Military Reliability ModelingWilliam P. Fox, Steven B. HortonIntroductionYouare an infantry rifle platoon leader.Yourplatoon is occupying a battleposition and has been ordered to establishan observationpost (OP)on ahilltopapproximatelyonekilometerforwardofyourposition.TheOPwill beoccupiedbythreesoldiersfor24hours.Hourlysituationreports mustbe madebyradio.All necessaryrations,equipment,andsuppliesforthe24hourperiod mustbecarried withthem. TheOPisineffectiveunlessitcancommunicatewithyouin a timelymanner.Therefore,radiocommunicationsmustbereliable.Theradiohasseveral componentswhichaffectitsreliabilityanessentialonebeingthebatteryBatterieshave a useful life whichis notdeterministic(wedo notknowexactlyhowlongabatterywill lastwhenweinstallit).Itslifetimeisavariablewhichmaydependonprevioususe,manufacturingdefects,weather,etc.Thebatterythatisinstalled intheradiopriortoleavingfortheOP could lastonlyafewminutesorfortheentire24hours.Sincecommunicationsaresoimportanttothismission,we are interested in modelingand analyzing the reliability of the battery.We will use the followingdefinition forreliability:If Tisthetimetofailure ofa component ofa system,and f(t)is theprobabilitydistributionfunction ofT,thenthe components'reliabilityattimetisR(t) = P(T >t) = 1 - F(t).R(t)iscalledthereliabilityfunctionandF(t)isthecumulativedistributionfunctionof f(t).Ameasure of this reliabilityis theprobabilitythata givenbattery will last morethan24hours.Ifweknowtheprobabilitydistributionforthebatterylife,wecanuseourknowledgeofprobabilitytheorytodeterminethereliability.Ifthebatteryreliability is below acceptable standards,one solution is to have the soldiers carryspares.Clearly,themore sparesthey carry,theless likelythereistobeafailureincommunicationsduetobatteries.Ofcoursethebatteryisonlyonecomponentof theradio.Others include the antenna,handset, etc.Failure of any one of theessential components causesthesystemtofail.153
153 Military Reliability Modeling William P. Fox, Steven B. Horton Introduction You are an infantry rifle platoon leader. Your platoon is occupying a battle position and has been ordered to establish an observation post (OP) on a hilltop approximately one kilometer forward of your position. The OP will be occupied by three soldiers for 24 hours. Hourly situation reports must be made by radio. All necessary rations, equipment, and supplies for the 24 hour period must be carried with them. The OP is ineffective unless it can communicate with you in a timely manner. Therefore, radio communications must be reliable. The radio has several components which affect its reliability, an essential one being the battery. Batteries have a useful life which is not deterministic (we do not know exactly how long a battery will last when we install it). Its lifetime is a variable which may depend on previous use, manufacturing defects, weather, etc. The battery that is installed in the radio prior to leaving for the OP could last only a few minutes or for the entire 24 hours. Since communications are so important to this mission, we are interested in modeling and analyzing the reliability of the battery. We will use the following definition for reliability: If T is the time to failure of a component of a system, and f(t) is the probability distribution function of T, then the components’ reliability at time t is R(t) = P (T >t) = 1 - F(t). R(t) is called the reliability function and F(t) is the cumulative distribution function of f(t). A measure of this reliability is the probability that a given battery will last more than 24 hours. If we know the probability distribution for the battery life, we can use our knowledge of probability theory to determine the reliability. If the battery reliability is below acceptable standards, one solution is to have the soldiers carry spares. Clearly, the more spares they carry, the less likely there is to be a failure in communications due to batteries. Of course the battery is only one component of the radio. Others include the antenna, handset, etc. Failure of any one of the essential components causes the system to fail
