U12-Time-Dependent-_22646

U12-Time-Dependent-_22646 - Time Dependent Failure Models...

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Unformatted text preview: Time Dependent Failure Models Unit 12 Risk Analysis in Safety Engineering Fall 2015 1 References •  Ebeling, C.E., Introduction to Reliability and Maintainability Engineering, 2nd ed, Waveland Press, 2010, Chapter 4 (Ebeling, 2010) •  Modarres, M., M. Kaminskiy, V. Krivtsov, Reliability Engineering and Risk Analysis, 2nd ed, Taylor & Francis, 2010 (Modarres, RERA) •  Jordaan, Ian, Decisions Under Uncertainty– Probabilistic Analysis for Engineering Decisions, Cambridge University Press, 2005 (Jordaan, 2005) •  Modarres, M., Risk Analysis in Engineering, Taylor&Francis, 2006 (Modarres, RAE) •  Modarres, M., Reliability Engineering and Risk Analysis in Engineering, Marcel Dekker, 1999 (Modarres, RE) •  O’Connor, P.D.T., Practical Reliability Engineering, 4th ed, Wiley, 2002 (O’Connor, PRE) •  Rausand, M. and Hoyland, A., System Reliability Theory, 2nd edition, Wiley, 2004 2 Time Dependent Failure Mode In general, λ (ROCOF) values in an engineering system are not constant, so a @me dependent failure rate model is needed that is [email protected] and can be approximated as @me independent under some [email protected] and periods of @me. λ (t) = at b , with a > 0 where a [email protected] rate [email protected] λ(t) is a power [email protected] with 2 parameters atb, which is linear for b = 1, and λ is constant ([email protected]) for b = 0. Rewrite the 2-­‐parameter atb to simplify the corresponding reliability [email protected]: where β ⎛ t⎞ λ (t) = ⎜ ⎟ θ ⎝θ ⎠ β -1 ; θ > 0, β > 0 ; t ≥ 0 β is the shape parameter. θ, the scale parameter, is called the characteris)c life, which = 1/λ for β = 1 . 3 Reliability Function Using this form for the hazard rate [email protected] λ(t), the Weibull [email protected] is derived from the basic expression for R(t). Recall! t R(t) = e ∫0 − λ (t) dt β ⎛ t⎞ 0θ ⎜θ ⎟ ⎝ ⎠ =e ∫ − t β −1 dt = e ⎛ t⎞ −⎜ ⎟ ⎝θ ⎠ β Simple R(t), which reduces to [email protected] for β = 1, λ = 1/θ Note for t = θ, [email protected] life: R(t = θ ) = e ⎛θ ⎞ –⎜ ⎟ ⎝θ ⎠ β –1 = e = 0.37 So the scale parameter of the Weibull is the @me t = θ at which 1–0.37 = 63 % of the failures have already occurred for all values of the shape parameter, β. For β = 1, 1/λ = MTTF, − MTTF MTTF We discussed this result for [email protected] in Unit 8. e = 0.37 4 Graph of the Reliability Function, R(t) R(t) 1.2 Beta Shape Parameter values θ = 2 R(t) 0.5 1 R(t) = e 0.8 ⎛ t⎞ −⎜ ⎟ ⎝θ ⎠ β 1.5 2.0 4.0 Same [email protected] life for all β 0.6 ⎛θ ⎞ –⎜ ⎟ ⎝θ ⎠ β θ = t, e = e-­‐1 = 0.37 0.4 [email protected] for β = 1 0.2 0 0.0 -0.2 1.0 2.0 3.0 4.0 5.0 6.0 Time, t t The effect of the shape parameter β [email protected] in a wide range of shapes for the Weibull reliability [email protected] Recall: [email protected] [email protected] has no shape parameter and only one shape. 