# MAINTAINED SYSTEMS. Harry G. Kwatny. Department of Mechanical Engineering & Mechanics Drexel University ENGINEERING RELIABILITY INTRODUCTION

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1 MAINTAINED SYSTEMS Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University

2 OUTLINE MAINTE MAINTE

3 MAINTAINED UNITS Maintenance can be employed in two different manners: Preventive maintenance employed before failure occurs goal is to improve reliability What is the effect of preventive maintenance on the reliability function? Corrective maintenance employed after failure occurs goal is to return item to service as quickly as possible How frequently do we anticipate repairs and how much inventory of replacement parts is required? Several concepts clarify maintenance issues. Notably: Availability A system is not always available to perform its function it has downtime. Uptime/downtime captures the essence of reliability/unreliability. Can downtime be minimized? Can downtime be avoided when the system is most needed?

4 IDEALIZED MAINTE Consider a system whose reliability without maintenance is R(t). Suppose maintenance is performed at regular time intervals T M. Denote the reliability of the maintained system by R M (t). Since no maintenance is performed for 0 t < T M, R M (t) = R (t), 0 t < T M Assume ideal maintenance so that the system is restored to as-good-as-new condition. Thus, R M (t) = R (T M ) R (t T M ),. R M (t) = R (T M ) n R (t nt M ), T M < t < 2T M nt M < t < (n + 1) T M

5 MTTF - MAINTAINED UNITS For a system with ideal preventive maintenance: MTTF = MTTF = n=0 0 (n+1)tm nt M R M (t) dt = n=0 (n+1)tm nt M R M (t) dt R (T M) n R (t T M) dt = TM R n=0 (TM)n R (τ) dτ The infinite series evaluates to R 1 n=0 (TM)n = 1 R (T M) Consequently, MTTF = TM 0 R (t) dt 1 R (T M ) 0

6 SYSTEMS WITH CONSTANT FAILURE RATE Ideal preventive maintenance has no affect on the reliability of systems with constant failure rate! Recall Thus, R M (t) = R (t) = e λt ( e λt M) n e λ(t nt M ) = e λt = R (t)

7 SYSTEMS WITH VARIABLE FAILURE RATE Consider a system with a Weibull distribution R (t) = e (βt)α With ideal preventive maintenance Compute R M (t) == e n(βt M) α e (β(t nt M)) α, R M (nt m) R (nt = α m) e n(βt M) +(nβtm ) α nt M < t < (n + 1) T M Preventive maintenance improves reliability if n α 1 1 > 0 and degrades reliability if n α 1 1 < 0 Reliability improves if α > 1 (aging) and degrades if α < 1 (burn-in).

8 MAINTE PROBLEM SETUP Consider a situation in which a failed unit can be repaired or replaced, so that operation can proceed indefinitely. We would like to anticipate the number of repairs that will be required in a given time period. Assume a constant failure rate λ 0. p (n t) probability of exactly n repairs at time = 0 : p (0 0) = 1, p (n 0) = 0, n = 1, = T > 0 : n=0 p (n T) = 1

9 MAINTAINED UNITS PROBABILITY OF 0 FAILURES The system operates for time period t + dt At most one failure can occur in the infinitesimal period dt 0 failures can occur at t + dt requires 0 failures at time t and 0 failures during dt p (0 t + dt) = p (0 t) (1 λ 0 dt) p (0 t + dt) p (0 t) = λ 0 dt dp (0 t) = λ 0 p (0 t) dt p (0 t) = e λ 0t

10 MAINTAINED UNITS PROBABILITY OF n FAILURES The system operates for time period t + dt At most one failure can occur in the infinitesimal period dt n failures can occur at t + dt in two ways n failures at time t and 0 failures during dt n 1 failures at time t and 1 failures during dt p (n t + dt) = p (n t) (1 λ 0 dt) + p (n 1 t) λ 0 dt dp (n t) dt p (n t) = λ 0 e λ 0t = λ 0 p (n t) + λ 0 p (n 1 t) t 0 p (n 1 τ)e λ 0τ dτ

11 MAINTAINED UNITS RECURSIVE SOLUTION We sequentially solve these equations for n = 0, 1, 2,... p (0 t) = e λ 0t p (1, t) = (λ 0 t) e λ 0t [ ] p (2 t) = (λ 0 t) 2 /2 e λ 0t. p (n t) = [(λ 0 t) n /n!] e λ 0t This is the Poisson distribution with λ = λ 0 t. p (n; λ) = λn n! e λ, n = 0, 1, 2,...

