Exercise set

Reliability Life Data, Weibull, and MTBF Bound Exercises

Worked reliability-data exercises for zero-failure tests, Weibull life, censored exposure, accelerated tests, Poisson bounds, MTBF and release gates.

Branch: Mathematical Engineering
Content: Exercise set
Updated: Jul 03, 2026
Revision: v1.0.0 · reviewed

These exercises practise reliability life-data analysis as release evidence. They cover zero-failure demonstrations, exponential mission reliability, Weibull reliability and B-life, censored exposure, accelerated tests, observed field failures, Poisson failure-rate bounds, demand-event reliability, warranty return rates, competing populations, confidence-bound margins, corrective-action evidence and release gates.

The focus is narrower than general statistical inference. Here the central question is whether exposure, failures, censoring and model assumptions support a reliability claim for a defined population, failure mode and operating environment.

How to Use These Exercises

For each calculation, define:

the item, function, failure definition and operating environment;
exposure basis: hours, cycles, starts, missions, demands or field units;
failure count, censoring rule and configuration identity;
model: exponential, Weibull, binomial, Poisson or accelerated-life screen;
confidence-bound decision, not only point estimate.

The common mistake is reporting MTBF as if it were a guarantee. Reliability evidence is only meaningful with exposure basis, confidence, failure definition and population comparability.

Release Evidence Notes

Failure definitions should be controlled before data are counted. Cosmetic defects, repairable interruptions, dangerous failures and user-induced events should not be mixed unless the release claim explicitly includes them.

Censored data should be visible. Units that survive to test end or leave service early still contribute exposure, but they do not contribute the same information as failures.

Confidence bounds should lead release decisions. A point MTBF can pass while the lower confidence bound fails badly when few failures or little exposure are available.

Accelerated tests need model justification. Acceleration factors are evidence only when the stress mechanism matches field failure physics.

Engineering Boundary Notes

These exercises use simplified reliability formulas. Real reliability demonstration can require life-data fitting, confidence intervals from exact methods, competing risks, repairable-system models, degradation analysis, Bayesian priors, environmental qualification, root-cause evidence and expert review.

A statistical reliability pass does not close a known failure mode. If failures share a root cause, corrective-action verification becomes the release gate.

Scenario Map

Scenario	Exercises	Primary calculation	Engineering decision
Demonstration and exposure	1-5, 9	zero-failure exposure, exponential reliability, demand reliability, censored exposure and acceleration	Decide whether the test earns enough confidence.
Life-data modelling	6-8, 10-13	Weibull reliability, B-life, field failures, warranty rate, mixed populations and confidence bounds	Decide whether the life claim is credible.
Release control	14-18	additional exposure, corrective action, configuration split, demand PFD and release gates	Decide whether reliability evidence can be released.

Exercise 1: Zero-Failure MTBF Demonstration

A test accumulates:

T=14{,}000\ \text{unit-h}

with zero failures. For a one-sided $90\%$ exponential lower MTBF bound, use:

MTBF_{90}\ge \dfrac{T}{2.303}

Solution

MTBF bound:

MTBF_{90}\ge \dfrac{14{,}000}{2.303}=6079\ \text{h}

Engineering Comment

Zero failures do not prove infinite reliability. Confidence is earned by exposure, and the bound depends on the assumed constant-rate model.

Plausibility Check

At $90\%$ confidence, zero-failure exposure must be about $2.3$ times the claimed MTBF.

Exercise 2: Required Zero-Failure Exposure

A product must demonstrate:

MTBF_{90}\ge 8000\ \text{h}

with zero failures under an exponential screen. Calculate required exposure.

Solution

Required exposure:

T=2.303(8000)=18{,}424\ \text{unit-h}

Engineering Comment

Parallel testing can earn exposure faster, but only if all units represent the same configuration and operating profile.

Plausibility Check

The required exposure is a little more than twice the target MTBF, as expected for a $90\%$ zero-failure demonstration.

