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RELIABILITY ENGINEERING – I SEM 2020-21 BITS Pilani Pilani Campus Rajesh P Mishra Probability Density Function : f(t) • Probability Density Function is the first derivative of the Failure Density Function and given by f(t) • F(t) = 1 – R(t) • f(t) = 𝑑𝐹(𝑡) 𝑑𝑡 • f(t) ≥ 0 • F(t) = =− and 𝑡 𝑓(𝑡) 0 𝑑𝑅(𝑡) 𝑑𝑡 ∞ 𝑓(𝑡) 0 𝑑𝑡 𝑑𝑡 = 1 and R(t) = ∞ 𝑓(𝑡) 𝑡 𝑑𝑡 BITS Pilani, Pilani Campus Design Life • A design life is defined to be the time to failure t R that corresponds to a specified / desired reliability R. • R (t R) = R • Failure occurring within some interval of time [a, b] = F(b) – F(a) = R(a) – R(b) = 𝑏 𝑓(𝑡) 𝑎 𝑑𝑡 BITS Pilani, Pilani Campus Mean Time to Failure (MTTF) MTTF is the first moment of probability density function f(t) MTTF = = ∞ 𝑡 0 𝑓(𝑡) 𝑑𝑡 ∞ 𝑅(𝑡) 0 𝑑𝑡 Integrating by parts, we get MTTF = ∞ 𝑅(𝑡) 0 𝑑𝑡 BITS Pilani, Pilani Campus Median Time to Failure (t med) • Median Time to Failure is the time with 50% of the failure occurring before the median time to failure and 50% occurring after the median • R (t med) = 0.5 • VARIANCE = ∞ 0 𝑡 − 𝑀𝑇𝑇𝐹 2 𝑓 𝑡 𝑑𝑡 = ∞ 𝑡 0 2 𝑓 𝑡 𝑑𝑡 − 𝑀𝑇𝑇𝐹 2 BITS Pilani, Pilani Campus Hazard Rate Function • Hazard Rate Function is defined as the limit of the failure rate as the interval approaches zero. • The hazard function is the instantaneous failure rate • ƛ (t) = Lim △t⤑0 ƛ (t) = ƛ (t) = - 𝑅 𝑡 −𝑅(𝑡+ △𝑡) 𝑅 𝑡 △𝑡 𝑑𝑅(𝑡) 𝑑𝑡 1 x 𝑅(𝑡) 𝑓(𝑡) 𝑅(𝑡) ƛ (t) dt = - R(t) = exp [ 𝑑𝑅(𝑡) 𝑅(𝑡) 𝑡 - 0ƛ Integrating both sides then 𝑡 ƛ 0 𝑡 𝑑𝑡 = - ln R(t) 𝑡 𝑑𝑡 ] BITS Pilani, Pilani Campus CFR and AFR • Cumulative Failure Rate (CFR) over a period of time t is defined by L(t) = 𝑡 ƛ 0 𝑡 𝑑𝑡 • Average Failure Rate (AFR) between two time periods t1 and t2 is defined by AFR (t1,t2) = If t1 = 0 and t 2 = t 𝑡2 1 ƛ 𝑡2−𝑡1 𝑡1 then AFR = 𝑡 𝑑𝑡 = ln 𝑅 𝑡1 −ln 𝑅 (𝑡2) 𝑡2 −𝑡1 ln 𝑅 0 −ln 𝑅 (𝑡) 𝑡 = 0 −ln 𝑅(𝑡) 𝑡 = 𝐿(𝑡) 𝑡 BITS Pilani, Pilani Campus Bath Tub BITS Pilani, Pilani Campus SOLUTION 2.1 Given R(t) = 1 0.001𝑡+1 t≥0 Find R(100) = 1 0.001 𝑥 100+1 R(1000) = = 1 0.001 𝑥 1000+1 1 1.1 = 0.9091 1 2 = = 0.5 BITS Pilani, Pilani Campus b. 𝑡 ƛ 0 𝑡 𝑑𝑡 = - ln R(t) ƛ 𝑡 = - 𝑑 ln(1/0.001𝑡+1) 𝑑𝑡 = 0.001 0.001𝑡+1 It is an increasing function as the value is positive. BITS Pilani, Pilani Campus Solution 2.2 ƛ 𝑡 = 0.4 t a. where t is in years R(t) = exp [ - 𝑡 ƛ 0 Hence when t = 1/12 year = 1 month b. 𝑡 𝑑𝑡 ] = exp [ - 𝑡 0.4𝑡 0 𝑑𝑡 ] = 𝑒^ − 0.2𝑡2 R(t) = 0.9986 R (tr) = R = 0.95 exp^ -0.2 t^2 = 0.95 t = 0.5064 years = 6 months BITS Pilani, Pilani Campus Solution 2.3 Given f(t) = 0.01 a. R(t) = 100 0.01 𝑑𝑡 𝑡 b. ƛ (t) = 𝑓(𝑡) 𝑅(𝑡) c. MTTF = = where 0 ≤ t ≤ 100 days = 0.01 (100 − 𝑡) 0.01 0.01 (100 −𝑡) 100 0.01 (100 0 − 𝑡) dt = 50 days 100 d. Variance = 0 0.01 𝑡2𝑑𝑡 − 𝑀𝑇𝑇𝐹 2 = 833.33 STD DEV = 28.87 e. R(t) = 1 – 0.01t = 0.5 tmed = 50 days BITS Pilani, Pilani Campus SOLUTION 2.4 Given f(t) = 3𝑡2 10 9 0 ≤ t ≤ 1000 hrs. a. Probability of failure within 100 hr warranty period F(t) = b. MTTF = 1000 𝑡 0 c. R(tr) = 0.99 = 𝑡 𝑓(𝑡) 0 𝑑𝑡 = 100 𝑓(𝑡) 0 𝑑𝑡 = 0.001 𝑓(𝑡) dt = 750 hrs. 1000 𝑓(𝑡) 𝑡 dt = 215.44 hrs. BITS Pilani, Pilani Campus EXPONENTIAL RELAIBILITY FUNCTION • A failure distribution that has a constant failure rate is called exponential probability distribution • If the failure rates of all failure modes of a component are constant and independent, then the overall failure rate of the component is also constant • Failures due to random or chance events will follow this distribution BITS Pilani, Pilani Campus CFR MODEL R(t) = exp [ - 𝑡 ƛ 0 𝑡 𝑑𝑡 ] If ƛ is constant then R(t) = e – ƛt where t ≥ 0 F(t) = 1 - R(t) = 1 - e – ƛt f(t) = 𝑑𝐹(𝑡) 𝑑𝑡 MTTF = =0 − ∞ 𝑅(𝑡) 0 Variance = ∞ 0 𝑑𝑅(𝑡) 𝑑𝑡 = ƛ e – ƛt 𝑑𝑡 = ∞ e 0 𝑡 − 𝑀𝑇𝑇𝐹 – 2 ƛt 𝑑𝑡 = 𝑓 𝑡 𝑑𝑡 = 1 = Standard Deviation ƛ ∞ 0 𝑡 − 1 2 ƛ ƛe – ƛt𝑑𝑡= 1 = (MTTF)2 ƛ2 BITS Pilani, Pilani Campus BITS Pilani, Pilani Campus BITS Pilani, Pilani Campus • Reliability of component with CFR has slightly more than one-third chance of surviving to its mean time to failure: R (MTTF) = e –MTTF/MTTF = e -1 = 0.368 • Design Life t R R(t R) = 0.5 = e – ƛ t R t R= - ln R / ƛ, if R 0.5 then, Median = - ln 0.5 MTTF = 0.69315 MTTF BITS Pilani, Pilani Campus Reliability in Series BITS Pilani, Pilani Campus Reliability in Parallel (Redundnacy) BITS Pilani, Pilani Campus Problem 3.1 Page 55 A component experiences chance (CFR) failures with an MTTF of 1100 hour. Find the following: i. Reliability for a 200-hr mission ii. The design life for a 0.9 reliability iii. The median time to failure iv. The reliability of 200 hr mission if a second, redundant (and independent) component is added. v. The reliability of 200 hr mission if a second, single component is serially added. BITS Pilani, Pilani Campus Solution 3.1 Given MTTF = 1100 hr hence ƛ = 1/1100 i. R(200) = exp – (1/1100* 200) = 0.9555 ii. 0.90 = exp – (1/1100*t) taking ln both side and solve td = 115.9 hours iii. 0.50 = exp –(1/1100 * t) taking ln both side and solve tmed = 762.46 hrs iv. R(t) = 1 – (1- exp – ƛt)2 = 1 – ( 1 – exp – 1/1100* 200)2 = 0.9723 v. R(t) = exp - 2ƛt = exp – 2*1/1100*200 = 0.9131 BITS Pilani, Pilani Campus Characterisitics • The time to failure of a component is not dependent on how long the component has been operating • The probability that the component will operate for next t hours is the same regardless of whether the component is brand new, has been operating for several hours • Using Condition Probability R (t I To) = R (t + To)/ R(To) = exp −ƛ (𝑡+𝑇𝑜) exp( − ƛ𝑇𝑜) = exp (-ƛt) BITS Pilani, Pilani Campus Problem 3.2 Page 55 • A CFR system with ƛ = 0.0004 has been operating for 1000 hrs. What is the probability that it will fail in the next 100 hr? the next 1000 hr? • Solution: a. F(t) = 1 - exp (– 0.0004 * 100) = 0.0392 b. F(t) = 1- exp (– 0.0004 * 1000) = 0.3297 BITS Pilani, Pilani Campus Failure Modes • Failure occurs due to different physical phenomena or different characteristics of individual components • If R(t) is the reliability function for the ith failure mode, assuming independence among the failure modes, then • System reliability is the product of these probabilities that none of thee modes occur before time t • System hazard rate function is the sum of the individual hazard rate of function of all the failure modes • ƛ s = ƛ1 + ƛ2 + ƛ3 +…….. R(t) = e - ƛ s t BITS Pilani, Pilani Campus MTTF = ƛs = 1 ƛ𝑠 1 𝑀𝑇𝑇𝐹1 = 1 ƛ1+ ƛ2+ ƛ3+ …. 1 + 𝑀𝑇𝑇𝐹2 = 1 1 1 + +⋯ 𝑀𝑇𝑇𝐹1 𝑀𝑇𝑇𝐹2 +⋯ If all n components have identical ƛ then ƛs = nƛ and MTTF = 1 𝑛ƛ BITS Pilani, Pilani Campus Bath-tub Failure Modes Failure modes in bath tub includes: Burn-in, Useful Life and Wear-Out The failure rates of these failure modes can be summed up to find ƛs Then we can find the reliability of the system for the mission hour BITS Pilani, Pilani Campus Problem 3.3 Page 55 • A gearbox has two independent failure modes: a constant failure rate of 0.0003 and a linearly increasing wear out failure given by ƛ (t) = t / 5 x 10 5, Find the reliability of the gear box for 100 hr of operation. • Solution: ƛs = 0.0003 + t / 5 x 10 5 R(t) = exp – (0.0003t + t 2 / 10 x 10 5) R(100) = exp – (0.0003*100 + (100)2 / 10 x 10 5) = exp -0.04 = 0.9608 BITS Pilani, Pilani Campus Failure on Demand • Components fail from switching from idle mode