Harry G. Kwatny S - Information Technologyhgk22/courses/MEM361/MEM361-Lecture5.pdf · Harry G....
Transcript of Harry G. Kwatny S - Information Technologyhgk22/courses/MEM361/MEM361-Lecture5.pdf · Harry G....
ENGINEERINGRELIABILITY
INTRODUCTION
PREVENTIVEMAINTE-NANCE
CORRECTIVEMAINTE-NANCE
AVAILABILITY
SUMMARY
ENGINEERING RELIABILITYMAINTAINED SYSTEMS
Harry G. Kwatny
Department of Mechanical Engineering & MechanicsDrexel University
ENGINEERINGRELIABILITY
INTRODUCTION
PREVENTIVEMAINTE-NANCE
CORRECTIVEMAINTE-NANCE
AVAILABILITY
SUMMARY
OUTLINE
INTRODUCTION
PREVENTIVE MAINTENANCE
CORRECTIVE MAINTENANCE
AVAILABILITY
SUMMARY
ENGINEERINGRELIABILITY
INTRODUCTION
PREVENTIVEMAINTE-NANCE
CORRECTIVEMAINTE-NANCE
AVAILABILITY
SUMMARY
MAINTAINED UNITS
Maintenance can be employed in two different manners:
I Preventive maintenanceI employed before failure occursI goal is to improve reliabilityI What is the effect of preventive maintenance on the reliability
function?
I Corrective maintenanceI employed after failure occursI goal is to return item to service as quickly as possibleI How frequently do we anticipate repairs and how much inventory of
replacement parts is required?
Several concepts clarify maintenance issues. Notably:
I AvailabilityI A system is not always available to perform its function – it has
downtime.I Uptime/downtime captures the essence of reliability/unreliability.I Can downtime be minimized? Can downtime be avoided when the
system is most needed?
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INTRODUCTION
PREVENTIVEMAINTE-NANCE
CORRECTIVEMAINTE-NANCE
AVAILABILITY
SUMMARY
IDEALIZED MAINTENANCE
I Consider a system whose reliability withoutmaintenance is R(t).
I Suppose maintenance is performed at regular timeintervals TM. Denote the reliability of the maintainedsystem by RM(t).
I Since no maintenance is performed for 0 ≤ t < TM,
RM (t) = R (t) , 0 ≤ t < TM
I Assume ideal maintenance so that the system isrestored to as-good-as-new condition. Thus,
RM (t) = R (TM) R (t − TM) , TM < t < 2TM...
RM (t) = R (TM)n R (t − nTM) , nTM < t < (n + 1) TM
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CORRECTIVEMAINTE-NANCE
AVAILABILITY
SUMMARY
MTTF - MAINTAINED UNITS
For a system with ideal preventive maintenance:
MTTF =
∫ ∞0
RM (t) dt =∑∞
n=0
∫ (n+1)TM
nTM
RM (t) dt
MTTF =∑∞
n=0
∫ (n+1)TM
nTM
R (TM)n R (t − TM) dt =∑∞
n=0R (TM)n
∫ TM
0R (τ) dτ
The infinite series evaluates to∑∞
n=0R (TM)n =
11− R (TM)
Consequently,
MTTF =
∫ TM0 R (t) dt
1− R (TM)
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INTRODUCTION
PREVENTIVEMAINTE-NANCE
CORRECTIVEMAINTE-NANCE
AVAILABILITY
SUMMARY
SYSTEMS WITH CONSTANT FAILURE RATE
I Ideal preventive maintenance has no affect on thereliability of systems with constant failure rate!
I RecallR (t) = e−λt
I Thus,
RM (t) =(
e−λTM)n
e−λ(t−nTM) = e−λt = R (t)
ENGINEERINGRELIABILITY
INTRODUCTION
PREVENTIVEMAINTE-NANCE
CORRECTIVEMAINTE-NANCE
AVAILABILITY
SUMMARY
SYSTEMS WITH VARIABLE FAILURE RATE
I Consider a system with a Weibull distribution
R (t) = e−(βt)α
I With ideal preventive maintenance
RM (t) == e−n(βTM)α
e−(β(t−nTM))α
, nTM < t < (n + 1) TM
I ComputeRM (nTm)
R (nTm)= e−n(βTM)α+(nβTM)α
I Preventive maintenance improves reliability if nα−1 − 1 > 0 anddegrades reliability if nα−1 − 1 < 0
I Reliability improves if α > 1 (aging) and degrades if α < 1 (burn-in).
