Replace existing machine now; AEC and ESL

MARR is not printed in the question; assume MARR = 10%. Study period = 5 years (the remaining life of the defender and the life of the challenger).

Defender (existing): market value now 10,000, falling by 1,600 per year to 2,000 at year 5 (market value = 10,000 − 1,600n). O&M = 4,000, 5,000, 6,000, 7,000, 8,000. Challenger (new): cost 25,000; O&M 2,000 each year; market value = 25,000 − 1,500n (17,500 at year 5).

a) Cash-flow approach (incremental: challenger − defender)

Replacing now means selling the old machine for 10,000 (a receipt of 10,000 against the 25,000 purchase), saving O&M each year, and giving up the old machine's salvage at year 5 for the new machine's.

Year 0: −25,000 + 10,000 = −15,000. Years 1–5: savings in O&M = defender O&M − 2,000 = 2,000, 3,000, 4,000, 5,000, 6,000. Year 5: also + (17,500 − 2,000) = +15,500.

Cash flow: 0: −15,000 | 1: +2,000 | 2: +3,000 | 3: +4,000 | 4: +5,000 | 5: +6,000 + 15,500 = +21,500
PW = −15,000 + 2,000(P/F,10%,1) + 3,000(P/F,10%,2) + 4,000(P/F,10%,3) + 5,000(P/F,10%,4) + 21,500(P/F,10%,5)
   = 9,068

PW of replacing = Rs 9,068 > 0. Yes, it is worthwhile to replace the existing machine now.

b) AEC and economic service life (ESL)

For a service life of n years: AEC(n) = [P + Σ O&M_k (P/F,10%,k) − MV_n (P/F,10%,n)] (A/P,10%,n), where P is the investment (for the defender, its present market value 10,000).

Defender

Challenger

Example (defender, n = 1): (10,000 + 4,000/1.1 − 8,400/1.1) × 1.1 = 6,600.

• Defender: AEC is lowest at n = 1, so ESL(defender) = 1 year, AEC = Rs 6,600.
• Challenger: AEC keeps falling to n = 5, so ESL(challenger) = 5 years, AEC = Rs 5,728.

The challenger's minimum AEC (5,728) is below the defender's (6,600), and at the full 5 years AEC(defender) = 8,121 against 5,728, which agrees with the cash-flow result.

Answer: Replace now. AEC and ESL: defender Rs 6,600 at 1 year; challenger Rs 5,728 at 5 years.