Thisisarelativelysimpleexampleofoneofmanymilitaryapplicationsofreliability.Thischapterwill showwecanuseelementaryprobabilitytogeneratemodels that can be used todeterminethe reliability of military equipmentModeling ComponentReliabilityInthis section,wewill discuss howtomodel component reliability.Recallthatthe reliability function, R(t) is defined as:R(t)= P(T > t) = P(component fails after time t).This canalsobe stated, using Tas thecomponent failure time,asR(t)= P(T> t) = 1- P(T ≤t)= 1- [f(x)dx = 1- F(t) Thus,if weknowtheprobabilitydensityfunction f(t)ofthetimetofailureT,wecan use probability theoryto determinethereliability function R(t).Wenormallythinkofthesefunctions asbeingtimedependent;however,thisisnotalways thecase.Thefunctionmightbe discrete such asthe lifetime ofa cannontube.It isdependentonthenumberofroundsfiredthroughit(adiscreterandomvariable)Auseful probabilitydistribution in reliabilityis theexponential distribution.Recallthat its densityfunction isgivenby[ae-st>0f(t) :otherwise'owhere the parameter a is such that Y, equals the mean of the random variableT. If T denotes the time to failure of a piece of equipment or a system, then isthemeantimetofailurewhichis expressed in units oftime.For applications ofreliability,we will usetheparameter α.Since/is themeantimetofailure,2istheaveragenumberoffailures perunit time orthefailurerate.Forexample, ifa light bulb has a time to failure thatfollows an exponential distribution with ameantimetofailureof50hours,thenitsfailurerateis1lightbulbper50hoursor1/50perhour,sointhiscase=0.02perhour.NotethatthemeanofT,themean time to failure of the component, is Ya.Example1:Let's considertheexamplepresented inthe introduction.Lettherandom variable T bedefined as follows:154
154 This is a relatively simple example of one of many military applications of reliability. This chapter will show we can use elementary probability to generate models that can be used to determine the reliability of military equipment. Modeling Component Reliability In this section, we will discuss how to model component reliability. Recall that the reliability function, R(t) is defined as: R(t) = P(T > t) = P(component fails after time t). This can also be stated, using T as the component failure time, as R(t) P(T t) P(T t) f ( x)dx F(t) t = > = - £ = - = - - ¥ ò 1 1 1 . Thus, if we know the probability density function f (t) of the time to failure T, we can use probability theory to determine the reliability function R(t). We normally think of these functions as being time dependent; however, this is not always the case. The function might be discrete such as the lifetime of a cannon tube. It is dependent on the number of rounds fired through it (a discrete random variable). A useful probability distribution in reliability is the exponential distribution. Recall that its density function is given by f t e t t ( ) = ì í î > - l l 0 0 otherwise , where the parameter l is such that 1 l equals the mean of the random variable T. If T denotes the time to failure of a piece of equipment or a system, then 1 l is the mean time to failure which is expressed in units of time. For applications of reliability, we will use the parameter l . Since 1 l is the mean time to failure, l is the average number of failures per unit time or the failure rate. For example, if a light bulb has a time to failure that follows an exponential distribution with a mean time to failure of 50 hours, then its failure rate is 1 light bulb per 50 hours or 1/50 per hour, so in this case l = 0.02 per hour. Note that the mean of T, the mean time to failure of the component, is 1 l . Example 1: Let's consider the example presented in the introduction. Let the random variable T be defined as follows:
T=timeuntilarandomlyselectedbatteryfailsSuppose radio batteries havea timetofailure that is exponentially distributedwithameanof3ohours.Inthiscase,wecouldwrite1T ~EXP =301Therefore,perhour,sothat301α930f(t) :t > 0 and F(t)dx-3030F(t),theCDFoftheexponential distribution,canbe integratedto obtain-1F(t) =1-e 30.t>oNowwecancomputethereliabilityfunctionforabattery:R(t) = 1- F(t) = 1-e30t>o1-Recall that intheearlierexample,thesoldiersmustoccupytheOPfor24hoursThereliabilityofthebatteryfor24hoursisR(24) = e%(24) =0.4493,so the probability that the battery lasts more than 24 hours is 0.4493.Example2:Wehavetheoptiontopurchaseanewnickelcadmiumbatteryforouroperation.Testinghas shownthatthedistribution ofthetimetofailurecanbe modeled using a parabolic function:XX0≤x≤4848384f(x):[0otherwiseLettherandomvariableTbedefined asfollows:155
