1. Reliability and survival - basic concepts

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1 . Reliability and survival - basic concepts. Books Wolstenholme, L.C. "Reliability modelling. A statistical approach." Chapman & Hall, 999. Ebeling, C. "An introduction to reliability & maintainability engineering." McGraw Hill, Applications This course is concerned with the study of components and systems that fail in time. e.g. Engineering mechanical and electrical equipment Medicine design and analysis of medical trials.3 The reliability function Suppose that an item (component, device etc.) is new at time t and fails at time T (a continuous random variable). T > - only takes positive values the item is assumed to be in one of two states - working or failed. The reliability function R (t) is de ned as R (t) Pr (T > t) (.) i.e. R (t) is the probability that the item is still working at time t Notice that R (t) F (t) R (t) d R (t) f (t) (.) dt where F (t) Pr (T t) cumulative distribution function (c.d.f.) of T and f (t) is the probability density function (p.d.f.) of T: Some properties of R (t) R (t) is a decreasing function of t: If t < t then R (t ) R (t ) : R () Pr (T > ) R () nothing survives for ever R (t)

2 In medical survival analysis, R (t) is also known as the survivor function S (t) : Example The failure time of a compressor (in hours) is a random variable with the probability density function (p.d.f.) ( : t > f (t) (:t+) otherwise Find the reliability for a hr. operating life. Solution Writing 3 we have R (t) Z t (u + ) du (u + ) t t (u + ) (t + ) Setting t gives R () : + :99:.4 Mean time before failure (MTBF) The mean time before failure is de ned as E (T ) Z tf (t) dt (.3) where E (:) denotes mathematical expectation (mean) of the random variable T: Result. is given by the area below the reliability curve. Proof. Use (.) (.3) and integration by parts, noting that tr(t) at t ; : Z tf (t) dt Z Z [ tr (t)] + R (t) dt Z tr (t) dt R (t) dt (.4)

3 The expectation (mean) of a distribution is not always nite. In such cases we can use the median lifetime as an alternative measure of a "typical" lifetime. The median life t m satis es and is the age reached by 5% of the population. Example R (t m ) :5 (.5) An electronic device has failure time density function (t measured in hours) :e :t t > f (t) otherwise Find a) the MTBF, b) the median time to failure of the device. Solution Observation: the density f (t) e t ; t > is known as the exponential distribution. MTBF: R (t) Z t Z t e u t e t f (u) du e u du MT BF E (T ) Z Z e t dt R (t) dt e t 5 hrs. Median life: R (t m ) :5 e tm t m ln 5 ln 346:6 hrs. 3

4 .5 The hazard function When an item has reached age t; its lifetime characteristics are usually di erent from new. The failure time density is f (t) : However when T > t we need to condition on this event and de ne the instantaneous failure rate of the item as lim t! Pr (Item fails in (t; t + t) jsurvives to t) t (.6a) lim Pr (t < T < t + t j T > t) t! t (.6b) showing that h (t) f (t) R (t) NB. Pr (t < T < t + t j T > t) Pr (t < T < t + t ) R (t) f (t) t R (t) h (t) t known as the hazard function is the instantaneous failure rate (.6). h (t) is not a probability density function, though for xed t we can de ne the failure time density (lifetime density) conditional on survival to time t f (tjt > t ) f (t) R (t ) which does satisfy the normalization lim t! t Pr (t < T < t + t j T > t ) (.7) Z t f (tjt > t ) dt Z t f (t) dt R (t ) Actuaries use a discrete form of h (t) with t year e.g. Pr (person reaching age 69 dies before 7th birthday) which is known as the "force of mortality" 4

5 .6 Relationship between R(t) and h(t) h (t) f (t) R (t) R (t) R (t) d ln R (t) dt Integrating and using ln R () ln () we obtain Z t ln R (t) h (u) du Z t R (t) exp h (u) du where H (t) is the cumulative hazard function. (.8a) (.8b) exp f H (t)g (.8c).7 Conditional reliability and monotone failure rates Given that an item has survived to a, the conditional reliability is R (tja) Pr (item survives further time t j survives to a) Pr (T > a + tjt > a) t > Pr (T > a + t) Pr (T > a) R (a + t) R (a) (.9) An item has monotone failure rate if either h(t) " or h (t) # for all t. The item is either IFR or DFR or both according to whether the failure rate h (t) is increasing, decreasing or constant. Result Conditional reliability improves as a function of a (for all t) if and only if the item is DFR. Proof i ) Suppose R (tja) R (a + t) R (a) is increasing as a function of a. h (a) lim Pr (fails in (a; a + t) jsurvives to a) t! t lim t! t [ R (tja)] Therefore h (a) is decreasing as a function of a 5

6 ii) Conversely, suppose h (a) is DFR, therefore decreasing as a function of a: exp ( H (a + t)) R (tja) exp ( H (a)) Z a+t exp h (u) du a (.) so R (tja) is increasing as a function of a: The above result shows that in certain circumstances aging can be bene cial to reliability. Certain metals increase in strength as they are work-hardened. New items can be prone to failure through manufacturing defects. c.f. "infant mortality" Aging more commonly has an adverse e ect on reliability, corresponding to a situation of increasing hazard (IFR). The following result can be proved in the same way as above. Result Conditional reliability decreases as a function of a (for all t) if and only if the item is IFR..8 The Bathtub curve Figure : Typical behaviour of h (t) throughout life cycle The bathtub curve represents the failure rate of a product during its lifecycle. function comprises three parts: The hazard. Early failures or infant mortality - DFR period.. Useful life - constant (CFR) failure rate period. Failures are "rare" events modelled by a Poisson process. 3. Old age - IFR period, failures due to wear-out. 6

