Reliability — MTTF, MTBF, exponential failure model: Question bank

This is a thinking gym, not a calculator drill. Every item below is a question ::: answer reveal. Read the question, commit to an answer out loud with a reason, then reveal. If your reason was wrong even when your yes/no was right, you failed the trap — that's the point.

Prerequisites live in the parent topic. Related traps hide in Bathtub Curve, Series vs. Parallel System Reliability, Redundancy Design, and Availability vs. Reliability.

True or false — justify

Answer true/false, then give the reason. A right verdict with a wrong reason still counts as a miss.

A component with MTTF = 1000 h is guaranteed to run for about 1000 h before failing

False — MTTF is a population average; only

R (1/ λ) = e^{- 1} \approx 37%

of units survive that long, so 63% die before their own MTTF.

If failure rate

λ

is constant, then a component that has already survived 500 h is "worn in" and more likely to keep going than a fresh one

False — the exponential model is memoryless; a survivor's future looks statistically identical to a brand-new unit, because

R (t + s) / R (s) = e^{- λ t} = R (t)

Doubling the failure rate

λ

halves the reliability

R (t)

at a fixed time

t

False —

R = e^{- λ t}

is exponential in

λ

, not linear; doubling

λ

squares

R

, since

e^{- 2 λ t} = (e^{- λ t})^{2}

, which is a much harsher drop than halving.

For an exponential model, MTBF and MTTF have the same numerical value

1/ λ

True — both equal

1/ λ

when repair time is negligible; the labels differ only in intent (MTTF for non-repairable, MTBF for repairable), not in arithmetic.

Reliability

R (t)

can equal exactly 1 for some

t > 0

under the exponential model

False —

e^{- λ t} = 1

only at

t = 0

(or the trivial

λ = 0

, meaning "never fails"); for any real

λ > 0

and

t > 0

R (t) < 1

strictly.

Adding a redundant parallel unit lowers the system failure rate to a new constant value

False — a parallel system's failure rate is not constant; it starts near zero and rises over time, so the neat "

λ =

const" exponential form is lost even if each unit is exponential.

MTBF includes repair time, so MTBF is always larger than MTTF

True in the strict definition

MTBF = MTTF + MTTR

, but for spacecraft we usually set MTTR to zero (can't repair in orbit), collapsing MTBF back to MTTF.

A higher MTTF always means higher reliability for a given mission

True for the exponential model at fixed

t

— larger MTTF means smaller

λ = 1/ MTTF

, so

e^{- λ t}

is larger; the two move together monotonically.

Spot the error

Each statement contains one flawed step. Name it.

"Two parallel units, each

λ = 0.001

, so the system has

λ = 0.002

The error is adding

λ

for parallel redundancy — that rule is for series; parallel reliability is

R_{sy s} = 1 - (1 - e^{- λ t})^{2}

, which improves reliability, not worsens it.

R (t) = e^{- λ t}

where

λ t

is in hours, so at

t = 1000

h the exponent is 1000."

The exponent

λ t

must be dimensionless;

λ

carries units of 1/hours, so failures/hour × hours cancels — you forgot to multiply by

λ

"The solar array MTTF is 20000 h, and the mission is 43830 h, so reliability is

20000/43830 = 46%

Reliability is never a ratio of times; use

R = e^{- λ t} = e^{- t / MTTF} = e^{- 2.19} \approx 11%

, far below the naive 46%.

"Failure rate is negative because reliability decreases, so

λ < 0

λ = - \frac{1}{R} \frac{d R}{d t}

; the minus sign cancels the negative slope

d R / d t

, leaving

λ > 0

— a rate is always positive.

f (t) = λ e^{- λ t}

is the reliability, so at

t = 0

reliability equals

λ

f (t)

is the failure density (a PDF), not reliability; reliability is

R (0) = 1

, while

f (0) = λ

describes how fast probability mass leaves at the start.

"Since MTTF

= \int_{0}^{\infty} t f (t) d t

diverges (t goes to infinity), MTTF is infinite."

The integrand

t λ e^{- λ t}

decays fast enough that the exponential kills the linear growth; the integral converges to the finite value

1/ λ

"Parallel redundancy makes reliability 100%, since it can't fail if there's a backup."

