Time to Failure: The Exponential Model

Main idea

If a part doesn't age, so its chance of failing in the next hour is the same whether it's new or old, then its lifetime is exponential: $R(t) = e^{-\lambda t}$ , mean life $1/\lambda$ , and a used part is exactly as good as a new one. A fleet of such parts fails as a Poisson process, which is how spares get planned.

{\htmlData{sym=0}{R(t)}} = P(T > t) = e^{-{\htmlData{sym=3}{\lambda}} t}, \qquad {\htmlData{sym=1}{F(t)}} = 1 - e^{-{\htmlData{sym=3}{\lambda}} t}, \qquad {\htmlData{sym=2}{f(t)}} = {\htmlData{sym=3}{\lambda}} e^{-{\htmlData{sym=3}{\lambda}} t}

What the symbols mean

$R(t)$
Reliability
The probability the part still works at time t.
$F(t)$
Failure CDF
The probability it has failed by time t.
$f(t)$
Failure density
How the failures are spread out over time.
$\lambda$
Failure rate
Failures per hour per surviving unit, the same at every age.

{\htmlData{sym=0}{\text{MTTF}}} = \int_0^\infty R(t)\,dt = \frac{1}{\lambda}

What the symbols mean

$\text{MTTF}$
Mean time to failure
The average lifetime. Only 36.8% of units survive past it.

Memoryless

P(T > {\htmlData{sym=0}{s}} + {\htmlData{sym=1}{t}} \mid T > {\htmlData{sym=0}{s}}) = e^{-\lambda {\htmlData{sym=1}{t}}}

What the symbols mean

$s$
Age so far
How long the part has already survived.
$t$
Extra time
How much longer you need it to last.

Failures as a Poisson process

P\big({\htmlData{sym=0}{N(t)}} = k\big) = \frac{(\lambda t)^ke^{-\lambda t}}{k!}

What the symbols mean

$N(t)$
Failures by time t
With instant replacement, the number of failures in [0, t] is Poisson with mean λt.

Estimate from a life test

\hat\lambda = \frac{\htmlData{sym=0}{{r}}}{\htmlData{sym=1}{{T}}}, \qquad \widehat{\text{MTTF}} = \frac{\htmlData{sym=1}{{T}}}{\htmlData{sym=0}{{r}}}

What the symbols mean

$r$
Failures observed
How many units failed during the test.
$T$
Total time on test
The hours of every unit added up, failed or still running.

Explanation

The exponential lifetime model is simply the right description for parts that fail from random accidents rather than from wearing out. The formulas look like a lot, but there's one idea underneath: $R(t) = e^{-\lambda t}$ (the fraction still working at time $t$ , which shrinks by the same percentage every hour). Everything else follows: $F$ (the fraction failed, $1 - R$ ), $f$ (how failures spread out over time), the MTTF $1/\lambda$ (the average life), "memoryless" (an old part is as good as new) and $r/T$ (how to estimate $\lambda$ from a test). One curve, viewed from several angles.

A part that doesn't age. An engine controller has no moving parts. It fails when a voltage spike or a stray particle hit happens to be big enough, and those events strike at random, with no regard for how long the controller has been running. So its chance of failing in the next hour is the same at hour 10 as at hour 10,000: a constant failure rate $\lambda$ .

A constant rate forces the exponential. Of the $R(t)$ units still working at age $t$ , a fraction $\lambda\,dt$ fails in the next $dt$ , so $R(t + dt) = R(t)(1 - \lambda\,dt)$ , which is

R'(t) = -\lambda R(t), \quad R(0) = 1 \implies R(t) = e^{-\lambda t},

the separable equation of radioactive decay. Then $F = 1 - R$ and $f = F' = \lambda e^{-\lambda t}$ : the exponential distribution from the intro course, now read as a lifetime.

Mean time to failure. Integrate by parts with $u = t$ and $dv = f(t)\,dt$ , so $v = -R(t)$ :

\text{MTTF} = \int_0^\infty t f(t)\,dt = \Big[-tR(t)\Big]_0^\infty + \int_0^\infty R(t)\,dt = \int_0^\infty e^{-\lambda t}\,dt = \frac1\lambda.

