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1、Probability multiplication,PFD=PD(AB)=PD(A)*PD(B),A,B,PSU,PSU,PSD,PSD,共因系数b,诊断覆盖率DC,PFD=PD(AB)=PD(A)+PD(B)PD(A B),PFS=PS(AB)=PS(A)+PS(B)-PD(A)*PD(B),Probability multiplication,PFD=P(A B)+P(B C)+P(A C),Functional safety Engineering-Probability,Probability,Rules of ProbabilityTypes of EventProbability
2、 multiplicationProbability additionFault Trees,Probability Assignment,Probability assigned by two methods:Physical property determinationGeometry,physical shapeExperimental outcome determinationNumber of occurrences/Number of TrialsProbability is a number:(0 P 1),Probability,P(Gold)=0.8,Functionalsa
3、fety,P(Mable)=0.75,Event Types,Independent Events that do not affect each otherCoin TossesDice ThrowsComplimentary When one outcome does not occur,the other will always occurCoinMutually Exclusive When one event occur the other can not happenDice,Independent Events,Events that do not effect each oth
4、erTossing of two coinsThrowing a pair of diceFailure of one component in a system?Normally considered independent,Complimentary Events,Complimentary When one outcome does not occur,the other will always occurTossing a coin Two events possible head and tailsSuccess/Failure?Probability of Complementar
5、y EventsP(A*)=1 P(A)Probability of successful operation for the next year is 0.8.What is the probability of failure the next year?(0,2),Mutually Exclusive Events,Mutually ExclusiveWhen one event occur the other can not happenToss of one dieOutcome of 1,2,3,4,5,6 are mutually exclusive,Probability mu
6、ltiplication,P(A AND B)=P(A)*P(B),Probability multiplication,For Independent eventsP(A AND B)=P(A)*P(B),Limit Switch,Solenoid Valve,In the next year,the probability of successful operation for a limit switch is 0.9 and the probability of successful operation for a solenoid valve is 0.98.What is the
7、probability of success for the system consisting of both elements?Limit switch AND solenoideP(A)*P(B)=0.9*0.98=0.882,Probability Addition,Mutually Exclusive EventsP(A or B)=P(A)+P(B),A,B,Probability Addition,Mutually Exclusive EventsP(A or B)=P(A)+P(B)Roll 1 die.What is the probability of getting a
8、4 or a 6?Probability of getting 4-1/6Probability of getting 6-1/6The probability of rolling 4 OR 6 P(A)+P(B)=1/6+1/6=2/6,Probability Addition,Independent EventsP(A or B)=P(A)+P(B)P(A AND B),A,B,Not Mutually Exclusive,Probability Addition,Independent EventsP(A or B)=P(A)+P(B)P(A AND B)A bag contains
9、100 objects.All are either round marbles or square blocksAll are either red or gold75%of objects are marbles80%of the objects are goldIf an object is randomly selected what is the probability that it will be either a marble OR gold?,Probability Addition,Independent EventsP(A or B)=P(A)+P(B)P(A AND B
10、)75%of objects are marbles(25%are square)80%of the objects are gold(20%are red)If an object is randomly selected what is the probability that it will be either a marble OR gold?The marble and gold are not mutually exclusive,since it is possible to pick an object that is both marble AND gold.P(A or B
11、)=P(A)+P(B)P(A AND B)=0,75+0,8(0,75*0,8)=1,55 0,6=0,95,Probability Addition,P(Gold)=0.8,P(Mable)=0.75,The probability of getting a gold or marble can also be calculated by using the rule of complementary events.The only way to NOT get the desired result is to get a red block.That possibility equals
12、0,2*0,25=0,05Therefor 1-0,05=0,95,Independent Events,Fault Tree Analysis,Fault Tree Graphical method to show the logical relationship of failure probabilities and frequencies,And GATE 1,OR GATE 2,TOP,Fault Tree Top Event,Fault Tree Analysis,Fault Tree Graphical method to show the logical relationshi
13、p of failure probabilities and frequenciesFault tree contains 3 partsInitiating EventsBranches or propagation step or escalating stepsOutcomes,Fault Tree AND Analysis,Battery dicharge,AND,Battery System Faiure,Charger fails,P=0,2,P=0,01,Quantitative analysis of Fault Trees combine probability using
14、probability multiplicationWhat is the probability of battery system failure?AND gates are solved using probability multiplicationPtop=0,2*0,01=0.002,Fault Tree OR Analysis,Solenoid fails to vent actuator,OR,Shutdown valve Fails to close,Valve stem stick,preventing closure,P=0,001,P=0,001,Quantitativ
