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Reliability Engineering ( Rekayasa Keandalan ). Y M Kinley Aritonang, Ph.D. Referen ces. Mitra , A. (1998), Fundamentals of Quality Control and Improvement 2 nd ed. , New Jersey: Prentice-Hall.
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Reliability Engineering(RekayasaKeandalan) Y M Kinley Aritonang, Ph.D
References Mitra, A. (1998), Fundamentals of Quality Control and Improvement 2nd ed., New Jersey: Prentice-Hall. Ebeling, C. E. (1997), An Introduction To Reliability And Maintainability Engineering, Singapore: The McGraw-Hill O’Connor, P. D. T., Newton, D., dan Bromley, R. (2002), Practical Reliability Engineering 4th ed., Chichester: John Wiley & Sons.
Purposes of study • Able to explain about Reliability. • Able to use the probability distribution in reliability modelling. • Able to calculate system reliability. • Able to perform the reliability testing.
Introduction Reliability → measure of quality of the product over the long run. Reliability → Probability of a productperforming its intended function for a stated period of time under certain specified condition(Mitra, 1998). → Probability of a system or component operating more than time t(Ebeling, 1997). Reliability implies successfull operation over a certain period of time
Introduction T : time to failure of a system or component. Reliability at time t(R(t)) → probability of the product that fail at ≥ t. Reliablity Function : R(t) = P(T ≥ t) where R(t) ≥ 0, R(0) = 1, danlimt→ R(t) = 0.
Introduction CDF (Cumulative Distribution Function): F(t) = P(T < t) F(t) = 1 – R(t) where F(0) = 0 danlimt→ F(t) = 1 So F(t) is the probability of the system or component that fail before time t.
Introduction PDF (Probability Density Function): f(t) = dF(t)/dt= -dR(t)/dt where f(t) ≥ 0 and 0 f(t) dt = 1 Reliability function andCDFcome from the probability with function f(t). So, reliability (R(t)) and thefailure probability (F(t)) are in the range [0,1].
Introduction Mean Time To Failure (MTTF) → The average of useful life until the product or component fail. PartialIntegral : u = t du = dt dv = -dR(t) v = -R(t)
Life-Cycle Curve 1 Based on thefailure rate there are 3 distinct phases of the product : • Debugging phase (infant – mortality phase / Burn-in) • The initial is high and then drop over time • The problem is identified and then is improved • Chance-failure phase (Useful life) • Failure occur randomly and independently. • is constant • Represents the Useful life of product • Wear-out phase • increases • The end of product useful live
Life-Cycle Curve 2 Debugging phase Chance-Failure phase Wear-out phase t (time) Life-cycle curve = bathtub curve
Probability Distribution to model Failure Rate (time to failure) : • EksponensialDistribution → is constant • WeibullDistribution → is notconstant
EksponensialDistribution1 T = time to failure Pdf : Cdf :
EksponensialDistribution2 Mean Time To Failure (MTTF) MTTF = E(T) For repairable equipment, MTTF = MTBF (Mean Time Between Failure) MTTF ≠ MTBF if there is a significant repair or replacement time upon failure of the product
EksponensialDistribution3 Reliabilityat time t (R(t)) → the probability of the product lasting up to at least timet. R(t) 1 t
EksponensialDistribution4 Failure Rate function(r(t)) Failure rate function→ ratio of the time-to-failure pdf to the reliability function. For the exponential failure distribution : So iftime-to-failureis exponential distribution, failure rateis equal to (constant failure rate)
EksponensialDistribution5 Example 11-1 An amplifier has an exponential time-to-failure with a failure rate of 8% per 1000h.. • What is the reliability at 5000h? • Find themean time to failure? Solution : • Reliability at 5000h = 0.08/1000 hour = 0.00008/hour t = 5000 hour → R(5000)= exp(-t) = exp(-0.00008*5000) = 0.6703 Mean time to failure MTTF = 1/ = 1 / 0.00008 jam = 12500 jam
EksponensialDistribution6 Example 11-2 What is the highest failure rate for a product if itis to have a probability of survival (that is, successful operation) of 0.95 at 4000 h. Assume that time-to-failure follows an exponential distribution. Soluiton : R(4000) = 0.95 exp(-x4000) = 0.95 = ln(0.95) / 4000 = 0.0000128 / hour (the highest failure rate)
WeibullDistribution1 Is used to model the time to failure of product that have a varying failure rate (in debugging or wear-out phase) T = time to failure Pdf : Parameters : : location parameter - < < : scale parameter > 0 : shape parameter > 0 If =0 and = 1, it becomes the exponential distribution.