5 Weibull, F(t): Effect of β F(t) θ = 2 θ = t, 1-­‐ e-­‐1 = 0.63 At t = θ, Reliability has dropped by a factor of 1/e = 2.72 t Both shape and scale parameters are required for failure distribution characterization, which is better understood from the parameter values based on observation and measurement of a 6 component during system monitoring or in field tests. Weibull Probability Density Function: Effect of Shape Parameter β on f(t) = λ(t)·∙R(t) f(t) 1.0 β < 1, similar to [email protected]: mode at t = 0 β = 1, [email protected], λ = 1/θ 1 < β < 3: skewed 3 < β, closer to symmetric 0.9 0.8 0.7 0.6 0.5 β 0.5 1.5 2.0 4.0 θ = 2 4.0 0.5 0.4 2.0 0.3 1.5 0.2 0.1 t 0.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 -0.1 dR(t) β ⎛ t⎞ − = f(t) = ⎜ ⎟ dt θ ⎝θ ⎠ λ(t) β -1 e ⎛ t⎞ –⎜ ⎟ ⎝θ ⎠ β R(t) = λ (t)⋅R(t) λ (t) = f(t) R(t) 7 Conditional Failure (Hazard) Rate Function, λ(t) Of special importance is [email protected] of the wide ranges of increasing failure rate behavior: [email protected] slope, decreasing [email protected] slope (concave), constant [email protected] slope (Rayleigh), or increasing [email protected] slope (convex). 6 λ(t) 5 θ = 2 β, shape parameter f(t) / R(t) 0.5 1.5 2.0 Hazard Rate 4 4.0 constant LFR, Rayleigh [email protected], β = 2 [email protected] slope Convex, β > 2 3 increasing [email protected] slope 2 Concave, 1 < β < 2 1 decreasing [email protected] slope [email protected] slope, β < 1 t 0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 IFR for β > 1; DFR for β < 1; [email protected], CFR, for β = 1 8 Weibull Shape Parameter Effect on λ(t) •  Value Property •  0 < β < 1 Decreasing Failure Rate (DFR) •  β = 1 [email protected] [email protected] (CFR), λ(t) = λ •  1 < β < 2 IFR-­‐Concave failure rate (decreasing + slope) •  β = 2 Rayleigh [email protected] (IFR, linear failure rate) •  β > 2 IFR – Convex failure rate (increasing + slope) •  3 < β < 4 IFR – close to Normal [email protected] -­‐ symmetrical β = 3.4393: Most closely approximates the normal [email protected] The shape of the [email protected] represents the failure process of the component or system as it develops through @me, so it is useful for failure behavior [email protected] 9 Weibull Characteristic Life: Effect of θ on f(t) 1.6 f(t) Basic Shape is set by β, which is unaffected by θ f(t) β = 1.5 1.4 θ 1.2 0.5 0.5 1 1.0 2.0 0.8 1.0 0.6 Variance increases as θ is increased 0.4 2.0 0.2 t 0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 t The [email protected] life = θ (scale parameter) value changes the mean, medium, mode, and variance or spread of the 10 [email protected] but not its basic shape, which is set by its β. Weibull Characteristic Life: Effect of θ on R(t) R(t) β ~ 2 R(t) broadens and reliability decreases more slowly as θ is increased but it has same basic shape set by β. 11 Weibull Characteristic Life: Effect of θ on λ(t) λ(t) β = 2, Rayleigh [email protected], linear IFR The [email protected] Life parameter influences the mean and the dispersion or spread of the Weibull [email protected] As θ is increased, the slope of λ(t) decreases and the [email protected] 12 broadens, but the linear shape set by β remains the same. Weibull MTTF from θ, β, and the Gamma Function MTTF = ∫ ∞ 0 ⎛ 1⎞ t f(t)dt = θ Γ ⎜ 1+ ⎟ ⎝ β⎠ Derived in IRME, App. 4A, p. 87 ∞ Γ(x) = the gamma function = ∫ 0 y x-1e-ydy IRME, Tab A.9, p. 531 Γ(x) = (x - 1)Γ(x - 1) for x >0 Γ(x) = (x - 1)! for x a positive integer; Γ(1) = (0)! = 1 1 lim MTTF = lim θ Γ(1+ ) = θΓ(1) = θ β→∞ β→∞ β As β increases, the mean approaches the [email protected] life, because the Weibull f(t) approaches symmetrical. 13 Gamma Function - Selected Values IRME, A.9. p. 531 x 1.53 1.54 1.55 1.56 1.57 1.58 1.59 1.6 1.61 1.62 1.63 1.64 1.67 1.68 1.69 Gamma(x) 0.88757 0.88818 0.88887 0.88964 0.89049 0.89142 0.89243 0.89352 0.89468 0.89592 0.89724 0.89864 0.9033 0.905 0.90678 x 2.23 2.24 2.25 2.26 2.27 2.28 2.29 2.3 2.31 2.32 2.33 2.34 2.35 Gamma(x) 1.12023 1.12657 1.133 1.13954 1.14618 1.15292 1.15976 1.16671 1.17377 1.18093 1.18819 1.19557 1.20305 14 1.2 R(t) 0.5 1 1.5 2.0 0.8 Example Problems 4.0 0.6 0.4 0.2 0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 t -0.2 Let T = a random variable, the time to failure of a circuit card. T has a Weibull distribution with β = 0.5 and θ = 500 (thousands of hours). Find R(50,000) and the MTTF. 0.5 R(50) = e ⎛ 50 ⎞ −⎜ ⎟ ⎝ 500 ⎠ = 0.729 ( ) MTTF = 500Γ 1+ 2 = 500(2) = 1000 Γ(x) = (x-­‐1)! = 2! = 2 Note DFR because β < 1 Let T = a random variable, the time to failure of a fuse. T has a Weibull distribution with beta = 1.5 and theta = 500 (K hours). Find R(50,000) and the MTTF. ⎛ 50 ⎞ R(50) = e− ⎜ ⎝ 500 ⎟ ⎠ 1.5 = 0.969 ⎛ 2⎞ MTTF = 500Γ ⎜ 1+ ⎟ = 500(0.903) = 451 ⎝ 3⎠ Note IFR because β > 1 15 The Variance and Standard Deviation 2 ⎧ ⎛ 2⎞ ⎡ ⎛ 1⎞ ⎤ 2 2 ⎪ σ = θ ⎨ Γ ⎜ 1+ ⎟ - ⎢ Γ ⎜ 1+ ⎟ ⎥ ⎝ β⎠ ⎣ ⎝ β⎠⎦ ⎪ ⎩ Γ(x) = (x − 1)! ⎫ ⎪ ⎬ ⎪ ⎭ for x = positive integer Variance and σ decrease as β increases, because Weibull f(t) becomes symmetrical as β increases to within (3,4). Derive the variance and σ for [email protected] with β = 1: σ2 = θ2(2 -­‐ 1) = θ2 = (1/λ)2 as is known for [email protected] 16 Example Problem - Standard Deviation θ = 500 θ = 500 β = 1/2 2.33 β = 1.5 σ 2 = 5002 [Γ(5) − 22 ] σ 2 = 5002 [Γ(1+ 4 / 3) − 0.90332 ] = 500 [24 − 4] = 5,000,000 σ = 2236 (thousands of hr.) = 5002 [1.18819 0.8160] = 93,048 σ = 305 (thousands of hr.) 2 ⎧ ⎛ 2 ⎞ ⎡ ⎛ 1 ⎞ ⎤2 ⎪ 2 = θ 2 ⎨ Γ ⎜ 1+ ⎟ - ⎢ Γ ⎜ 1+ ⎟ ⎥ σ ⎪ ⎝ β⎠ ⎣ ⎝ β⎠⎦ ⎩ ⎫ ⎪ ⎬ ⎪ ⎭ 17 Design Life and Median Set then and R(t) = e t -⎛ θ ⎞ ⎜ ⎟ ⎝ ⎠ β =R tR = θ(- ln R) a targeted value 1 β t 0.50 = t med = θ(-ln0.5) 1 β Note that when β = 1, the expressions for [email protected] appear, showing again that Weibull is an extension of [email protected]! 