18 RELIABILITYSystems ReliabilityA system consists of components which determine whether or not itwill work. T here are various types of configurations o f the componentsin different systems.• Series SystemThis is a system in which all the components are in series and theyall have to work for the system to work. If one component fails,the system fails.• Parallel SystemThis is a system that will fail only if they all fail.• Series-Parallel SystemThis is a system where some of the compo nents in series are repli-cated in parallel.8 RELIABILITY 28.1 Reliability of Series SystemExamples1. A simple computer consists of a processor, a bus and a memory.The computer will work only i f all three are functioning correctly.The probability that the processor is functioning is 0.99, tha t thebus is functioning 0.95, and that the memory is functioning is 0.99 .P rocessor Bus M emoryinput −→0.99−→0.95−→0.99−→outputThe probability that the computer will work is:Rel = .99 × .95 × .99 = 0.893475So even though all the components have abo ve 95% or more reli-ability, the overall Reliability of the computer is less that 90%.2. A system consists of 5 components in series each hav ing a relia-bility of 0.97. What is the reliability of the system?C1C2C3C4C5input −→0.97−→0.97−→0.97−→0.97−→0.97−→ outputRel = 0.975= 0.86With series systems, reliability decreases as the number of comp o-nents i ncreases.With 6 components in series;Rel = 0.976= 0.832972With 7 components; Rel = 0.977= 0.80798288 RELIABILITY 33. An electronic product contains 100 integrated circuits. The prob-ability that any integrated circuit is defective is .001 and the in-tegrated circuits are independent. The product operates only ifall the integrated circuits are operational. What is the probabilitythat the product is operational?Solution:The probability that any component is functioning is .999. Sincethe product operates only if all 100 components are operational,the probability that the 100 components are functioning is:Rel = .999100obtained in R with.999**100[1] 0.9047921The reliability is just over 90% even though each component hasa reliability of 99.9%.Bearing in mind that computing and electrica l systems have hundredsor thousands of components, a series formation on its own will neverbe sufficiently reliable, no matter how high the individual componentreliability is. The components need to be backed up in parallel.8 RELIABILITY 48.2 Reliability of a Parallel SystemSystems with parallel structure ha ve built in redundancy. Componentsare backed up, and the system will work even if some of these cease tofunction.Examples1. A system consists of 5 components in parallel. If each componenthas a reliability of 0.97, what is the overall reliability of the system?input−→0.970.970.970.970.97−→outputSystem will function provided at least one of the 5 componentsworks:Rel = P(At least one component is functioning)Taking the complementary approach,P (at least one component functioning ) = 1- P (all componentsfail).ThereforeRel = 1 − (0.03)5≈ 1.000008 RELIABILITY 5With parallel systems the “law of diminishing returns” operates:With 2 components Rel = 1 − 0.032= 0.9991With 3 components Rel = 1 − 0.033= 0.999973With 4 components Rel = 1 − 0.034= 0.9999992Reliability of a parallel system with increasing number ofcomponents: component reliability = .972 4 6 8 100.970 0.975 0.980 0.985 0.990 0.995 1.000Number of components Reliabilityk<-1:10p <- .97plot (k, 1-(1-p)^k , xlab = "Number of components ",ylab = "Reliability")abline(1,0)8 RELIABILITY 6Series-Parallel SystemsExamples1. Consider a system with 5 kinds of component, with reliabilities• component 1 : 0.95,• component 2 : 0.95,• component 3 : 0.70,• component 4 : 0.75,• component 5 : 0.90.Because of the low reliability of the third a nd fourth components, theyare replicated; the system contains 3 of the third comp onent and 2 ofthe fourth component.The System:0.95 −→ 0.95 −→0.700.700.70−→0.750.75−→ 0.90C1−→ C2−→ C3−→ C4−→ C5P (C1) × P(C2) × P( at least 1 C3) × P(at least 1 C4) × P(C5)Rel = .95 × .95 × (1 − .33) × (1 − .252) × .9In R.95*.95*(1-.3^3) * (1- .25^2) *.9[1] 0.74092438 RELIABILITY 7Examples2. The following system operates only if there is a path of functionaldevices from left to right. The probability that each device functions isshown on the g raph. Assume the devices fail independently.C1C2C3input −→0.900.90−→0.850.850.85−→ 0.99 −→ outputReliability: P (at least 1 of 2 C1) × P (at least 1 of 3 C2) × P (C3)1 − .12× 1 − .153× .99In R> (1-.2^2)* (1-.15^3)*.99[1] 0.94719248 RELIABILITY 8Examples3. The following diagram is a representation of a system that com-prises two subsystems operating in parallel.input1PPPPPPPPPPqSubsystem ASubsystem B0.9-0.90.90.9-PPPPPPPPPPq1outputThe system operates if at least one of subsystem A operates.Rel = P (at least one subsystem works)EquivalentlyRel = 1- P (both subsystems fail)P (Subsystem A fails) = 1 − .92.P (Subsystem B fails) = 1 − .92.Overall system will operate if at least one of these subsystems works.Rel = P (at least one subsystem works)EquivalentlyRel = 1- P (both subsystems fail)Rel = 1 − (1 − .92)2= 0.96398 RELIABILITY 9Generally with k subsystems, the reliability isRel = 1 − (1 − .92)kWe can ex amine this with different values of kk<-1:10plot(k, 1-(1-.9^2)^k, xlab = "Number of Subsystems", ylab = "Reliability")abline(.99, 0)calculates and plots the reliability for various levels of backup:2 4 6 8 100.85 0.90 0.95 1.00Number of SubsystemsReliability8 RELIABILITY 10SummaryThe reliability of a system, i.e. the probability that it is functioningproperly depends on• the reliability of each of its the components• the type of systemReliability with Series SystemsThe problem with series systems is that reliability quickly decreases asthe number of components increases.Reliability with Parallel SystemsThe problem with parallel systems is that the ‘law of diminishing re-turns’ operates. The rate of increase in reliability with each additionalcomponent decreases as the number of components increases.Most systems are combinations of series and parallel
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