WeibullDistribution2 For reliability modelling, the location parameter will be zero(Mitra, 1998) if : < 1, PDF will close to exponential distr. = 1, PDF is eksponentialdistribution 1 < < 3, PDF is skewed ≥ 3, PDF is symetrically distributed (close to normal distr.) f(t) = 0.5 2 - = 4 = 1 = 2 1 - t
DistribusiWeibull3 Reliability function : Failure rate :
WeibullDistribution4 r(t) vst r(t) = 0.5 = 3.5 = 1 = 1 → r(t)is constant Then follows the exp. distr t
WeibullDistribution5 Mean Time To Failure : Ifr I (Integer) :
WeibullDistribution6 Example 11-3 Capacitors in an electrical circuit have a time-to-failure distr. that can be modelled by the weibull distribution with a scale parameter of 400h and a shape parameter of 0.2. . • What is the reliability of the capacitor after 600h of operation? • Find the mean time to failure ? • Is the failure rate increasing or decreasing with time? Solution : 1. R(600) = exp(-(600/400)0.2) = 0.3381
WeibullDistribution8 2. MTTF = 400 (1/0.2 + 1) = 400 (5 + 1) = 400 (6) = 400 (6 – 1)! = 400 * 120 = 48000 hours 3 r(t) = (0.2 t0.2-1)/(4000.2) = 0.0603 t-0.8 This function decreases with time. It would model component in the debugging phase
System Reliability 1 Illustration: SYSTEM
System Reliability 2 • Most products are made up of a number of components. The reliability of each component and the configuration of the system consisting of these components determines the system reliability . • Component configuration: • Series • Parallel • Combination between series and parallel
System Reliability - Series1 Systems with components in series: System reliability (Rs) is : Rs = R1 * R2 * R3 * … * Rn → The more components in series the smaller Rs . 1 2 3 n
System Reliability - Series2 If thetime to failure for each component follow the exponential distr. with the failure rate 1, 2, 3, … , nConstant, then : Rs = R1 * R2 * R3 * … * Rn = e-1*t e-2*t e-3*t … e-n*t = exp[-(∑ i)t]
System Reliability - Series3 Time to failure for the system withexponensially distributedwith thefailure rate : s = ∑i → If thecomponent isidentical failure rate then: s = n Generally the MTTF of the system is : MTTFs = 1 / s Example 11-5.page 520 Example 11-6.page 520
System Reliability – Parallel1 Systems with components in parallel: • All components operate at the same time. • System will operate at least one component operate. • System will fail if all component fail. • Purpose : To increase the system reliability. 1 2 … n
System Reliability – Parallel2 System has ncomponents : Reliability forcomponenti = Ri ; i = 1, 2, 3, … , n Failure probability of componenti = Fi = 1 – Ri System will fail with the probability : Fs = (1 – R1) (1 – R2) (1 – R3) … (1 – Rn) The System reliability is : Rs = 1 - Fs
System Reliability – Parallel3 If the time to failure of each component can be modelled by the exponential distr. With constantfailure rate1, 2, 3, … , n : → The time to failuredistribution of the system is not exponentially distributed.
System Reliability – Parallel4 The Mean Time To Failure of system with n identically components (assuming that each failed component is immediately replaced by an identical component) : Example 11-7 page 522 Example 11-8 page 522
System Reliability – Combination(Series-Parallel) The steps to calculate reliability: • Devide system into sub-system • Calculate the reliability for each sub-system • Calculate the reliability for the total system
CONTOH SOAL: System Reliability 5 ContohSoal 8 Determine the system reliability for the following figure : RA1 = 0.92 RA2 = 0.90 RA3 = 0.88 RA4 = 0.96 RB1 = 0.95 RB2 = 0.90 RB3 = 0.92 RC1 = 0.93 Page 523 B1 A1 A2 B2 C1 A3 A4 B3
CONTOH SOAL: System Reliability 6 ContohSoal 9 If the time to failure for each component is exponentially distributed with the following failure rate (in units/hour) : A1 = 0.0006 A2 = 0.0045 A3 = 0.0035 A4 = 0.0016 B1 = 0.006 B2 = 0.006 B3 = 0.006 C1 = 0.005 Page 523-524 B1 A1 A2 B2 C1 A3 A4 B3
System Reliability:Standby Component1 System with standby components : • It is assumed that only one component in the parallel configuration is operating at any given time. • One or more components wait to take over operation upon failure of the currently operating component. • The system reliability is higher than the systems with component in parallel. Basic Component Standby Comp. … Standby Comp.
