Availability and Maintainability...• The Laplace transform of the restoration density is g (t)...

11Piero Baraldi

Availability and Maintainability

22Piero Baraldi

Introduction: reliability and availability

• Non maintained systems: theycannot be repaired after a failure(a telecommunication satellite, aF1 engine, a vessel of a nuclearpower plant).

• Maintained systems: they can berepaired after the failure (pump ofan energy production plant, acomponent of the reactoremergency cooling systems)

System types

33Piero Baraldi

Introduction: reliability and availability

• Non maintained systems: theycannot be repaired after a failure(a telecommunication satellite, aF1 engine, a vessel of a nuclearpower plant).

• Maintained systems: they can berepaired after the failure (pump ofan energy production plant, acomponent of the reactoremergency cooling systems)

Reliability (𝑹𝑹(𝑻𝑻𝑴𝑴))It quantifies the system capabilityof satisfying a specified missionwithin an assigned period of time(𝑇𝑇𝑀𝑀): 𝑅𝑅 𝑇𝑇𝑀𝑀 = 𝑃𝑃 𝑇𝑇 > 𝑇𝑇𝑀𝑀

Availability (𝑨𝑨(𝒕𝒕))It quantifies the system ability to fulfill the assigned mission at any specific moment of the lifetime

Performance parametersSystem types

44Piero Baraldi

Definition of Availability

• X(t) = indicator variable such that:• X(t) = 1, system is operating at time t

• X(t) = 0, system is failed at time t

• Instantaneous availability p(t) and unavailability q(t)

X(t)

1

0

F F F

R R R t

F = FailedR = under Repair

[ ] [ ]( ) ( ) 1 ( )p t P X t E X t= = = [ ]( ) ( ) 0 1 ( )q t P X t p t= = = −

55Piero Baraldi

Contributions to Unavailability

• RepairA component is unavailable because under repair after a failure

66Piero Baraldi


• RepairA component is unavailable because under repair

• Testing / preventive maintenanceA component is removed from the system because:

a. it must undergo preventive maintenance

TRANSMISSION BELT

https://context.reverso.net/traduzione/inglese-italiano/belts

77Piero Baraldi





b. it has to be tested (safety system, standby components)

EMERGENCY CORE COOLING SYSTEM

88Piero Baraldi





b. it has to be tested (safety system, standby components)

• Unrevealed failureA stand-by component fails unnoticed. The system goes on without noticing the component failure until a test on the component is made or the component is demanded to function

99Piero Baraldi

Average availability descriptors

• How to compare different maintenance strategies?

1010Piero Baraldi



• We need to define quantities for an average description of the system probabilistic behavior:• If after some initial transient effects, the instantaneous availability assumes a time

independent value → Limiting or steady state availability:

)(lim tppt ∞→

=

1111Piero Baraldi



• We need to define quantities for an average description of the system probabilistic behavior:• If after some initial transient effects, the instantaneous availability assumes a time

independent value → Limiting or steady state availability:

• If the limit does not exist → Average availability over a period of time Tp:

∫

∫

===

===

p

p

p

p

T

ppT

T

ppT

TDOWNTIMEdttq

Tq

TUPTIMEdttp

Tp

0

0

...)(1

...)(1

)(lim tppt ∞→

=

1212Piero Baraldi

Average Availability12

p(t)

p(t)

p(t)

t

t

1

1

1

t

𝑇𝑇𝑝𝑝

𝑇𝑇𝑝𝑝

𝑇𝑇𝑝𝑝

∫pT

dttp0

)(

p

p

Tp

T

pT

dttp=∫0 )(

∫pT

dttp0

)(

UPTIME DOWNTIME

2𝑇𝑇𝑝𝑝

∫pT

dttp0

)(

∫pT

dttp0

)(

∫pT

dttp0

)(

1313Piero Baraldi

Maintenability

• How fast a system can be repaired after failure?

• Repair time TR depends from many factors such as:

13

1414Piero Baraldi

Maintenability

• How fast a system can be repaired after failure?

