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Lesson 1: Lifetimes

In this lesson

You will learn:

  • how to describe when a part fails: its reliability \(R(t)\), unreliability \(F(t)\), density \(f(t)\) and hazard rate \(h(t)\);
  • what the mean time to failure (MTTF) does, and does not, tell you;
  • how the exponential and Weibull distributions describe random failure and wear-out, and why the Weibull shape is the most useful single number in maintenance;
  • how to read a lifetime model in Python, and how RePyability treats one part as the simplest possible system.

Before you start: basic probability (the probability of an event, complements, conditional probability). About 30 minutes.

The question

A supplier tells you their pump has a mean time to failure of 887 hours. You need a pump to run a 200-hour campaign without stopping. How likely is a new one to make it? And the spare in the store has already run 500 hours on another job: is it as good as a new one?

"887 hours" cannot answer either question. Pumps do not all last 887 hours: some fail in their first week and some run for years. To answer, you need the whole distribution of the time to failure. This lesson builds that picture, and by the end of it you will answer both questions.

The time to failure is a random variable

Imagine putting a large number of new pumps to work at time zero and writing down when each one fails. The time to failure, \(T\), varies from pump to pump; its probability distribution is the pump's lifetime model. Two functions describe it.

The unreliability \(F(t)\) is the probability that a unit has failed by age \(t\): the fraction of the population that has failed by then.

\[ F(t) = \Pr(T \le t) \]

The reliability, also called the survival function, is its complement: the probability that a unit is still working at age \(t\), or the fraction of the population still working.

\[ R(t) = \Pr(T > t) = 1 - F(t) \]

Every reliability curve starts at \(R(0) = 1\) (everything works when new) and falls towards 0 (eventually everything fails). Here is the supplier's pump, whose model you will meet properly in a moment:

xychart-beta
    title "Reliability of the pump"
    x-axis "age (hours)" ["0", "250", "500", "750", "1000", "1250", "1500", "1750", "2000"]
    y-axis "R(t)" 0 --> 1
    line [1, 0.969, 0.838, 0.614, 0.368, 0.174, 0.064, 0.017, 0.003]

Almost every pump survives its first 250 hours, most are gone by 1250, and hardly any see 2000. Most of this course is about computing curves like this one, first for parts and then for whole systems.

A third function says when failures happen. The density \(f(t) = dF/dt\) is the fraction of the original population that fails per hour around age \(t\). The area under \(f\) between two ages is the fraction that fails between them.

The hazard rate: the risk right now

The person running the pump asks a different question: given that it is still working now, how likely is it to fail in the next hour? That is the hazard rate (engineers also say failure rate):

\[ h(t) = \frac{f(t)}{R(t)} \]

Among the units still working at age \(t\), a fraction of about \(h(t)\,\Delta t\) fails in the next short interval \(\Delta t\). The density counts failures against the whole original population; the hazard counts them against the survivors only.

The difference, in numbers

Start 1000 pumps. At 500 hours, \(R(500) = 0.838\), so 838 are still running, and the density is \(f(500) = 0.00074\) per hour: about 0.74 of the original 1000 pumps fail in the next hour. As a share of the 838 survivors that is \(0.74 / 838 = 0.00088\) per hour, which is \(h(500)\).

The shape of \(h(t)\) is the most useful single thing to know about a part:

  • a falling hazard means early failures (infant mortality): weak units fail first, and the survivors are the strong ones;
  • a constant hazard means failures from outside causes (shocks, overloads) that do not care how old the part is;
  • a rising hazard means wear-out: the older the part, the more likely it is to fail soon.

The hazard and the reliability determine each other. Since \(f = -dR/dt\), the hazard is \(h = -\frac{d}{dt}\ln R(t)\), and integrating,

\[ R(t) = \exp\!\left(-H(t)\right), \qquad H(t) = \int_0^t h(u)\,du, \]

where \(H(t)\) is the cumulative hazard. Keep this in mind: it explains the bathtub curve at the end of the lesson.

The mean time to failure

The MTTF is the average life, \(\operatorname{E}[T]\). For a lifetime it equals the area under the reliability curve:

\[ \text{MTTF} = \int_0^\infty R(t)\,dt \]

One way to see this: follow the whole population through time. At each age \(t\), a fraction \(R(t)\) of the units is still alive and "earning" life, so the total life earned per unit is the area under \(R\).