155 T = time until a randomly selected battery fails. Suppose radio batteries have a time to failure that is exponentially distributed with a mean of 30 hours. In this case, we could write T ~ EXP = 1 30 l æ è ç ö ø ÷ . Therefore, l = 1 30 per hour, so that f(t) e t = - 1 30 1 30 , t > 0 and F(t) e dx x t = - ò 1 30 1 30 0 . F(t), the CDF of the exponential distribution, can be integrated to obtain F(t) e t = - - 1 1 30 , t > 0. Now we can compute the reliability function for a battery: R(t) F(t) e e t t = - = - - æ è ç ö ø ÷ = - - 1 1 1 1 30 1 30 , t > 0 . Recall that in the earlier example, the soldiers must occupy the OP for 24 hours. The reliability of the battery for 24 hours is ( ) ( ) R 24 e 0 4493 1 30 24 = = - . , so the probability that the battery lasts more than 24 hours is 0.4493. Example 2: We have the option to purchase a new nickel cadmium battery for our operation. Testing has shown that the distribution of the time to failure can be modeled using a parabolic function: f x x x x ( ) = æ è ç ö ø ÷ - æ è ç ö ø ÷ £ £ ì í ï ï î ï ï 384 1 48 0 48 0 otherwise Let the random variable T be defined as follows:
T=time until a randomly selected batteryfailsInthiscase,wecouldwritef(t)=(t / 384)(1-t / 48), 0 ≤t ≤ 48 and F(t)= [(x/384)(1-x/48)dxRecall that intheearlierexamplethesoldiersmustmantheOPfor24hoursThereliabilityofthebatteryfor24hoursistherefore24R(24) = 1- F(24) = 1- [(t / 384)(1-t / 48)dx = 0.500Cwhichisanimprovementoverthebatteriesfromexample1ModelingSeriesSystemsNowweconsiderisasystemwithncomponentsC,,C,...,C,whereeachoftheindividual componentsmustwork inorderforthesystemtofunction.Amodelofthis type of system is shown in figure 1.ccc...Figure 1:Series systemIf we assume these components are mutually independent, the reliability of thistype of system iseasyto compute.Wedenotethe reliability of componentiattime t by R,(t). In other words, R,(t) is simply the probability that component iwill function continuously fromtime O through until time t.We are interested inthe reliability of the entire system of n components, but since these componentsaremutuallyindependent,thesystemreliabilityisR(t)= R,(t).R(t). ...-R,(t) .Example 3: Our radio has several components. Let us assume that there arefourmajor components--they are (in order)thehandset,the battery,thereceiver-transmitter,andthe antenna.Sincetheyall mustfunction properlyfortheradioto operate, we can model the radio with the diagram shown in figure2.156
156 T = time until a randomly selected battery fails. In this case, we could write f (t) = (t / 384)(1- t / 48), 0 £ t £ 48 and F(t) ( x )( x )dx t ò = - 0 / 384 1 / 48 . Recall that in the earlier example the soldiers must man the OP for 24 hours. The reliability of the battery for 24 hours is therefore R(24) 1 F(24) 1 (t 384)(1 t 48)dx 0 500 0 24 = - = - - = ò / / . which is an improvement over the batteries from example 1. Modeling Series Systems Now we consider is a system with n components C1 C2 Cn , ,., where each of the individual components must work in order for the system to function. A model of this type of system is shown in figure 1. C . 1 C2 C3 Cn Figure 1: Series system If we assume these components are mutually independent, the reliability of this type of system is easy to compute. We denote the reliability of component i at time t by Ri (t). In other words, Ri (t) is simply the probability that component i will function continuously from time 0 through until time t. We are interested in the reliability of the entire system of n components, but since these components are mutually independent, the system reliability is R(t) = R1 (t)×R2 (t)× ××× ×Rn (t) . Example 3: Our radio has several components. Let us assume that there are four major components - they are (in order) the handset, the battery, the receiver-transmitter, and the antenna. Since they all must function properly for the radio to operate, we can model the radio with the diagram shown in figure 2