The middle step, MTTF as the area under $R(t)$ , holds for any lifetime distribution with $tR(t) \to 0$ , and it comes back in reliability and hazard functions. The tools are integration by parts and improper integrals.

A common mistake. It's tempting to read the MTTF as a typical life, the age a part will usually reach. It isn't. At $t = \text{MTTF}$ , $R = e^{-1} = 0.368$ : 63.2% of units have already failed. The median life is $\ln 2/\lambda \approx 0.693\,\text{MTTF}$ . An MTTF of 50,000 hours doesn't promise that a controller lasts 5.7 years. It says that a fleet of 1,000 controllers loses about one every 50 hours.

Memoryless. For a part that has already survived to age $s$ ,

P(T > s + t \mid T > s) = \frac{P(T > s + t)}{P(T > s)} = \frac{e^{-\lambda(s + t)}}{e^{-\lambda s}} = e^{-\lambda t},

exactly the chance that a new part survives $t$ . Used is as good as new, so replacing exponential parts on a schedule wastes money: you swap a part for one that's just as likely to fail. Preventive replacement pays only for parts that wear out, which takes the Weibull model with $\beta > 1$ .

The Poisson process. Run one socket forever, replacing the part the instant it fails. Failures then arrive at rate $\lambda$ , independently in separate stretches of time: a Poisson process, and the number of failures by time $t$ is Poisson with mean $\lambda t$ . The link runs both ways. The first failure comes after $t$ exactly when there are no failures in $[0, t]$ , so

P(T > t) = P\big(N(t) = 0\big) = e^{-\lambda t}.

For a fleet of $m$ identical units the rates add, and failures arrive at $m\lambda$ . That's how spare parts are stocked: choose $k$ so that $P(N \le k)$ is high enough.

Estimating λ from a life test. You can't wait for every unit to fail, so a life test stops early. The units still running are censored: all you know is that their lives exceed the test time. The likelihood of what you saw is a density $\lambda e^{-\lambda t_i}$ for each failure and a survival probability $e^{-\lambda c_j}$ for each survivor, which multiplies out to $\lambda^r e^{-\lambda T}$ , where $T$ adds up every unit's hours. Setting the derivative of $r\ln\lambda - \lambda T$ to zero gives

\hat\lambda = \frac{r}{T}, \qquad T = \text{total hours of all units, failed or not},

an optimization problem. Averaging only the failure times ignores the survivors and badly underestimates the MTTF.

When the model fits: electronics in their useful life, failing from random overloads and shocks. It fails for anything that wears: bearings, belts, seals, batteries. Electronic failure rates are often quoted in FIT, failures per $10^9$ device-hours.

See it

A controller with an MTTF of 2,000 hours (time in thousands of hours). As the age s moves along, the dashed curve shows the survival curve of the units still working at that age, and it's an exact copy of the original, just restarted at s. That's memorylessness: the chance of lasting another 1,000 hours stays 0.61 at every age. The shaded area under R(t) is the MTTF, and only 36.8% of units make it that far.

One equipment socket with a constant failure rate of 0.5 per 1,000 hours, its part replaced at every failure. Notice that random failures don't arrive evenly: two pairs come only 400 and 300 hours apart, and elsewhere there's a 4,400-hour lull. Yet the running count N(t) stays close to the expected λt, ending at 10 failures by 20,000 hours, because the failures form a Poisson process: the count in any window is Poisson with mean λ times its length.

Worked examples

Example 1

An engine controller has a constant failure rate of $2 \times 10^{-5}$ per hour. Find its MTTF, its reliability at 10,000 hours, its median life and the age by which 10% of controllers have failed.

1
$\text{MTTF} = 1/\lambda = 1/(2 \times 10^{-5}) = 50{,}000$ hours.

Example 2

Twenty power modules run on a life test for 1,000 hours. Four fail, at 220, 450, 610 and 880 hours; the other 16 are still running when the test ends. Estimate the failure rate and the MTTF.