to operating mode like bulbs or air-conditioners • If constant failure rate is assumed for each failure mode and a constant probability of failure on demand is assumed, then the component will have an effective failure rate on a clock hour basis • ƛi is the average failure rate when idle • ƛo is the average failure rate while operating • p is the probability of failure on demand • ti is the average length of the idle time period per cycle • to is the average length of the operating time period per cycle • ƛ eff = 𝑡𝑖 𝑡𝑖+𝑡𝑜 ƛi + 𝑡𝑜 𝑡𝑖+𝑡𝑜 ƛo + 𝑝 𝑡𝑖+𝑡𝑜 BITS Pilani, Pilani Campus Problem 3.18 Page 57 • A 60 watt outdoor lightbulb is advertised as having an average life (i.e. MTTF) of 1000 operating hours. However, experience has shown that it will also fail on demand an average of once every 120 cycles. A particular bulb is turned once each evening for an average of 10 hr. If it is desired to have a reliability of 90 percent, what is the design life. ƛ eff = ƛ eff = 𝑡𝑖 𝑡𝑜 𝑝 ƛi + ƛo + 𝑡𝑖+𝑡𝑜 𝑡𝑖+𝑡𝑜 𝑡𝑖+𝑡𝑜 14 10 ∗0+ (1/1000) 14+10 14+10 0.90 = exp (-0.000764 t) + 1/120 14+10 = 0.000764 therefore td = 137.9 hrs BITS Pilani, Pilani Campus Renewal Process • A failed component is immediately replaced with a new one, will cause the system to reach a steady state constant number of failures per unit of time • For a renewal process in which failed components are replaced as they fail, a steady state constant failure rate is obtained such that : ƛ = 𝑛𝑖 1/𝑀𝑇𝑇𝐹𝑖 BITS Pilani, Pilani Campus Problem 3.7 Page 56 • A system contains 20 identical and critical components that will be replaced on failure (renewal process). As a result, a constant failure rate for the system will be observed. If a design life of 10 year with a reliability of 0.99 is required, what should the system MTTF and median time to failure? If each component has CFR, what will the component MTTF and median time to failure be? • Solution: 0.99 = exp ( - ƛ * 10) therefore ƛs = 0.001/ year MTTF = 1 0.001 = 1000 years BITS Pilani, Pilani Campus Component ƛi = ƛs / n = 0.001/20 = 0.00005 / year Component MTTF = 1/ ƛi = 1/0.00005 = 20000 years System Median time to failure : 0.5 = exp (– 0.001 * tmed) tmed = 693.15 years Component Median time to failure : 0.5 = exp (- 0.00005 * tmed) t med = 13863 years BITS Pilani, Pilani Campus Repetitive Loading • Let p be the probability of failure as a result of a load or stress on the system and if independent loads are applied at constant, fixed interval of time, an approximate reliability function results R = 1 – p for single load Rn = (1-p) n for n loads = exp ln (1-p) n = exp n ln (1-p) but as ln (1-p) = - p = exp (– np) As n = t / △t , △t being the fixed time between loads, so R(t) = exp (- p/ △t) t = exp – ƛ t , where ƛ = p/ △t, a constant failure rate. BITS Pilani, Pilani Campus Solution 3.6 Page 56 • A landing gear system has repetitive stresses placed on it twice a day as a result of loadings. The probability of a failure during landing is 0.0028. Determine the reliability of the landing gear system over a 30-day contingency operation. What is the probability of a failure occurring between days 10 and 20 of the operation? • Solution: p = 0.0028 △t = ½ day t = 30 days • i. R(t) = exp (- 0.0028 *2) * 30 = exp – 0.168 = 0.8454 • ii. F (t) = F(b) – F(a) = R(a) – R(b) = exp (-0.0028*2*10) – exp (-0.0028*2*20) = 0.9455 – 0.8940 = 0.0515 BITS Pilani, Pilani Campus Two-Parameter Exponential Distribution Guaranteed Lifetime : to :The minimum time prior to which no failure occurs. • It is a location parameter that shifts the distribution an amount equal to to to the right on the time axis. • f(t) = ƛ exp –ƛ (t – to) where 0 ≤ to ≤ t ≤ ∞ • R(t) = exp –ƛ (t – to) 𝑡 1 • MTTF = t ƛ t exp –ƛ (t – to) dt = to+ ƛ o • R (tmed)= exp –ƛ (tmed – to) = 0.5 • tmed = to+ 0.69315 ƛ BITS Pilani, Pilani Campus • Design life t R = to+ ln 𝑅 −ƛ • Variance (1/ ƛ 2) and standard deviation (1/ƛ) of the two parameter exponential distribution are not affected by the location parameter. BITS Pilani, Pilani Campus Problem 3.15 • Consider two identical and redundant CFR components having a guaranteed life of 2 months and a failure rate of 0.15 failure per year. What is the system reliability for 10,000 hr of continuous operation? • Solution: to = 2/12 = 0.167 ƛ = 0.15 t = 10000/365*24 = 1.14 years • R(t) = 1- {1- exp – 0.15 ( 1.14 – 0.167)}2 = 0.982 BITS Pilani, Pilani Campus Poisson’s Distribution • If a component having a constant failure rate is immediately repaired or replaced upon failing, then the number of failures observed over a period of time t has a Poisson Distribution • The probability of observing n failures in time t is given by e – ƛt ƛ𝑡 𝑛 P (t) = 𝑛ǃ BITS Pilani, Pilani Campus Problem 3.13 • An electronic circuit board with ƛ (t) = 0.00021 per hour is replaced on failure. What is the probability that the third failure will occur by 10000 hours? • Solution: ƛ = 0.00021 ƛ t = 0.00021* 10000 = 2.1 𝑒 −2.1 2.1 0 p0= = 0.122 0 ǃ 𝑒 −2.1 2.1 1 P1 = = 0.257 1 ǃ 𝑒 −2.1 2.1 P2 = 2 ǃ 2 = 0.270 P3 = 1- p0-p1-p2= 0.351 BITS Pilani, Pilani Campus Redundancy and CFR Model Two Components in Parallel: The system will fail when both the components fail F(t) = (1- exp –ƛt) * (1- exp –ƛt) = (1- exp –ƛt) 2 R ( t) = 1 - (1- exp –ƛt) 2 = 2 exp –ƛt – exp -2 ƛt ƛ(t) = 𝑓(𝑡) 𝑅(𝑡) MTTF = = 2ƛ𝑒−ƛ𝑡 − 2ƛ exp − 2ƛ𝑡 2 exp –ƛt – exp −2 ƛt 1.5 ƛ BITS Pilani, Pilani Campus BITS Pilani Pilani Campus MODULE 1: TIME DEPENDENT FAILURE MODELS: Weibull Distribution Lecture No. 3 Chapter 4 T1 WEIBULL DISTRIBUTION • Used to model both increasing and decreasing failure rates • Characterized by the hazard rate function • ƛ (t) = ß Ɵ 𝑡 Ɵ ƛ (t) = a t b ß−1 • ß is the shape parameter • • Ɵ is the scale parameter and also called the characteristic life BX Life : The time at which X% of units will fail. BITS Pilani, Pilani Campus Weibull Shape Parameter: ß or k • The Weibull shape parameter, β, is also known as the Weibull slope • The value of β is equal to the slope of the line in a probability plot • Different values of the shape parameter can have marked effects on the behaviour of the distribution • Some values of the shape parameter will cause the distribution equations to reduce to those of other distributions. For example, when β = 1, the pdf of the three-parameter Weibull reduces to that of the two-parameter exponential distribution. • The parameter β is a pure number (i.e., it is dimensionless). BITS Pilani, Pilani Campus BITS Pilani, Pilani Campus BITS Pilani, Pilani Campus Interpretation • Weibull distributions with β < 1 have a failure rate that decreases with time, also known as infantile or early-life failures • Weibull distributions with β close to or equal to 1 have a fairly constant failure rate, indicative of useful life or random failures • Weibull distributions with β > 1 have a failure rate that increases with time, also known as wear-out failures • These comprise the three sections of the classic "bathtub curve." A mixed Weibull distribution with one subpopulation with β < 1, one subpopulation with β = 1 and one subpopulation with β > 1 would have a failure rate plot that was identical to the bathtub curve. An example of a bathtub curve is shown