UNIT 15 RELIABILITY EVALUATION OF k-out-of-n AND STANDBY SYSTEMS

Transcription

1 UNIT 1 RELIABILITY EVALUATION OF k-out-of-n AND STANDBY SYSTEMS Structure 1.1 Introduction Objective 1.2 Redundancy 1.3 Reliability of k-out-of-n Sytem 1.4 Reliability of Standby Sytem 1. Summary 1.6 Solution/Anwer 1.1 INTRODUCTION In Unit 14, we have dicued reliability evaluation of erie, parallel and mixed ytem. You have learnt that for the ucceful operation of a ytem having n component: it i neceary that all n component mut perform their intended function uccefully in the cae of erie ytem, and only one need to work uccefully in the cae of parallel ytem. However, in practice, there exit ytem, which require the ucceful operation of at leat k, (1 k n), component out of n identical component. Such ytem are known a k-out-of-n ytem or partially redundant ytem. Further, in parallel ytem dicued in Unit 14, all n component work imultaneouly, while for ucceful operation of a parallel ytem only one need to work uccefully. Thi i known a active redundancy. We explain what i meant by redundancy in Sec In Sec. 1.3, we dicu how to evaluate the reliability of k-out-of-n ytem of identical component. The redundancy complementary to active redundancy i known a the tandby (paive) redundancy, which i dicued in Sec In Unit 16, we hall dicu three technique to evaluate the reliability of complex ytem. Objective After tudying thi unit, you hould be able to: define redundancy, active redundancy and tandby (paive) redundancy; evaluate the reliability of a k-out-of-n ytem; and evaluate the reliability of a tandby ytem under certain aumption. 1.2 REDUNDANCY Redundancy in a ytem mean that there exit an alternative parallel path for the ucceful operation of the ytem. For example, uppoe a ytem ha n component connected in parallel and at leat k, (1 k n), component are needed for the ucceful operation of the ytem. In uch a ytem, we ay that

2 Reliability Theory the number of baic component i k and the remaining (n k) component are known a redundant component. In fact, thi ytem i known a a k-outof-n ytem and i dicued in the next ection of thi unit. The baic purpoe of introducing redundant component in a ytem i to improve the reliability of the ytem. A we have een in Example 6 in Unit 14, the reliability of a ytem increae a we introduce redundant component in a ytem. Redundancy can be claified a hown in Fig Fig. 1.1: Claification of redundancy. Let u explain the term ued in Fig Fully Redundant Sytem: A k-out-of-n ytem i known a a fully redundant ytem if k = 1. But for k = 1, the k-out-of-n ytem i the ame a the parallel ytem, which we have dicued in Sec. 14. of Unit 14. Thu, the parallel ytem i nothing but a 1-out-of-n ytem and i alo known a a fully redundant ytem. Partially Redundant Sytem: Before defining a partially redundant ytem, let u firt define a non-redundant ytem. A k-out-of-n ytem i aid to be a non-redundant ytem if k = n. But for k = n, the k-out-of-n ytem i imply the erie ytem, which we have dicued in Sec of Unit 14. Thu, a erie ytem i nothing but an n-out-of-n ytem and i alo known a a non-redundant ytem. Now, we define a partially redundant ytem. A k-outof-n ytem i aid to be a partially redundant ytem if 1 < k < n. A far a conditional redundancy i concerned, it i beyond the cope of the coure. On the other hand, what we mean by tandby redundancy i dicued in Sec. 1.4 of thi unit. Let u now determine the reliability of the k-out-of-n ytem. 1.3 RELIABILITY OF k-out-of-n SYSTEM A ytem having n component i aid to be k-out-of-n ytem if and only if at leat k component out of n are required to function uccefully for the ucceful operation of the ytem. In k-out-of-n ytem, although all n component are in operation, the k component that are eential for the ytem to operate are called the baic component. The (n k) component that are added with the objective to improve the reliability of the ytem are known a redundant component. Thu, a erie ytem i an n-out-of-n ytem and a parallel ytem i a 1-outof-n ytem. That i, the erie and parallel ytem are particular cae of k- out-of-n ytem when k = n and k = 1, repectively. The poible path for a 2- out-of-3 ytem having component 1, 2 and 3 are hown in Fig. 1.2.

