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Reliability calculation models

Distinguish lifetime reliability, failure distributions, redundancy, availability and statistical evidence through twenty worked examples.

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Matter pathway: atom → solid → liquid → gas → plasma. Quantum mechanics and quantum field theory provide foundations across the pathway; they are not additional phases. This is a connected modeling route, not a universal heating curve. Actual phases depend on pressure, composition, and kinetics.

1. Lifetime and hazard

Definitions & inputs. T random lifetime,R(t) survival probability,F(t) failure CDF,f density,h hazard.

  1. Hazard is a conditional instantaneous failure rate, not a probability by itself.

    R(t)=P(T>t)=1−F(t),h(t)=f(t)/R(t)R(t)=P(T>t)=1-F(t),\quad h(t)=f(t)/R(t)

  2. Differentiate survival and substitute the hazard definition.

    dR/dt=−h(t)R(t)dR/dt=-h(t)R(t)

  3. Integrate with R(0)=1.

    R(t)=exp⁡[−∫0th(u)du]R(t)=\exp[-\int_0^t h(u)du]

Interpretation. Constant hazard gives exponential survival; wearout and infant mortality do not generally have constant hazard.

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2. Lifetime distributions

Definitions & inputs. λ constant hazard,η Weibull scale,β Weibull shape,MTTF expected lifetime.

  1. Integrate exponential survival over time.

    Rexp=e−λt,MTTF=1/λR_{exp}=e^{-\lambda t},\quad MTTF=1/\lambda

  2. Differentiate the Weibull cumulative hazard.

    RW=e−(t/η)β,hW=(β/η)(t/η)β−1R_W=e^{-(t/\eta)^\beta},\quad h_W=(\beta/\eta)(t/\eta)^{\beta-1}

  3. Mean lifetime averages over the distribution; it is not a guaranteed life.

    MTTF=∫0∞R(t)dtMTTF=\int_0^\infty R(t)dt

Interpretation. Weibull shape below, equal to, or above one represents decreasing, constant, or increasing hazard.

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3. System structure

Definitions & inputs. Ri component survival at the same mission time; n identical units with reliability R.

  1. All series components must survive.

    Rs=∏iRiR_s=\prod_iR_i

  2. Parallel success is the complement of all components failing.

    Rp=1−∏i(1−Ri)R_p=1-\prod_i(1-R_i)

  3. Count mutually exclusive cases with two or three surviving units.

    R2/3=3R2(1−R)+R3R_{2/3}=3R^2(1-R)+R^3

Interpretation. Common causes, load sharing, standby aging and imperfect switching invalidate simple independence formulas.

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4. Repair and evidence

Definitions & inputs. MTBF mean operating time between failures,MTTR mean repair time,Ttest total exposure,α one-sided tail probability.

  1. Long-run uptime fraction is mean up duration divided by the full cycle.

    A=MTBF/(MTBF+MTTR)A=MTBF/(MTBF+MTTR)

  2. Poisson counting under a constant failure rate gives the zero-event likelihood.

    P(0 failures)=e−λTtestP(0\text{ failures})=e^{-\lambda T_{test}}

  3. Solve the zero-event probability for a one-sided upper rate bound.

    λU=−ln⁡α/Ttest\lambda_U=-\ln\alpha/T_{test}

Interpretation. A test bound is model-dependent statistical evidence; zero observed failures does not establish perfect reliability.

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Graphical worked example

Constant failure hazard λ=10⁻⁵ per hour; nonrepairable mission survival. X axis: Mission time (hours). Y axis: Survival probability (dimensionless).
Constant failure hazard λ=10⁻⁵ per hour; nonrepairable mission survival. Related worked calculation · Download SVG · Plot data

Twenty worked examples

Open a problem to see its defined inputs, assumptions, equation, numerical substitution, result, and interpretation. Values are illustrative analytical exercises.

Example 01. Exponential mission reliability

Definitions & inputs. λ=10⁻⁵/h,t=1000 h.

  1. Choose the governing model and isolate the requested quantity.

    R=e−λtR=e^{-\lambda t}

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    R=e−0.01R=e^{-0.01}

  3. Evaluate the expression; the result uses the units shown.

    Result=0.9900498 \mathrm{Result}=0.9900498\ {}

Interpretation. Constant hazard is assumed over this mission interval.

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Example 02. Mission failure probability

Definitions & inputs. Same λ=10⁻⁵/h,t=1000 h.

  1. Choose the governing model and isolate the requested quantity.

    F=1−e−λtF=1-e^{-\lambda t}

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    F=1−e−0.01F=1-e^{-0.01}

  3. Evaluate the expression; the result uses the units shown.

    Result=0.009950166 \mathrm{Result}=0.009950166\ {}

Interpretation. Reliability and failure probability sum to one.