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INTRODUCTION
PREVENTIVEMAINTE-NANCE
CORRECTIVEMAINTE-NANCE
AVAILABILITY
SUMMARY
CORRECTIVE MAINTENANCE – PROBLEM SETUP
I Consider a situation in which a failed unit can berepaired or replaced, so that operation can proceedindefinitely.
I We would like to anticipate the number of repairs thatwill be required in a given time period.
I Assume a constant failure rate λ0.
p (n|t) , probability of exactly n repairs at time t
@t = 0 : p (0|0) = 1, p (n|0) = 0, n = 1, 2, . . .
@t = T > 0 :∑∞
n=0p (n|T) = 1
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CORRECTIVEMAINTE-NANCE
AVAILABILITY
SUMMARY
MAINTAINED UNITS – PROBABILITY OF ‘0’FAILURES
I The system operates for time period t + dtI At most one failure can occur in the infinitesimal period
dtI 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− λ0dt)p (0|t + dt)− p (0|t) = λ0dt
dp (0|t)dt
= −λ0p (0|t)
p (0|t) = e−λ0t
ENGINEERINGRELIABILITY
INTRODUCTION
PREVENTIVEMAINTE-NANCE
CORRECTIVEMAINTE-NANCE
AVAILABILITY
SUMMARY
MAINTAINED UNITS – PROBABILITY OF nFAILURES
I The system operates for time period t + dtI At most one failure can occur in the infinitesimal period
dtI n failures can occur at t + dt in two ways
I n failures at time t and 0 failures during dtI n− 1 failures at time t and 1 failures during dt
p (n|t + dt) = p (n|t) (1− λ0dt) + p (n− 1|t)λ0dt
dp (n|t)dt
= −λ0p (n|t) + λ0p (n− 1|t)
p (n|t) = λ0e−λ0t∫ t
0p (n− 1|τ)e−λ0τdτ
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CORRECTIVEMAINTE-NANCE
AVAILABILITY
SUMMARY
MAINTAINED UNITS – RECURSIVE SOLUTION
We sequentially solve these equations for n = 0, 1, 2, . . .
p (0|t) = e−λ0t
p (1, t) = (λ0t) e−λ0t
p (2|t) =[(λ0t)2 /2
]e−λ0t
...p (n|t) = [(λ0t)n /n!] e−λ0t
This is the Poisson distribution
p (n;λ) =λn
n!e−λ, n = 0, 1, 2, . . .
with λ = λ0t.
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PREVENTIVEMAINTE-NANCE
CORRECTIVEMAINTE-NANCE
AVAILABILITY
SUMMARY
MEAN TIME BETWEEN FAILURES
From the Poisson distribution p (n;λ), we can compute theexpected number of failures in the time period (0, t]
µN = E {N (t)} =∑∞
n=0np (n;λ0t) = λ0t
The variance is
σ2N = E
{(N − µN)2
}= λ0t
The mean time between failures is
MTBF =tµN
=1λ0
Finally, we may wish to compute the probability that thenumber of failures in time t is greater than the number n
P (N > n) =∑∞
k=n+1
(λ0t)k
k!e−λ0t = 1−
∑n
k=0
(λ0t)k
k!e−λ0t
ENGINEERINGRELIABILITY
INTRODUCTION
PREVENTIVEMAINTE-NANCE
CORRECTIVEMAINTE-NANCE
AVAILABILITY
SUMMARY
DOWNTIME
I The downtime due to a failure is the sum of severalelements: diagnosis time, part(s) access time, repairtime, etc. Thus, we consider downtime, TD to be arandom variable.
I The mean down time is
MDT =∫ ∞
0τ f TD
(τ) dτ
Mean downtime is also called mean time to repair,MTTR.
I Three distributions are commonly used for downtime:I ExponentialI NormalI Lognormal
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INTRODUCTION
PREVENTIVEMAINTE-NANCE
CORRECTIVEMAINTE-NANCE
AVAILABILITY
SUMMARY
EXAMPLE: DOWNTIME
I The downtime associated with a particular failure isassumed to have an exponential distribution.