7 .9 Expected future life If the hazard function h (t) is increasing, then we would expect the expected future life conditional on reaching age a to decrease with a and conversely. It turns out to be true. We show this by linking the conditional mean life (i.e. life expectancy) to the conditional reliability. Result De ne the (future) life expectancy at age a by L (a) E (T ajt > a) (.) Z tr (tja) dt since the conditional p.d.f. of future life T f T a is f (tja) R (tja) (t > ). (We have already made extensive use of f (t) R (t)). Integrating by parts gives Z L (a) [ tr (tja)] + R (tja) dt Z R (a) R (a) R (tja) dt Z Z a R (a + t) dt R (t) dt (.a) (.b) We see that future life expectancy is the area below the conditional reliability function R (tja) : Example Show that the reliability function R (t) ( b t b < t < b elsewhere is IFR and nd the residual mean life (MTTF) at age t : (i) Solution h (t) which is an increasing function of t: d (ln R (t)) dt ; < t < b b t 7

8 (ii) At any age a the conditional life expectancy at age t is MT T F (t ) L (t ) Z b t R (t) R (t ) dt b b t b t Z b b t (b t ) t dt b (b t) t b which is a decreasing function of t. In fact the expected life is half the time interval from t to b:. Summary of reliability basics. Knowing any one of the following functions uniquely characterizes a lifetime distribution: lifetime density f (t) reliability (survival) function R (t) hazard function h (t) and we can go from one to another using formulae presented above.. The bathtub curve is an important concept characterizing a product life cycle or human lifespan (if we are lucky!). BUT we have not yet considered the possibility of repairs and maintenance. 3. The hazard function h (t) is also known as the failure rate or force of mortality. Certain h (t) representing monotone or constant failure rates can be used to model di erent sections of the bathtub curve. Such h (t) also imply monotone (future) life expectancies.. 8

9 . Some common failure time distributions In reliability studies, because failure times are de ned only for T >, some families of distributions are commonly used in place of the normal distribution (truncated at t ). In particular the exponential E () the Weibull distribution W (; ) the Gamma distribution (k; ) the Gumbel (extreme value) distriburion the lognormal distribution. The exponential distribution The exponential distribution E () ( > ) is the characterized by a constant failure rate (CF R) ; namely h (t) (.) It is often used for electronic components (failures are "rare events") and can be used to model items during their useful working life (the at section of the bathtub curve). We easily obtain and R (t) exp f H (t)g e t (.) f (t) R (t) The e ect of changing on the pdf is shown here. Ex. Verify that e t (.3) Mean Median Variance ln :693 9

10 .. Memoryless property At any age t ; the conditional reliability is R (tjt ) R (t + t) R (t ) exp f (t + t)g exp f t g e t Thus R (tjt ) R (t) so a component age t has the same reliability characteristics as a new component. (Called the non-aging or memoryles property.) In particular, future life expectancy is constant at : These properties follow from the previous section considering that E () is both IFR and DFR... Poisson shock model Suppose a device is subject to a stream of shocks arriving in a Poisson process with rate per unit time. If the device fails at the rst shock Pr fn shocks in time tg (t)n e t (.4) n! n ; ; ; ::: R (t) Pr fno shocks in (; t)g e t

11 . The Gamma family Consider a device subject to shocks arriving in a Poisson process (.4) with rate : Suppose the device fails if and only if it receives k shocks. This model leads to a reliability function belonging to the Gamma family (k; ) with integer k. NB.. The exponential distribution is a special case of the gamma distribution with k :. The rate is entirely dependent on the time scale (seconds, minutes etc.) So is known as the scale parameter and k the shape parameter. Recall that:. The sum of k independent exponential random variables with parameter has the gamma distribution (k; ) : In this case (k integer) it is also called the Erlang distribution.. The gamma integral (gamma function) is de ned as (z) Z Two important properties of the gamma integral: u z e u du (.6) (i) When z k; integer, (k) (k )! the factorial function. Otherwise we have to refer to tables of the gamma function. (ii) The gamma integral satis es the recursion (z + ) k (k). 3. The p.d.f. of (k; ) is f (t) (t)k e t t > for k > (k) (t)k (k )! e t t > for integer k ( ) (.7).. Reliability function for (k; ) from shock model Consider a xed time interval (; t) and a device new at time. From the Poisson process (.4) P n (t) Pr fexactly n shocks in time (; t)g (t)n e t n!

12 Therefore R (t) Pr ffewer than k shocks in time (; t)g P (t) + ::: + P k (t) e t " + t + (t)! + ::: + (t)k k! e t S k (t) (.8) Notice that S k (t) is the partial sum to k terms of the exponential series expansion of e t and satis es the recursion : d dt S k Sk S k (.9) for k > : #.. Lifetime density and hazard function The lifetime density f (t) can be obtained from R (t) using (:8) and (:9) f (t) R (t) d dt [e t S k (t)] [e t (S k ) + e t S k ] which is of the required form (.7). The hazard function is e t [S k S k ] (t)k (k )! e t (.) h (t) f (t) R (t) (t) k (k )!S k (t) (.) Now [h (t)] is a polynomial in t consisting of entirely negative powers t ; t ; :::; t (k ). So [h (t)] is a decreasing function of t, i.e. h (t) is an increasing function of t: Hence Result The gamma failure time distribution is IFR for integer k:

13 .3 The Weibull distribution The Weibull family provides a very exible life distribution model. It is characterized by the reliability function h R (t) exp (t) i ; t > (.) where > is a scale parameter and > is a dimensionless shape parameter. Notice that has dimensions Time and some books use instead, known as characteristic life, as the scale parameter. The lifetime density (p.d.f.) is f (t) R (t) h (t) exp (t) i (.3) leading to h (t) f (t) R (t) (t) (.4) a simple power of t: This property characterizes the Weibull distribution. As a check, we see that (.) has the form R (t) exp [ H (t)] where H (t) R t h (u) du and (.4) is substituted for h (t) :.3. The hazard function and special cases Since h (t) _ t which is a positive power of t for > ; therefore W (; ) is IFR for >. Similarly for < ; W (; ) is DFR and for the distribution is CFR. gives the exponential distribution h (t) gives the Rayleigh distribution h (t) t 3:44 gives a good approximation to a normal p.d.f. 3

