SYSTEM RELIABILITY EVALUATION 1 System reliability evaluation using

SYSTEM RELIABILITY EVALUATION 1

System reliability evaluation using probability distributions Series systems The reliability of a 2 -component series system for an n-component series system 2

In the special case of the exponential distribution, For a 2 -component series system for an n-component series system 3

for the general case and for the exponential case The effective failure rate of a series system containing components whose reliabilities follow exponential distributions is 4

Example A simple electronic circuit consists of 6 transistors each having a failure rate of 1 O-6 f/hr, 4 diodes each having a failure rate of O. 5 x 1 O-6 f/hr. 3 capacitors each having a failure rate of 0. 2 x 10 -6 f/hr, 10 resistors each having a failure rate of 5 x 10 -6 f/hr and 2 switches each having a failure rate of 2 x 10 -6 f/hr. Assuming connectors and wiring are 100% reliable (these can be included if considered significant), evaluate the equivalent failure rate of the system and the probability of the system surviving 1000 hr and also 10 000 hr if all components must operate for system success. 5

Parallel systems The special case of a constant hazard rate, i. e. , the exponential distribution, we obtain, for a 2 -component parallel system 6

for an n-component parallel system 7

Reconsider the previous Example and evaluate the probability of surviving 1000 hr and 10 000 hr if (a) two, (b) three identical circuits to those described in the previous Example are used in parallel and it is assumed that the system operates successfully if only one of the circuits is successful. In this example, all the parallel components (each being equivalent to one of the series circuits) are identical and each having a failure rate given by λe in the previous Example, i. e. , 8

Partially redundant systems The binomial expression 9

Consider a system comprising 4 identical units each having a failure rate of 0. 1 f/yr. Evaluate the probability of the system surviving 0. 5 yr and 5 yr if at least two units must operate successfully. Using the binomial expression for n =4 gives 10

Number of units required for success Probability of system success 4 3 2 1 11

In the more general case of non-identical units, the probability of each system state Mean time to failure (MTTF) 12

is the distribution of times to failure, this expected value was the mean time to failure or MTTF, which can be designated as m Integration by parts gives The MTTF of any series or parallel system can therefore be evaluated by using the appropriate system expression for R(t) and integrating between the limits (0, ∞ ). 13

Consider the particular case of exponential distributions. For a series system For a 2 parallel system 14

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For an n-component parallel system 16

Standby systems B S A A simple and most basic standby system is shown. A represents the main operating component, B the standby component and S the sensing and changeover switch. One assumption which is used through out is that the components, both normally operating and standby, have a constant hazard rate, i. e. , that failures are described by exponential distributions. 17

Perfect switching (a) 2 - component system Therefore, the reliability of the system is 18

(b) 2 standby components (c) n standby components 19

(d) Mean time to failure and for n standby components (case (c)), Example Compare the reliability of a 2 -component system each having a failure rate of 0. 02 f/hr after a time of 10 hr if they are (a) parallel redundant and, (b) standby redundant with a 100% reliable sensing and changeover device. Also, compare the MTTFs of 20 the two systems,

(a) Parallel system. (b) Standby system. 21

Imperfect switching The value of PS can be established in practice from a data collection scheme which records the number of successful and failed operations of the device since Returning to the 2 -component standby system in which the switching and sensing device is less than 100% reliable, system success requires that either no failures occur or one failure occurs and the switching device operates. 22

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Effect of spare components Consider a system comprising N identical components all of which must operate for system success and consider that there are it spares available to the operating personnel as standby components. It follows that n failures in the system can be tolerated and only the (n + 1) failure causes system failure. This logic also indicates that the spare components are themselves not replaced following their use as a normally operating component. Again, consider the case of exponential distributions and assume that the failure rate of each component is λ. 24

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A system contains 50 identical components each of which has a failure rate of 0. 001 f/hr. Assuming that system failure occurs when any one component fails, evaluate the system reliability for an operating period of 20 hr and the MTTF when no spares are available and also when a varying number of spares between 1 and 6 are carried as immediate replacements. If the system is to have a minimum reliability of 0. 9950, what is the minimum number of spares that must be carried as immediate replacements? 26

Number of Spare System Reliability R(20) MTTF m, hr 0 0. 367879 20 1 0. 735759 40 2 0. 919699 60 3 0. 981012 80 4 0. 996340 100 5 0. 999406 120 6 0. 999917 140 27