CCFigure2:RadioSystemSuppose we know that the probability that the handset will work for at least 24hoursis0.6703,andthereliabilitiesfortheothercomponentsare0.44930.7261,and0.9531,respectively.Ifweassumethatthecomponentsworkindependentlyofeachother,thentheprobabilitythat theentiresystemworksfor24 hours is:R(24) = R;(24)·R2(24)·R(24)·R4(24) = (.6703)(.4493)(.7261)(.9531) = 0.2084Recall that two events A and B are independent if P(A|B)= P(A)ModelingParallelSystems(TwoComponents)Nowweconsiderasystemwithtwocomponentswhereonlyoneofthecomponentsmustworkforthesystemtofunction.Asystemofthistypeisdepicted in figure 3.Figure3:ParallelSystemofTwoComponentsNotice that in this situation the two components are both put in operation at timeO;theyareboth subjecttofailurethroughout theperiod of interest.Onlywhenbothcomponentsfail beforetimetdoesthesystemfail.Againwealsoassumethat thecomponentsare independent.Thereliability of thistype of system canbe found using the following well known model:P(AU B) = P(A) + P(B)- P(An B)Inthis case,Aistheeventthatthefirst componentfunctionsforlongerthansometime,t,andBistheeventthatthesecond componentfunctions longerthanthe sametime,t.Since reliabilities are probabilities,we can translate theaboveformula into the following:R(t) = R,(t)+R,(t)- R,(t)R2(t)157
157 C1 C2 C3 C4 Figure 2: Radio System Suppose we know that the probability that the handset will work for at least 24 hours is 0.6703, and the reliabilities for the other components are 0.4493, 0.7261, and 0.9531, respectively. If we assume that the components work independently of each other, then the probability that the entire system works for 24 hours is: R(24) = R1 ( 24)×R2 (24)×R3 (24)×R4 (24) = (.6703)( .4493)(.7261)( .9531) = 0.2084 . Recall that two events A and B are independent if P( A|B) = P(A). Modeling Parallel Systems (Two Components) Now we consider a system with two components where only one of the components must work for the system to function. A system of this type is depicted in figure 3. C1 C2 Figure 3: Parallel System of Two Components Notice that in this situation the two components are both put in operation at time 0; they are both subject to failure throughout the period of interest. Only when both components fail before time t does the system fail. Again we also assume that the components are independent. The reliability of this type of system can be found using the following well known model: P( AÈ B) = P(A) + P(B) - P( AÇ B). In this case, A is the event that the first component functions for longer than some time, t, and B is the event that the second component functions longer than the same time, t. Since reliabilities are probabilities, we can translate the above formula into the following: R(t) = R1 (t) + R2 (t) - R1 (t)R2 (t)
Example4:Supposeyourbattalioniscrossingariverthathastwobridges inthearea.Itwill take3hourstocompletethecrossing.Thecrossingwill besuccessful as long as at least one bridgeremains operational during the entirecrossing period.Youestimatethat enemyguerrllas with mortars havea one-third chance of destroying bridge 1 and a one-fourth chance of destroying bridge2 in the next 3hours.Assume the enemy guerrillas attacking eachbridgeoperateindependently.Whatistheprobabilitythatyourbattalioncancompletethecrossing?Solution:First we computethe individual reliabilities:R,(3) = 1- 1_ 233andR(3)=1_ 1_ 344Now it is easyto compute the system reliability:R(3) = R(3) + R,(3)-R(3)R,(3) = +3311=0.9167.343412ModelingActiveRedundantSystemsConsiderthesituation inwhichasystemhasncomponents,all ofwhichbeginoperating (areactive)attimet=o.The systemcontinuestofunction properlyaslong as atleast k of the components do notfail.In otherwords, if n-k+1componentsfail, the systemfails.Thistypeof component systemis called anactiveredundant system.The active redundant system can bemodeled asaparallelsystemofcomponentsasshowninfigure4below:158