• Repair time TR depends from many factors

• Repair time TR varies statistically from one failure to another, depending on the conditions associated to the particular maintenance events

• Let TR denotes the downtime random variable, distributed according to the pdf 𝑔𝑔𝑇𝑇𝑅𝑅(𝑡𝑡)

• Maintenability:

14

( ) ( )∫=≤t

TR dgtTPR0

ττ

1515Piero Baraldi

Failure classification

• Failure can be:

• revealed

• unrevealed (stanby components, safety systems)

15

1616Piero Baraldi

REVEALED FAILURE

• Hypotheses:• Constant failure rate 𝜆𝜆

• Restoration starts immediately after the component failure

• Probability density function of the random time duration TR of the repair process = 𝑔𝑔𝑇𝑇𝑅𝑅(𝑡𝑡)

• Objective:• Computation of the availability p(t) of a component

continuously monitored

• Method• Conceptual Experiment with𝑁𝑁0 identical components that start

at time t = 0

Balance equation for the number N(t) of components working between time t and time t+∆t

Availability of a continuously monitored component (1)

working at t+Δt -

Failurebetween tand t+ Δt

working at t

∫ ∆⋅−⋅∆⋅⋅⋅+∆⋅⋅⋅−⋅=∆+⋅t

ttgpNttpNtpNttpN0

)()()()()( ττλτλ

Repair between t and t+ Δt

+=

(1) (2) (3) (4)

1) Number of items UP at time t+∆t

2) Number of items UP at time t

3) Number of items failing during the interval ∆t

4) Number of items that had failed in (τ, τ+∆τ) and whose restoration terminates in (t, t+∆t);


0 0 0 0

• The integral-differential form of the balance

• The solution can be obtained introducing the Laplace transforms

∫ ⋅−⋅⋅+⋅−=t

dtgptpdt

tdp

0

)()()()( τττλλ 1)0( =p


𝑝𝑝 𝑡𝑡

𝑑𝑑 𝑝𝑝 𝑡𝑡𝑑𝑑𝑡𝑡

𝑝𝑝 𝑡𝑡 ∗ 𝑔𝑔 𝑡𝑡 = �0

𝑡𝑡𝑝𝑝 𝜏𝜏 𝑔𝑔 𝑡𝑡 − 𝜏𝜏 𝑑𝑑𝜏𝜏

𝐿𝐿 𝑝𝑝 𝑡𝑡 = �0

+∞𝑒𝑒−𝑠𝑠𝑡𝑡𝑝𝑝 𝑡𝑡 𝑑𝑑𝑡𝑡 = �𝑝𝑝 (𝑠𝑠)

𝐿𝐿𝑑𝑑 𝑝𝑝 𝑡𝑡𝑑𝑑𝑡𝑡

= 𝑠𝑠 � �𝑝𝑝 𝑠𝑠 − 𝑝𝑝 0

𝐿𝐿 𝑝𝑝 𝑡𝑡 ∗ 𝑔𝑔 𝑡𝑡 = �𝑝𝑝 (𝑠𝑠) � �𝑔𝑔 (𝑠𝑠)

Time domain Laplace Transform

• Applying the Laplace transform we obtain:

which yields:

• Inverse Laplace transform → p(t)

• Limiting availability:

)(~)(~)(~1)(~ sgspspsps ⋅⋅+⋅−=−⋅ λλ

( ))(~11)(~

sgssp

−⋅+=

λ

[ ] ( )

−⋅+

=⋅==→→∞→∞ )(~1

lim)(~lim)(lim00 sgs

sspstppsst λ


• As s → 0, the first order approximation of �𝑔𝑔(𝑠𝑠) is:

Rs sdgsdgsdgesg ττττττττττ ⋅−=⋅−≅+⋅−== ∫∫∫

∞∞∞− 1)(1)()...1()()(~

000

[ ]R g RE Tτ =

cycle""pair failure/rea of timeaverageUP iscomponent the timeaverage

=+

=τ+λλ

=τ⋅λ+

=τ⋅⋅λ+

=→∞ MTTRMTTF

MTTF1

11

1ss

slimpRRR

0s

General result!


MTTR

MTTR=

Exercise 1

• Find instantaneous and limiting availability for an exponential component whose restoration probability density is

tetg ⋅−⋅= µµ)(

Exercise 1: Solution


• The Laplace transform of the restoration density is


[ ]µ

µ+

==s

tgLsg )()(~

)(1)(~

λµµ

µλ ++⋅

+=

+⋅+

=sss

sss

sp 𝑝𝑝 𝑡𝑡 = 𝐿𝐿−1 �𝑝𝑝(𝑠𝑠) = 𝐿𝐿−1𝑠𝑠 + 𝜇𝜇

𝑠𝑠(𝑠𝑠 + 𝜆𝜆 + 𝜇𝜇)