14 .3. Mean time before failure (MTBF) The mean life or MTBF for W (; ) is + where (z) R u z e u du is the gamma function previously de ned. Proof (.5a) (.5b) Make the substitution Z Z R (t) dt h exp u (t) t u dt u (t) i dt Then we arrive at (.5b) Z u The equivalence of (.5a) (.5b) follows from the e u du function recursion. An online calculator for (z) can be found at ndgamma.cfm Example Consider a device with lifetime distributed as W (; ) where 3 ; 6; in hours : Find the MTTF and the median life t m : Which is the better measure of "typical life"? Solution i) The MTTF is + 6; (4) 6; 3! 96; hours 4

15 ii) The median life is given by h exp R (t m ) :5 (t m ) i :5 (t m ) ln t m (ln ) Given values of ; give t m 6 (ln ) 3 538: 4 Note: The median life is a better measure than the mean life since the p.d.f. is highly skewed. Since < the distribution is DFR - suitable for modelling initial "infant mortality" portion of bathtub curve..3.3 Further properties. Approximately 63% of failures will have occurred by t, the characteristic life of W (; ) : h R (t) exp (t) i R exp [ ] :368 :63 All Weibull survival curves pass through a common point.. A series system of k identical Weibull distributed components also has the Weibull distribution of failure with the same shape parameter but a di erent scale parameter. Let T ; T ; :::; T k be the individual times to failure. Let Y be the system failure time. for a series system and let R S (t) be the system reliability Y min ft ; T ; :::; T k g (.6) R S (t) Pr (Y > t) Pr (T > t \ T > t \ ::: \ T k > t) ky R i (t) i assuming independent failures. For a system of identical components T i s W (; ) h R i (t) exp (t) i for each i h R S (t) exp k (t) i exp k t Hence system time to failure Y s W ; k. This is known as the self-reproducing property of the Weibull distribution. 5

16 .4 Systems of components Many systems can be modelled as an network of components joined in series or in parallel..4. Series systems A pure series system fails if and only if any component fails. Assume (usually) that T i are independent random variables. Then R S (t) ky R i (t) (.7) as obtained previously. Notice that if R min (t) min i fr i (t)g then R S (t) R min (t) : A series system is less reliable than the weakest link! When components are identical R i (t) R (t) each i then i R S (t) [R (t)] k The hazard function of a series system is additive Proof h S (t) kx h i (t) (.8) i h S (t) f S (t) R S (t) kx i d dt ln R i (t) kx h i (t) i d dt ln R S (t) This result explains why the Weibull (and in particular the exponential) failure time distribution has the self-reproducing property: when h i (t) are a simple power of t, the sum h S (t) will be also be a multiple of that power of t:.4. Parallel systems A pure parallel system fails if and only if all components fail. R P (t) R P (t) ky ( R i (t)) i ky ( R i (t)) (.9) In particular letting R max (t) max i fr i (t)g we have R P (t) R max (t) : Thus a parallel system improves on the reliability of the most reliable component (common sense result). i 6

17 .4.3 k-out of -n systems Such systems have n identical components of which k is the minimum number of working components for the system to function. The system reliability is given by R S (t) nx Pr (exactly r working components) rk nx rk n r R (t) r ( R (t)) n r (.) Assuming independent failures, the number of working components at time t has the binomial distribution Bin (n; p) with p R (t) : Ex. Show that setting k n and k respectively in (.) results in expressions (.7) (.9) for the system reliability of a pure series and a pure parallel system..4.4 Examples. Two components with T s E ( ) ; T s E ( ) : Find the system reliability if joined (a) in series and (b) in parallel. Show that the mean times to failure are respectively S + P + +. Two identical components with CFR are joined in parallel operation. What should the component failure rate be to achieve a system reliability R P () :95? [Ans.:53]. Show that the corresponding system MTTF is 597 hours, a 5% improvement on MTTF for a single component. 3. Find the reliability at hours of a -out of-3 system comprising two components each with constant failure rate 3 5 : [Ans..9974] 7

18 3. Replacement policies Consider a renewable system in which components that fail in service are replaced immediately on failure by an identical new component. A light tting is an example of such a system. Assume also a new component at time t. The intervals T ; T ;... between renewals are a sequence of non-negative random variables (RV s) that are independently and identically distributed (IID). Suppose that f (t) is the common failure time density The time to the k th renewal Y k is the sum of IID RV s Y k T + T + ::: + T k (3.) and the density function (p.d.f.) f k (t) of Y k is related to f (t) through the moment generating function (m.g.f.) or through the Laplace transform (LT) in the following way. De ne the LT of f (t) by ef (s) Z e st f (t) dt where T stands for any one of the T i (i ; :::; k) : Then ef k (s) E e sy k E e s(t +T +:::+T k ) (3.a) E e st (3.b) ky E e st i i [ e f (s)] k (3.3) i.e. the LT raised to the k th power. E.g. for k the density of T + T is given by the convolution integral and the LT s related by f (t) Z t f (t h i ef (s) ef (s) u) f (u) du 3. Renewal theory De nitions A renewal process is a sequence of IID non-negative random variables X ; X ;... The renewal function M (t) is the expected number of renewals in [; t] M (t) E [N (t)] (3.4) where fn (t) : t g is a discrete RV known as the renewal counting process The renewal density m (t) is the intensity of renewals at time t m (t) lim Pr fa renewal occurs in time (t; t + t)g (3.5) t! t 8