158 Example 4: Suppose your battalion is crossing a river that has two bridges in the area. It will take 3 hours to complete the crossing. The crossing will be successful as long as at least one bridge remains operational during the entire crossing period. You estimate that enemy guerrillas with mortars have a onethird chance of destroying bridge 1 and a one-fourth chance of destroying bridge 2 in the next 3 hours. Assume the enemy guerrillas attacking each bridge operate independently. What is the probability that your battalion can complete the crossing? Solution: First we compute the individual reliabilities: R1 (3) 1 1 3 2 3 = - = and R2 (3) 1 1 4 3 4 = - = . Now it is easy to compute the system reliability: R(3) R ( 3) R ( 3) R (3)R ( 3) 2 3 3 4 2 3 3 4 11 12 = 1 + 2 - 1 2 = + - 0 9167 æ è ç ö ø ÷ æ è ç ö ø ÷ = = . . Modeling Active Redundant Systems Consider the situation in which a system has n components, all of which begin operating (are active) at time t = 0. The system continues to function properly as long as at least k of the components do not fail. In other words, if n - k + 1 components fail, the system fails. This type of component system is called an active redundant system. The active redundant system can be modeled as a parallel system of components as shown in figure 4 below:
C1Figure4:ActiveRedundantSystemWeassumethatall n componentsare identical and will fail independentlyIfweletT,bethetimetofailureoftheithcomponent,thentheT,terms areindependent and identically distributed for i=1,2,3,..,n. Thus R,(t), thereliabilityattimetforcomponenti,is identicalforall componentsRecallthatoursystemoperates ifatleastkcomponentsfunctionproperlyNowwedefinetherandomvariablesXandTasfollows:X = number of components functioning at time t, andT =time tofailureof the entire system.ThenwehaveR(t) = P(T >t) = P(X ≥k) .It iseasyto see that wenowhaven identical and independent components withthe same probability offailure bytime t.This situation corresponds to a binomialexperiment and we can solve for the system reliabilityusing thebinomialdistribution with parameters n and p = R,(t).Example5:ThreesoldiersonanOPhavebeeninstructedtoputout15sensorsforward oftheir OP todetect movement ina wooded area.Theyestimate thatanymovement throughthearea canbedetectedas longas atleast 12of thesensors areoperating.Sensorsareassumedtobe inparallel (activeredundant)i.e.:theyfail independently.Ifweknowthateach sensor has a 0.6065probabilityofoperatingproperlyforat least24hours,wecan computethereliabilityoftheentiresensorsystemfor24hours.Definethe randomvariable:X=numberof sensors workingafter24hours.159
159 Cn C2 C1. Figure 4: Active Redundant System We assume that all n components are identical and will fail independently. If we let Ti be the time to failure of the ith component, then the Ti terms are independent and identically distributed for i = 1,2,3,.,n . Thus Ri (t), the reliability at time t for component i, is identical for all components. Recall that our system operates if at least k components function properly. Now we define the random variables X and T as follows: X = number of components functioning at time t, and T = time to failure of the entire system. Then we have R(t) = P(T > t) = P( X ³ k). It is easy to see that we now have n identical and independent components with the same probability of failure by time t. This situation corresponds to a binomial experiment and we can solve for the system reliability using the binomial distribution with parameters n and p = Ri (t). Example 5: Three soldiers on an OP have been instructed to put out 15 sensors forward of their OP to detect movement in a wooded area. They estimate that any movement through the area can be detected as long as at least 12 of the sensors are operating. Sensors are assumed to be in parallel (active redundant), i.e.: they fail independently. If we know that each sensor has a 0.6065 probability of operating properly for at least 24 hours, we can compute the reliability of the entire sensor system for 24 hours. Define the random variable: X = number of sensors working after 24 hours