15、e analysis of Fault Trees combine probability using probability additionWhat is the probability the valve fails to close?OR gates are solved using probability addition(non-mutual exclusive)Ptop=0,001+0,001(0.001*0.001)=0.001999,System Reliability Engineering,Reliability block diagramMarcovMultiple F
16、ailure rateCommon Cause,Quantitative system Analysis Techniques,System Modeling We know the Reliability(failure rate)of the components,what is the Reliability of the system?Reliability Block DiagramSimplified EquationsFault Tree DiagramsMarkov model,Power Supply,Power Supply,Controller,Controller,A,
17、B,A,B,Ra=m2Rb=(m2+m2)-m4=m2(2-m2)Rc=(2m-m2)(2m-m2)=m2(2-m)2Rc-Rb=m2(4-4m+m2-2+m2)=2m2(1-2m+m2)=2m2(m-1)20,Quantitative system Analysis Techniques,Simplified equations derived from one of the techniques listed belowReliability Block Diagram(RBD)Best for Reliability/Availability analysis.Probability c
18、ombination methods.Takes the success view.Confusing when in multiple failure mode modelingFault Tree Diagram(FTD).Takes the best failure view.Probability combination methods.Multiple drawings can be used for multiple failure mode.Easy to understand the drawingMarkow Model(MM),Looks at the success an
19、d failure on one drawing.Flexible,solved for probability as a function of time interval.Few educated in the method,Reliability Block diagram,Series systemAvailabilityProbability of SuccessUnavailabilityProbability of failure,Power Supply,Controller,A,B,System operate only if all components operate,R
20、eliability Block diagram,Parallel systemAvailabilityProbability of SuccessUnavailabilityProbability of failure,Power Supply,Power Supply,A,B,System operate if any component operate,Reliability Block Diagram,Series/ParallelExAPS=0.6AC=0.8(for a 1 year interval)ASystem=(APS*AC)+(APS*AC)-(APS*AC)2=(0,6
21、*0,8)+(0,6*0,8)(0,6*0,8)2=0,7296,Power Supply,Power Supply,Controller,Controller,A,B,A,B,Multiple Failure Modes,Typically categorized asSafe(System causes a false trip)Dangerous(System can not resond)Ex Failure mode from a pressure transmitterOutput saturated HI(S/D)Output saturated LO(S/D)Frozen ou
22、tput(D)Indication Error(HI)DIndication Error(LO)D,Common Cause,Common Cause can be used when use of same type of equipmentCommon stress failed both units in a redundant systemStress combination of temperature,humidity,corrosion,shock vibration,electrical surge,and more,Reducing Common Cause,Physical
23、 Separation redundant units are less likely to use common stressDiverse ArchitectureRedundant units respond differently to common stress,Basic Reliability Engineering,Failure rateReliability/UnrealiabilityReliability Enginering TermsConstanf Failure RatePF avg,Failures,Random failureA failure occuri
24、ong at a random time,which result from one or more degradation mechanismCan be influenced by temperature,humidity,vibrationsSystematic failureA failure related in a deterministic way to a certain cause,which can only be eliminated by:A modification of the design or of the manufactoring process,Opera
25、tor proceduresDocumentationOther relevant factors,Failure rate,Failure rate number of failures per operating hoursFailure rate that varied with timeConstant failure rateAverage failure rate over a long periodEx:100 solenoids are placed into operationDuring first year 7 failWhat is the avg failure ra
26、te during the year?f=7/(100 units*8760 hrs/year)=7,99E-6 failure/yearMost conservative:=7/(93 units*8760 hrs/year)=8,6 E-6 failure/year,Failure rate Equation,=Nf/(Ns*t)=f/Rf=Nf/(N*t)F=Nf/NR=Ns/(Ns+Nf)Ns=Number of successfull units at the end of the time periodNf=Number of failed units at the end of
27、the timeperiodNf=Number of failed units during a time period.t=time period,Terms,Failure rate the number of failures per unit of timeFailure rate that varies with timeConstant failure rateAverage failure rate over along period of timeProbability of success the chance that a system will perform its i
28、ntended function when operated within its specified limits.,Reliability/Safety engineering terms,Reliabilitythe probability of success during an interval of timeR=Ns/(Ns+Nf)ExampleThe probability of successful operation for 1 hour=0,999What is the probability of successful operation in 1 day?PS(24)=
29、PS(1hour)*PS(1hour).24PS(24)=(1 hour)24=(0,999)24=0,976,Reliability/Safety Engineering Terms,Failure Rate Failure per unit time per deviceMean Time To Fail(MTTF)The average successful operating time of a systemMTTF=R(t)dt,Constant Failure Rate,Failure rate:(t)=Reliability:R(t)=e-t Unreliability:F(t)