1
Total time on test: $T = 220 + 450 + 610 + 880 + 16(1000) = 18{,}160$ hours.

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

A pressure sensor has a constant failure rate of $4 \times 10^{-6}$ per hour. What is its MTTF in hours?

Problem 2+5 XP

A pump controller with an MTTF of 40,000 hours runs around the clock. What's the probability it survives its first year (8,760 hours)? Give 4 decimal places.

Problem 3+5 XP

A pressure transmitter has a constant failure rate and an MTTF of 10 years. It has already run 3 years without failing. What's the probability it lasts at least 2 more years? Give 4 decimal places.

Problem 4+5 XP

For a part with an exponential lifetime, what fraction of units fail before reaching the MTTF? Give 4 decimal places.

Problem 5+5 XP

Twelve cooling fans run on a test for 2,000 hours. Three fail, at 600, 1,100 and 1,700 hours; the other nine are still running when the test stops. Assuming a constant failure rate, estimate the MTTF in hours.

Problem 6+5 XP

A telecom site runs 50 identical line cards, each with a constant failure rate of $1 \times 10^{-5}$ per hour, and a failed card is swapped right away. Spares are restocked every 30 days (720 hours). What's the smallest number of spares to keep so that the chance of running out within a restock period is at most 1%?

Problem 7+5 XP

A drive belt's failure rate grows with age: $h(t) = 0.02t$ per thousand hours, with $t$ in thousands of hours. The surviving fraction obeys $R'(t) = -h(t)R(t)$ with $R(0) = 1$ . Solve for $R(t)$ .

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

A satellite transmitter has a constant failure rate of $1 \times 10^{-5}$ per hour. A second, identical transmitter rides along switched off as a cold spare: it can't fail until it's switched on, and the switch-over is perfect. What's the probability the satellite can still transmit at the end of a 10-year mission (87,600 hours)? Give 4 decimal places.

Where it's used

Electronics makers publish failure rates in FIT for every component, and system designers add them up for a whole product, which is legitimate exactly because the rates are constant. Data centers and telecom operators size spare-part stocks from Poisson counts of failures across fleets of identical units. Life tests that stop early are standard practice, and the time-on-test estimate is how a few weeks of testing turns into an MTTF claim on a datasheet.

Waiting for the Bus: Schedule vs RandomStats · Economics & Finance Counting Clicks on a Geiger CounterStats · Physics

Counting Defects: Binomial and Poisson The Weibull Distribution

Time to Failure: The Exponential Model

Main idea

{\htmlData{sym=0}{R(t)}} = P(T > t) = e^{-{\htmlData{sym=3}{\lambda}} t}, \qquad {\htmlData{sym=1}{F(t)}} = 1 - e^{-{\htmlData{sym=3}{\lambda}} t}, \qquad {\htmlData{sym=2}{f(t)}} = {\htmlData{sym=3}{\lambda}} e^{-{\htmlData{sym=3}{\lambda}} t}

What the symbols mean

$R(t)$
Reliability
The probability the part still works at time t.
$F(t)$
Failure CDF
The probability it has failed by time t.
$f(t)$
Failure density
How the failures are spread out over time.
$\lambda$
Failure rate
Failures per hour per surviving unit, the same at every age.

{\htmlData{sym=0}{\text{MTTF}}} = \int_0^\infty R(t)\,dt = \frac{1}{\lambda}

What the symbols mean

$\text{MTTF}$
Mean time to failure
The average lifetime. Only 36.8% of units survive past it.

Memoryless

P(T > {\htmlData{sym=0}{s}} + {\htmlData{sym=1}{t}} \mid T > {\htmlData{sym=0}{s}}) = e^{-\lambda {\htmlData{sym=1}{t}}}

What the symbols mean

$s$
Age so far
How long the part has already survived.
$t$
Extra time
How much longer you need it to last.