3 Reliability Evaluation of k-out-of-n and Standby Sytem Fig. 1.2: Poible path for 2-out-of-3 ytem. So, for reliability evaluation of k-out-of-n ytem, we have to find a technique, which erve our purpoe for all value of k varying from 1 to n. For implicity, uppoe all the n component of the ytem are identical and independent. In fact, we impoe thee two retriction on the component of the ytem intentionally o that we can apply binomial ditribution for evaluating the reliability of the ytem. Recall from Unit 9 of MST-003 that the probability ma function of the binomial ditribution i given by n x nx p(x) P X x C p q, where x n = number of trial in the random experiment, x = number of uccee out of n trial, p = probability of ucce in each trial, q = 1 p = probability of failure in each trial. n Cx binomial coefficient which repreent number of way of occurrence of x uccee out of n trial. You may like to refer to Unit 4 of MST-001 if needed. (1) We now conider the main objective of thi ection: To obtain an expreion for the reliability of k-out-of-n ytem. Let u firt ee whether we can apply binomial ditribution or not. Let u undertand the above lited meaning of n, x, p, q in term of our k-out-of-n ytem. So we firt define ucce and probability of ucce in our cae a follow. Succeful operation of a component in k-out-of-n ytem repreent a ucce. Since there are n component, the number of uccee for a given time may be 0, 1, 2,, n. A far a the probability of ucce i concerned, reliability of each component repreent the probability of ucce. But for applying the binomial ditribution, the probability of ucce hould be the ame for each trial. However, we have already aumed that all the n component are identical. Therefore, the reliability of each component i aumed to be the ame. Hence, the requirement of contant probability in each trial i atified. Further, to apply binomial ditribution, the probability of ucce in each trial hould be independent. But we have already aumed that n identical component in the ytem are independent. Thu, all the requirement for applying binomial diribuition are atified by the k-out-of-n ytem of identical and independent component. Therefore, if then R denote the reliability of k-out-of-n ytem for a given time, R = Probability of ucce of at leat k component n n x nx x xk or R C R Q, (2)

4 Reliability Theory where n = number of component in the ytem, k = minimum number of component required for the ucce of the ytem, R = reliability of each component, Q = 1 R = unreliability of each component. (3) Let u explain the procedure of evaluating R with the help of an example. Example 1: Conider a piping ytem having pipe connected in parallel a hown in Fig Aume that all pipe are identical and independent. If reliability of mooth flow of the liquid from each pipeline i 0.80 for a miion of 1 year, evaluate the reliability of the ytem working uccefully. The ytem i aid to work uccefully if at leat 3 pipeline perform their intended function uccefully. Fig. 1.3: Sytem of pipeline connected in parallel. Solution: Since it i given that for ucceful operation of the ytem, at leat 3 pipeline hould work uccefully, it i a k-out-of-n ytem, where k = 3 and n =. If R denote the reliability of each pipeline and R that of the 3-out-of- ytem, then applying the binomial ditribution given in equation (2), we get k k k3 k R C R 1R C3 R 1 R C4 R 1 R C R 1 R (0.8) (1 0.8) (0.8) (1 0.8) (0.8) = for a miion of 1 year Let u explain the procedure of handling the cae of non-identical component in k-out-of-n ytem with the help of an example. Example 2: For the ytem given in Example 1, aume that intead of pipeline, we have 4 pipeline, which are independent but not identical. It i given that the reliabilitie of 4 pipeline for a miion of 1000 hour are R 1 = 0.60, R 2 = 0.70, R 3 = 0.80 and R 4 = 0.7, repectively. Evaluate the reliability of 2-out-of-4 ytem.