in the following chart BITS Pilani, Pilani Campus BITS Pilani, Pilani Campus Weibull Scale Parameter: Ɵ or ƞ • A change in the scale parameter, Ɵ, has the same effect on the distribution as a change of the abscissa scale. • Increasing the value of Ɵ while holding β constant has the effect of stretching out the pdf. Since the area under a pdf curve is a constant value of one, the "peak" of the pdf curve will also decrease with the increase of Ɵ, as indicated in the following figure. • If Ɵ is increased, while β and γ are kept the same, the distribution gets stretched out to the right and its height decreases, while maintaining its shape and location. • If Ɵ is decreased, while β and γ are kept the same, the distribution gets pushed in towards the left (i.e., towards its beginning or towards 0 or γ), and its height increases. • Ɵ has the same unit as T, such as hours, miles, cycles, actuations, etc. BITS Pilani, Pilani Campus Effect of varying Scale parameter (Ɵ or ƞ) BITS Pilani, Pilani Campus Location Parameter: Ƴ or To • Frequently, the location parameter is not used, and the value for this parameter can be set to zero • When this is the case, the pdf equation reduces to that of the two-parameter Weibull distribution • There is also a form of the Weibull distribution known as the one-parameter Weibull distribution. This in fact takes the same form as the two-parameter Weibull pdf, the only difference being that the value of β is assumed to be known beforehand • This assumption means that only the scale parameter needs be estimated, allowing for analysis of small data sets. It is recommended that the analyst have a very good and justifiable estimate for β before using the one-parameter Weibull distribution for analysis BITS Pilani, Pilani Campus Effect of ß on Weibull PDF i.e. f (t) BITS Pilani, Pilani Campus Effect of ß on CDF i.e. F(t) BITS Pilani, Pilani Campus Effect of ß on Weibull Reliability i.e. R (t) BITS Pilani, Pilani Campus Effect of ß on Weibull Hazard Rate BITS Pilani, Pilani Campus Reliability and Failure Rate • R(t) = exp - • f(t) = • • • • 𝑑 𝑅(𝑡) 𝑑𝑡 = 𝑡ß 0 Ɵ ß Ɵ 𝑡 Ɵ 𝑡 Ɵ ß−1 ß−1 𝑑𝑡 = exp – exp – 𝑡 Ɵ 𝑡 Ɵ ß R(Ɵ) = exp (-1) = 0.368 ß For ß = 1, ƛ (t) is constant and equal 1/Ɵ For ß < 1, PDF is similar in shape to the exponential For ß ≥ 3, PDF is symmetrical like normal distribution For 1 < ß < 3 PDF is skewed BITS Pilani, Pilani Campus MTTF and Variance • MTTF = ∞ 𝑡 0 Let y = 𝑡 Ɵ MTTF = ∞ 𝑡 0 ∞ ß 𝑓 𝑡 𝑑𝑡 = dy = ß Ɵ ∞ß 0 Ɵ 𝑡 Ɵ ß−1 𝑡 Ɵ ß−1 exp – 𝑡 Ɵ ß t dt dt 1 𝑒 − 𝑦 𝑑𝑦 = Ɵ ∞ 𝑦ß 0 𝑒 − 𝑦 𝑑𝑦 = Ɵ Γ (1 + 1/ß) − Γ(x) = 0 𝑦 𝑥 1 𝑒 − 𝑦 𝑑𝑦 Variance = σ2 = Ɵ2 [Γ (1 + 2/ß) – {Γ (1 + 1/ß)}2] For x > 0, Γ(x) = (x-1)ǃ or Γ(x) = (x-1) Γ(x-1) Γ1/2 = √ᴫ BITS Pilani, Pilani Campus Design Life, Median and Mode Design Life R(t) = exp – 𝑡 Ɵ ß =R Design Life t R = Ɵ (- ln R) 1/ß Median Life : R = 0.5 therefore t med = Ɵ (- ln 0.5) 1/ß= Ɵ (0.69315) 1/ß Mode: when ß > 1, t mode = Ɵ ( 1- 1/ß) 1/ß when ß ≤ 1 t mode = 0 BITS Pilani, Pilani Campus Problem 1 Page 79 • For a system having a Weibull failure distribution with a shape parameter of 1.4 and a scale parameter of 550 days. Find the following: i. R(100 days) ii. The B1 Life iii. MTTF iv. The standard deviation v. tmed vi. tmode vii. Design life for a reliability of 0.90 BITS Pilani, Pilani Campus Solution i. R(t) = exp – 𝑡 Ɵ ß ii. R = 0.99 = exp – ( = exp – ( 𝑡 1.4 ) 550 100 1.4 ) 550 = 0.912 t = 20.58 days iii. MTTF = Ɵ Γ (1 + 1/ß) = 550 Γ(1 + 1/1.4) = 500.81 days iv. σ2 = Ɵ2 [Γ (1 + 2/ß) – {Γ (1 + 1/ß)}2] Standard Deviation σ = 363.95 days BITS Pilani, Pilani Campus v. tmed R = 0.5 0.5 = exp – (t/550) 