I The MDT is known to be 5 hours. Hence the repair rateis
µ =1
MDT= 0.2 hr−1
I The probability that the repair will take longer than 5hours is
P (TD > 5) = e−0.2×5 = 0.368
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INTRODUCTION
PREVENTIVEMAINTE-NANCE
CORRECTIVEMAINTE-NANCE
AVAILABILITY
SUMMARY
AVAILABILITY
I Availability, A(t), is the probability that the system isperforming properly at time t. This is sometimes called‘point’ availability.
I The (average) interval availability over the time interval(t1, t2), Aav(t1, t2), is
Aav (t1, t2) =1
t2 − t1
∫ t2
t1A (t)dt
This is sometimes referred to as ‘mission’ availability.I The long run average availability is
Alr = limT→∞
1T
∫ T
0A (t)dt
ENGINEERINGRELIABILITY
INTRODUCTION
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CORRECTIVEMAINTE-NANCE
AVAILABILITY
SUMMARY
NON-REPAIRABLE SYSTEMS
If a system is not repairable, thenI availability is simply reliability,
A(t) = R(t)
I interval availability is
Aav (t1, t2) =1
t2 − t1
∫ t2
t1R (t)dt
I long run availability is
Alr = 0
ENGINEERINGRELIABILITY
INTRODUCTION
PREVENTIVEMAINTE-NANCE
CORRECTIVEMAINTE-NANCE
AVAILABILITY
SUMMARY
CONSTANT REPAIR RATE
Consider an item that has a constant failure rate λ andconstant repair rate µ.
I λ∆t is the conditional probability of failure during ∆t,given availability at t.
I µ∆t is the conditional probability of repair during ∆t,given unavailability at t.
I 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
ddt
A (t) = − (λ+ µ) A (t) + µ
I assuming A(0) = 1,
A (t) =µ
λ+ µ+
λ
λ+ µe−(λ+µ)t
ENGINEERINGRELIABILITY
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CORRECTIVEMAINTE-NANCE
AVAILABILITY
SUMMARY
CONSTANT REPAIR RATE, CONT’D
I Integrate the last expression to obtain
Aav (t1, t2) =µ
λ+ µ+
λ
(λ+ µ)2 (t2 − t1)
(e−(λ+µ)t1 − e−(λ+µ)t2
)I The long run availability is obtained with t1 → 0, t2 →∞:
Alr =µ
λ+ µ
I ordinarily µ >> λ, so that
Alr ≈ 1− λ
µ=
MTTFMTTF + MTTR
ENGINEERINGRELIABILITY
INTRODUCTION
PREVENTIVEMAINTE-NANCE
CORRECTIVEMAINTE-NANCE
AVAILABILITY
SUMMARY
SYSTEM AVAILABILITY
I Consider a system composed of n components.I Suppose the components are independent with
availabilities Ai(t).I The system availability, A(t), can be computed from
RBD’s using the same formulas as for computingsystem reliability.
I For a serial structure, all components must be availablefor the system to be available, so the system availabilityis
A (t) =∏n
i=1Ai (t)
I For a parallel structure all components must beunavailable for the system to be unavailable, so thesystem availability is
A (t) = 1−∏n
i=1(1− Ai (t))
ENGINEERINGRELIABILITY
INTRODUCTION
PREVENTIVEMAINTE-NANCE
CORRECTIVEMAINTE-NANCE
AVAILABILITY
SUMMARY
SYSTEM AVAILABILITY – CONSTANT RATE
COMPONENTS
I The component long run availabilities are
Alr,i =µi
λi + µi≈ 1− µi
λifor µi >> λi
I For a serial configuration
Alr =∏n
i=1
(1− µi
λi
)≈ 1−
∑n
i=1
µi
λi
I For a parallel configuration
Alr = 1−∏n
i=1
(1− µi
λi + µi
)≈ 1−
∏n
i=1
(µi
λi
)
ENGINEERINGRELIABILITY
INTRODUCTION
PREVENTIVEMAINTE-NANCE
CORRECTIVEMAINTE-NANCE
AVAILABILITY
SUMMARY
SUMMARY
I Preventive maintenanceI periodic, ideal maintenanceI improves reliability only when failure rate is increasing
(aging period).I Corrective maintenance
I maintenance is performed upon failureI For constant failure rate λ0, MTBF = 1/λ0I The probability of more than n failures in a a given time
period T is
1−∑n
k=0
(λ0T)k
k!e−λ0T
I AvailabilityI downtime and uptimeI point and interval availabilityI computing MDT (MTTR) and system availability