Pump Set Redundancy: Two, Three or Four Pumps
A process plant needs a cooling-water pumping set that delivers full flow with two pumps running. A RAM study gives each identical pump a one-year mission reliability of 0.95, with failures assumed independent. Installed cost is P1.2 million per pump; headers, switchgear and spares are common to any number of pumps and can be excluded from the comparison. Production loss while full flow is unavailable is costed at P25,000 per hour over 8 760 operating hours per year. Capital is annualized at 12% over 15 years (capital recovery factor 0.1468). The project engineer proposes three pumps on the grounds that one standby is enough; the maintenance superintendent wants four. Management wants the choice settled on expected annual cost.
1. Compute the one-year reliability and expected unavailable hours for two, three and four pumps under the capacity rule.
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The capacity rule is decisive: two pumps must be available, so a set fails whenever fewer than two are up.
Two pumps (0.95² = 0.9025): Unavailability = 1 − 0.9025 = 0.0975 → 0.0975 × 8 760 = 854 h/yr Note that 1 − (1−0.05)² = 0.9975 is the wrong model here. It answers 'at least one pump', which does not meet the flow requirement. This is the classic trap in this question.
Three pumps (need ≥2 of 3): R = 3R²(1−R) + R³ = 3(0.9025)(0.05) + 0.857375 = 0.135375 + 0.857375 = 0.99275 Unavailability = 0.00725 → 63.5 h/yr
Four pumps (need ≥2 of 4): R = 6R²q² + 4R³q + R⁴ = 6(0.9025)(0.0025) + 4(0.857375)(0.05) + 0.814506 = 0.013538 + 0.171475 + 0.814506 = 0.999519 Unavailability = 4.81 × 10⁻⁴ → 4.2 h/yr
2. Convert each option to expected annual cost and recommend a configuration. Check whether a fifth pump is worth adding.
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Annualized capital per pump = 1.2 M × 0.1468 = P176 200.
Three pumps: capital 528 600 + unreliability 63.5 × 25 000 = 1 587 500 → P2 116 100/yr Four pumps: capital 704 700 + unreliability 4.2 × 25 000 = 105 000 → P809 700/yr Five pumps (need ≥2 of 5): unavailability = 5R q⁴ + q⁵ = 5(0.95)(5.0625 × 10⁻⁵) + 3.125 × 10⁻⁷ = 2.41 × 10⁻⁴ → 2.1 h/yr. Capital 880 900 + 52 500 = P933 400/yr
Recommendation: four pumps. The fourth pump cuts expected annual cost by more than P1.3 million against the three-pump scheme, and pays for its own P176 200 capital charge many times over. The fifth pump is past the knee of the curve - it adds P176 200 of capital to save only about P52 000 of production loss, so it fails on value even though it is more reliable. Diminishing returns, not 'more is better', decides this.
3. What single assumption most threatens this recommendation, and what would you do about it?
Read the worked answer
Independence of failures. The whole calculation multiplies (1−R) terms as if each pump were an isolated coin flip. In reality the pumps share a suction header, a discharge header, one MCC bus, one control system and one operating crew - a single common-cause event (header rupture, bus fault, operator error, contaminated water supply) takes out all four simultaneously and is completely invisible in the model. For a pump set of this type, common cause can dominate the true unavailability.
Actions:
- Re-run the RAM study with an explicit beta factor (often 0.05–0.10 for identical pumps on a shared header) and see whether the four-pump case still beats three.
- Compare spending the fourth pump's P1.2 million on segmenting the header, splitting the electrical supply, or adding strainer and condition monitoring instead - availability per peso is often far higher there.
- Verify the 0.95 figure: if it was taken from the vendor rather than plant records, the entire ranking is built on an unverified input, and the P25 000/h downtime figure deserves the same scrutiny before a P4.8 million decision is signed.