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Non-identical components In order to illustrate the joint density function approach, consider two components A and B in a standby system and having failure rates λa and λb respectively. Let A be the normal operating component and B be the standby one. Assume A fails at time t 1 when B takes over instantaneously. Assume B fails at time t. Time to failure of B is then t 2 = t - t 1 The joint density function of both components operating is given by 29

There are two time functions, t 1 and t. In order to obtain the joint density function in terms of t only, f(t) is integrated with reference to t 1 This gives The system reliability can now be evaluated from 30

By adding and subtracting exp (-λat) and rearranging to give 31

The MTTF is given by If the sensing and changeover device is not 100% reliable, This approach can be extended to include any number of standby components. Consider a standby system consisting of one normally operating component having a density function f 1(t 1) and (n-1) components in standby having density functions f 2(t 2), f 3(t 3), ……. fn(tn). 32 The joint density function is

where t 1 represents the time of failure of component 1 t 2 represents the time of failure of component 2 tn represents the time of failure of component n and tn = t - tn-1 33

Failures in the standby mode (a) Case 1 Consider the case of two non-identical components A and B forming the standby system described in Section 7 in which the standby component B cannot fail in the standby mode. The events leading to the system success are either (i) Component A does not fail for an interval of time 0 to t, or (ii) Component A fails at time t 1< t and component B does not fail in the interval t 1 to t. Let R 1 and R 2 be the reliabilities associated with these two events respectively, then 34

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Since the events (i) and (ii) are mutually exclusive 36 Graphical representation of Case 1. (a) Contribution R 1. (b) Contribution R 2

The contribution R 1 is simply given by the area under the exponential curve associated with component A for times greater than t, i. e. , R 1 = exp (-λat) The contribution R 2 for a given value of t 1 is the area under the exponential curve associated with component B [RB(t 1 -t 2)] weighted by the probability that component A fails in the time interval dt, at t 1. That is, The value R 2 can then be determined by integrating R 2(t 1) over all values of t 1 37

(b) Case 2 Consider now the case of 2 components in a standby system for which λ 1 is the failure rate of the normal operating component, A is the failure rate of the standby component when operating and λ 2 is its failure rate when in a standby mode. Using a similar logic to that used in Case 1, the mutually exclusive events leading to system success are (i) Component 1 does not fail during the time (0 -t), (ii) Component 1 fails at time t 1 and component 2 is not failed at time t 1 (failure rate λ 3) and component 2 does not fail in the time (t-t 1) (failure rate λ 2). Event Mode of operation in time domain of Component 1 1 Good for time t 1 2 Fail at time t 1 Component 2 Good for time t 1 to t 38

Using the same logical derivation as in case 1 39

And the system reliability is (c) Case 3 Consider the system shown below in which components 1 and 2 operate as a parallel redundant system and component 3 is used when both 1 and 2 have failed. 40 System used for Case 3

In this case assume that the following system data is available: Øall components operate in their useful life period and wear-out can be neglected Øfailure rate of component 1 when energized= λ 1 e Øfailure rate of component 2 when energized = λ 2 e Øfailure rate of component 3 when energized = λ 3 e Øfailure rate of component 3 when on standby = λ 3 s Øfailure rate of the sensing device = λs Øfailure rate of changeover device when on standby = λcs Øfailure rate of changeover device after switching = λce Øprobability of successful changeover = Ps 41

Mode of Operation in time domain of component Event 1 2 1 Good/t 2 Good/t Bad/t 3 Bad/t Good/t 4 Bad/t 1 Bad/t 2 3 3 Sensing Changeover Device Standby Energized device Standb y Energized After Switching Good/t 2 Good/t-t 2 Good/t Good/one cycle Good/t-t 2 2 5 Bad/t 2 Bad/t 3 Good/t 2 Good/t-t 2 Good/t 2 42

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Problems 1. 4 S 3 2 1 1. For the standby system shown, mission success requires at least two components. Components 1, 2, 3 are in parallel and component 4 is in stand-by. Assuming 100% reliable sensing and changeover arrangement, develop the expression for the system reliability. Assume constant failure rates λ 1, λ 2, λ 3 and λ 4 for components 1, 2, 3, and 4 respectively. 45

2. 1 out of 2 requirement 2 2 out of 3 requirement 4 5 1 2 6 1 Redundant paths 3 7 Calculate the reliability of the system shown for a 100 hr mission. 46

3. 1 2 3 4 input output input 7 5 6 8 9 10 Consider the system shown. Assume that the signal can flow only in the directions shown. (a). Develop an equation for the reliability of this system. (b). Use your equation to calculate the system reliability if all components have a reliability of 0. 9. 47