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