18 Weibull Mode f(t mode ) = MAX t≥0 1 ⎧ ⎛ 1⎞ β ⎪ θ 1⎪ t mod e = ⎨ ⎜ β⎟ ⎝ ⎠ ⎪ ⎪0 ⎩ f(t) for β > 1 for β ≤ 1 Derived in Appendix 4B [email protected] Recall tmode = 0 for [email protected], for β = 1 19 Example - Design Life, Median, & Mode θ= 500 1/β β = 0.5 t0.9 = 500 (-ln 0.90)2 = 5.55 tmed = 500 (0.69315)2 = 240 tmode = 0, (β<1) θ= 500 β = 1.5 = 3/2 t0.9 = 500 (-ln 0.90)2/3 = 112 tmed = 500 (0.69315)2/3 = 392 tmode = 500 [1-2/3)]2/3 = 240 Sameθ, but notice the major difference due to different β 20 Weibull Conditional Reliability ⎛ t+T ⎞ −⎜ 0 ⎟ ⎝ θ ⎠ R(T0 + t) e R(t | T0 ) = = β ⎛ T0 ⎞ R(T0 ) −⎜ ⎟ e ⎝θ⎠ β ⎡ ⎢ ⎛ t+T0 ⎞ = exp ⎢ − ⎜ θ ⎟ ⎠ ⎢⎝ ⎣ β ⎛ T0 ⎞ + ⎜ ⎟ ⎝θ⎠ β ⎤ ⎥ ⎥ ⎥ ⎦ Note that when β = 1, the [email protected] R(t) = exp(-­‐λt) appears due to lack of memory of the [email protected]! Find R(50|50) for Weibull with β = 0.5, θ = 500 21 Example - Conditional Reliability 1.2 R(t) 0.8 Different failure behavior shown by change in shape, β 0.5 1.5 1 1.5 2.0 4.0 0.6 0.5 0.4 0.2 θ= 500 β = 0.5; R(50) = 0.7289 θ= 500 β = 1.5; R(50) = 0.969 R(50|50) = R(100)/R(50) = R(50|50) = R(100)/R(50) = exp[-(100/500)0.5] / 0.7289 exp[-(100/500)1.5] / 0.969 = 0.6394 /0.7289 = 0.8772 0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 t -0.2 = 0.9144 / 0.969 = 0.9437 Sameθ, but notice the significant shift due to different β 22 Failure Modes Recall that a system composed of several independent failure modes in series each with a Weibull failure [email protected] has a system [email protected] failure rate [email protected] equal to the sum of the [email protected] failure mode failure rate [email protected] If all failure modes have the [email protected] shape parameter, then the system also has a Weibull [email protected] β⎛ t⎞ λS (t) = ∑ ⎜ ⎟ i=1 θ i ⎝ θ i ⎠ n β -1 ⎡ n ⎛ 1⎞β⎤ β β -1 = β t β -1 ⎢ ∑ ⎜ ⎟ ⎥ = t ⎢ i=1 ⎝ θ i ⎠ ⎥ θ S ⎣ ⎦ which is a Weibull [email protected] failure rate [email protected] with a shape parameter β and a [email protected] life for the 1 − system, θS ⎡ n ⎛ 1⎞β ⎤ β System θS t = θS = ⎢ ∑ ⎜ ⎟ ⎥ ⎢ i=1 ⎝ θ i ⎠ ⎥ 23 ⎣ ⎦ MTTF and tmed for Jet Engine A jet engine is constructed from 5 modules each [email protected] Weibull [email protected] failure behavior with β = 1.5 (IFR, concave). Is the engine system also Weibull. Why or why not? The [email protected] life values ([email protected] cycles) of 3,600, 7,200, 5,850, 4,780, and 9,300. Calculate the, θS , MTTFS, median TTF, and the reliability of the engine. ⎡ n ⎛ 1⎞ θS = ⎢ ∑ ⎜ ⎟ ⎢ i=1 ⎝ θ i ⎠ ⎣ β ⎤ ⎥ ⎥ ⎦ − 1 β ⎡⎛ 1 ⎞ 1.5 ⎛ 1 ⎞ 1.5 ⎛ 1 ⎞ 1.5 ⎛ 1 ⎞ 1.5 ⎛ 1 ⎞ 1.5 ⎤ ⎥ θS = ⎢ ⎜ +⎜ +⎜ +⎜ +⎜ 3,600 ⎟ 7,200 ⎟ 5,850 ⎟ 4,780 ⎟ 9,300 ⎟ ⎥ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎢⎝ ⎣ ⎦ ⎛ ⎛ 2⎞ 1⎞ MTTFS = θ S Γ ⎜ 1+ ⎟ = 1,842.7Γ ⎜ 1+ ⎟ = 1,664.5 cycles ⎝ β⎠ ⎝ 3⎠ − 1 / 1.5 = 1842.7 Note effect of skew to higher t values. 1 β 1/1.5 = 1,443.2 cycles t 0.50 = t med = θ(-ln0.5) = 1,842.7(0.69315) t R(t) = e ∫0 − λ (t) dt ⎛ t⎞ 0θ ⎜θ ⎟ ⎝ ⎠ =e ∫ − tβ β −1 dt = e ⎛ t⎞ −⎜ ⎟ ⎝θ ⎠ β =e ⎛ t ⎞ −⎜ ⎝ 1,842.7 ⎟ ⎠ 1.5 24 Identical Weibull Components If n independent components in series have iden8cal [email protected] failure rate [email protected] with the same scale and β -1 shape parameters: β⎛ t⎞ λ (t) = ⎜ ⎟ θ ⎝θ ⎠ then β⎛ t⎞ λS (t) = ∑ ⎜ ⎟ θi ⎝ θi ⎠ i=1 n and ⎡ θS = ⎢ ⎢ ⎣ t β -1 = , nβ θ β (t ) β -1 ∫0 λ (t) dt = −n⎛ θt ⎞ ⎜ ⎟ RS (t) = e e ⎝ ⎠ − ⎛ 1⎞ ⎤ ∑ ⎜ θi⎟ ⎥ ⎠ ⎥ i=1 ⎝ ⎦ n β − 1 β θ = 1/ β n β ⎛ 1⎞ MTTFS = θ S Γ ⎜ 1+ ⎟ ⎝ β⎠ 25 Student Exercise - Weibull •  A certain brand lightning arrester has a Weibull failure distribution with a shape parameter of 2.4 and a characteristic life of 10 years. Find: •  a. R(5 yrs) •  b. MTTF •  c. Standard deviation •  d. Median and Mode •  e. 99% (B1) and 95% B(5) design life •  f. R(5|5) 26 Student Exercise - Solution a. R(5) = e ⎛ 5⎞ −⎜ ⎟ ⎝ 10 ⎠ 2.4 = 0.827 b. MTTF = 10 Γ(1+ 1/ 2.4) = 10 Γ(1.42) 10(0.88636) = 8.86 yrs { c. σ 2 = 10 2 Γ(1+ 2 / 2.4) − 0.88636 2 } or σ = 3.93 yr. 27 Student Exercise - Solution (continued) ( ) = 10 (1− 1/ 2.4 ) d. t med = 10 0.69315 t mode t 0.95 = 8.6 yr. 1/2.4 ( ) = 10 ( − ln.95 ) e. t.0.99 = 10 − ln.99 1/2.4 = 8.0 yr. 1/2.4 = 1.5 yr. 1/2.4 = 2.9 yr. 28 Student Exercise - Solution (continued) R(t + T0 ) f. Recall: R(t | T0 ) = R(T0 ) ⎛ 10 ⎞ −⎜ ⎟ ⎝ 10 ⎠ 2.4 R(10) e 0.3679 R(5 | 5) = = = = 0.445 R(5) 0.8274 0.8274 29 Time-­‐Dependent 2-­‐Parameter Reliability Models •  Weibull [email protected]: A [email protected] reliability model with a shape parameter that represents the progression of a [email protected] or failure process. Defined for [email protected] values of t. •  Normal (Gaussian) [email protected] for failures that are primarily the result of small (not dominant), [email protected] effects. Defined for [email protected] and [email protected] values of t. Represents only the IFR region where the [email protected] failure rate increases with @me. •  Lognormal [email protected]: Defined for [email protected] values of t and especially suitable when failures are the result of [email protected]@ve effects. Similar to the Weibull in [email protected] Because of a skewed tail to larger t, ideal for modeling repair frequency and other skewed [email protected] [email protected] •  Gamma Dist[email protected]: A [email protected] reliability model similar to Weibull, including defined for [email protected] values of t. In the same family of [email protected] as [email protected] and Poisson, so gamma is useful also as a prior [email protected] for Bayesian parameter [email protected] with [email protected] or Poisson as the likelihood [email protected] 30 The Normal Probability Density Function(PDF) f(t) = 1 2π σ e 1 (t-µ ) − 2 σ2 2 ,- ∞ < t < ∞ MTTF = µ, location parameter Std Dev = σ, scale or spread parameter There is no shape parameter, because the Normal pdf always has a symmetric shape but with a wide range of scale parameter, σ. 31 Normal Conditional Failure Rate, Effect of σ σ = 0.5 λ(t) = f(t) / R(t) IFR σ = 1 The Normal [email protected] failure rate [email protected] is always increasing, so it represents only the IFR region of component life and cannot be used for modeling defects that are gradually removed in the DFR region. 