Clearly,therandomvariableXisbinomiallydistributedwithn=15andp=0.6065.Inthelanguageofmathematics,wewritethissentenceasX ~b(15,0.6065) or X~BINOMIAL(15,0.6065)Weknowthatthereliabilityofthesensorsystemfor24hours isR(24)= P(X ≥12) = P(12≤X ≤15) = 0.0990 Thus the reliability of the system for 24 hours is only 0.0990.ModelingStandbyRedundant SystemsActiveredundantsystemscansometimesbeinefficient.Thesesystemsrequireonlykofthencomponentstobeoperational,butall n components are initially in operationandthus subjecttofailure.Analternativeistheuseofsparecomponents.Suchsystems have onlyk components initiallyinoperation,exactlywhatweneedforthewhole systemto be operational.Whenacomponent fails,we have a spare"standingby"whichisimmediatelyputintooperation.For this reason,wecall these StandbyRedundant Systems.Supposeoursystemrequireskoperationalcomponentsandweinitiallyhave n-k spares available.When a component inoperationfails,adecision switch causesa spareor standby componenttoactivate(becominganoperationalcomponent).Thesystemwillcontinuetofunctionuntiltherearelessthank operational componentsremaining.In otherwords,the systemworksuntiln-k+1componentshavefailed.Wewillconsideronlythecasewhereoneoperational component isrequired (thespecial casewherek=1)and there aren-1 standby (spare) components available. We will assume that a decisionswitch(DS)controlstheactivationofthestandbycomponentsinstantaneouslyand100%reliably.Weusethemodelshowninfigure5belowtorepresentthissituation.160
160 Clearly, the random variable X is binomially distributed with n = 15 and p = 0.6065. In the language of mathematics, we write this sentence as X ~ b(15,0.6065) or X ~ BINOMIAL(15,0.6065). We know that the reliability of the sensor system for 24 hours is R(24) = P( X ³ 12) = P(12 £ X £ 15) = 0.0990 . Thus the reliability of the system for 24 hours is only 0.0990. Modeling Standby Redundant Systems Active redundant systems can sometimes be inefficient. These systems require only k of the n components to be operational, but all n components are initially in operation and thus subject to failure. An alternative is the use of spare components. Such systems have only k components initially in operation; exactly what we need for the whole system to be operational. When a component fails, we have a spare “standing by” which is immediately put in to operation. For this reason, we call these Standby Redundant Systems. Suppose our system requires k operational components and we initially have n - k spares available. When a component in operation fails, a decision switch causes a spare or standby component to activate (becoming an operational component). The system will continue to function until there are less than k operational components remaining. In other words, the system works until n - k + 1 components have failed. We will consider only the case where one operational component is required (the special case where k = 1) and there are n - 1 standby (spare) components available. We will assume that a decision switch (DS) controls the activation of the standby components instantaneously and 100% reliably. We use the model shown in figure 5 below to represent this situation
CiC2OS.Figure5:StandbyRedundantSystemIf we letT;bethetimetofailureofthethcomponent,thentheT's areindependent and identically distributed for i=1,2,3..,n.Thus R,(t) is identicalfor all components. Let T= time to failure of the entire system. Since the systemfails onlywhen all n componentshavefailed,and component i+1isput intooperation onlywhen componentifails,itis easyto see thatT = T, +T,+.+T,.Inotherwords,wecancomputethesystemfailuretimeeasily ifweknowthefailure timesoftheindividual components.Finally,wecandefinearandomvariableX=number of components that fail beforetimet ina standby redundantsystem.Nowthereliabilityofthesystemissimplyequaltotheprobabilitythatlessthanncomponents fail during the time interval (O, t). In other wordsR(t) = P(X <n).It canbe shownthatXfollowsaPoissondistributionwithparameter=αtwhereαisthefailurerate,sowewriteX~POissON(a)Forexample,iftimeismeasured inseconds,thenα isthenumberoffailurespersecond.Thereliabilityforsomespecifictimetthenbecomes:R(t)= P(X<n)= P(0≤X ≤n-1) .Example6:Considerthereliabilityofaradiobattery.Wedeterminedpreviouslythat one batteryhasa reliabilityfor24hours of 0.4493.Inlightofthe importanceofcommunications,youdecidethatthis reliabilityisnotsatisfactory.Suppose161