5 Solution: Here we are given that k = 2 and n = 4. Since the component (pipeline) are non-identical, we cannot apply the binomial ditribution in thi cae. Thi i becaue one of the requirement for applying the binomial ditribution i that the probability of ucce in each trial hould be the ame. Thi requirement i not met in thi cae due to the preence of non-identical component. The cae of non-identical component i handled by conidering all poible mutually excluive and exhautive cae for each poible value of k. Here the value of k may be 2, 3 or 4. Let u find the mutually excluive and exhautive cae for each value of k one at a time. Thi i explained below: For k = 2, the number of mutually excluive cae the number of way of exactly 2 uccee out of C Reliability Evaluation of k-out-of-n and Standby Sytem Thee ix mutually excluive cae are R1R 2R 3R 4, R1R 2R 3R 4, R1R 2R 3R 4, R1R 2R 3R 4, R1R 2R 3R 4, R1R 2R 3R 4,where R i denote the unreliability of the component i, i = 1, 2, 3, 4. Thee mutually excluive cae are hown in Fig. 1.4a. For k = 3, the number of mutually excluive cae = the number of way of exactly 3 uccee out of C Thee four mutually excluive cae are R1R 2R 3R 4, R1R 2R 3R 4, R1R 2R3R 4, R1R 2R 3R 4 a hown in Fig 1.4b. For k = 4, the number of mutually excluive cae the number of way of exactly 4 uccee out of C 4 1a There i only one poibility of 4 uccee out of 4, which i R1R 2R3R 4.Thi cae of 4 uccee i hown in Fig. 1.4c. Fig. 1.4: Mutually excluive path for (a) k = 2; (b) k = 3 and (c) k = 4.

6 Reliability Theory Let R a,r b and R c denote the reliabilitie for the cae when k = 2, k = 3 and k = 4, repectively. Now, to evaluate reliability for k = 2, we imply have to add the reliabilitie of the ix mutually excluive cae a hown in Fig. 1.4a (by addition law of probability for mutually excluive event). R R R R R R R R R R R R R R R R R a R R R R R R R R (1 0.8)(1 0.7) 0.6 (1 0.7)0.8(1 0.7) 0.6 (1 0.7)(1 0.8) 0.7 (1 0.6) (1 0.7) (1 0.6) 0.7(1 0.8)0.7 (1 0.6)(1 0.7) = 0.24 R b R1R 2R 3R 4 R1R 2R 3R 4 R1R 2R 3R 4 R1R 2R 3R (1 0.7) (1 0.8) (1 0.7) (1 0.6) = = Rc R1R 2R3R The reliability of the whole ytem i given by R R a R b R c You may like to try the following exercie to evaluate the reliability of a k-out-of-n ytem. E1) For the ytem given in Example 1, evaluate the reliability of the ytem for k = 1, 2, 3, 4, and comment on the reult. 1.4 RELIABILITY OF STANDBY SYSTEM Refer to Fig. 1.1, where we have claified redundancy broadly into two categorie, namely, active redundancy and tandby redundancy. By active redundancy, we mean that all the component connected in parallel are turned on at the beginning of operation of the ytem, and continue to perform until they fail. Thu, in active redundancy, all component are imultaneouly in the operating mode. We have dicued two uch ytem, namely, the parallel ytem and k-out-of-n ytem, where all the component of the ytem are turned on at the beginning of the operation of the ytem. So all the component are imultaneouly in the operating mode. On the other hand, in tandby redundancy, the component are connected in parallel but do not tart operating imultaneouly from the beginning of the operation of the ytem. In tandby redundancy, the component() in operating mode, i/are known a normally operating component(). The

7 component() kept in tandby or reerve mode i/are known a tandby component(). Other than thi, there i a changeover device. The function of the changeover device i to ene the failure of normally operating component and in cae of a failure, to bring a tandby component into the normal operating mode. To explain the concept of tandby ytem, let u firt conider the implet cae of the two component (ay, A and B) tandby ytem. A witch in the tandby ytem can put any component into operation. In the 2-component tandby ytem, initially thi witch i connected to component A and turn it on, a hown in Fig. 1.. In Fig. 1., the witch i repreented by S. In thi etup, component B will remain in tandby (reerve or inoperative) mode till uch time a component A perform it function uccefully. A oon a component A fail, the witch ene the failure and put the component B into operation. Reliability Evaluation of k-out-of-n and Standby Sytem Fig. 1.: A typical two component tandby ytem. In general, if we have a tandby ytem having n component, namely, 1, 2, 3,, n, then there exit a witch, ay S, which can put any one of the n component into operation. The ytem work in the ame way a the two-component tandby ytem. Initially the witch i connected to component 1 and turn it on (ee Fig. 1.6). Here we are auming that only one component i in the normally operating mode. Until the component 1 perform it intended function uccefully, the remaining (n 1) component, namely 2 to n, remain in the tandby (reerve or inoperative) mode. A oon a component 1 fail, the witch ene the failure and turn component 2 on. Till uch time a the component 2 perform it intended function uccefully, the remaining (n 2) component, namely 3 to n, remain in the tandby (reerve or inoperative) mode. A oon a component 2 fail, the witch ene the failure and turn the component 3 on. Thi proce continue until the failure of the n th component. When the n th component fail, the ytem will go into failure mode. If there are more than one component in the normally operating mode, the proce will work in the ame way a it work for one normally operating component. Fig. 1.6: A typical n component tandby tytem for the cae of only one component in normally operating mode. In the dicuion o far, we have not conidered a key point of the tandby ytem: that the witch itelf may fail during the operation of a component or during the time of changeover of a normally operating component to a tandby