System Reliability:Standby Component2 If the Time to failure of the component is to be exponential with failure rate . → Than the number of failure in certain time t adheres to the Poisson distribution with parameter = t. P(xfailures in time t) =
System Reliability:Standby Component3 System with one basic component and one standby component: Rs = P(system operate at time t) = P(no more than one failure at time t) = P(x = 0) + P(x = 1) = e-t + e-t(t)
System Reliability:Standby Component4 System with one basic component and two standby component : Rs = P(system operate at time t) = P(no more than two failure at time t) = P(number of failure ≤ 2) = P(x ≤ 2) = P(x = 0) + P(x = 1) + P(x = 2) = [e-t]+ [e-t(t)] + [e-t(t)2 / 2!]
System Reliability:Standby Component5 Generally, system with one basic component and n standby component will have reliability : The Mean Time To Failure of the system : Example 11-11, page 525
Operating Characteristic Curve 1 • OC curve shows the probability of acceptance of a lot from the life and reliability testing plan • In the OC curve, the probability of acceptance of a lot, Pa (ordinate Y) is a function of lot quality shown by The Mean Time To Failureof the item in the lot, (ordinate X)
Operating Characteristic Curve 2 • Life and reliability sampling plan : • Taking sample from a batch/lot • Observe the sample for a certain predetermined time • If the number of failures exceeds a stipulated acceptance number, the lot is rejected; if the number of failure is less than or equal to the acceptance number, the lot is accepted. • Two options are possible : 1. an item that fails is replaced immediately by an identical item. 2. failed items are not replaced
Operating Characteristic Curve 4 Making the OC curve (from a testing plan) : • Assume : time to failure is exponential distributed () → The number of failure at time t follows a Poisson distribution(t). • Parameters of testing plan : • Test time (T) • Sample size (n) • Acceptance number (c) • Probability to accept lot is calculated by using Poisson distribution.
example: OC Curve1 Example 11-12 It is known the parameter T = 800 hour, n = 12, and c = 2. The failure Item is replaced immediately by an identical item. Construct the OC curve! Expectation of the failure number in time T = nT For = 8000 hours (it is chosen) = (1/8000) per hour E(Failure) = (12)(800)(1/8000) = 1.2 Pa = P(lot acceptance) = P(failure ≤ 2 | = 1.2) = 0.879 For = 1000 hours = (1/1000) per hour E(failure) = (12)(800)(1/1000) = 9.6 Pa = P(lot acceptance) = P(failure ≤ 2 | = 9.6) = 0.0042 See table 11-1 See Figure 11-8
Operating Characteristic Curve 5 This OC curve is valid for other life testing plan as long as the total number of item hours tested is 9600 (item hour = nT) and acceptance number = 2. Example : Plan with n = 10, T = 960, c = 2 or n = 8, T = 1200, c = 2
Operating Characteristic Curve 6 Producer’s risk () : The risk of rejecting a good lot (products with a satisfactory mean life of ) = P(rejecting a good product) = 1 - Pa Consumer’s risk () : The risk of accepting a poor lot (products with an unsatisfactory mean life of ) = P(accepting a poor product) = Pa
Operating Characteristic Curve 7 With testing plan n = 12, T = 800, c = 2 : If the lot with a mean life of 20000 hours are satisfactory ( = 20000), then probability to reject this lot is 1 – Pa = 1 - 0.9872 = 0.0128 (Producer’s risk) If the lot with a mean life of 2000 hours are undesirable, then the probability to accept this lot is Pa = 0.1446 (Consumer’s risk)
Reliability And Life Testing Plans 1 Plans for reliability and life testing are usually destructive in nature . The testing time increase → The cost also increase The testing is usually performed at prototype stage.
Reliability And Life Testing Plans 2 Types of test Failure – Terminated test • The tests are terminated when a preassigned number of failure occurs in the chosen sample. • Parameters : sample size (n), preassigned number of failure (r), and the stipulate mean life by C.