Exercise 3: Exponential Mission Reliability

A component has claimed MTBF:

MTBF=12{,}000\ \text{h}

Mission duration is:

t=500\ \text{h}

Use:

R(t)=e^{-t/MTBF}

Solution

Mission reliability:

R=e^{-500/12000}=e^{-0.04167}=0.959

Engineering Comment

Mission reliability depends on mission time and model shape. Exponential MTBF assumes constant failure rate, which may be wrong for wear-out.

Plausibility Check

The mission is only about $4.2\%$ of MTBF, so reliability near $96\%$ is plausible.

Exercise 4: Demand-Event Reliability

A safety function is demanded:

n=240

times during validation with zero dangerous failures. A one-sided $90\%$ upper bound for probability of failure per demand is approximated by:

PFD_U=\dfrac{2.303}{n}

Calculate the bound.

Solution

Upper bound:

PFD_U=\dfrac{2.303}{240}=0.00960

Therefore:

PFD_U=0.96\%

Engineering Comment

Demand-based reliability is different from hour-based MTBF. The exposure unit must match how the function is used.

Plausibility Check

Several hundred successful demands can support a one-percent-scale upper bound, not a parts-per-million claim.

Exercise 5: Censored Field Data Exposure

Five field units have observed operating hours:

Unit	Hours	Result
A	$900$	failed
B	$1200$	censored
C	$700$	failed
D	$1500$	censored
E	$1100$	censored

Calculate total exposure, failures and point MTBF.

Solution

Total exposure:

T=900+1200+700+1500+1100=5400\ \text{h}

Failures:

N_f=2

Point MTBF:

MTBF=\dfrac{5400}{2}=2700\ \text{h}

Engineering Comment

Censored units contribute exposure but not failures. The censoring reason should be recorded so weak or removed units are not hidden.

Plausibility Check

Two failures over a little over five thousand hours should give a point MTBF of a few thousand hours.

Exercise 6: Weibull Reliability at Mission Time

A fitted Weibull model has shape:

\beta=2.1

and scale:

\eta=4200\ \text{h}

Calculate reliability at:

t=2000\ \text{h}

using:

R(t)=e^{-(t/\eta)^\beta}

Solution

Exponent:

\left(\dfrac{2000}{4200}\right)^{2.1}=0.211

Reliability:

R=e^{-0.211}=0.810

Engineering Comment

Because $\beta>1$ , failure rate increases with time. A constant-rate MTBF interpretation would hide wear-out.

Plausibility Check

The mission time is less than half the scale parameter, so reliability above $80\%$ is plausible.

Exercise 7: Weibull B10 Life

Use the Weibull model:

\beta=2.1,\qquad \eta=4200\ \text{h}

Calculate B10 life, where $R=0.90$ :

t_{10}=\eta[-\ln(0.90)]^{1/\beta}

Solution

Substitute:

t_{10}=4200[-\ln(0.90)]^{1/2.1}

t_{10}=4200(0.1053)^{0.476}=1438\ \text{h}

Engineering Comment

B10 is not a warranty guarantee. It depends on fit quality, confidence bounds and whether the data represent the field population.

Plausibility Check

B10 should be far below the scale parameter because only $10\%$ cumulative failure is allowed.

Exercise 8: Observed Field Failures and Poisson MTBF Bound

A fleet accumulates:

T=18{,}400\ \text{h}

and observes:

k=2

relevant failures. A one-sided $90\%$ upper failure-rate factor for two failures is:

6.296

Use:

\lambda_U=\dfrac{6.296}{T}

and calculate lower MTBF bound.

Solution

Point MTBF:

MTBF_{point}=\dfrac{18{,}400}{2}=9200\ \text{h}

Upper failure rate:

\lambda_U=\dfrac{6.296}{18{,}400}=3.422\times10^{-4}\ \text{h}^{-1}

Lower MTBF bound:

MTBF_L=\dfrac{1}{3.422\times10^{-4}}=2922\ \text{h}

Engineering Comment

The point estimate passes many targets, but the confidence bound is much lower because only two failures were observed.

Plausibility Check

The confidence factor is more than three times the observed failure count, so the lower bound should be much lower than point MTBF.