The MTTF is a single average over the whole population, and it hides the shape. You will meet two consequences again and again:

  • It is not a typical life. For many parts, half of them or more fail before the MTTF.
  • It is not a failure rate. \(1/\text{MTTF}\) equals the hazard only for the exponential distribution, next.

The exponential: failures that ignore age

The exponential distribution has a constant hazard \(\lambda\):

\[ R(t) = e^{-\lambda t}, \qquad h(t) = \lambda, \qquad \text{MTTF} = \frac{1}{\lambda} \]

Its defining property is that it has no memory. The probability that a unit which has survived to age \(s\) survives a further \(t\) is

\[ \Pr(T > s + t \mid T > s) = \frac{R(s + t)}{R(s)} = \frac{e^{-\lambda (s + t)}}{e^{-\lambda s}} = e^{-\lambda t} = R(t), \]

exactly the probability that a new unit survives \(t\). A used exponential part is as good as new.

Take a part with MTTF 1000 hours, so \(\lambda = 0.001\) per hour. Lifetime models in Python come from surpyval, the library that fits them to data; RePyability takes them as they are.

import numpy as np
import surpyval as surv

unit = surv.Exponential.from_params([0.001])   # rate: 0.001 per hour
unit.mean()                    # -> 1000.0   MTTF = 1 / rate
unit.sf(200)                   # -> 0.8187   R(200) = exp(-0.2)
unit.sf(700) / unit.sf(500)    # -> 0.8187   200 more hours, after 500
unit.hf(300)                   # -> 0.001    the hazard at every age
unit.sf(unit.mean())           # -> 0.3679   only 37% reach the MTTF

sf is the reliability (survival function), ff the unreliability, df the density, hf the hazard and qf the inverse of \(F\) (quantiles). The last line is worth remembering: only \(e^{-1} = 36.8\%\) of exponential parts live as long as their MTTF.

The exponential suits parts that fail from outside causes rather than from age, such as electronics in their useful life. Its lack of memory has a practical consequence you will prove in Lesson 8: replacing an exponential part before it fails buys nothing.

The Weibull: the shape tells the story

The Weibull distribution adds a shape \(\beta\) to a scale \(\alpha\):

\[ R(t) = \exp\!\left[-\left(\frac{t}{\alpha}\right)^{\beta}\right], \qquad h(t) = \frac{\beta}{\alpha}\left(\frac{t}{\alpha}\right)^{\beta - 1} \]
  • The scale \(\alpha\) is the characteristic life: \(R(\alpha) = e^{-1}\) for every shape, so 63.2% of units have failed by age \(\alpha\).
  • The shape \(\beta\) sets the hazard's direction: falling for \(\beta < 1\) (early failures), constant for \(\beta = 1\) (the Weibull is then the exponential), rising for \(\beta > 1\) (wear-out). The larger \(\beta\), the more tightly the lives cluster around \(\alpha\).

Here are three parts with the same scale, \(\alpha = 1000\) hours, and different shapes:

for beta in (0.5, 1, 3):
    part = surv.Weibull.from_params([1000, beta])
    print(beta, part.sf([200, 1000]).round(3), (1000 * part.hf([200, 1000])).round(3))
# 0.5 [0.639 0.368] [1.118 0.5  ]
# 1 [0.819 0.368] [1. 1.]
# 3 [0.992 0.368] [0.12 3.  ]
\(\beta = 0.5\) \(\beta = 1\) \(\beta = 3\)
\(R(200)\) 0.639 0.819 0.992
\(R(1000)\) 0.368 0.368 0.368
\(h(200)\), per 1000 h 1.118 1.0 0.12
\(h(1000)\), per 1000 h 0.5 1.0 3.0
MTTF (h) 2000 1000 893

With \(\beta = 0.5\), 36% of the parts are gone within 200 hours, but the survivors get better: their hazard halves between 200 and 1000 hours. With \(\beta = 3\) almost none fail early, but by 1000 hours the hazard is 25 times what it was at 200. Same \(\alpha\), three very different parts, and the MTTF (\(\alpha\,\Gamma(1 + 1/\beta)\) for a Weibull) says little about the difference.