8 Reliability Theory component. Due to thi feature of the tandby ytem, we divide the dicuion of thi ection under the following two head: Standby ytem with perfect witching, and Standby ytem with imperfect witching. Let u dicu thee one at a time, in brief. Standby Sytem with Perfect Switching By perfect witching, we mean that the changeover device/witch will neither fail during the operation of a component nor during the changeover of a failed component to the next available tandby component. In reliability term, we are auming that the changeover device i 100% reliable. To evaluate the reliability of the tandby ytem with perfect witching, we aume that the tandby component() will not fail in it/their tandby poition. Now, let E i be the event that component i perform it intended function uccefully, where i = 1, 2, 3,, n. Let Qi denote the unreliability of the component i, given that component 1to(i 1) have failed. Further, if R and Q denote the reliability and unreliability of the tandby ytem, then PE1 PE2 E1 PE3 E1E 2...P En E1E 2...E n 1 Q P E E E... E, where E denote complement of event E n i i Q1Q 2Q 3...Q n [uing the concept of conditional probability] (4a) (4b) reliability (R) of the tandby ytem i given by R = 1 Q () In particular, in the cae of one normally operating component and one tandby component, the unreliability of the ytem i given by Q Q Q (6) 1 2 and reliability (R) of the tandby ytem i given by equation (). Note1: Here the unreliability Q 2 of the component 2 i not the ame a it i in the cae of parallel ytem becaue in a tandby ytem, component 2 i ued for a horter duration a compared to the cae of parallel ytem. Let u conider an example. Example 3: A tandby ytem ha three component 1, 2, 3, where component 1 i normally operating and component 2, 3 are tandby component. The reliability of component 1 i 0.9. The reliability of component 2 given that component 1 ha failed i 0.96 and that of component 3 given that component 1 and 2 have failed i Evaluate the reliability of the ytem under the aumption that the witch i perfect. Solution: We know that for an n-component tandby ytem (with one normally operating component), if Qi i the unreliability of component i given that component 1 to (i 1) have failed, then reliability (R) of the ytem i given by R 1 Q Q Q...Q n In thi cae, Q , Q , Q R 1 Q Q Q = 1 (0.0) (0.04) (0.02) = =

9 We now dicu the reliability evaluation of the tandby ytem in the cae of imperfect witching. Standby Sytem with Imperfect Switching By imperfect witching, we mean that the changeover device/witch may fail either during the operation of a component or during the time of bringing a tandby component in place of a normally operating failed component. To explain the concept, let u conider a imple two-component tandby ytem with one normally operating component (ay, A) and one tandby component (ay, B). To evaluate the reliability of the ytem, let u aume that the witch fail only at the time of changeover. Therefore, the unreliability of the ytem i given by Q = P(ucceful changeover) P(ytem failure given ucceful changeover) + P(unucceful changeover) P(ytem failure given unucceful changeover) Reliability Evaluation of k-out-of-n and Standby Sytem in cae of ucceful changeover,ytem failure A fail and then component Bfail. Q P Q Q P Q,...(7) In cae of unucceful changeover,ytem failure failure of component A. A B A where, P probability of ucceful changeover P 1 P probability of unucceful changeover QA QB unreliability of component A unreliability of component B given that component A ha failed Let u conider an example. Example 4: Conider a two-component tandby ytem with A a normally operating component and B a tandby component. The reliability of component A i 0.90 while the reliability of component B given that A ha failed i 0.9. Aume that the witch can fail only at the time of changeover with a probability of failure Evaluate the reliability of the ytem. Solution: We know that if P probability of ucceful changeover P 1 P probability of unucceful changeover QA QB unreliability of component A unreliability of component B given that component A ha failed then from equation (7), the unreliability (Q) of the two-component tandby ytem under the aumption that the witch can only fail during the time of changeover i given by Q P QAQB P QA (1 0.03) (1 0.90)(1 0.9) 0.03 (1 0.90) [uing given reliabilitie] =

