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Reliability Engineering ( Rekayasa Keandalan )

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 ( Rekayasa Keandalan )

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  1. Reliability Engineering(RekayasaKeandalan) Y M Kinley Aritonang, Ph.D

  2. 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.

  3. 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.

  4. 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

  5. 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.

  6. 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.

  7. 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].

  8. 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)

  9. 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

  10. Life-Cycle Curve 2 Debugging phase Chance-Failure phase Wear-out phase  t (time) Life-cycle curve = bathtub curve

  11. Probability Distribution to model Failure Rate (time to failure) : • EksponensialDistribution →  is constant • WeibullDistribution →  is notconstant

  12. EksponensialDistribution1 T = time to failure Pdf : Cdf :

  13. 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

  14. EksponensialDistribution3 Reliabilityat time t (R(t)) → the probability of the product lasting up to at least timet. R(t) 1 t

  15. 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)

  16. 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

  17. 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)

  18. 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.

  19. 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

  20. DistribusiWeibull3 Reliability function : Failure rate :

  21. WeibullDistribution4 r(t) vst r(t)  = 0.5  = 3.5  = 1  = 1 → r(t)is constant Then follows the exp. distr t

  22. WeibullDistribution5 Mean Time To Failure : Ifr I (Integer) :

  23. 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

  24. 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

  25. System Reliability 1 Illustration: SYSTEM

  26. 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

  27. 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

  28. 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]

  29. 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

  30. 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

  31. 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

  32. System Reliability – Parallel3 If the time to failure of each component can be modelled by the exponential distr. With constantfailure rate1, 2, 3, … , n : → The time to failuredistribution of the system is not exponentially distributed.

  33. 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

  34. 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

  35. 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

  36. 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

  37. 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.

  38. 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) =

  39. 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)

  40. 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!]

  41. 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

  42. 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)

  43. 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

  44. 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.

  45. 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

  46. 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

  47. 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

  48. 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)

  49. 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.

  50. 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.

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