Replacement models

Why it matters

Machines get costlier to run as they age, and many small parts (lamps, bearings, filters, tool inserts) fail without warning. Replacing too early wastes capital; replacing too late pays for breakdowns, repairs and lost output. Replacement models give the age at which a deteriorating machine should be replaced and decide whether low-cost items should be replaced one by one on failure or all together at fixed intervals.

Key ideas

Two kinds of replacement problems.

1. Items that deteriorate gradually (machine tools, vehicles): running and maintenance cost R(t) rises with age, resale (scrap) value S falls. Replace when the average annual cost is lowest.
2. Items that fail suddenly (bulbs, fuses, small bearings): the item works until it fails completely. The question is individual replacement on failure versus group (preventive) replacement of all units every k periods, plus individual replacement of failures in between.

Deteriorating items — money value ignored. If a machine costing C is kept n years, the total cost is C − S(n) + Σ R(t) (t = 1 to n), and the average annual cost is A(n) = [C − S(n) + Σ R(t)]/n. Tabulate A(n) and pick the n where it is minimum. Equivalent rule when the resale value is negligible or does not change from year to year: keep the machine while next year's running cost R(n + 1) is less than the current average A(n); replace when R(n + 1) exceeds it. When resale value keeps falling, rely on the A(n) table itself.

Deteriorating items — money value considered. With interest rate i, use the discount factor v = 1/(1 + i). Compare the weighted average of discounted costs: replace at year n when the discounted total cost divided by Σ vᵗ⁻¹ is minimum. Costs further in the future count less, so the optimal life generally changes.

Sudden failure — individual vs group.

• Given the probability pₖ that a new item fails in period k, the expected number of failures in period k is found by the renewal recursion Nₖ = N₀pₖ + N₁pₖ₋₁ + … + Nₖ₋₁p₁.
• Mean life = Σ k·pₖ. In the long run, individual replacement causes N₀/(mean life) failures per period.
• Group replacement every k periods costs [N₀·c_g + c_i·(N₁ + … + Nₖ)]/k per period. Choose the k with the lowest value and compare it with the individual policy.
• Rule: group-replace at the end of period k if the cost of individual replacements in period k + 1 exceeds the average cost per period up to k.

Other factors. Technological obsolescence, downtime and lost production, safety, and the availability of spares can justify replacement earlier than the pure cost rule suggests.

Formulas

• Total cost of keeping a machine n years: T(n) = C − S(n) + Σ R(t), t = 1…n
• Average annual cost: A(n) = T(n) / n
• Replacement rule (resale value constant or negligible): replace at end of year n when R(n + 1) > A(n) and R(n) < A(n − 1)
• Discount factor: v = 1 / (1 + i)
• Expected failures in period k: Nₖ = Σ (j = 0 to k − 1) Nⱼ · p₍ₖ₋ⱼ₎
• Mean life: t̄ = Σ k · pₖ
• Individual policy cost per period: (N₀ / t̄) · c_i
• Group policy cost per period (interval k): [N₀ · c_g + c_i · Σ (j = 1 to k) Nⱼ] / k

Symbols: C = purchase price (₹); S(n) = resale value after n years (₹); R(t) = running and maintenance cost in year t (₹/year); i = interest rate per year; N₀ = number of items in the group; pₖ = probability of failure in period k; c_i, c_g = cost per item for individual and group replacement (₹).

Worked examples

Example 1 (machine replacement). A machine costs ₹6000. Running costs in years 1–8 are ₹1000, 1200, 1400, 1800, 2300, 2800, 3400 and 4000; resale values at the end of years 1–8 are ₹3000, 1500, 750, 375, 200, 200, 200 and 200. When should it be replaced?

1. n = 1: T = 6000 − 3000 + 1000 = 4000; A = 4000.
2. n = 2: Σ R = 2200; T = 6000 − 1500 + 2200 = 6700; A = 3350.
3. n = 3: Σ R = 3600; T = 6000 − 750 + 3600 = 8850; A = 2950.
4. n = 4: Σ R = 5400; T = 6000 − 375 + 5400 = 11,025; A = 2756.25.
5. n = 5: Σ R = 7700; T = 6000 − 200 + 7700 = 13,500; A = 2700.
6. n = 6: Σ R = 10,500; T = 16,300; A = 2716.67 (rising again).
7. The resale value is constant (₹200) from year 5 on, so the running-cost rule also applies: R(6) = 2800 > A(5) = 2700.
8. Replace at the end of the 5th year; minimum average cost ₹2700 per year.

Example 2 (GATE-type group replacement). 1000 bulbs; probability that a new bulb fails in week 1, 2, 3, 4, 5 is 0.10, 0.15, 0.25, 0.30, 0.20. Individual replacement costs ₹12 per bulb; group replacement costs ₹3 per bulb. Find the best policy.

1. Expected failures: N₁ = 1000(0.10) = 100; N₂ = 1000(0.15) + 100(0.10) = 160; N₃ = 1000(0.25) + 100(0.15) + 160(0.10) = 281; N₄ = 300 + 25 + 24 + 28.1 = 377.1.
2. Mean life = 1(0.10) + 2(0.15) + 3(0.25) + 4(0.30) + 5(0.20) = 3.35 weeks.
3. Individual policy: (1000/3.35) × 12 = ₹3582 per week.
4. Group policy, interval k: k = 1: (3000 + 12 × 100)/1 = ₹4200; k = 2: (3000 + 12 × 260)/2 = ₹3060; k = 3: (3000 + 12 × 541)/3 = ₹3164; k = 4: (3000 + 12 × 918.1)/4 = ₹3504.
5. Minimum at k = 2 (₹3060). Rule check: failures in week 3 would cost 12 × 281 = ₹3372 > 3060.
6. Group-replace all bulbs every 2 weeks and replace failures individually in between: ₹3060 per week, saving about ₹522 per week over individual replacement.

Common mistakes

• Forgetting to subtract the resale value, or subtracting the resale value of the wrong year.
• Using constant maintenance cost: then A(n) keeps falling and there is no optimum — the model needs rising running costs.
• Comparing R(n + 1) with the total cost instead of the average cost A(n).
• In group replacement, using N₀·pₖ alone and ignoring failures of replacement bulbs (the renewal terms).
• Comparing the group cost with individual cost per bulb rather than per period.
• Ignoring the interest rate when the problem gives one.

For GATE PI

• Optimal replacement age from a table of running costs and resale values (NAT).
• Average annual cost at the optimum; the R(n + 1) > A(n) rule.
• Group replacement: expected failures per period, mean life, best group interval.
• Practise laying out the cumulative table neatly — most errors are in the running totals.

Quick check

1. C = ₹10,000, resale after 3 years ₹4000, running costs 1000, 1500, 2000. Find A(3).
2. A(4) = ₹3000 and R(5) = ₹3400. Keep the machine for year 5?
3. Failure probabilities 0.2, 0.3, 0.5 in weeks 1–3. What is the mean life?
4. Why is group replacement attractive for street lamps?

Answers: 1. (10,000 − 4000 + 4500)/3 = ₹3500. 2. No — replace at the end of year 4 (provided A was still falling up to year 4). 3. 0.2 + 0.6 + 1.5 = 2.3 weeks. 4. The cost per lamp is much lower when all are changed together than when a crew is sent for each failure.