Reliability: failure rate, bathtub curve and MTBF

Why it matters

Quality asks whether a product is right when it leaves the factory; reliability asks whether it keeps working afterwards. Warranty budgets, spare-parts stocking, maintenance intervals and the design of safety systems all rest on failure-rate data. The bathtub curve, the exponential model and MTBF are the vocabulary every maintenance, quality and design engineer uses.

Key ideas

Reliability. The probability that an item performs its required function, under stated conditions, for a stated period of time. All four parts matter: a pump's reliability for 1000 h at rated head is not its reliability for 5000 h at overload.

The lifetime functions. Let T be the time to failure of an item.

• Failure density f(t): the probability per unit time of failing at t (a probability density).
• Unreliability F(t) = P(T ≤ t), the fraction failed by time t.
• Reliability R(t) = 1 − F(t), the fraction still working at t.
• Hazard rate (instantaneous failure rate) h(t) = f(t)/R(t): the probability per unit time that an item that has survived to t fails in the next instant. It is a conditional rate, which is why it is the right quantity to plot over life.

The bathtub curve. For many populations h(t) plotted against age has three regions:

1. Infant mortality (early failure, burn-in): h decreasing. Causes: manufacturing defects, weak components, assembly and installation errors. Remedies: burn-in or run-in testing, better process control and inspection, supplier quality.
2. Useful life (random or chance failures): h roughly constant. Causes: random overloads and stress peaks. Preventive replacement does not help here, because an old item is as good as a new one; derating and redundancy do.
3. Wear-out: h increasing. Causes: fatigue, wear, corrosion, ageing. Remedy: preventive replacement or overhaul before wear-out begins. Electronic components tend to have long flat regions; mechanical parts often show little flat region and a gradual wear-out.

Constant failure rate: the exponential model. If h(t) = λ is constant, R(t) = e^(−λt). The mean time to failure is 1/λ. The model is memoryless: the probability of surviving the next 100 h is the same for a new item and for one that has already run 5000 h. Only 36.8 % of items survive to the mean life (R = e⁻¹), so MTBF is not "the time by which most items will still be working".

Weibull model. h(t) = (β/η)(t/η)^(β−1). The shape parameter β describes the bathtub region: β < 1 decreasing hazard (infant mortality), β = 1 constant (exponential), β > 1 increasing (wear-out). η is the characteristic life, at which 63.2 % have failed. Parameters are fitted to test or field data.

MTTF, MTBF, MTTR, availability.

• MTTF (mean time to failure) is used for non-repairable items: MTTF = ∫₀^∞ R(t) dt.
• MTBF (mean time between failures) is used for repairable systems: total operating (up) time divided by number of failures. In the constant-λ region, MTBF = 1/λ.
• MTTR (mean time to repair) measures maintainability.
• Inherent availability A = MTBF / (MTBF + MTTR): the long-run fraction of time the system is up.

Estimating the failure rate from data. From a test of N₀ items with failures recorded in time intervals, the hazard in an interval is the number failing in it divided by the number surviving at its start and the interval length. From field data on repairable equipment with constant λ, λ = number of failures / total operating time (all units added together).

Formulas

R(t) = 1 − F(t), h(t) = f(t) / R(t)

• R, F dimensionless; f and h in failures per hour (h⁻¹).

R(t) = exp[−∫₀ᵗ h(t) dt]

• General link between hazard and reliability.

R(t) = e^(−λt), F(t) = 1 − e^(−λt), f(t) = λ·e^(−λt)

• Constant failure rate λ (h⁻¹), time t (h).

MTTF = ∫₀^∞ R(t) dt = 1/λ (exponential)

λ̂ = r / ΣTᵢ and MTBF = ΣTᵢ / r

• r = number of failures; ΣTᵢ = total operating time of all units (h).

h(interval) = n_f / (N_s · Δt)

• n_f = failures in the interval; N_s = survivors at the start of the interval; Δt = interval length (h).

A = MTBF / (MTBF + MTTR)

• Inherent availability (dimensionless).

R(t) = exp[−(t/η)^β]

• Weibull reliability; η = characteristic life (h); β = shape parameter (dimensionless).

Worked examples

Example 1 (standard): exponential reliability Given: a control valve in its useful-life period has λ = 2 × 10⁻⁴ h⁻¹. Find (a) MTBF, (b) reliability for a 1000 h mission, (c) the longest mission with reliability at least 0.95, (d) the probability of surviving to the MTBF.

1. (a) MTBF = 1/λ = 1 / (2 × 10⁻⁴) = 5000 h.
2. (b) R(1000) = e^(−2 × 10⁻⁴ × 1000) = e^(−0.2) = 0.8187.
3. (c) e^(−λt) = 0.95 gives t = −ln(0.95)/λ = 0.05129 / (2 × 10⁻⁴) = 256.5 h.
4. (d) R(5000) = e^(−1) = 0.368.

MTBF = 5000 h; R(1000 h) = 0.819; maximum mission for 95 % reliability ≈ 256 h; only 36.8 % survive to the MTBF.

Example 2 (GATE level): hazard rate from test data Given: 1000 identical components are put on test. Failures: 60 in 0–100 h, 40 in 100–200 h, 30 in 200–300 h. Find the hazard rate in each interval, the reliability at 300 h, and the region of the bathtub curve.

1. Interval 1: survivors at start = 1000. h₁ = 60 / (1000 × 100) = 6.0 × 10⁻⁴ h⁻¹.
2. Interval 2: survivors = 1000 − 60 = 940. h₂ = 40 / (940 × 100) = 4.26 × 10⁻⁴ h⁻¹.
3. Interval 3: survivors = 940 − 40 = 900. h₃ = 30 / (900 × 100) = 3.33 × 10⁻⁴ h⁻¹.
4. Survivors at 300 h = 900 − 30 = 870, so R(300) = 870 / 1000 = 0.87.
5. The hazard rate falls with age.

h = 6.0, 4.26 and 3.33 × 10⁻⁴ h⁻¹; R(300 h) = 0.87; decreasing hazard means infant mortality, so a burn-in period is worth considering.

Common mistakes

• Dividing failures in an interval by the original population instead of the survivors at the start of the interval.
• Treating MTBF as a guaranteed life; with constant λ, 63 % of items fail before the MTBF.
• Applying R = e^(−λt) in the infant-mortality or wear-out regions, where λ is not constant.
• Expecting preventive replacement to help in the constant-λ region; it only helps for wear-out.
• Mixing hours and years, or failures per 10⁶ h with failures per hour.
• Using MTBF for non-repairable parts (MTTF is the correct term).

For GATE PI

Expect numericals on R(t) = e^(−λt), MTBF/MTTF, the mission time for a target reliability, λ from total operating time, hazard rates from tabulated failure data, and availability with MTTR. Conceptual MCQs ask which phase of the bathtub curve has which hazard trend and cause, and what the Weibull shape parameter indicates.

Quick check

1. λ = 0.001 h⁻¹. What is R(100 h)?
2. What fraction of exponential items survive to the MTTF?
3. A Weibull fit gives β = 0.6. Which region of the bathtub curve is this?
4. MTBF = 400 h and MTTR = 20 h. What is the availability?

Answers: 1. e^(−0.1) = 0.905. 2. 36.8 %. 3. Infant mortality (decreasing hazard). 4. 0.952.