19 We may show the following Result m (t) d dt M (t) M (t) (3.6) Proof For small t the probability of more than one renewal in (t; t + t) is negligible so w.p. m (t) t N (t) N (t + t) N (t) w.p. m (t) t Hence applying the formula P xp (x) for expectation of a random variable Therefore in the limit as t! : Fundamental renewal theorem E [N (t)] :m (t) t + : ( m (t) t) m (t) t m (t) t E [N (t)] E [N (t)] t d dt M (t) The renewal function satis es an integral equation (a convolution) M (t) F (t) + Z t M (t u) f (u) du (3.7) Proof N (t) is a discrete RV taking values in the range f; ; ; :::g : Consider a xed t. Let U be the time to the rst failure.conditional on u the expected number of renewals in [; t] is E [N(t)jU u] The density function of U is clearly f (t) so if u t + M (t u) if u < t as required. M (t) E [N(t)] : Pr (U t) + Z t Z t F (t) + Z t [ + M (t u)] f (u) du f (u) du + Z t Z t M (t E [N(t)jU < t] f (u) du M (t u) f (u) du u) f (u) du 9

20 Relationship between Laplace transforms The fundamental renewal equation can be solved by using the convolution theorem for Laplace transforms. From (3.7) we obtain fm (s) F e (s) + M f (s) f e (s) (3.8) dy Apply the LT identity L sey y to y M (t) gives dt em (s) sm f (s) fm (s) em (s) s noting that M () { no. of renewals in time interval of length }. Also F e f (s) e (s) s F () ; hence because em (s) f e (s) + em (s) f e (s) s s s em (s) f e (s) + em (s) f e (s) em (s) ef (s) f e (s) (3.9) (3.9) can be inverted for a speci c lifetime density f (t) to give m (t) and M (t) : Example Components having an exponential distribution of failure p.d.f. f (t) e t ef (s) Thus em (s) Z Z s + ef (s) e f (s) e st f (t) dt e (s+)t dt s (3.) By taking the inverse LT we nd (by inspection or from tables) m (t) (3.a) M (t) Z t m (u) du t (3.b) Example

21 For components with lifetime distributed as (; ) p.d.f. f (t) te t ef (s) Z e st f (t) dt Z te (s+)t dt s + Z s + : (s + ) (s + ) te (s+)t dt s + since R t:e t dt may be veri ed (using integration by parts or observing that the integral is E (T ) where T s E ()): Thus em (s) ef (s) f e (s) s (s + ) To invert this Laplace transform use partial fractions em (s) A s + B s + We nd (Ex.) that A ; B so em (s) s s + which can be inverted by observing that L () s and L e kt (LT shift theorem). s + k m (t) e t (3.a) The renewal function is found by integration and using M () M (t) t e t (3.b) 4 We now examine two examples of replacement policies. Block replacement. Age replacement

22 3. Block replacement Under a block replacement policy the unit is replaced on failure and at xed times T; T; 3T; ::: In practice we have a block or set of components (e.g. lights on a motorway). Two types of replacement are carried out:. units as they fail (in-service failures). all units (irrespective of which have failed and been replaced) at the xed times T; T; 3T; ::: The former are sometimes termed emergency replacements and the latter are planned replacements. Suppose that the unit cost for planned replacement is c p and the unit cost for in-service replacement is c s with c s > c p What is an optimal choice for T? [n.b. T is not a random variable.] Consider a cycle between kt and (k + ) T Cost incurred per cycle of length T is c p +c s N (T ) where N (:) is the renewal counting function. Mean cost per cycle of length T is C (T ) c p + c s E [N (T )] c p + c s M (T ) : We choose T to minimize the mean cost per unit time Di erentiate w.r.t. T and set to zero (T ) (T ) c p + c s M (T ) T c p T + c s m (T ) T M (T ) T (3.3) Hence Example T m (T ) M (T ) c p c s (3.4) For components with constant hazard rate (that do not age) M (t) t (T ) c p + c s T T c p T + c s We see that (T ) decreases monotonically with T so for such components it is not worthwhile to perform block replacement. In fact block replacement is only worthwhile for IFR components since otherwise (for DFR components) replacing old with new doesn t improve the failure rate. Example

23 Substitute (3.) into (3.4) gives (Ex.) after simplifying e T (T + ) 4 c p c s (3.5) This is a nonlinear equation for T of the form (T ). Since (T ) < for T > (Ex.) LHS is monotonic decreasing from () and (T )! as T! : Therefore (3.5) has a unique solution T provided < 4 c p c s < and block replacement is therefore worthwhile for components with the failure provided c p < c s 4 (; ) distribution of i.e. the cost of an in-service replacement is greater than four times the cost of a planned replacement. Notes on general solution In general, for components with other lifetime distributions, (T ) leading to (3.4) may have multiple solutions giving rise to local minima. We examine (3.3) for T # and as T! :. As T # we have M (T ) # so (T ) s c p T showing that the cost of planned replacement dominates as the time interval for planned replacements reduces.. As T! we have c p T! so (T ) s c s M (T ) T and the cost of in-service replacements dominates. In fact as T! where is the mean life of the component so (T ) s c s M (T ) s T cost of an in-service replacement MT BF 3

24 3.3 Age replacement Block replacement policy has advantage of ease of administration, since records do not have to be kept of when items are replaced. All items are replaced at times kt irrespective of the length of time in service. However it has the disadvantage that almost new items (those replaced upon failure just before a time kt ) are replaced at planned times kt: We may instead replace components either at a predetermined age T or on failure if it occurs earlier. The costs are respectively c p, c s with c p < c s : Let the time interval between consecutive replacements (of either type) de ne a cycle. We nd T to minimize Expected replacement cost per cycle (T ) Expected cycle length E [C] say E [L] (3.6) The cost per cycle is a random variable (RV) cp w.p. R (T ) C c s w.p. R (T ) The numerator of (3.6) is therefore E [C (T )] c p R (T ) + c s [ R (T )] c s (c s c p ) R (T ) (3.7) The length of a cycle is another RV having both discrete and continuous components T w.p. R (T ) L t w.p. f (t) t Strictly we should write Pr [L (t; t + t)] f (t) t for t < T: Therefore the denominator of (3.6) is Z T E [L] T R (T ) + Z T R (t) dt tf(t)dt (3.8a) (3.8b) Let I (T ) R T R (t) dt: Recall that R (t) f (t) and use integration by parts Z T tf(t)dt Z T tr (t)dt Z T [ tr (t)] T + R (t) dt T R (T ) + I (T ) 4