Failures as a Poisson process

P\big({\htmlData{sym=0}{N(t)}} = k\big) = \frac{(\lambda t)^ke^{-\lambda t}}{k!}

What the symbols mean

$N(t)$
Failures by time t
With instant replacement, the number of failures in [0, t] is Poisson with mean λt.

Estimate from a life test

\hat\lambda = \frac{\htmlData{sym=0}{{r}}}{\htmlData{sym=1}{{T}}}, \qquad \widehat{\text{MTTF}} = \frac{\htmlData{sym=1}{{T}}}{\htmlData{sym=0}{{r}}}

What the symbols mean

$r$
Failures observed
How many units failed during the test.
$T$
Total time on test
The hours of every unit added up, failed or still running.

Explanation

A constant rate forces the exponential. Of the $R(t)$ units still working at age $t$ , a fraction $\lambda\,dt$ fails in the next $dt$ , so $R(t + dt) = R(t)(1 - \lambda\,dt)$ , which is

R'(t) = -\lambda R(t), \quad R(0) = 1 \implies R(t) = e^{-\lambda t},

the separable equation of radioactive decay. Then $F = 1 - R$ and $f = F' = \lambda e^{-\lambda t}$ : the exponential distribution from the intro course, now read as a lifetime.

Mean time to failure. Integrate by parts with $u = t$ and $dv = f(t)\,dt$ , so $v = -R(t)$ :

\text{MTTF} = \int_0^\infty t f(t)\,dt = \Big[-tR(t)\Big]_0^\infty + \int_0^\infty R(t)\,dt = \int_0^\infty e^{-\lambda t}\,dt = \frac1\lambda.

Memoryless. For a part that has already survived to age $s$ ,

P(T > s + t \mid T > s) = \frac{P(T > s + t)}{P(T > s)} = \frac{e^{-\lambda(s + t)}}{e^{-\lambda s}} = e^{-\lambda t},

P(T > t) = P\big(N(t) = 0\big) = e^{-\lambda t}.

For a fleet of $m$ identical units the rates add, and failures arrive at $m\lambda$ . That's how spare parts are stocked: choose $k$ so that $P(N \le k)$ is high enough.

\hat\lambda = \frac{r}{T}, \qquad T = \text{total hours of all units, failed or not},

an optimization problem. Averaging only the failure times ignores the survivors and badly underestimates the MTTF.

See it

Worked examples

Example 1

An engine controller has a constant failure rate of $2 \times 10^{-5}$ per hour. Find its MTTF, its reliability at 10,000 hours, its median life and the age by which 10% of controllers have failed.

1
$\text{MTTF} = 1/\lambda = 1/(2 \times 10^{-5}) = 50{,}000$ hours.

Example 2

Twenty power modules run on a life test for 1,000 hours. Four fail, at 220, 450, 610 and 880 hours; the other 16 are still running when the test ends. Estimate the failure rate and the MTTF.

1
Total time on test: $T = 220 + 450 + 610 + 880 + 16(1000) = 18{,}160$ hours.

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

A pressure sensor has a constant failure rate of $4 \times 10^{-6}$ per hour. What is its MTTF in hours?

Problem 2+5 XP

A pump controller with an MTTF of 40,000 hours runs around the clock. What's the probability it survives its first year (8,760 hours)? Give 4 decimal places.

Problem 3+5 XP

A pressure transmitter has a constant failure rate and an MTTF of 10 years. It has already run 3 years without failing. What's the probability it lasts at least 2 more years? Give 4 decimal places.

Problem 4+5 XP

For a part with an exponential lifetime, what fraction of units fail before reaching the MTTF? Give 4 decimal places.

Problem 5+5 XP

Problem 6+5 XP

Problem 7+5 XP

A drive belt's failure rate grows with age: $h(t) = 0.02t$ per thousand hours, with $t$ in thousands of hours. The surviving fraction obeys $R'(t) = -h(t)R(t)$ with $R(0) = 1$ . Solve for $R(t)$ .

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Where it's used

Waiting for the Bus: Schedule vs RandomStats · Economics & Finance Counting Clicks on a Geiger CounterStats · Physics