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Example 03. Mean exponential life

Definitions & inputs. λ=2×10⁻⁵/h.

  1. Choose the governing model and isolate the requested quantity.

    MTTF=1/λMTTF=1/\lambda

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    MTTF=1/(2×10−5)MTTF=1/(2\times10^{-5})

  3. Evaluate the expression; the result uses the units shown.

    Result=50000 h\mathrm{Result}=50000\ {\rm h}

Interpretation. This is a mean, not a lower guaranteed lifetime.

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Example 04. Median exponential life

Definitions & inputs. λ=2×10⁻⁵/h.

  1. Choose the governing model and isolate the requested quantity.

    t50=ln⁡2/λt_{50}=\ln2/\lambda

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    t50=ln⁡2/(2×10−5)t_{50}=\ln2/(2\times10^{-5})

  3. Evaluate the expression; the result uses the units shown.

    Result=34657.36 h\mathrm{Result}=34657.36\ {\rm h}

Interpretation. Half of the modeled population survives past the median.

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Example 05. Required hazard for target

Definitions & inputs. R≥.99 over 1000 h,exponential model.

  1. Choose the governing model and isolate the requested quantity.

    λmax=−ln⁡R/t\lambda_{max}=-\ln R/t

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    λmax=−ln⁡0.99/1000\lambda_{max}=-\ln0.99/1000

  3. Evaluate the expression; the result uses the units shown.

    Result=1.005034×10−5 h−1\mathrm{Result}=1.005034\times10^{-5}\ {\rm h}^{-1}

Interpretation. The requirement applies to the modeled system hazard.

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Example 06. Two-component series

Definitions & inputs. Independent R1=.99,R2=.98.

  1. Choose the governing model and isolate the requested quantity.

    Rs=R1R2R_s=R_1R_2

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    Rs=0.99(0.98)R_s=0.99(0.98)

  3. Evaluate the expression; the result uses the units shown.

    Result=0.9702 \mathrm{Result}=0.9702\ {}

Interpretation. Both components are necessary for success.

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Example 07. Four-component series

Definitions & inputs. Independent identical R=.99.

  1. Choose the governing model and isolate the requested quantity.

    Rs=R4R_s=R^4

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    Rs=0.994R_s=0.99^4

  3. Evaluate the expression; the result uses the units shown.

    Result=0.960596 \mathrm{Result}=0.960596\ {}

Interpretation. More required components reduce series reliability.

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Example 08. Parallel pair

Definitions & inputs. Independent active R1=R2=.9.

  1. Choose the governing model and isolate the requested quantity.

    Rp=1−(1−R)2R_p=1-(1-R)^2

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    Rp=1−0.12R_p=1-0.1^2

  3. Evaluate the expression; the result uses the units shown.

    Result=0.99 \mathrm{Result}=0.99\ {}

Interpretation. Either unit can satisfy the full mission requirement.

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Example 09. Three parallel units

Definitions & inputs. Independent R=.8.

  1. Choose the governing model and isolate the requested quantity.

    Rp=1−(1−R)3R_p=1-(1-R)^3

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    Rp=1−0.23R_p=1-0.2^3

  3. Evaluate the expression; the result uses the units shown.

    Result=0.992 \mathrm{Result}=0.992\ {}

Interpretation. Common power or software faults are excluded.

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Example 10. Two-out-of-three voting

Definitions & inputs. Independent R=.9,ideal voter.

  1. Choose the governing model and isolate the requested quantity.

    R=3r2(1−r)+r3R=3r^2(1-r)+r^3

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    R=3(0.9)2(0.1)+(0.9)3R=3(0.9)^2(0.1)+(0.9)^3

  3. Evaluate the expression; the result uses the units shown.

    Result=0.972 \mathrm{Result}=0.972\ {}

Interpretation. Voter failures must be added separately for a real architecture.

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Example 11. Series constant hazards

Definitions & inputs. λ1=10⁻⁵/h,λ2=2×10⁻⁵/h,independent.

  1. Choose the governing model and isolate the requested quantity.

    λs=λ1+λ2\lambda_s=\lambda_1+\lambda_2

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    λs=10−5+2×10−5\lambda_s=10^{-5}+2\times10^{-5}

  3. Evaluate the expression; the result uses the units shown.

    Result=3×10−5 h−1\mathrm{Result}=3\times10^{-5}\ {\rm h}^{-1}

Interpretation. Exponential survival factors combine by adding hazards.

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Example 12. Weibull survival

Definitions & inputs. η=1000 h,β=2,t=500 h.

