keep it at least until year 2. Since the horizon is 8 years and tables stop at 6, each plan is the defender for n years followed by challengers covering the remaining years. PW of a plan = AEC_D(n)(P/A,10%,n) + AEC_C(m)(P/A,10%,m)(P/F,10%,n) (+ the same for a second challenger, if used).

Worked example, defender 2 years then challenger 3 + 3 years:

PW = 5,200(P/A,10%,2) + [5,700(P/A,10%,3)](P/F,10%,2) + [5,700(P/A,10%,3)](P/F,10%,5)
   = 5,200(1.7355) + 5,700(2.4869)(0.8264) + 5,700(2.4869)(0.6209)
   = 9,025 + 11,714 + 8,802 = 29,541
Defender keptThen challenger(s) forPW of costEquivalent AEC (8 yr)
Defender 2 yr3 + 3 yr29,5415,537
Defender 1 yr4 + 3 yr29,8485,595
Defender 3 yr5 yr29,8555,596
Challenger only4 + 4 yr29,8765,600
Defender 4 yr4 yr30,1935,659
Defender 5 yr3 yr32,3046,055
Defender 6 yr2 yr34,8196,527

The lowest PW is Rs 29,541 (equivalent to Rs 5,537 per year).

Answer: Keep the defender for 2 years (its economic life), then replace it by the challenger and use challengers (each replaced after 3 years) for the remaining 6 years. The three plans "defender 1 yr, then challengers 4 + 3", "defender 3 yr, then challenger 5" and "challenger replaced after 4 yr each" are within about 1% of this, so the choice is not very sensitive.

✓ Answer checked by Claude Opus 5.5 (high effort)

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