25 so, as required, Hence Z T T R (T ) + tf(t)dt I (T ) (T ) c s (c s c p ) R (T ) I (T ) (3.9) Set (T ) at the optimal (minimum cost) time to planned replacement T (c s c p ) f (T ) I (T ) R (T ) [c s (c s c p ) R (T )] f (T ) I (T ) R (T ) + R (T ) R (T ) + h (T ) I (T ) c s (c s c p ) c s (c s c p ) (3.) We can solve (3.) given a speci c lifetime distribution. Example For components with a (; ) failure time distribution We need to evaluate I (t) : f (t) te t t > R (t) ( + t) e t h (t) f (t) R (t) t + t Substitute into (3.) I (t) Z t Z t R (u) du ( + u) e u du e t te t by parts (verify) R (T ) + I (T ) h (T ) ( + T ) e T + T + T c s (c s c p ) e T T e T Simplifying, we obtain (Ex.) e T + T + T e T c s (c s c p ) c s ( + T ) T (c s c p ) e T + ( ) T (3.) 5

26 c s where (c s c p ) > : The solution T to (3.) can be found numerically. Graphically T is the intersection of the curve y e T with the straight line y + ( ) T: This has a unique solution if and only if < : Therefore age replacement is worthwhile if and only if c s (c s c p ) < or. c p < c s (3.) i.e. the cost of in-service replacement on failure should be greater than twice the cost of a planned replacement at age T determined by (3.). General solution (for any failure distribution) Let us examine (3.9) for the limiting cases T # and as T! : (T ) c s (c s c p ) R (T ) I (T ) As T # I (T )! and R (T )! so (T )! as expected As T! R (T )! I (T )! (since MTBF R R (t) dt) so (T )! c s which is the same limit (of course) as we found for block replacement. 3.4 Preventive maintenance (PM) Reliability and therefore mean time to failure (which is the integral of the reliability function) can be increased by preventive maintenance. Suppose that a system is restored to "as good as new" periodically at intervals of time T: So the system renews itself at times T; T; :::: De ne system reliability under PM as R M (t) Pr ffailure has not occurred by time tg We decompose an interval of time (; t) into k PM cycles of length T and a remainder t kt. The system survives to time t if and only if it survives every PM cycle f; ; :::; kg and a further time t T: Assuming independence we have Interval containing t Formula for R M (T ) < t < T R (t) T < t < T R (T ) :R (t T ) T < t < 3T [R (T )] :R (t T ) kt < t < (k + ) T [R (T )] k :R (t kt ).. 6

27 System mean time to failure (mean time to rst failure) MTTFF But R (t Z Z T R M (t) dt Z T Z 3T R (t) dt + R (T ) R (t T ) dt + R (T ) R (t T ) dt + ::: T T kt ) R (t) if PM restores system to "as good as new". MTTFF X Z T R (T ) k R (t) dt k R T R (t) dt R (T ) I (T ) R (T ) (3.3) Exercise Find R M (9) and MTTFF for a component with Weibull distributed time to failure with and : (a characteristic life of days) if PM is performed every days. 3.5 PM with minimal repairs With PM we assumed a system was restored to "as good as new". With block replacement we also assumed in-service failures were replaced by a new component. In a policy of minimal repair for in-service failures, the component is brought back to "working order" but its age (hazard function) is only restored to what it was immediately before failure. Suppose we carry out PM, e ectively a "full service" or replacement by a new component, every T time units, and carry out a minimal repair for failures in-service. We can use the theory of block replacement to determine an optimal interval T for PM. Let De ne the renewal function c p cost of PM c r cost of minimal repair M (t) E [N (t)] to be the mean (expected) number of restorations (minimal repairs) in time (; t) : With minimal repairs and no PM we show that M (t) H (t) (3.4) where H (t) R t h (u) du is the cumulative hazard function of the component. Proof 7

28 Consider a small time interval (t; t + t) and let N (t) N (t + t) small t N (t) : For su ciently Hence in the limit t! we obtain N (t) w.p. h (t) t w.p. h (t) t E [N (t)] h (t) t (3.5) is obtained by integration using M () : The mean cost per PM cycle is therefore dn (t) E dt h (t) d E [N (t)] dt h (t) E [N (t)] H (t) (3.5) (t) c p + c r H (T ) T Hence from previous block replacement theory T is a root of T h (T ) H (T ) c p c r (3.6) 8

29 4.State dependent models 4. Markov processes Markov processes can model a system which is considered to be in any one of a discrete set of states fs ; S ; :::; S k g at time t (continuous time). The fundamental Markov assumptions are. The probability that a system will undergo a transition from one state to another state depends only on the current state and not on the previous state history - system is "memoryless".. Transition probabilities (instantaneous rates) are constant over time - system is "stationary": Throughout this section, components are assumed to have the exponential lifetime distribution. As an example, consider a system consisting of two components each with constant failure rate (i.e. lifetimes are exponential) We may de ne the system states is as follows: State Component Component S Working Working S Failed Working S 3 Working Failed S 4 Failed Failed For a series system, the system failed or "down" states are D fs ; S 3 ; S 4 g : For a parallel system the system failed state is D fs 4 g : The corresponding system "up" states are respectively U fs g and U fs ; S ; S 3 g. Let P i (t) Pr (System is in state S i at time t) (4.) System reliability is de ned as the probability of being "up", i.e. not being in a failed state R S (t) X iu P i (t) (4.) Let the instantaneous failure rate for component i be i (i ; ) : A rate diagram shows the rates of transition between the states: S. & S S 3 &. S 4 Pr [Transition S! S in (t; t + t)] t Pr [Transition S! S 3 in (t; t + t)] t Pr [Transition S 3! S 4 in (t; t + t)] t Pr [Transition S! S 4 in (t; t + t)] t 9