�𝑝𝑝 𝑠𝑠 =𝐴𝐴𝑠𝑠

+𝐵𝐵

(𝑠𝑠 + 𝜇𝜇 + 𝜆𝜆)=𝐿𝐿−1

𝑠𝑠 + 𝜇𝜇


𝐿𝐿 1 =1𝑠𝑠

𝐿𝐿1

𝑠𝑠 + 𝑎𝑎= 𝑒𝑒−𝑎𝑎𝑡𝑡

𝑝𝑝 𝑡𝑡 = 𝐴𝐴𝐿𝐿−11𝑠𝑠

+ 𝐵𝐵𝐿𝐿−11

𝑠𝑠 + 𝜆𝜆 + 𝜇𝜇= 𝐴𝐴 + 𝐵𝐵𝑒𝑒− 𝜆𝜆+𝜇𝜇 𝑡𝑡




�𝑝𝑝 𝑠𝑠 =𝐴𝐴𝑠𝑠

+𝐵𝐵

(𝑠𝑠 + 𝜇𝜇 + 𝜆𝜆)=𝐴𝐴 𝑠𝑠 + 𝜇𝜇 + 𝜆𝜆 + 𝐵𝐵𝑠𝑠𝑠𝑠(𝑠𝑠 + 𝜆𝜆 + 𝜇𝜇)

=

𝑠𝑠(𝐴𝐴 + 𝐵𝐵) + 𝐴𝐴𝜇𝜇 + 𝐴𝐴𝜆𝜆𝑠𝑠(𝑠𝑠 + 𝜆𝜆 + 𝜇𝜇)

=𝑠𝑠 + 𝜇𝜇


� 1 = 𝐴𝐴 + 𝐵𝐵𝜇𝜇 = 𝐴𝐴𝜆𝜆 + 𝐴𝐴𝜇𝜇

𝐴𝐴 =𝜇𝜇

𝜇𝜇 + 𝜆𝜆

𝐵𝐵 =𝜆𝜆

𝜇𝜇 + 𝜆𝜆𝑝𝑝 𝑡𝑡 = 𝐴𝐴 + 𝐵𝐵𝑒𝑒− 𝜇𝜇+𝜆𝜆 𝑡𝑡 =

𝜇𝜇𝜇𝜇 + 𝜆𝜆

+𝜆𝜆

𝜇𝜇 + 𝜆𝜆𝑒𝑒− 𝜇𝜇+𝜆𝜆 𝑡𝑡

𝐿𝐿−1𝑠𝑠 + 𝜇𝜇


𝐿𝐿 1 =1𝑠𝑠

𝐿𝐿1

𝑠𝑠 + 𝑎𝑎= 𝑒𝑒−𝑎𝑎𝑡𝑡

0 20 40 60 80 100 120 140 160 180 2000.9

0.91

0.92

0.93

0.94

0.95

0.96

0.97

0.98

0.99

1

t [day]

p(t)

constant failure rate =1/50 [day-1]constant repair rate = 1/5 [day-1]


UNREVEALED FAILURE

Safety Systems: Examples

Fire extinguisher

Risk: Ignition of the gases in the oil tank, lightning strike or generation of sparks due to electrostatic charges.

Fire protection

Exercise 2: Instantaneous Availability of an Unattended Component

• Find the instantaneous unavailability of an unattended component (no repair is allowed) whose cumulative failure time distribution is F(t)

• Find the instantaneous availability of an unattended component (no repair is allowed) whose cumulative failure time distribution is F(t)

SOLUTION

The istantaneous unavailability, i.e. the probability q(t) that at time t the component is not functioning is equal to the probability that it fails before t

)()( tFtq ≡


3434Piero Baraldi

Availability of a safety system under periodic maintenance (1)

• Safety systems are generally in standby until accident and their components must be periodically tested (interval between two consecutive test = Tp)

• The instantaneous availability is a periodic function of time

p(t)

R(t)1

τ 2τ 3τ tTp 2Tp 3Tp

3535Piero Baraldi

Availability of a safety system under periodic maintenance (1)

• Safety systems are generally in standby until accident and their components must be periodically tested

• The instantaneous unavailability is a periodic function of time

• The performance indicator used is the average unavailability over a period of time Tp

complete maintenance cycle

Average time the system is not working

pp

T

T TDOWNtime

T

dttqq

p

p== ∫0

)(

3636Piero Baraldi

EX. 3: Availability of a component under periodic maintenance

• Unavailability is due to unrevealed random failures, e.g. with constantfailure rate λ

• Assume also instantaneous and perfect testing and maintenanceprocedures performed every Tp hours

• Draw the qualitative time evolution of the instantaneous availability

• Find the average unavailability of the component

3737Piero Baraldi37

3838Piero Baraldi

Ex 3. Solution (1)

• Unavailability is due to unrevealed random failures, e.g. with constantrate λ

• Assume also instantaneous and perfect testing and maintenanceprocedures

• The instantaneous availability within a period Tp coincides with thereliability

p(t)

R(t)1

τ 2τ 3τ tTp 2Tp 3Tp

3939Piero Baraldi

• The average unavailability and availability are then:

For different systems, we can compute 𝑞𝑞𝑇𝑇𝑝𝑝 and 𝑝𝑝𝑇𝑇𝑝𝑝 by first computing their failure probability distribution FT(t) and reliability R(t) and then applying the above expressions.