Answering the question

The supplier's pump is a Weibull with \(\alpha = 1000\) hours and \(\beta = 2.5\): it wears out. That is where "887 hours" came from.

pump = surv.Weibull.from_params([1000, 2.5])   # alpha (hours), beta
pump.mean()                     # -> 887.26   the supplier's MTTF
pump.sf(200)                    # -> 0.9823   a new pump survives the campaign
pump.sf(700) / pump.sf(500)     # -> 0.792    the spare that has run 500 h
pump.ff(pump.mean())            # -> 0.5236   52% fail before the MTTF
np.exp(-200 / pump.mean())      # -> 0.7982   what "exponential, MTTF 887" says

A new pump makes the campaign 98.2% of the time; the spare that has already run 500 hours only 79.2% of the time. Age matters because the pump wears out. If you had assumed an exponential life with the same MTTF, you would have said 79.8% for both: far too pessimistic about the new pump, and slightly optimistic about the used one. The MTTF alone could not have told you either.

Percentiles and B-lives

Warranty and maintenance planning often use the B-life: \(B_{10}\) is the age by which 10% of units have failed, \(F(B_{10}) = 0.1\). For a Weibull, solving \(F(t) = p\) gives

\[ B_p = \alpha\left(-\ln(1 - p)\right)^{1/\beta} \]

so the pump's \(B_{10}\) is \(1000 \times (-\ln 0.9)^{0.4} = 406.5\) hours: plan on one pump in ten failing by then. The median life (\(B_{50}\)) is 863.6 hours, less than the MTTF: a few long-lived pumps pull the mean up.

pump.qf(0.10)    # -> 406.51   B10: 10% have failed
pump.qf(0.50)    # -> 863.63   the median life

The bathtub curve

Real populations often mix failure modes: early failures from manufacturing defects, random failures from outside causes, and wear-out. A part survives to age \(t\) only if no mode has struck by then. If the modes act independently, their reliabilities multiply:

\[ R(t) = R_{\text{early}}(t)\,R_{\text{random}}(t)\,R_{\text{wear}}(t) \]

and because \(R = e^{-H}\), multiplying reliabilities means adding hazards: \(h(t) = h_{\text{early}}(t) + h_{\text{random}}(t) + h_{\text{wear}}(t)\).

"The part fails at the first of several causes" is exactly the logic of a series system, which Lesson 2 builds on. So RePyability can compute the combined hazard by putting the three modes in series, as three blocks between an input and an output:

flowchart LR
    s((in)) --> e[early defects] --> r[random shocks] --> w[wear-out] --> t((out))
from repyability import NonRepairableRBD

modes = {
    "early": surv.Weibull.from_params([2000, 0.5]),       # falling hazard
    "random": surv.Exponential.from_params([1 / 5000]),   # constant hazard
    "wear": surv.Weibull.from_params([1500, 5]),          # rising hazard
}
life = NonRepairableRBD(
    [("in", "early"), ("early", "random"), ("random", "wear"),
     ("wear", "out")],
    modes,
)
ages = np.arange(100, 1501, 100)
(1000 * life.hf(ages)).round(3)   # failures per 1000 hours
# array([1.318, 0.992, 0.851, 0.776, 0.741, 0.742, 0.781, 0.865, 1.005,
#        1.212, 1.501, 1.888, 2.391, 3.028, 3.822])
1000 * life.hf(500)               # -> 0.741   the bottom of the bathtub

xychart-beta
    title "Hazard of a part with three failure modes"
    x-axis "age (hours)" ["100", "200", "300", "400", "500", "600", "700", "800", "900", "1000", "1100", "1200", "1300", "1400", "1500"]
    y-axis "failures per 1000 hours" 0 --> 4
    line [1.318, 0.992, 0.851, 0.776, 0.741, 0.742, 0.781, 0.865, 1.005, 1.212, 1.501, 1.888, 2.391, 3.028, 3.822]

This is the classic bathtub curve: a falling hazard early on, a flat stretch of useful life, and a rising hazard as wear-out takes over. Each part of the tub calls for a different response: burn-in or better quality control for early failures, nothing age-based for random failures, and preventive replacement for wear-out (Lesson 8).

One part as the simplest system

A block diagram with a single block is a system that is exactly its part. It is a useful first step in RePyability, because every system method then works on it:

from repyability import NodeState

single = NonRepairableRBD([("in", "pump"), ("pump", "out")], {"pump": pump})
single.sf(200)                     # -> 0.9823   the same as pump.sf(200)
single.bx_life(10)                 # -> 406.51   B10
single.time_to_reliability(0.9)    # -> 406.51   when R falls to 0.9
single.sf_given_state(200, {"pump": NodeState(age=500)})   # -> 0.792   the spare
single.mean_time_to_failure()   # -> 887.26   the pump's own mean life

sf_given_state conditions on each part's current age: the spare's 79.2%, computed for you. The system's MTTF is the area under its reliability curve, which RePyability integrates numerically; with one part it is that part's mean life.