161 Cn C2 C1. DS Figure 5: Standby Redundant System If we let Ti be the time to failure of the ith component, then the Ti’s are independent and identically distributed for i = 1,2,3,.,n . Thus Ri (t) is identical for all components. Let T = time to failure of the entire system. Since the system fails only when all n components have failed, and component i + 1 is put into operation only when component i fails, it is easy to see that T T T Tn = + + + 1 2 . . In other words, we can compute the system failure time easily if we know the failure times of the individual components. Finally, we can define a random variable X = number of components that fail before time t in a standby redundant system. Now the reliability of the system is simply equal to the probability that less than n components fail during the time interval (0, t). In other words, R(t) = P( X < n). It can be shown that X follows a Poisson distribution with parameter l = a t where a is the failure rate, so we write X ~ POISSON( l). For example, if time is measured in seconds, then a is the number of failures per second. The reliability for some specific time t then becomes: R(t) = P( X < n) = P(0 £ X £ n - 1) . Example 6: Consider the reliability of a radio battery. We determined previously that one battery has a reliability for 24 hours of 0.4493. In light of the importance of communications, you decide that this reliability is not satisfactory. Suppose
wecarrytwosparebatteries.Theadditionofthesparesshould increasethebattery system reliability.Later inthe course,youwill learn howto calculatethefailure rate α for a battery given the reliability (0.4493 in this case).For now, wewill give this toyou:α =1/30 per hour.Weknowthat n=3total batteriesTherefore: lso - -(24) -0a)andR(24) = P(X <3) = P(0 ≤X ≤2) = 0.9526Thereliabilityof thesystemwithtwosparebatteriesfor24hours isnow0.9526.Example7:IftheOPmuststayoutfor48hourswithoutresupply,howmanysparebatteriesmust betakentomaintain a reliability of0.95?Wecan use trialand errortosolvethisproblem.Westartbytryingour current load of2spares.Wehave(48) = 1.6)X~POISSON=αt=30andwecannowcomputethesystemreliabilityR(48) = P(X <3)= P(0 ≤X ≤2) = 0.7834 <0.95whichisnotgoodenough.Therefore,wetryanother spareso n=4(3spares)and we compute:R(48)= P(X <4)= P(0 ≤X ≤3) = 0.9212<0.95which is still not quite good enough, but we aregetting close!Finally,we tryn=5whichturnsouttobesufficient:R(48) = P(X <5)= P(0 ≤X ≤4) = 0.9763 ≥ 0.95.Therefore, we conclude that the OP shouldtakeout at least 4 spare batteries fora48hourmission.Modelsof LargeScaleSystemsInourdiscussionofreliabilityuptothispoint,wehavediscussed seriessystems,active redundant systems, and standby redundant systems. Unfortunatelythingsarenotalwaysthis simple.Thetypesofsystems listedaboveoftenappearassubsystems inlargerarrangements ofcomponentsthatweshall call"large scalesystems".Fortunately,ifyouknowhowtodeal with series systems,162
162 we carry two spare batteries. The addition of the spares should increase the battery system reliability. Later in the course, you will learn how to calculate the failure rate a for a battery given the reliability (0.4493 in this case). For now, we will give this to you: a = 1 30 per hour. We know that n = 3 total batteries. Therefore: X ~ POISSON l = a t = ( ) = . æ è ç ö ø ÷ 1 30 24 0 8 and R(24) = P( X < 3) = P(0 £ X £ 2) = 0.9526. The reliability of the system with two spare batteries for 24 hours is now 0.9526. Example 7: If the OP must stay out for 48 hours without resupply, how many spare batteries must be taken to maintain a reliability of 0.95? We can use trial and error to solve this problem. We start by trying our current load of 2 spares. We have X ~ POISSON l = a t = ( ) = . æ è ç ö ø ÷ 1 30 48 16 , and we can now compute the system reliability R(48) = P( X < 3) = P(0 £ X £ 2) = 0.7834 < 0.95 which is not good enough. Therefore, we try another spare so n = 4 (3 spares) and we compute: R(48) = P( X < 4) = P(0 £ X £ 3) = 0.9212 < 0.95 which is still not quite good enough, but we are getting close! Finally, we try n = 5 which turns out to be sufficient: R(48) = P( X < 5) = P(0 £ X £ 4) = 0.9763 ³ 0.95. Therefore, we conclude that the OP should take out at least 4 spare batteries for a 48 hour mission. Models of Large Scale Systems In our discussion of reliability up to this point, we have discussed series systems, active redundant systems, and standby redundant systems. Unfortunately, things are not always this simple. The types of systems listed above often appear as subsystems in larger arrangements of components that we shall call “large scale systems”. Fortunately, if you know how to deal with series systems