32 Normal Distribution - Applications •  Useful for random stresses over time, such as the additive effects of temperature variation, material wear, and friction leading to increasing conditional failure rate, λ(t), along with Weibull β>1 (concave, linear, or convex behavior cases) •  Tool failures •  Brake lining wear •  Tire tread wear 33 Finding Normal Cumulative Probabilities If T is normally distributed, transform T to Z, which is the standard normal variable T−µ z= σ Then Z has a standard normal [email protected] with a mean of 0 and a standard [email protected] of 1. The pdf for Z is given by φ(z) = 1 2π 2 e −z 2 The cdf for failure = F(t), [email protected] probability of failure to @me = t, is then given by z -∞ P{Z ≤ z} = Φ(z) = ∫ φ(z) dz 34 Standard Normal Probability Tables •  •  •  •  •  •  •  •  •  •  •  •  •  Z -0.55000 -0.54000 -0.53000 -0.52000 -0.51000 -0.50000 -0.49000 -0.48000 -0.47000 -0.46000 -0.45000 -0.44000 F(Z) 1-F(Z) 0.29116 0.29460 0.29806 0.30153 0.30503 0.30854 0.31207 0.31561 0.31918 0.32276 0.32636 0.32997 0.70884 0.70540 0.70194 0.69847 0.69497 0.69146 0.68793 0.68439 0.68082 0.67724 0.67364 0.67003 P{Z < -­‐ 0.5} = 0.30854 Area under pdf curve up to Z = -­‐ 0.5 P{Z > -­‐ 0.46 = 0.67724 Area under pdf curve > Z = -­‐ 0.46 35 Normal Reliability Function R(t) = ∫ ∞ t 1 2π σ e − (t−µ)2 2σ 2 dt Standardize to Z: Pr of fail for T ≥ t Z T Z t ⎧T − µ t − µ ⎫ R(t) = P{T ≥ t} = P ⎨ ≥ ⎬ = P(Z T ≥ Z t ) σ ⎭ ⎩ σ ⎧ ⎛ t − µ⎞ t−µ⎫ = P ⎨Z T ≥ ⎬ = 1− Φ ⎜ ⎟ = 1− Φ Z t σ ⎭ ⎝ σ ⎠ ⎩ ( ) Tables: A.1, pp. 514-­‐519 36 Example Problem - Normal •  The time to failure of a fan belt is normally distributed with a MTTF = 220 (in hundreds of vehicle miles) and a standard deviation of 40 (in hundreds of vehicle miles). t μ σ •  R(100) = 1 - F[ (100-220)/40] = 1 - F(-3) = 0.9987 •  R(200) = 1 - F[ (200-220)/40] = 1 - F(-0.5) = 0.6915 •  R(300) = 1 - F[ (300-220)/40] = 1 - F(2) = 0.02275 R(t|T0) •  R(100|200) = R(300) / R(200) = 0.02275 / 0.6915 = 0.0329 •  Note: the median = mode = MTTF = 22,000 miles 37 Normal Example problem - Design Life •  A new fan belt is developed from a higher grade of material. The belt has a time to failure distribution that is normal with a mean of 35,000 vehicle miles and a standard deviation of 7,000 vehicle miles. Find its design life if a 0.97 reliability is desired. •  •  •  •  R(t) = 1 – F[(t - 350)/70] = 0.97; find t ! From the normal CDF table, 1 - F(-1.88) = 0.970 Therefore; (t - 350 ) / 70 = -1.88 = Zt and t0.97 = 350 - 1.88 (70) = 218.4 or 21,840 vehicle miles Design life 38 Student Exercise - Normal •  The operating hours until failure of a halogen headlamp is normally distributed with a mean of 1200 hr. and a standard deviation of 450 hr. •  Find: •  a. The 5 year reliability if normal driving results in the use of the headlamp an average of 0.2 hr a day. •  b. The 0.90 design life in years. 39 Student Exercise - Solution •  a. t = 0.2 hr./da. x 365 day/yr x 5 yr = 365 hr. •  R(365) = 1 – F[(365 - 1200)/450] = 1 – F[-1.86] = 0.969 •  b. R(t0.90) = 0.90 or 1 – F[(t0.90 - 1200)/450] = 0.90 •  (t0.90 - 1200) / 450 = -1.28 •  t0.90 = 1200 - 1.28 (450) = 624 hr. or t0.90 = 624 / (0.2 x 365) = 8.5 yr. 40 some normal logs The Lognormal Failure Process Let T = a random variable, the time to failure. If T has a lognormal distribution, then the logarithm of T has a normal distribution. 