1.4 tmed = 423.32 days vi. tmode = Ɵ ( 1- 1/ß) 1/ß tmode = 550 ( 1 – 1/1.4) ^1/1.4 tmode = 224.77 days vii. 0.90 = exp – (t/550) 1.4 t = 110.22 days BITS Pilani, Pilani Campus Burn-In screening for Weibull If To is Burn-in Period, then conditional reliability R(T| To) = exp – [(t + To)/Ɵ] ß exp - (To/Ɵ) ß = exp [ - 𝑡+𝑇𝑜 ( Ɵ ) ß + 𝑇0 ß ( ) ] Ɵ BITS Pilani, Pilani Campus Problem 4.2 • A turbine blade has demonstrated a Weibull failure pattern with a decreasing failure characteristics by a shape parameter of 0.6 and a scale parameter of 800 hr. a. Compute the reliability for a 100 – hr mission b. If there is a 200 – hr burn-in of the blades, what is the reliability for a 100-hr mission BITS Pilani, Pilani Campus Solution 4.2 Given: ß = 0.6 a. R(t) = exp – = exp – ( Ɵ = 800 𝑡 Ɵ ß 100 0.6 ) = 800 b. R(t) = exp [- ( t = 100 𝑡+𝑇𝑜 Ɵ 0.75 𝑇0 Ɵ ) ß + ( ) ß ] = exp [- ( 100+200 800 ) 0.6 + ( 200 0.6 ) ] 800 = 0.887 BITS Pilani, Pilani Campus Failure Modes • For a system having n serially related components or having n independent failure modes, each having an independent shape parameter ß and scale parameter Ɵi: • Failure rate is summation from i = 1 to i = n ƛ(t) = ∑ ß Ɵ𝑖 𝑡 ß-1 ( ) Ɵ𝑖 1 Ɵ𝑖 = ß t ß-1∑ ( ) ß 1 Ɵ𝑖 Characteristic Life = [ ∑ ( ) ß ] -1/ß • If the failure modes are all Weibull but with differing shape parameter, then the system failure distribution will not be Weibull. BITS Pilani, Pilani Campus Identical Weibull Components • If a system of a serially related and independent components have identical hazard rate function • ƛ(t) = ∑ = ß Ɵ𝑖 𝑛ß (t Ɵß 𝑡 Ɵ𝑖 ( ) ß-1 ) ß-1 R(t) = exp {– n 𝑡 ß ( ) } Ɵ This follows Weibull distribution with shape parameter ß and scale parameter Ɵ 𝑛 ^1/ß BITS Pilani, Pilani Campus Problem 4.7 • What is the maximum number of identical and independent Weibull components having a scale parameter of 10,000 operating hours and a shape parameter of 1.3 that can be put in series if a reliability of 0.95 at 100 operating hours is desired? What is the resulting system MTTF? BITS Pilani, Pilani Campus Solution 4.7 Given : Ɵ = 10,000 R(t) = exp 0.95 = exp ƛ(t) = 𝑛ß (t Ɵß 𝑡 ß {– n ( ) } Ɵ 100 1.3 –n( ) 10000 ß = 1.3 R(t) = 0.95 t = 100 Solving we get n = 21 ) ß-1 ƛ (t) = 0.0006857 MTTF = 1/ƛ (t) = 1458 hours BITS Pilani, Pilani Campus The Three Parameter Weibull Minimum life : to such that T > to The parameter to is called the location parameter The distribution assumes that no failure will take place prior to time to R(t) = exp [ - ( ƛ (t) = 𝑡 −𝑡0 Ɵ ) ß] t ≥ to ß 𝑡 −𝑡0 ß-1 ( ) Ɵ Ɵ 1 ß MTTF = to + Ɵ Γ(1 + ) tmed = to + Ɵ (0.69315) 1/ß Td = to + Ɵ (-ln R) 1/ß BITS Pilani, Pilani Campus Problem 4.4 • For a three parameter Weibull distribution with ß = 1.54 Ɵ = 8500 hr to = 50 hr, find the following: a. R(150) b. MTTF c. tmed d. The standard deviation e. The design life for a reliability of 0.98 BITS Pilani, Pilani Campus Solution 4.4 Given t = 150 to = 50 ß = 1.54 a. R(t) Ɵ = 8500 𝑡 −𝑡0 ß = exp [ - ( ) ] t ≥ to Ɵ 150−50 1.54 = exp [ - ( ) = 0.9989 8500 MTTF = 50 + 8500 Γ ( 1 + 1/1.54) = 50 + 8500 x 0.9 = 7700 hours BITS Pilani, Pilani Campus t med = to + Ɵ (0.69315) 1/ß= 50 + 8500 (0.69315) 1/1.54= 6750 hours Variance σ 2 = (8500)2 [ Γ(1 + 2/1.54) – { Γ( 1 + 1/1.54)}2 ] Standard Deviation = 5070 hours Design Life td = to + Ɵ (- ln R) 1/ß = 50 + 8500 ( - ln 0.98 ) 1/1.54=725 hours BITS Pilani, Pilani Campus Redundancy with Weibull Failures Let us assume two identical and independent components used to form redundant system Rs = 1 – [ (1 – R(t) ] 2 If R(t) = exp – (t / Ɵ ) ß Rs = 2 exp – (t / Ɵ ) ß - exp – 2 (t / Ɵ ) ß MTTF = 2 Ɵ Γ( 