30 In order for the system to be in state S at time t + t ) the system must be in state S at time t, and ) no transition occurs from state S in time (t; t + t) Pr [system is in state S at time t + t] Pr [system is in state S at time t] Pr [no transition during (t; t + t)] P (t + t) P (t) ( t t) P (t + t) P (t) ( + ) P (t) t In the limit as t! d dt P (t) P (t) ( + ) P (t) (4.3) For the system to be in state S at time t + t; either ) system is in state S at time t and no transition occurs during (t; t + t) ; or ) system is in state S at time t and a transition S! S occurs in (t; t + t) P (t + t) P (t) ( t) + P (t) t () P (t + t) P (t) t P (t) + P (t) In the limit as t! P (t) P (t) + P (t) (4.4) By symmetry in the state transition diagram P3 (t) P 3 (t) + P (t) (4.5) Since the system must be in one of the four possible states 4X P i hence i P 4 (t) [P (t) + P (t) + P 3 (t)] Note that S 4 is an absorbing state. We solve the st order di erential equation (4.3) for P (t) : Z Z dp ( + ) dt P ln (P (t)) ( + ) t + C The constant of integration C since P () ; hence ln () + C: P (t) e ( + )t (4.6) Now substitute (4.6) into (4.4) 3

31 An integrating factor is e t. Then P (t) + P (t) e ( + )t e t P (t) + P (t) d h i e t P (t) e t dt Z t e t P (t) e u du e t using the initial condition P () : Therefore P (t) e t e t (4.7) By symmetry P 3 (t) e t e t (4.8) and P 4 (t) may be found by subtraction. Inserting these expressions for P i (t) into the de nition of R S (t) given above in (4.) we can verify the reliability expressions for series and parallel systems: Series system R s (t) X P i (t) P (t) iu e ( + )t e t :e t R (t) :R (t) as before Parallel system R p (t) X P i (t) P (t) + P (t) + P 3 (t) iu e t + e t e ( + )t e t e t ( R (t)) ( R (t)) as before 3

32 4. Redundancy and load sharing Standby systems are a way of increasing reliability. e.g. hospital electricity generator electrical systems on an aircraft (in triplicate). Standby systems are also known as systems with redundancy. They may involve a switching process that itself can fail. The standby system(s) may be subject to a small probability of failure through deterioration (even when on standby) and so fail to operate when switched in. As an alternative to "cold" standby systems or "passive redundancy", reliability is also increased by units in "active redundancy" where units operate in parallel e.g. multiple tyres on a lorry kidneys in a human body Such systems are "load-sharing". On failure of the primary unit, the secondary unit operates under an increased load and may have a higher rate of failure as a result. Example Consider two identical components in active parallel operation. When both are working, the failure rates are (combined failure rate is ). When one component fails the remaining component has increased rate of failure. Find the system reliability and MTBF. We de ne a Markov process on a system with three states: S both units operating S one failed, one working S 3 both failed and transition rates rates as follows S! S! S 3 Probability arguments as before lead to the following: Hence P (t + t) P (t) [ t] P (t + t) P (t) :t + P (t) [ t] P (t) P (t) P (t) P (t) P (t) with initial conditions P () ; P () P 3 () : 3

33 then As before we obtain Use e t as integrating factor d dt Integrate and use P () to obtain assuming that 6 ; hence System reliability is P (t) e t (4.9) P (t) + P (t) e t e t P (t) P (t) R S (t) X iu P i (t) System mean time before failure is MT BF recalling that R e t dt : e.g. suppose that h i e tp (t) e ( )t n e ( )t o n o e t e t (4.) P (t) + P (t) e t + ne o t e t n o e t e t (4.) Z R S (t) dt Z e t dt Z e t dt + (4.) : failures per day : failures per day system reliability over a day period is R S () :e : :8 :934 and MTBF is : + 6 days : :e 33

34 Notes on solution. The system MTBF has a natural interpretation. Under load sharing the time to rst failure of either component is n.b. hazard function (failure rate) of a series system is : When either fails the system operates as a single unit with (increased) failure rate > : The system lifetime is the sum of the lifetimes T ; T under each phase (MT T F ) S E (T ) + E (T ) +. We can also verify the expression for R S (t) by a direct probability argument conditioning on the time to rst failure of either component. Note that the event "no system failure by time t" occurs if and only one of the following occurs: (a) rst failure occurs after time t, or (b) rst failure occurs at time u, then no subsequent failure. The time to rst failure T is an exponentially distributed with constant failure rate : Therefore the p.d.f. of time to rst failure is e t : Pr fsystem survives to time tg Pr f rst failure occurs after time tg X + lim Pr f rst failure occurs in (u; u + u)g u! u<t Pr fno subsequent failure in time t Let R (t) ; R (t) be the reliability function of each phase of sytem life, f (t) be the p.d.f. of time to rst failure. ug R S (t) R (t) + e t + Z t Z t e t + e t f (u) R (t e u :e (t Z t e ( u) du u) du )u du e t + n e t e ( )t n o e t e t o con rming (4.). 34

