Presentation on theme: "Mitra, A. (1998), Fundamentals of Quality Control and Improvement 2 nd ed., New Jersey: Prentice- Hall. Ebeling, C. E. (1997), An Introduction To Reliability."— Presentation transcript:
Mitra, A. (1998), Fundamentals of Quality Control and Improvement 2 nd 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 4 th ed., Chichester: John Wiley & Sons.
1.Able to explain about Reliability. 2.Able to use the probability distribution in reliability modelling. 3.Able to calculate system reliability. 4.Able to perform the reliability testing.
Reliability → measure of quality of the product over the long run. Reliability → Probability of a product performing 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
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, dan lim t → R(t) = 0.
CDF (Cumulative Distribution Function): F(t) = P(T < t) F(t) = 1 – R(t) where F(0) = 0 dan lim t → F(t) = 1 So F(t) is the probability of the system or component that fail before time t.
PDF (Probability Density Function): f(t) = dF(t)/dt= -dR(t)/dt where f(t) ≥ 0 and 0 f(t) dt = 1 Reliability function and CDF come from the probability with function f(t). So, reliability (R(t)) and the failure probability (F(t)) are in the range [0,1].
Mean Time To Failure (MTTF) → The average of useful life until the product or component fail. Partial Integral : u = tdu = dt dv = -dR(t)v = -R(t)
Based on the failure rate there are 3 distinct phases of the product : 1.Debugging phase (infant – mortality phase / Burn-in) The initial is high and then drop over time The problem is identified and then is improved 2.Chance-failure phase (Useful life) Failure occur randomly and independently. is constant Represents the Useful life of product 3.Wear-out phase increases The end of product useful live
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
Reliability at time t (R(t)) → the probability of the product lasting up to at least time t. R(t) t 1
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 if time-to-failure is exponential distribution, failure rate is equal to (constant failure rate)
Example 11-1 An amplifier has an exponential time-to-failure with a failure rate of 8% per 1000h.. 1.What is the reliability at 5000h? 2.Find the mean time to failure? Solution : 1.Reliability at 5000h = 0.08/1000 hour = 0.00008/hour t = 5000 hour → R(5000)= exp(- t) = exp(-0.00008*5000) = 0.6703 2.Mean time to failure MTTF= 1/ = 1 / 0.00008 jam = 12500 jam
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)
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.
For reliability modelling, the location parameter will be zero (Mitra, 1998) 1 - 2 - f(t) = 0.5 = 4 = 1 = 2 t if : < 1, PDF will close to exponential distr. = 1, PDF is eksponential distribution 1 < < 3, PDF is skewed ≥ 3, PDF is symetrically distributed (close to normal distr.)
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.. 1.What is the reliability of the capacitor after 600h of operation? 2.Find the mean time to failure ? 3.Is the failure rate increasing or decreasing with time? Solution : 1. R(600)= exp(-(600/400) 0.2 ) = 0.3381
2. MTTF= 400 (1/0.2 + 1) = 400 (5 + 1) = 400 (6) = 400 (6 – 1)! = 400 * 120 = 48000 hours 3 r(t)= (0.2 t 0.2-1 )/(400 0.2 ) = 0.0603 t -0.8 This function decreases with time. It would model component in the debugging phase
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
Systems with components in series: System reliability (Rs) is : Rs = R1 * R2 * R3 * … * Rn → The more components in series the smaller Rs. 1 1 2 2 3 3 n n
If the time to failure for each component follow the exponential distr. with the failure rate 1, 2, 3, …, n Constant, then : Rs = R1 * R2 * R3 * … * Rn = e - 1*t e - 2*t e - 3*t … e - n*t = exp[-( ∑ i )t]
Time to failure for the system with exponensially distributed with the failure rate : s = ∑ i → If the component is identical failure rate then: s = n Generally the MTTF of the system is : MTTF s = 1 / s Example 11-5.page 520 Example 11-6.page 520
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 1 2 2 n n …
System has n components : Reliability for component i = R i ; i = 1, 2, 3, …, n Failure probability of component i = F i = 1 – R i System will fail with the probability : F s = (1 – R 1 ) (1 – R 2 ) (1 – R 3 ) … (1 – R n ) The System reliability is : R s = 1 - F s
If the time to failure of each component can be modelled by the exponential distr. With constant failure rate 1, 2, 3, …, n : → The time to failure distribution of the system is not exponentially distributed.