1 + 1/ß ) ƛs (t) = ß Ɵ ( 𝑡 Ɵ Ɵ 2 ^ 1/ß Γ( 1 + 1/ß ) = Ɵ Γ (1 + 1/ß) (2 – 2 -1/ß) 2−2 𝑒 −(t /Ɵ)ß ) ß-1 2− 𝑒 −(t / Ɵ ) ß BITS Pilani, Pilani Campus Problem 4.5 • For the components defined in Exercise 4.1. determine the reliability for 100 hr if there are two redundant components. • Given : t = 100 Ɵ = 550 ß = 1.4 • Rs = 2 exp – (t / Ɵ ) ß - exp – 2 (t / Ɵ ) ß = 2 exp – (100/550) 1.4 – exp –2 (100/550) 1.4 = 0.992 BITS Pilani, Pilani Campus BITS Pilani Pilani Campus Reliability of Systems Chapter – 5 RL 2.1-2.5 Introduction • A system consist of number of different components • Components within a system may be related to one another in two primary ways -Series configuration or -Parallel configuration • In series configuration all components must function for the system to function. • However in a parallel or redundant configuration at least one component must function for the system to function. 76 BITS Pilani, Pilani Campus Serial Configuration • In a series configuration, all assets are considered critical Reliability block diagram for components in series • • • • • E1=Event that component 1 does not fail E2=Event that component 2 does not fail Then P(E1)=R1 and P(E2)=R2 R1,R2=Reliabilities Rs=P(E1∩E2)=P(E1) P(E2)= R1 R2 Therefore generelizing system reliability asRs= (R1) x (R2) x (R3) x …………….x (Rn) 77 BITS Pilani, Pilani Campus Serial Configuration 𝑛 𝑅𝑠 = 𝑅𝑖 (𝑡) 𝑖=1 If component has constant failure rate 𝜆𝑖 𝑛 𝑅𝑠 = 𝑒𝑥𝑝(−𝜆𝑖 𝑡) 𝑖=1 𝑛 𝑅𝑠 = exp − 𝜆𝑖 𝑡 = exp −𝜆𝑠t 𝑖=1 MTTF=1/𝜆𝑠 = 1/ 𝑛 𝑖=1 𝜆𝑖 78 BITS Pilani, Pilani Campus Example • Consider a four component system of which the components are independent and identically distributed with CFR. If Rs(100)=0.95 is the specified reliability , Find individual component MTTF. 79 BITS Pilani, Pilani Campus Example • A system is comprised of four serially related components each having a Weibull time to failure distribution with parameters as shown in table. Determine the R(10). Component Scale parameter Shape parameter 1 100 1.20 2 150 0.87 3 510 1.80 4 720 1.00 80 BITS Pilani, Pilani Campus 81 BITS Pilani, Pilani Campus Parallel Configuration • In parallel system all units must fail to for system to fail • Parallel units are represented by block diagram • System reliability is found by taking 1 minus probability that all n components fail 82 BITS Pilani, Pilani Campus Parallel Configuration • Rs(t) = 1 - [(1 - R1) (1 - R2) (1 - R3) …………. (1 - Rn)] 𝑛 𝑅𝑠 𝑡 = 1 − 1 − 𝑅𝑖 (𝑡) 𝑖=1 • For a redundant system consisting of all CFR components 𝑛 1 − 𝑒 −𝜆𝑖𝑡 𝑅𝑠 𝑡 = 1 − 𝑖=1 EXAMPLE- For two component parallel system and having CFR derive expression for MTTF 83 BITS Pilani, Pilani Campus 84 BITS Pilani, Pilani Campus Combined Series Parallel Systems • Generally systems contain assets in both series and parallel relationships. Consider, the example below • To compute the system reliability, the network may be broken into series or parallel subsystems • The system reliability may be calculated on the basis of the relationship among the subsystems 85 BITS Pilani, Pilani Campus Combined Series Parallel Systems • • • • • In the network the subsystems have the following reliabilities RA= [1 – (1 - R1) (1 - R2)] RB= (RA) x (R3) RC= (R4) x (R5) RS= [1 – (1 - RB) (1 - RC)] (R6) If R1= R2= 0.90, R3= R6= 0.98, and R4= R5= 0.99, then RB= [1 – (0.10)2] (0.98) = 0.9702 RC= (0.99)2= 0.9801 And RS= [1 – (1 – 0.9702) (1 – 0.9801)] (0.98) = 0.9794 86 BITS Pilani, Pilani Campus Example • Given the system, calculate its useful life reliability for a 100-hr mission. All given failure rates are in failures per million hours of operation. 