41 41 Lognormal Density Function^, Effect of Shape Parameter, s f(t) = 1 2π s t s2 = ln(1+ δ 2 ), δ t = t e 1 ⎛ t ⎞ − 2 ⎜ ln ⎟ 2 s ⎝ tMED ⎠ σt = cov µt 2 ;t ≥ 0 Defined for only [email protected] value of t. More symmetrical for smaller s values s2 ~ δ 2 , δ t ≤ 0.3 t tmed = median @me to failure, [email protected] parameter s = shape parameter s = 0.1 s = 1 s = 0.5 ^Lognormal pdf is [email protected] represented in terms of the mean and std. dev. of the underlying normal [email protected] Here the parameters are median (more [email protected] than mean for skewed) and shape parameter s (~ σ). 42 Lognormal F(t), Effect of Shape Parameter, s F(t) s = 0.1 s = 0.5 s = 1.0 As s increases, the variance increases and the [email protected] is more asymmetric. Data represented by Weibull will o|en be represented [email protected] by lognormal. 43 Lognormal for Repair Frequency Majority of repair @mes near Mode t med e Short repair @mes s2 / 2 Long repair @mes NASA, Lesson 0840 The lognormal [email protected] is widely used to describe the frequencies of system repair, because it reflects normal [email protected] repair-­‐@mes, a large number of repairs closely grouped about a modal value, and long repair @me data points with decreasing frequency in the tail. 44 Lognormal/Normal Relationship •  Given T is a lognormal random variable, then Log T space T space •  [email protected] Lognormal Normal •  Mean •  Variance •  Mode s2 /2 t med e 2 s2 s2 med t e [e −1] ln tmed s2, σ = S t = t med ln tmed mode s2 e 45 Lognormal Failure and Reliability Distribution Pr of failure up to t F(t) = P {T ≤ t} = P {lnT ≤ lnt} logic expression [email protected] parameter ⎧ lnT - ln tMED lnt - ln tMED ⎫ = P⎨ ≤ ⎬ s ⎩ Z s ⎭ Z Standardize: T t ⎧ ⎛1 1 t ⎫ t ⎞ = P ⎨z ≤ ln ⎬ = Φ ⎜ ln s tMED ⎭ ⎝ s tMED ⎟ ⎠ ⎩ Values in Std. Normal tables Pr of failure < t ⎛1 t ⎞ R(t) = 1− Φ ⎜ ln ⎝ s tMED ⎟ ⎠ Pr of failure > t 46 Lognormal Design Life and Median Specify R Find Z in the normal table Calculate tR ⎛1 t ⎞ 1- Φ ⎜ ln R ⎟ = R ⎝ s t med ⎠ ⎛1 t ⎞ Φ ⎜ ln R ⎟ = 1- R ⎝ s t med ⎠ Find z1-­‐R = such that: Φ(z 1−R ) = 1- R 1 tR ln = Z1−R s t med Solve for tR t R = t med e s z1−R 47 Lognormal Conditional Failure Rate Function, λ(t) Effect of Shape Parameter, s 0.30 0.25 S = 0.4 tmed = 10 λ(t) HAZARD RATE 0.20 S = 0.6 0.15 S = 0.8 0.10 0.05 S = 1.0 0.00 1 6 11 16 21 S=.4 26 S=.6 31 S=.8 36 41 46 TIME S=1 λ(t) increases until it reaches a max. and then it gradually decreases. Smaller the s, the smaller the variance and longer time until the max. 48 Lognormal Conditional Failure Rate Function, λ(t) for tmed = 10 in previous slide Lognormal Mode, MTTF, λ(t) with several shape parameters: s t mode = t med / e 1.0 0.8 0.6 5.3 13.8 10 7.0 12.0 16 0.4 s2 /2 Mode 3.7 f(t):...
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