35 4.3 Matrix form of Markov Equations The system of linear di erential equations for a load sharing system can be written in matrix-vector form. De ne Then p (t) T (P (t) ; P (t) ; P 3 (t)) may be written where P (t) P (t) P (t) P (t) P (t) P 3 (t) P (t) d dt p (t)t p (t) T M (4.) M Notice that each row of M sums to zero. This can be seen by postmultiplying (4.) by the column vector of ones (; ; ) T d dt p (t)t p (t) T M (4.3) Now p (t) T P 3 j P j (t) and is constant for all t: So the derivative w.r.t. t is zero. Therefore M. A simpli ed procedure for calculating MTBF: We can calculate MTBF (MTTF) without explicitly determining R S (t) R S (t) X ju P j (t) MT T F Z Z X ju R S (t) dt X P j (t) dt ju Z P j (t) dt (4.4) Let q j R P j (t) dt then MT BF P ju q j: The vector q T (q ; q ; q 3 ) can be found by integrating (4.). Z Z d dt p (t)t dt p (t) T M dt p () T p () T q T M (4.5) 35

36 Now p () T (; ; ) and p () T (; ; ) since S is the initial state at t and S 3 is the failed state at which is achieved with probability as t! : Therefore 3 ( ; ; ) (q ; q ; q 3 ) 4 5 which represents the 3 equations in unknowns q ; q q q q q having solution Hence q q as we obtained previously. MT T F q + q + (4.6) 4.4 Further examples 4.4. Standby system with two non-identical components A system consists of a primary unit and a secondary standby unit. The main unit has failure rate and the secondary has a failure rate when working as the primary. When on standby the secondary has a low probability of failure : Find the system MTTF assuming perfect switching and no repairs to either unit. The system states are S S S 3 S 4 both working primary failed, secondary working primary working, secondary failed both failed From the rate diagram we obtain the set of di erential equations S. & S S 3 &. S 4 P (t) + P (t) P (t) P (t) P (t) P 3 (t) P (t) P 3 (t) 36

37 or d dt p (t)t p (t) T M where p (t) T (P (t) ; P (t) ; P 3 (t)) is a reduced probability vector (omitting P 4 (t)) and M 4 is the correspondingly reduced matrix. p () T p () T (; ; ) (; ; ) ; hence ( ; ; ) (q ; q ; q 3 ) q q q q q 3 Hence and therefore q + q q q 3 q MT T F q + q + q Standby system with identical components + A special case of the above arises when : Then the system mean time to failure is MT T F + + (4.7) (see Examples 4, Q). In this case states S and S 3 can be combined as whichever state results, the subsequent rate of transition to S 4 is : So the state transition diagram is as follows + S! S! S 3 Notice that the combined rate of transition into S is + (the time of transition is that of rst failure of either component c.f. a series system) and the corresponding set of di erential equations! P (t) + P (t) P (t) + P (t) P (t) 37

38 d dt p (t)t p (t) T M where p (t) T (P (t) ; P (t)) and M + + so and hence (4.7) results. p () T p () T q T M ( ; ) (q ; q ) q q (4.8) Switch subject to failure Consider a standby system that has a probability p (< ) of successful operation when required. The rate diagram is then as follows The di erential equation system is S p. & ( p)! S S 3 P (t) P (t) P (t) pp (t) P (t) Hence we nd P (t) e t (using P () ) and using P () so P (t) + P (t) pe t d P (t) e t p dt P (t) e t pt P (t) pte t R S (t) P (t) + P (t) ( + pt) e t (4.9) 38

39 4.5 Systems with repair We consider now the possibility of repairs to a system described by a set of states. To maintain the Markov property we will assume that times to repair follow an exponential distribution. Let MT T R denote the component mean time to repair where is the repair rate. As before let MT T F be the component mean time to failure Availability The life of a repairable system consists of alternating periods of being in working and failed states. The mean time to rst failure is now denoted MT T F F and we de ne the availability of a system at time t by A (t) Pr [System is working at time t] For a system without repair A (t) R S (t) : Example ( state Markov model) Consider a single component with failure rate and repair rate : The states are S : working and S : failed.! Here S A (t) P (t) S Pr [System in state S at time t] Since P (t) P (t) the sytem can be described by the single di erential equation dp (t) dt P (t) + ( P (t)) An integrating factor is e (+)t dp (t) dt + ( + ) P (t) (4.) d h P (t) e (+)ti e (+)t dt P (t) e (+)t + e(+)t + C Using the initial condition P () we nd C P (t) e (+)t (4.) 39

40 This is the availability A (t) : Notice that A () lim t! A (t) + (4.) A () is known as the limiting or steady state availability. It may be calculated as A () + MT T F MT T F + MT T R which is the up-time ratio (UTR).. Note the limiting cases: As!, MT T R! and A ()! (instant repairs). As!, MT T R! and A ()! (slow repair).. The steady state availability is also obtained by setting d dt P (t) in (4.) Active parallel redundancy with repair We consider the e ect on the system MTTF of adding a repair facility to an active parallel system of two identical units, each with failure rate. De ne the following states: S : units working S : unit failed under repair S 3 : both units failed S 3 is system failed state The rate diagram is as follows: S S! S 3 Note that S 3 is an absorbing state. At this stage we assume no return from S 3. The system di erential equations are dp (t) dt dp (t) dt dp 3 (t) dt P (t) + P (t) P (t) ( + ) P (t) (4.3) P (t) To determine system MTTF we write the system as 4

41 d dt p (t)t p (t) T M where p (t) T (P (t) ; P (t) ; P 3 (t)) and M 4 3 ( + ) 5 Write q R p (t) dt then as before by integration we obtain p () T p () T q T M Solving the system ( ; ; ) q T M q + q q ( + ) q q gives Hence q + q MT T F q + q 3 + (4.4) The solution has two components. The MTTF without repair of an active parallel system of two identical components is 3 : The extra life of the system due to repair is : Reliability function calculation We can verify this result by solving the di erential equation system (4.3) to obtain the reliability function R S (t) which we then integrate. Recall the following properties of Laplace transforms:. L dy dt Z e st y (t) dt e st y (t) Z + se st y (t) dt s~y (s) y (). L e at Z s + a e (s+a)t dt shift theorem 4

