Series, parallel and standby system reliability

Why it matters

Real machines and plants are built from many components, and the way they are connected decides how reliable the whole is. A transfer line of twenty stations is far less reliable than any one station; a boiler-feed system with a standby pump can be far more reliable than either pump. Block-diagram calculations let an engineer decide where redundancy pays and how much, and they are among the most common reliability numericals in GATE.

Key ideas

Reliability block diagram (RBD). The system is drawn as blocks connected by the logic of success, not by the physical layout. Two valves physically in a pipe in series are in series in the RBD if either one failing open-circuit stops flow, but in parallel if the failure mode is "fails to close" and either can isolate. Always draw the RBD for the failure mode in question. Component failures are assumed independent.

Series system. The system works only if every component works. Reliability is the product of component reliabilities, so it is always lower than the weakest component and falls quickly as components are added: twenty components of 0.99 each give only 0.818. With constant failure rates, the system failure rate is the sum of component rates.

Parallel (active redundant) system. All units run together and the system works if at least one works. The system fails only if all fail, so unreliabilities multiply. Gains diminish: going from one to two units of 0.9 raises R from 0.90 to 0.99, the third unit adds only 0.009. Active redundancy has a cost: both units age and wear together, and a common cause (power loss, contamination, fire) can fail them all at once.

k-out-of-n system. The system needs at least k of n identical units (e.g. 2 of 3 engines, voting logic in safety systems). Reliability is a binomial sum.

Standby (passive) redundancy. The second unit is idle (cold standby) and is switched in only when the first fails. For two identical units with constant λ, a perfect switch and no failure while idle, the system survives to t if the first unit survives, or if it fails and the standby survives the remaining time. Because the standby does not age while waiting, standby is more reliable than active parallel with the same units, provided the switch (sensing and changeover) is reliable. An imperfect switch with reliability R_sw reduces the benefit.

Mixed systems. Reduce the RBD step by step: replace each parallel group by an equivalent block, then multiply blocks in series (and vice versa). For networks that are neither (bridge networks), use the decomposition (key-component) method or enumerate states.

Where to put redundancy. For the same number of spare components, redundancy at component level (each component duplicated, then in series) gives higher reliability than duplicating the whole series chain, but needs more switches and connections. Redundancy is applied first to the least reliable or most critical element.

Formulas

R_s = R₁ · R₂ · … · Rₙ

• Series system; Rᵢ = reliability of component i for the same mission time (dimensionless).

λ_s = λ₁ + λ₂ + … + λₙ and MTTF_s = 1 / λ_s

• Series system with constant failure rates λᵢ (h⁻¹).

R_p = 1 − (1 − R₁)(1 − R₂)…(1 − Rₙ)

• Active parallel system.

MTTF_p = (1/λ)(1 + 1/2 + … + 1/n)

• n identical active parallel units with constant λ; for n = 2, MTTF = 3/(2λ).

R_k/n = Σ (i = k to n) C(n, i) Rⁱ (1 − R)ⁿ⁻ⁱ

• k-out-of-n system of identical units; for 2-out-of-3, R = 3R² − 2R³.

R_sb(t) = e^(−λt) (1 + λt)

• Two identical units, cold standby, perfect switch, constant λ; MTTF = 2/λ.

R_sb(t) = e^(−λt) (1 + R_sw·λt)

• Same, with switch reliability R_sw.

R_sb(t) = e^(−λt) Σ (k = 0 to n − 1) (λt)ᵏ / k!

• One operating unit plus (n − 1) identical cold spares, perfect switching; MTTF = n/λ.

Worked examples

Example 1 (standard): series–parallel system Given: a hydraulic system has a motor A (R = 0.95) in series with two identical pumps B in active parallel (R = 0.90 each) and a control valve C (R = 0.98). Find the system reliability.

1. Parallel pumps: R_B = 1 − (1 − 0.90)² = 1 − 0.01 = 0.99.
2. Series chain: R_s = R_A · R_B · R_C = 0.95 × 0.99 × 0.98.
3. R_s = 0.9217.

System reliability = 0.922. The motor (0.95) is now the weakest block; duplicating it would help more than adding a third pump.

Example 2 (GATE level): active parallel versus cold standby Given: two identical pumps, each with constant λ = 0.001 h⁻¹, mission time t = 500 h. Compare (a) a single pump, (b) both pumps in active parallel, (c) one running with one in cold standby and a perfect switch, (d) as (c) with switch reliability 0.95. Also find the MTTF for (a)–(c).

1. λt = 0.001 × 500 = 0.5; R = e^(−0.5) = 0.6065.
2. (a) R = 0.6065; MTTF = 1/λ = 1000 h.
3. (b) R_p = 1 − (1 − 0.6065)² = 1 − 0.1548 = 0.8452; MTTF = 3/(2λ) = 1500 h.
4. (c) R_sb = e^(−0.5)(1 + 0.5) = 0.6065 × 1.5 = 0.9098; MTTF = 2/λ = 2000 h.
5. (d) R_sb = e^(−0.5)(1 + 0.95 × 0.5) = 0.6065 × 1.475 = 0.8946.

R(500 h): single 0.607; active parallel 0.845; cold standby 0.910 (0.895 with a 0.95 switch). MTTF: 1000, 1500 and 2000 h.

Common mistakes

• Multiplying reliabilities for a parallel group; it is the unreliabilities that multiply.
• Adding MTTFs of series components instead of adding failure rates.
• Using the parallel formula for standby redundancy (it gives the active-redundancy value, which is lower).
• Ignoring the switch in standby systems, or common-cause failures in parallel systems.
• Combining reliabilities computed for different mission times.
• Drawing the RBD from the physical layout instead of the success logic.

For GATE PI

Expect series, parallel and series–parallel reductions, 2-out-of-3 systems, exponential component reliabilities combined into system reliability, series failure rates and MTTF, and standby versus active redundancy with constant λ. Practise reducing diagrams quickly and keep four significant figures until the final answer.

Quick check

1. Three components of reliability 0.9 in series: system reliability?
2. Same three in active parallel: system reliability?
3. Failure rates 1 × 10⁻⁴, 2 × 10⁻⁴ and 0.5 × 10⁻⁴ h⁻¹ in series: MTTF?
4. 2-out-of-3 system with R = 0.9 per unit: system reliability?

Answers: 1. 0.729. 2. 0.999. 3. 2857 h. 4. 0.972.