30、=1-e-t Mean time to Failure:MTTF=R(t)dt=1/Approximation:ex=1+x+x2/2!+x3/3!.1+xR(t)e-t 1 tF(t)t or PF=t,Repairable Systems,What about repairable systems?The measurement“reliability”requires that a system be successful for an interval of time.What is needed for a repairable system is a measure that gi
31、ves us the probability that it will work successfully in the situation where repair can be done,MTTR Mean Time To Restore,Mean Time To Failure(MTTF)The average successful operating time interval of a systemMean Time to Restore(MTTR)The average failure time interval of a system.Applies only to repair
32、able systemsAverage time to detect that a failure has occurred+average time to actually make the repairRestore Rate()-Number of restores per time period,Mean Time to Restore,Average time to detect that a failure has occurred+average time to actually make the repairEx:If failures only are detected by
33、 periodic inspection and testTI=Test interval,RT=Repair TimeMTTR=TI/2+RT,Mean Time Between Failures(MTBF),Mean Time Between FailuresThe average time interval of one failure/restore cycle of a systemApplies only to repairable systemsMTBF=MTTF+MTTR,Availability/Unavailability,Probability of Success th
34、e chance that a system will perform its intended function when operational within its specified limitsAVAILABILITYThe probability of success at a moment of time(allows for past failures)UNAVAILABILITYThe probability of failure at a moment in timeA=1-U,Availability vs reliability,Availability the pro
35、bability of success at a moment of timeReliability the probability of success for an interval of time(no failure and repair during interval)0 or 1,Reliability/Safety engineering terms,Reliabilitythe probability of success during an interval of timeExampleThe probability of successful operation for 1
36、 hour=0,999What is the probability of successful operation in 1 day?PS(24)=PS(1hour)*PS(1hour).24PS(24)=(1 hour)24=(0,999)24=0,976,Available vs reliabilty(Safety reliable),A system or component is reliable even if it has hade a safe shutdown(PFS)or spurious trip.But a system is only available if it
37、is up and running and performs the tasks it is intended to do.The relation between availability and reliability is as follow:Reliability is the probability of a component,or system,functioning correctly over a given period of time under a given set of operating conditions.Reliability example:A SIF i
38、s operating in a PSD system on an oil installation.This SIF is controlling a pressure relief valve on a gas vessel.It is critical that this SIF is able to respond on an action.It is only possible to do maintenance once a year.Then the proof test interval is 1 year.The reliability for this SIF is the
39、 probability that the SIF will relief the pressure if a high pressure should occur,or that the SIF is in a failsafe mode.The availability of a system is the probability that the system will be functioning correctly at any given time.Availability example:The same SIF as mentioned in the reliability e
40、xample also have an availability figure.The availability for this SIF is the probability that the SIF is operational.Operational mean that the SIF continuously performs its intended functionality,i.e.supervising the pressure including relief of the pressure upon a high pressure demand.This is not po
41、ssible when the SIF is in the fail safe mode.A:Availability,R:Reliability(0 or 1),PFS:Probability of Failing Safely(spurious trio);A=R PFS(0&1),IEC61508 Part 6举了好几个系统结构例子。但描述得很粗略和含糊，加上其名字和体系结构图非常相近，所以很多人没法理解其区别。个人认为，其用意不在系统结构设计，而是在于突出不同的硬件体系结构设计会有不同的系统可靠性结果，但似乎很多人把它们作为安全相关系统设计的典范。以下我按IEC原文意思解释一下这三个体
42、系结构：1oo2结构包含了两个并联通道，每个通道都可以处理同一个安全功能，很明显这两个通道是冗余设计（尽管IEC原文没有说），一通道可以看作是另外一通道的备份。当这双通道均失效，其安全功能便失效。该结构还含有在线诊断，但其诊断只作检测错误和记录，不能改变系统状态和输出，所以对系统可靠性和安全都没有帮助。2oo2结构和1oo2相似，都含有两个并联通道和在线诊断，但不同的是2oo2的双通道是一起合作完成一个安全功能，所以如果其中一个通道失效会导致整个安全功能失效。我理解，其并联结构安排是因为该两通道都需要同一输入数据，但负责安全功能的不同部分，因此每个通道的输出结果需要捆绑在一起，才算完成其安全功
43、能。很明显，2oo2的可靠性肯定远低于1oo2，因为前者是单点失效模型。IEC原文没有提到这双通道是否冗余，照理应该不是，既然每个通道负责不同的功能。当然也可以有冗余部分，只能看具体应用。1oo2D结构和2oo2几乎一样，但不同的是1oo2D的在线诊断在必要时可以改变系统状态：一旦发现其一通道有问题，可以把系统输出转换为以另一通道的输出为准，或者如果该双通道都失效，把系统转换为安全模式（fail safety）。另外IEC原文暗示其在线诊断是通过互相比较这双通道的状态来检测是否双通道均失效。如果这样话，该双通道应该是冗余，这样却跟2oo2有点矛盾，况且1oo2D这个名字应该是基于1oo2而不是2oo2，真是想不明白。IEC原文也没有交待如何检测单一通道错误，我估计应该是sanity checking/consistency checking/etc。,