12 MEAN TIME BETWEEN FAILURES From the Poisson distribution p (n; λ), we can compute the expected number of failures in the time period (0, t] The variance is µ N = E {N (t)} = n=0 np (n; λ 0t) = λ 0 t { σn 2 = E (N µ N ) 2} = λ 0 t The mean time between failures is MTBF = t = 1 µ N λ 0 Finally, we may wish to compute the probability that the number of failures in time t is greater than the number n P (N > n) = k=n+1 (λ 0 t) k e λ0t = 1 n k! k=0 (λ 0 t) k k! e λ 0t

13 DOWNTIME The downtime due to a failure is the sum of several elements: diagnosis time, part(s) access time, repair time, etc. Thus, we consider downtime, T D to be a random variable. The mean down time is MDT = 0 τf TD (τ) dτ Mean downtime is also called mean time to repair, MTTR. Three distributions are commonly used for downtime: Exponential Normal Lognormal

14 EXAMPLE: DOWNTIME The downtime associated with a particular failure is assumed to have an exponential distribution. The MDT is known to be 5 hours. Hence the repair rate is µ = 1 = 0.2 hr 1 MDT The probability that the repair will take longer than 5 hours is P (T D > 5) = e = 0.368

15 Availability, A(t), is the probability that the system is performing properly at time t. This is sometimes called point availability. The (average) interval availability over the time interval (t 1, t 2 ), A av (t 1, t 2 ), is A av (t 1, t 2 ) = 1 t2 A (t)dt t 2 t 1 t 1 This is sometimes referred to as mission availability. The long run average availability is 1 T A lr = lim A (t)dt T T 0

16 NON-REPAIRABLE SYSTEMS If a system is not repairable, then availability is simply reliability, interval availability is A(t) = R(t) A av (t 1, t 2 ) = 1 t2 R (t)dt t 2 t 1 t 1 long run availability is A lr = 0

17 CONSTANT REPAIR RATE Consider an item that has a constant failure rate λ and constant repair rate µ. λ t is the conditional probability of failure during t, given availability at t. µ t is the conditional probability of repair during t, given unavailability at t. Consequently, A (t + t) = A (t) λ t A (t) + µ t (1 A (t)) A(t+ t) A(t) t = (λ + µ) A (t) + µ d A (t) = (λ + µ) A (t) + µ dt assuming A(0) = 1, A (t) = µ λ + µ + λ λ + µ e (λ+µ)t

18 CONSTANT REPAIR RATE, CONT D Integrate the last expression to obtain A av (t 1, t 2) = µ λ + µ + λ (λ + µ) 2 (t 2 t 1) ( ) e (λ+µ)t 1 e (λ+µ)t 2 The long run availability is obtained with t 1 0, t 2 : A lr = µ λ + µ ordinarily µ >> λ, so that A lr 1 λ µ = MTTF MTTF + MTTR

19 SYSTEM Consider a system composed of n components. Suppose the components are independent with availabilities A i (t). The system availability, A(t), can be computed from RBD s using the same formulas as for computing system reliability. For a serial structure, all components must be available for the system to be available, so the system availability is A (t) = n A i (t) i=1 For a parallel structure all components must be unavailable for the system to be unavailable, so the system availability is A (t) = 1 n i=1 (1 A i (t))

20 SYSTEM CONSTANT RATE COMPONENTS The component long run availabilities are A lr,i = µ i λ i + µ i 1 µ i λ i for µ i >> λ i For a serial configuration A lr = ( n 1 µ ) i 1 n µ i i=1 λ i i=1 λ i For a parallel configuration A lr = 1 ( n 1 µ ) i 1 n i=1 λ i + µ i i=1 ( µi λ i )

21 Preventive maintenance periodic, ideal maintenance improves reliability only when failure rate is increasing (aging period). Corrective maintenance maintenance is performed upon failure For constant failure rate λ 0, MTBF = 1/λ 0 The probability of more than n failures in a a given time period T is 1 n (λ 0T) k e λ 0T k=0 k! Availability downtime and uptime point and interval availability computing MDT (MTTR) and system availability

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