Ex. 3: Solution (2)

p

T

T

p

T

T T

dttF

T

dttqq

pp

p

∫∫ === 00)()(

p

T

p

T

T T

dttR

T

dttpp

pp

p

∫∫ == 00)()(

pp TT qp −=1

4040Piero Baraldi

• For a system with constant faiure rate:

Ex. 3: Solution (3)

pTT Tqppp

λ2111 −=−=

Good approximationwhen 𝜆𝜆𝑡𝑡 < 0.1

pp

p

p

T

p

T t

p

T

T

p

T

T TTT

T

dtt

T

dte

T

dttF

T

dttqq

pppp

pλ

λλλ

21

2

)1()()( 20000 ==≈

−==== ∫∫∫∫ −

4141Piero Baraldi

Ex.4: A more realistic case 41

• The test is performed after a time 𝜏𝜏 from the end of the previous test

• Assume a finite test time τR

• Unavailability is due to unrevealed random failures, e.g. with constant rate λ


• Estimate the average unavailability and availability over the complete maintenance cycle period Tp =τ + τR

4242Piero Baraldi

42Safety System under periodic maintenance: a more realistic case

4343Piero Baraldi

Ex.4: Solution (1) 43

• Assume a finite test time τR:


t

p(t)

1

τR τR τR

τ

4444Piero Baraldi

• Assume a finite test time τR

• The average unavailability and availability over the complete maintenance cycle period Tp = τ +τR are:

[ ]0 0

( ) ( )R T R T

RR

F t dt F t dtq

τ τ

τ ττ τ

τ τ τ

+ += ≈ << ≈

+

∫ ∫

[ ]0 0

( ) ( )

RR

R t dt R t dtp

τ τ

τ ττ τ τ

= ≈ << ≈+

∫ ∫

Ex. 4: Solution (2)

4545Piero Baraldi

Safety System under periodic maintenance: a more realistic case

• Objective: computation of the average unavailability over the lifetime [0, TM]

• Hypoteses:• The safety system is initially working: q(0) = 0; p(0) = 1

• 𝜏𝜏 time interval between two consecutive maintenance interventions

• 𝜏𝜏𝑟𝑟 duration of the maintenance intervention

• Failure causes:

• random failure at any time T ∼ FT(t)• on-line switching failure on demand ∼ Q0

• maintenance disables the component ∼ γ0 (human error during inspection, testing or repair)

MT T

DOWNtimeqM=],0[

4646Piero Baraldi46

𝑞𝑞[0,𝑇𝑇𝑀𝑀] = ∫0𝑇𝑇𝑀𝑀 𝑞𝑞 𝑡𝑡 𝑑𝑑𝑡𝑡

𝑇𝑇𝑀𝑀= ∫0

𝐴𝐴 𝑞𝑞 𝑡𝑡 𝑑𝑑𝑡𝑡+∫𝐴𝐴𝑇𝑇𝑀𝑀 𝑞𝑞 𝑡𝑡 𝑑𝑑𝑡𝑡

𝑇𝑇𝑀𝑀

=∫0𝐴𝐴 𝑞𝑞 𝑡𝑡 𝑑𝑑𝑡𝑡+𝐾𝐾 ∫𝐴𝐴

𝐵𝐵 𝑞𝑞 𝑡𝑡 𝑑𝑑𝑡𝑡+∫𝐵𝐵𝐶𝐶 𝑞𝑞 𝑡𝑡 𝑑𝑑𝑡𝑡

𝑇𝑇𝑀𝑀=

= 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝑇𝑇𝐷𝐷𝑀𝑀𝐸𝐸0𝐴𝐴+𝐾𝐾 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝑇𝑇𝐷𝐷𝑀𝑀𝐸𝐸𝐴𝐴𝐵𝐵+𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝑇𝑇𝐷𝐷𝑀𝑀𝐸𝐸𝐵𝐵𝐶𝐶𝑇𝑇𝑀𝑀

=𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝑇𝑇𝐷𝐷𝑀𝑀𝐸𝐸0𝑇𝑇𝑀𝑀

𝑇𝑇𝑀𝑀

A B τ C Ft

T0 τ τ+τR 2τ+2τR2τ+τR

maintenance maintenance

𝑘𝑘 =𝑇𝑇𝑀𝑀 − 𝜏𝜏𝜏𝜏 + 𝜏𝜏𝑟𝑟


4747Piero Baraldi

A B τ C Ft



(1) (2) (3) (4)