42 Take Laplace transforms of (4.3) s ~ P ~ P + ~ P s ~ P ~ P ( + ) ~ P (4.5) s ~ P 3 ~ P Since R S (t) P 3 (t) we need only solve for ~ P 3. We nd that ~P 3 s s + (3 + ) s + (4.6) A s + B + C say s s where ; (3 + ) q are the roots of the quadratic in the denominator of (4.6). By partial fractions we nd Inverting (4.6) A ; B ( ) ; C ( ) P 3 (t) A + Be t + Ce t e t + ( ) e t Observe that (product of roots of quadratic) R S (t) P 3 (t) e t e t (4.7) Hence MT T F Z Z R S (t) dt e t ( + ) e t dt 3 + as obtained in (4.4). 4

43 4.5.4 Steady state availability Consider now the component active parallel system with the possibility of a repair when the system reaches the failed state S 3 : The repair restores the system to the state S it was in immediately preceding failure. The rate diagram and di erential equations are now so the revised M dp (t) dt dp (t) dt dp 3 (t) dt matrix is S S S 3 P (t) + P (t) P (t) ( + ) P (t) + P 3 (t) P (t) P 3 (t) M 4 The equilibrium distribution is found using Hence letting and p : 3 ( + ) 5 d dt pt p T M (4.8) p + p p ( + ) p + p 3 p p 3 p ; p ; p 3 Now p + p + p 3 ) + + Therefore p p p 3 ( + + ) ( + + ) ( + + ) A () p + p (4.9) 43

44 4.5.5 Alternative derivation We may verify (4.9) using the expression [see (4.) above] A () MT T F MT T F + MT T R (4.3) known as the uptime ratio (UTR). However (4.4) which is MT T F F is not the appropriate MT T F to use because when the system is restored (after failed state S 3 ) it begins a new working cycle starting from state S not from the new state S as we assumed before! S S S 3 Therefore MT T F q + q where q ; q satisfy p () T p () T q T M with p () (; ) and p () (; ) : Thus giving q + q q ( + ) q q q where, as before, so MT T F + MT T R and as before. A () MT T F MT T F + MT T R

45 4.5.6 Further example ( out of 3 system) [see Q. of Examples 5] Consider a system of 3 identical units of which are in active parallel operation. The 3 rd system is either in working order on standby or failed and under repair. If only one unit is working the system is shut down to prevent further failure and is non-operational. The repair rate is (single repair facility). The system becomes operational again when two units are available. State Working Failed S 3 S (under repair) S 3 ( under repair) S S S 3 To nd the MT T F F we assume no return from S 3 (an absorbing state). Then d dt p (t)t p (t) T M where p (t) T (P (t) ; P (t) ; P 3 (t)) and 3 M 4 ( + ) 5 Write q R p (t) dt then p () T p () T q T M with p () (; ; ) and p () (; ; ) gives the system q + q q ( + ) q q and We nd q q + 4 MT T F F q + q + 4 System availability can be determined from the steady state condition p T M together with normalization of probabilities + A () (Ex.) 45

46 5. Analysis of survival data In real life we are faced with the problem of estimating a life distribution in the form of a hazard or a reliability function. The data may come from a trial of n components where a set of times to failure is recorded. In engineering this is is known as life testing. In clinical trials the data may be a set of observed times to remission (cure) or a set of survival data, often collected to compare survival rates between two drugs. 5. Censoring A practical problem with survival data is that it is often too costly or otherwise impractical to collect a full set of data. Examples. items on a 5 hour lifetime trial. 3 items are unfailed after 5 hours when the trial ends. T > 5 for unfailed items. (T is the lifetime random variable). There are are 7 uncensored observations.. In a medical trial a number of patients are lost to the study because they move out of the area, walk out or die from a cause unrelated to the disease being monitored. Other patients are still alive at the end of the trial. Both these examples produce incomplete data known as (right-) censored observations. Two forms of censoring in particular are: Type I We have n units on test at the start of the trial and terminate the test at a xed time T c : The data consist of r ordered failure times t () ::: t (r) T c and n r surviving units for which the failure time T i is unobserved but we have a lower bound T i > T c : Here the number of failures r is a random variable. Type II The test is terminated after a xed number r of failures have been observed (out of the n items initially on trial). The length of the trial T c t (r) is now a random variable. Again there are n r observations for which T i is unobserved but we have a lower bound T i > T c : 5. Estimation for the exponential distribution Example (Wolstenholme, P59) Given the following data from the exponential distribution E () estimate and provide a 95% con dence interval (CI). + ; 3; 6; ; ; 7; ; 7; 5; 8 + A + denotes a right-censored observation. 46

47 Divide the observations into two sets: U funcensored observationsg C fcensored observationsg The likelihood function can be modi ed to include censored observations.. For t i U the contribution to the likelihood function L () is f (t i ) where f (:) is the p.d.f. of E ().. For t i C the contribution to L () is Pr (T i > t i ) R (t i ) Take logs L () Pr observations j f (t) e t Y f (t i ) Y R (t i ) iu ic Y Y e t i e t i iu ic r e P n i t i log L () r log n P d d log L () r t i i np t i Hence ^ r T where T P n i t i is known as the "total time on test". The reliability function may also be estimated as ^R (t) e ^t using the invariance property of maximum-likelihod estimators. In our example n ; r 8 T i ^ 8 :8 To compute a CI we can use the asymptotic theory of mle s. ^ is approximately normal with E ^ d V ar ^ log L () d ^ 47

48 Note that V ar ^ is the reciprocal of observed Fisher information matrix. For the exponential distribution d d log L () r V ar ^ ^ s.e. ^ r ^ pr A 95% CI for is ^ :96 ^ :8 pr :8 :96 p 8 (:5; :35) 48