Both units can still fail; parallel reliability is

1 - (1 - R_{1})^{2} < 1

— redundancy raises reliability but never reaches certainty.

Why questions

Explain the reason, not just the fact.

Why does the exponential model appear in the flat middle of the bathtub curve but not the ends?

Only the middle "useful life" region has a roughly constant

λ

; infant mortality (falling

λ

) and wear-out (rising

λ

) need the Weibull Distribution since a constant rate can't bend.

Why does

R (t) = e^{- λ t}

look identical to radioactive decay?

Both obey

d R / d t = - λ R

— a quantity whose loss rate is proportional to how much remains — and that single differential equation forces an exponential solution regardless of physical context.

Why do engineers design missions to run well below the MTTF rather than up to it?

t = MTTF

reliability is only

e^{- 1} \approx 37%

; to reach 90% survival you need

t \approx 0.1/ λ

, i.e. about one-tenth of the MTTF.

Why does taking a logarithm appear when solving for

λ

from a target reliability?

The unknown sits inside an exponential (

R = e^{- λ t}

); the natural log is the exact inverse operation that "unwraps"

e^{(\cdot)}

to expose

- λ t

Why is the failure rate defined as

- \frac{1}{R} \frac{d R}{d t}

and not simply

- \frac{d R}{d t}

Dividing by

R

normalizes against the surviving population — it asks "what fraction of those still alive fail per unit time," which is what an engineer replacing parts actually cares about.

Why does adding a second parallel unit help even when each unit is more likely to fail than survive?

System death requires both to fail, and multiplying two sub-one probabilities gives a smaller number:

(0.63)^{2} \approx 0.40 < 0.63

, so joint failure is rarer than single failure.

Why can't we describe a series system's overall reliability by its worst single component alone?

A series system fails if any link fails, so reliabilities multiply:

R_{sy s} = \prod R_{i}

, which is always

\leq

the smallest

R_{i}

— every weak part drags the whole chain down further. See Series vs. Parallel System Reliability.

Edge cases

Push the model to its limits and see what survives.

What is

R (t)

when

λ = 0

R (t) = e^{0} = 1

for all

t

— a hypothetical component that never fails; MTTF

= 1/ λ

becomes infinite, consistent with "immortal."

What happens to reliability as

t \to \infty

for any

λ > 0

R (t) \to 0

— given infinite time every real component eventually fails, so long-term survival probability drains to zero no matter how small

λ

is.

What is the reliability at exactly

t = MTTF

R (1/ λ) = e^{- 1} \approx 0.368

— a fixed universal number for any exponential system, independent of the specific

λ

If MTTR (repair time) is not negligible, does MTBF still equal

1/ λ

No — then

MTBF = MTTF + MTTR = 1/ λ + MTTR

, and the gap between uptime and cycle-time matters for Availability vs. Reliability.

For a parallel system, what is the reliability at

t = 0

R_{sy s} (0) = 1 - (1 - 1)^{n} = 1

— every unit starts working, so the system starts perfectly reliable, exactly like a single unit at launch.

What does MTTF mean for a truly non-repairable, single-use item like a pyrotechnic bolt?

It's the average time-to-failure over a batch; there's no "between failures" because the item fires once and is done, so MTBF is meaningless and only MTTF applies.

If two parallel units are not independent (a shared power bus can kill both), is

R_{sy s} = 1 - (1 - R_{1})^{2}

still valid?

No — that formula assumes independence; a common-cause failure correlates the two, so real

R_{sy s}

is lower, which is exactly what Fault Tree Analysis is built to catch.

Recall Fast self-check before you close this page

One-line answers; force yourself before revealing. Survivor fraction at $t =$ MTTF? ::: About 37% ( $e^{- 1}$ ). Series reliability rule? ::: Multiply: $R_{sy s} = \prod R_{i}$ . Parallel reliability rule? ::: $R_{sy s} = 1 - \prod (1 - R_{i})$ . Is exponential $λ$ constant or time-varying? ::: Constant — that's the whole assumption. What property makes "already survived" irrelevant? ::: Memorylessness.