10 Reliability Theory Hence, the reliability of the ytem = = You can try the following exercie to evaluate the reliability of a tandby ytem. E2) If in Example 2, the probability of unucceful changeover i 0.2 intead of 0.03, evaluate the reliability of the ytem. Let u ummarie the main point that we have covered in thi unit. 1. SUMMARY 1. Redundancy in a ytem mean that there exit an alternative parallel path for the ucceful operation of the ytem. If a ytem ha n component connected in parallel and at leat k, (1 k n) component are needed for the ucceful operation of the ytem, we ay that the number of baic component i k and the remaining (n k) component are known a redundant component. 2. Fully Redundant Sytem: A k-out-of-n ytem i known a a fully redundant ytem if k = A k-out-of-n ytem i aid to be a non-redundant ytem if k = n. 4. A k-out-of-n ytem i aid to be a partially redundant ytem if 1 < k < n.. Reliability of k-out-of-n ytem i obtained by uing binomial ditribution in the cae of identical component. In the cae of nonidentical component, it i calculated by firt liting out all mutually excluive and exhautive cae for different value of k and then by applying the addition law of probability. 6. Reliability of tandby ytem having n component, namely 1, 2,, n, in the cae of perfect witching i given by Q Q Q Q...Q, n whereqi denote the unreliability of the component i, given that component 1to(i 1) have failed. 7. Reliability of tandby ytem having n component, namely 1, 2,, n, in the cae of imperfect witching under the aumption that the changeover device (witch) fail only at the time of changeover (in cae of two component) i given by R 1 Q where Q P QAQB P Q A, and P probability of ucceful changeover P 1 P probability of unucceful changeover QA QB unreliability of component A unreliability of component B given that component A ha failed 1.6 SOLUTIONS/ANSWERS E1) If R denote the reliability of the ytem and R that of each component, where R = 0.80, then for different value of k, R i calculated below: For k = 1,

11 k k k k1 R C R (1 R) C R(1 R) C R (1 R) C R (1 R) C4R (1 R) CR 0.8(1 0.8) 10(0.8) (1 0.8) 10(0.8) (1 0.8) 4 (0.8) (1 0.8) (0.8) = For k = 2, Reliability Evaluation of k-out-of-n and Standby Sytem k k k k2 R C R (1 R) C R (1 R) C R (1 R) C R (1 R) C R (1 R) = For k = 3, k k k k3 R C R (1 R) [uing already calculated value] C R (1 R) C R (1 R) C R (1 R) = For k = 4, [Uing already calculated value] k k 4 k 4 0 k4 R C R (1 R) C R (1 R) C R (1 R) [uing already calculated value] = For k =, k k k 0 k R C R (1 R) C R (1 R) Let u put all the calculated value in Table 1.1 a follow. Table 1.1: Reliability of the Sytem for k = 1, 2, 3, 4, Value of k Reliability of the Sytem for a Miion of 1 year % Decreae in Reliability of the Sytem Compared to the Value for k = % % % %

12 Reliability Theory Comment: From Table 1.1, we oberve that the reliability of the ytem decreae a k increae. The maximum decreae occur when k = = n, i.e., in the cae when k-out-of-n ytem coincide with the erie ytem. E2) We know that if P probability of ucceful changeover P 1 P probability of unucceful changeover QA QB unreliability of component A unreliability of component B given that component A ha failed then unreliability (Q) of the two-component tandby ytem under the aumption that the witch can only fail during the time of changeover i given by Q P QAQB P QA (1 0.2) (1 0.90)(1 0.9) 0.2 (1 0.90) [uing given reliabilitie] = Hence, the reliability of ytem = =