Exercise 9: Accelerated Reliability Exposure

Twenty units test for:

300\ \text{h}

each at an acceleration factor:

AF=4.5

with zero failures. Calculate equivalent field exposure.

Solution

Physical exposure:

T_p=20(300)=6000\ \text{unit-h}

Equivalent exposure:

T_e=6000(4.5)=27{,}000\ \text{unit-h}

Engineering Comment

Acceleration is valid only when the stress accelerates the same failure mechanism expected in the field. Otherwise equivalent exposure is misleading.

Plausibility Check

The acceleration factor multiplies the physical exposure, so $6000$ h becomes $27{,}000$ h.

Exercise 10: Accelerated MTBF Bound Shortfall

Use the equivalent exposure:

T_e=27{,}000\ \text{h}

with zero failures. The target is:

MTBF_{95}\ge 12{,}000\ \text{h}

For a zero-failure $95\%$ screen, use factor $2.996$ .

Solution

Demonstrated bound:

MTBF_{95}=\dfrac{27{,}000}{2.996}=9012\ \text{h}

Required exposure:

T_{req}=12{,}000(2.996)=35{,}952\ \text{h}

Shortfall:

\Delta T=35{,}952-27{,}000=8952\ \text{h}

The test does not meet the target.

Engineering Comment

Zero failures can still fail a reliability target when exposure is insufficient. The correct action is more exposure, a lower claim or stronger model evidence.

Plausibility Check

$27{,}000$ h is about three quarters of the required exposure, so the demonstrated MTBF bound is about three quarters of the target.

Exercise 11: Additional Test Time Needed

The accelerated test in Exercise 10 needs:

\Delta T=8952\ \text{equivalent h}

more exposure. With $20$ units and $AF=4.5$ , calculate extra physical hours per unit.

Solution

Equivalent exposure per physical hour across all units:

20(4.5)=90\ \text{equivalent h/h}

Extra hours:

t=\dfrac{8952}{90}=99.5\ \text{h}

Round up:

100\ \text{h per unit}

Engineering Comment

This assumes no failures occur during the added exposure. A failure would change the model and the release decision.

Plausibility Check

One hundred extra hours on twenty units at $4.5$ acceleration adds $9000$ equivalent hours.

Exercise 12: Warranty Return Rate Screen

A field population has:

N=3200

units in service for a one-year warranty window. There are:

26

confirmed reliability returns. Calculate return rate.

Solution

Return rate:

p=\dfrac{26}{3200}=0.008125

Therefore:

p=0.813\%

Engineering Comment

Warranty returns mix exposure, usage, detection, logistics and classification. They are useful but weaker than controlled life data unless failure modes are coded.

Plausibility Check

About $32$ returns would be $1\%$ , so $26$ returns is a little below $1\%$ .

Exercise 13: Mixed Population Failure-Rate Trap

Two configurations are combined:

Configuration	Exposure	Failures
A	$12{,}000$ h	$1$
B	$4{,}000$ h	$3$

Calculate point failure rates separately and combined.

Solution

Configuration A:

\lambda_A=\dfrac{1}{12{,}000}=8.33\times10^{-5}\ \text{h}^{-1}

Configuration B:

\lambda_B=\dfrac{3}{4{,}000}=7.50\times10^{-4}\ \text{h}^{-1}

Combined:

\lambda_C=\dfrac{4}{16{,}000}=2.50\times10^{-4}\ \text{h}^{-1}

Engineering Comment

Combining configurations hides that B has much higher failure rate. Reliability release should preserve configuration identity.

Plausibility Check

The combined rate lies between A and B, but it does not represent either configuration well.

Exercise 14: Corrective-Action Exposure After Fix

A failure mode is corrected. The post-fix validation plan runs:

12

units for:

600\ \text{h}

with zero recurrence. Calculate post-fix exposure and $90\%$ zero-failure MTBF bound.

Solution

Exposure:

T=12(600)=7200\ \text{unit-h}

MTBF bound:

MTBF_{90}=\dfrac{7200}{2.303}=3126\ \text{h}

Engineering Comment

Corrective-action validation should target the failed mechanism, not only accumulate generic operating hours.