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
The steps to calculate reliability: 1.Devide system into sub-system 2.Calculate the reliability for each sub- system 3.Calculate the reliability for the total system
Contoh Soal 8 Determine the system reliability for the following figure : R A1 = 0.92 R A2 = 0.90 R A3 = 0.88 R A4 = 0.96 R B1 = 0.95 R B2 = 0.90 R B3 = 0.92 R C1 = 0.93 Page 523 A1 A2 A3 A4 B1 B2 B3 C1
Contoh Soal 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 A1 A2 A3 A4 B1 B2 B3 C1
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. …
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(x failures in time t) =
System with one basic component and one standby component: R s = 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 with one basic component and two standby component : R s = 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!]
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
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 Failure of the item in the lot, (ordinate X)
Life and reliability sampling plan : 1.Taking sample from a batch/lot 2.Observe the sample for a certain predetermined time 3.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. 4.Two options are possible : 1. an item that fails is replaced immediately by an identical item. 2. failed items are not replaced
Making the OC curve (from a testing plan) : 1.Assume : time to failure is exponential distributed ( ) → The number of failure at time t follows a Poisson distribution( t). 2.Parameters of testing plan : – Test time (T) – Sample size (n) – Acceptance number (c) 3.Probability to accept lot is calculated by using Poisson distribution.
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
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 withn = 10, T = 960, c = 2 or n = 8, T = 1200, c = 2
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
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)
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.
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.
From the test result, let’s suppose the accumulated test time of the item is, from which an estimate of average life is if : → the lot is accepted → the lot is rejected
2. Time – Terminated test Testing is terminated when a preassigned time T is reached. Acceptance of the lot is based on the observed number of failure during the test time. If > r, the lot is rejected Parameter : sample size (n), testing time (T), and number of failure (r).
3. Sequential Reliability Testing (no prior decision is made as to the number of failures or the to conduct the test, instead, the accumulated results of the test are used to decide whether to accept or to reject the lot) sequential reliability testing : ordinate X : Testing accumulated time Ordinate Y : cumulative failure number Rejection line and acceptance line are calculated based on : Acceptance mean life o dan producer’s risk for o Minimum mean life 1 dan consumer’s risk for 1
When testing is performed, the cumulative failure number and accumulated time are plotted Plot in between two lines: Continue the testing Plot is in the acceptance region: the testing is stopped and the lot is accepted Plot is in the rejection region : the testing is stopped and the lot is rejected Look figure 11-9, page 528
Performing the reliability testing and product lifetime. With replacement The failure product is changed by the other product chosen randomly from the batch. Without replacement The failure product is not changed The three type of testing can be performed with or without replacement.
Assume : time to failure is exponentially distributed with parameter [~exp( )] 1. FAILURE – TERMINATED TEST Sample size : n Preassigned number of failure : r Suppose the failures occur at t 1 ≤ t 2 ≤... ≤ t r
Accumulated life for the test items until the r th failures (T r ) Without replacement With replacement
Estimated mean life Estimated failure rate 100(1- )% confidence interval for : Example 11-13, page 529 Example 11-14, page 530
2. TIME – TERMINATED TEST Sample size : n Preassigned time to terminate : T Observed number of failure : x (Random variable) t i : the time of failure of the i th item
Accumulated life for the test items Without replacement With replacement
Estimated mean life Estimated failure rate 100(1- )% confidence interval for : Example 11-15, page 531
“Failure is success if we learn from it.” -- Malcolm S. Forbes “I am only one; but still I am one. I cannot do everything, but still I can do something; I will not refuse to do the something I can do.” -- Helen Keller