87 BITS Pilani, Pilani Campus 88 BITS Pilani, Pilani Campus Example • Determine the reliability of the given system. 89 BITS Pilani, Pilani Campus 90 BITS Pilani, Pilani Campus 91 BITS Pilani, Pilani Campus Example • Find the reliability of the following system which is shown in the figure if the reliability of components are RA =0.96, RB = 0.92, RC = 0.94, RD = 0.89, RE = 0.95, RF = 0.88, RG = 0.90. 92 BITS Pilani, Pilani Campus Solution • The reliability of the subsystem with components A and B is R1 =[1- (1-0.96) (1-0.92)]= 0.9968 • The reliability of the subsystem with components E, F and G is R2 = [1- (1-(0.95)(0.88))(1-0.90)]= 0.9836 • System Reliability = Rs = 0.9968 X 0.94 X 0.89 X 0.9836 = 0.8202 93 BITS Pilani, Pilani Campus High – level Vs Low-level Redundancy • System redundancy may be obtained in two ways i. Each component may have parallel component ii. Entire system may placed in parallel with another identical system High level redundancy Low level redundancy Let A and C are identical B and D are identical components If both the component has same reliability R then 94 BITS Pilani, Pilani Campus High – level vs Low-level Redundancy • 𝑅𝑙𝑜𝑤 = 1 − 1 − 𝑅 2 = 2 𝑅 − 𝑅2 2 • 𝑅ℎ𝑖𝑔ℎ = 1 − 1 − 𝑅2 2 = 2𝑅2 − 𝑅4 • By Comparing two reliabilities, it can be shown that reliability of low level redundancy is greater than reliability of high level redundancy 95 BITS Pilani, Pilani Campus 96 BITS Pilani, Pilani Campus Example • A radio set consists of three major components: a power supply, a receiver, and an amplifier, having reliabilities of 0.8,0.9 and 0.85 respectively. Compute system reliabilities for both high level and low level redundancy for systems with two parallel components. 97 BITS Pilani, Pilani Campus Solution 98 BITS Pilani, Pilani Campus k-out of –n Redundancy • A requirement exist for k out of n identical and independent components to function for the system to function. • If each component is viewed as independent trial with R as a constant probability of success, then probability of exactly x components operating is𝑛 𝑃 𝑥 = 𝑅 𝑥 1 − 𝑅 𝑛−𝑥 𝑥 𝑛 𝑛! = 𝑥! (𝑛 − 𝑥)! 𝑥 𝑛 𝑅 𝑠 = 𝑃(𝑥) 𝑥=𝑘 Example- A vehicle require more than its 4 engines to operate to work. If each engine has a reliability of 0.97, determine the reliability of working. 99 BITS Pilani, Pilani Campus 100 BITS Pilani, Pilani Campus Complex configuration • Certain systems cannot be decomposed into series and parallel systems • However such networks can be analyzed by using either decomposition or enumeration 101 BITS Pilani, Pilani Campus Complex configuration • Decomposition• Two sub networks are created : one in which component E is assumed to be functioning and one in which E has failed. • Reliability of each sub network is determined separately • Then the system reliability is given by- 102 BITS Pilani, Pilani Campus 103 BITS Pilani, Pilani Campus Common Mode Failures • Assumption of independence of failure among n components may be violated. For example some components may share common power source, environmental conditions may affect etc • A common mode failure can be depicted in series with those common components sharing failure mode • Rs=[1-(1-R1)(1-R2)(1-R3)]R’ • In order to redundancy network to have an effect, common mode failure must have high reliability 104 BITS Pilani, Pilani Campus Example • Two parallel, identical and independent components have CFR. If it is desired that Rs(1000) = 0.95, Find the MTTF. Assume a common-mode CFR of 0.00001 in addition to the components’ independent failure rates. 105 BITS Pilani, Pilani Campus THANK YOU BITS Pilani, Pilani Campus