Plausibility Check

Exposure just over $7000$ h supports a $90\%$ bound just over $3000$ h.

Exercise 15: Repairable-System Event Rate

A repairable fleet operates for:

T=44{,}000\ \text{h}

and records:

N=11

relevant service interruptions. Calculate event rate and mean time between events.

Solution

Event rate:

\lambda=\dfrac{11}{44{,}000}=2.50\times10^{-4}\ \text{h}^{-1}

Mean time between events:

MTBE=\dfrac{44{,}000}{11}=4000\ \text{h}

Engineering Comment

Repairable systems may need trend, repair quality and recurrence analysis. Treating events as independent can be weak if the same unit repeatedly fails.

Plausibility Check

Eleven events over $44{,}000$ hours gives exactly one event per $4000$ hours.

Exercise 16: Lower MTBF Bound Release Margin

A lower confidence bound is:

MTBF_L=5200\ \text{h}

The release target is:

5000\ \text{h}

Calculate margin and percentage margin.

Solution

Margin:

M=5200-5000=200\ \text{h}

Percentage margin:

M_{\%}=\dfrac{200}{5000}=4.0\%

Engineering Comment

A small reliability margin should trigger a review of exposure accounting, failure classification and configuration coverage before final release.

Plausibility Check

The bound barely exceeds the target, so the percentage margin should be small.

Exercise 17: Failure-Mode-Specific Release Split

A test records $5$ failures over:

30{,}000\ \text{h}

Three failures are cosmetic and two are functional. The release claim is for functional reliability. Calculate functional point MTBF.

Solution

Functional failure count:

N_f=2

Functional point MTBF:

MTBF=\dfrac{30{,}000}{2}=15{,}000\ \text{h}

If all failures were incorrectly mixed:

MTBF_{mixed}=\dfrac{30{,}000}{5}=6000\ \text{h}

Engineering Comment

Failure classification must match the claim. Excluding failures is acceptable only when the exclusion rule is pre-defined and technically justified.

Plausibility Check

Counting fewer relevant failures increases MTBF, but the classification decision must be defensible.

Exercise 18: Reliability Release Gate

A reliability release package has five gates:

Gate	Weight	Result
exposure accounting	$0.25$	$0.96$
failure classification	$0.20$	$0.91$
confidence-bound margin	$0.25$	$0.87$
model and acceleration validity	$0.20$	$0.93$
configuration traceability	$0.10$	$0.95$

The weighted release threshold is:

S\ge 0.92

and confidence-bound margin may not be below $0.90$ . Calculate the decision.

Solution

Weighted score:

\begin{aligned} S&=0.25(0.96)+0.20(0.91)+0.25(0.87)+0.20(0.93)+0.10(0.95)\\ &=0.240+0.182+0.2175+0.186+0.095\\ &=0.9205 \end{aligned}

The weighted score is:

92.05\%

The score passes, but confidence-bound margin fails:

0.87<0.90

Release is held.

Engineering Comment

Reliability claims should not be released from a weighted average when the confidence-bound gate fails. The bound is the claim.

Plausibility Check

The total score barely passes, but the mandatory confidence floor fails, so the hold decision follows the rule.

Validation Package Checklist

Failure definition, exposure basis and configuration identity are frozen before counting data.
Censored units, removed units and partial exposures are recorded explicitly.
Point MTBF is separated from lower confidence bound.
Weibull claims include shape, scale, fit quality, censoring treatment and confidence notes.
Accelerated exposure is justified by a failure-physics model.
Field failures are classified by root cause and recurrence relevance.
Release decisions state what additional exposure or corrective action is required when the bound fails.

Common Release Mistakes

Treating zero failures as proof of no failures.
Reporting point MTBF without confidence, exposure and failure definition.
Mixing configurations or duty cycles into one reliability number.
Using accelerated-life factors without proving the same failure mechanism.
Ignoring censored observations or removed units.
Excluding failures after seeing the data rather than from a controlled classification rule.
Averaging release gates when the confidence-bound gate fails.

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