Where do the models come from? From failure data: the times at which units failed, and the ages of units still running (censored observations). Fitting them is surpyval's job; the Tutorial shows a fit with censored data in its first step.

Pitfalls

Common misreadings

  • The MTTF is not a guaranteed life. 52% of the pumps fail before it, and 63% of exponential parts do.
  • "Failure rate" has two meanings. Datasheets often quote \(1/\text{MTTF}\) as a failure rate. It equals the hazard only for the exponential; for a part that wears out, the hazard of a new unit is far lower, and that of an old one far higher.
  • A constant failure rate is an assumption, not a fact. Mechanical parts usually wear out (\(\beta > 1\)). Assuming an exponential makes a used part look as good as new, and hides the case for preventive maintenance.
  • A fitted model describes the data it came from. Extrapolating it far beyond the ages observed, or to a harsher duty, is risky.

Summary

Key ideas

  • The time to failure is random. \(R(t)\) is the fraction of units still working at age \(t\), and \(F(t) = 1 - R(t)\) the fraction failed.
  • The hazard \(h(t) = f(t)/R(t)\) is the risk of failing now, for a unit that has survived so far. Its shape (falling, flat or rising) matters more than any single number.
  • \(\text{MTTF} = \int_0^\infty R(t)\,dt\) is an average that hides the shape; often half the units or more fail before it.
  • The exponential has a constant hazard and no memory: an old unit is as good as new.
  • The Weibull's shape \(\beta\) says whether failures are early (\(\beta < 1\)), random (\(\beta = 1\)) or wear-out (\(\beta > 1\)); its scale \(\alpha\) is the age by which 63.2% have failed.
  • Independent failure modes act like blocks in series: reliabilities multiply and hazards add.

Exercises

1. An exponential part has an MTTF of 2000 hours. What is the probability that it survives its first 1000 hours?

Answer

\(\lambda = 1/2000\) per hour, so \(R(1000) = e^{-1000/2000} = e^{-0.5} = 0.6065\).

surv.Exponential.from_params([1 / 2000]).sf(1000)   # -> 0.6065

2. The same part has already run 3000 hours. What is the probability that it survives another 1000?

Answer

Still 0.6065. The exponential has no memory: \(R(4000)/R(3000) = e^{-0.5}\). Its past tells you nothing about its future.

part = surv.Exponential.from_params([1 / 2000])
part.sf(4000) / part.sf(3000)   # -> 0.6065

3. A bearing's life is Weibull with \(\alpha = 1000\) hours and \(\beta = 2\). What is its \(B_{10}\) life?

Answer

\(B_{10} = 1000 \times (-\ln 0.9)^{1/2} = 1000 \times 0.3246 = 324.6\) hours.

surv.Weibull.from_params([1000, 2]).qf(0.1)   # -> 324.6

4. For the same bearing, compare the hazard at 100 hours and at 1000 hours. What does the ratio tell you?

Answer

\(h(t) = \frac{2}{1000}\cdot\frac{t}{1000}\), so \(h(100) = 0.0002\) and \(h(1000) = 0.002\) per hour: ten times higher. With \(\beta = 2\) the hazard grows in proportion to age, so an old bearing is far riskier than a young one: a candidate for preventive replacement.

bearing = surv.Weibull.from_params([1000, 2])
bearing.hf(1000) / bearing.hf(100)   # -> 10.0

5. A component's life is Weibull with \(\alpha = 1000\) hours and \(\beta = 0.5\). Would replacing it with a new one every 500 hours make it more reliable? Compare a new unit and a 500-hour-old unit over the next 100 hours.

Answer

No: it would make things worse. With \(\beta < 1\) the hazard falls with age, so a unit that has survived 500 hours is more reliable than a new one. A new unit survives the next 100 hours with probability 0.729; a 500-hour-old unit with probability 0.935. Replacing it would throw away a proven unit for an unproven one. What helps here is burn-in: running new units for a while before service, so the weak ones fail on the test bench.

early = surv.Weibull.from_params([1000, 0.5])
early.sf(100)                   # -> 0.729
early.sf(600) / early.sf(500)   # -> 0.935

Where next