  1. Choose the governing model and isolate the requested quantity.

    R=e−(t/η)βR=e^{-(t/\eta)^\beta}

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    R=e−(0.5)2R=e^{-(0.5)^2}

  3. Evaluate the expression; the result uses the units shown.

    Result=0.7788008 \mathrm{Result}=0.7788008\ {}

Interpretation. The increasing hazard models a wearout trend.

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Example 13. Weibull hazard

Definitions & inputs. Same η=1000 h,β=2,t=500 h.

  1. Choose the governing model and isolate the requested quantity.

    h=(β/η)(t/η)β−1h=(\beta/\eta)(t/\eta)^{\beta-1}

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    h=(2/1000)(0.5)h=(2/1000)(0.5)

  3. Evaluate the expression; the result uses the units shown.

    Result=0.001 h−1\mathrm{Result}=0.001\ {\rm h}^{-1}

Interpretation. This instantaneous rate varies with age.

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Example 14. Weibull B10 life

Definitions & inputs. η=1000 h,β=2; 10% cumulative failures.

  1. Choose the governing model and isolate the requested quantity.

    t10=η[−ln⁡0.9]1/βt_{10}=\eta[-\ln0.9]^{1/\beta}

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    t10=1000−ln⁡0.9t_{10}=1000\sqrt{-\ln0.9}

  3. Evaluate the expression; the result uses the units shown.

    Result=324.5928 h\mathrm{Result}=324.5928\ {\rm h}

Interpretation. Ninety percent of the modeled population survives beyond this time.

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Example 15. Steady availability

Definitions & inputs. MTBF=1000 h,MTTR=10 h.

  1. Choose the governing model and isolate the requested quantity.

    A=MTBF/(MTBF+MTTR)A=MTBF/(MTBF+MTTR)

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    A=1000/1010A=1000/1010

  3. Evaluate the expression; the result uses the units shown.

    Result=0.990099 \mathrm{Result}=0.990099\ {}

Interpretation. Availability includes repair, unlike nonrepairable mission reliability.

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Example 16. Expected annual downtime

Definitions & inputs. Availability .999,year=8760 h.

  1. Choose the governing model and isolate the requested quantity.

    D=(1−A)TD=(1-A)T

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    D=0.001(8760)D=0.001(8760)

  3. Evaluate the expression; the result uses the units shown.

    Result=8.76 h\mathrm{Result}=8.76\ {\rm h}

Interpretation. This is a long-run mean, not a bound on any particular year.

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Example 17. Repair-rate parameter

Definitions & inputs. Exponential repair with MTTR=4 h.

  1. Choose the governing model and isolate the requested quantity.

    μ=1/MTTR\mu=1/MTTR

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    μ=1/4\mu=1/4

  3. Evaluate the expression; the result uses the units shown.

    Result=0.25 h−1\mathrm{Result}=0.25\ {\rm h}^{-1}

Interpretation. Real repair distributions often are not exponential.

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Example 18. Zero-failure rate upper bound

Definitions & inputs. Total exposure 10000 h,no failures,one-sided 95% confidence.

  1. Choose the governing model and isolate the requested quantity.

    λU=−ln⁡0.05/T\lambda_U=-\ln0.05/T

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    λU=−ln⁡0.05/10000\lambda_U=-\ln0.05/10000

  3. Evaluate the expression; the result uses the units shown.

    Result=0.0002995732 h−1\mathrm{Result}=0.0002995732\ {\rm h}^{-1}

Interpretation. This bound relies on constant hazard and appropriate independent exposure.

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Example 19. Required zero-failure exposure

Definitions & inputs. Demonstrate λ≤10⁻⁴/h at one-sided 95% confidence.

  1. Choose the governing model and isolate the requested quantity.

    T=−ln⁡0.05/λtargetT=-\ln0.05/\lambda_{target}

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    T=−ln⁡0.05/10−4T=-\ln0.05/10^{-4}

  3. Evaluate the expression; the result uses the units shown.

    Result=29957.32 h\mathrm{Result}=29957.32\ {\rm h}

Interpretation. This is the required total exposure if no failures occur.

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Example 20. Binomial zero-failure survival bound

Definitions & inputs. n=100 independent units complete the same mission; all pass; one-sided 95%.

  1. Choose the governing model and isolate the requested quantity.

    RL=α1/nR_L=\alpha^{1/n}

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    RL=0.051/100R_L=0.05^{1/100}

  3. Evaluate the expression; the result uses the units shown.

    Result=0.970487 \mathrm{Result}=0.970487\ {}

Interpretation. The exact all-success binomial lower bound is below one despite zero observed failures.

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Symbols and units

Each derivation and problem defines its own symbols and inputs. Symbols may be reused with different meanings in other subjects. Keep units consistent, retain sufficient precision during calculation, and apply the stated validity limits.