Period from O to A

Possible failure causes: random failure (𝐸𝐸1) on-line switching failure on demand (𝐸𝐸2)

∀𝑡𝑡 ∈ O, A → 𝑞𝑞0𝐴𝐴 𝑡𝑡 = 𝑃𝑃 𝐸𝐸1 + 𝑃𝑃 𝐸𝐸2 − 𝑃𝑃 𝐸𝐸1 ∩ 𝐸𝐸2 = 𝐹𝐹 𝑡𝑡 + 𝑄𝑄0 − 𝐹𝐹 𝑡𝑡 𝑄𝑄0

[ ](0 ) 0 0 0 0 00 0 0

( ) (1 ) ( ) (1 ) ( )D A A T TT q t dt Q Q F t dt Q Q F t dtτ τ τ

τ= = + − ⋅ = ⋅ + − ⋅∫ ∫ ∫],0[ ADOWNtime

4848Piero Baraldi

A B τ C Ft



(1) (2) (3) (4)


Period (2) From A to B

System under maintenance:

∀𝑡𝑡 ∈ A, B → 𝑞𝑞𝐴𝐴𝐴𝐴 𝑡𝑡 = 1

𝐷𝐷𝐷𝐷𝐷𝐷𝑁𝑁𝑇𝑇𝐷𝐷𝐷𝐷𝐸𝐸[A,B] = ∫𝐴𝐴𝐴𝐴 𝑞𝑞𝐴𝐴𝐴𝐴 𝑡𝑡 𝑑𝑑𝑡𝑡 = ∫𝜏𝜏

𝜏𝜏+𝜏𝜏𝑅𝑅 1 𝑑𝑑𝑡𝑡 = 𝜏𝜏𝑟𝑟

4949Piero Baraldi

A B τ C Ft



(1) (2) (3) (4)


Period (3) From B to C Possible failure causes: random failure (𝐸𝐸1) on-line switching failure on demand (𝐸𝐸2) maintenance disables the component (𝐸𝐸3)

∀𝑡𝑡 ∈ B, C → 𝑞𝑞𝐴𝐴𝐵𝐵 𝑡𝑡 = 𝑃𝑃 𝐸𝐸1 + 𝑃𝑃 𝐸𝐸2 + 𝑃𝑃 𝐸𝐸3 − 𝑃𝑃 𝐸𝐸1 ∩ 𝐸𝐸2 − 𝑃𝑃 𝐸𝐸2 ∩ 𝐸𝐸3 −−𝑃𝑃 𝐸𝐸1 ∩ 𝐸𝐸3 + 𝑃𝑃 𝐸𝐸1 ∩ 𝐸𝐸2 ∩ 𝐸𝐸3 =

=𝛾𝛾0 + (1 − 𝛾𝛾0)(𝑄𝑄0 + 1 − 𝑄𝑄0 𝐹𝐹(𝑡𝑡))

( ) 0 0 0 00 0

( ) (1 ) (1 ) ( )D BC BC TT q t dt Q Q F t dtτ τ

γ τ γ τ

= = ⋅ + − ⋅ ⋅ + − ⋅

∫ ∫],[ CBDOWNtime

5050Piero Baraldi

(0 ) 0 0 0 0 0 00 0

(1 ) ( ) (1 ) (1 ) ( )MM

D T T R TR

TT Q Q F t dt Q Q F t dtτ τ

τ τ γ τ γ ττ τ

= + − ⋅ + ⋅ + + − ⋅ ⋅ + − ⋅ + ∫ ∫

(0 ) 0 00 0 0 0 0

0 0

1 1( ) (1 ) (1 ) ( )M

M

D TT T R T

M M M R

T Q Qq F t dt Q Q F t dtT T T

τ ττ τ γ τ γ ττ τ

− = = + ⋅ + ⋅ + + − ⋅ ⋅ + − ⋅ + ∫ ∫

DT

MTq

DOWNtime

MT T

DOWNtimeqM=

−𝜏𝜏


5151Piero Baraldi

(0 ) 0 00 0 0 0 0

0 0

1 1( ) (1 ) (1 ) ( )M

M

D TT T R T

M M M R



− = = + ⋅ + ⋅ + + − ⋅ ⋅ + − ⋅ + ∫ ∫MTq

• 𝜏𝜏 ≪ 𝑇𝑇𝑀𝑀

• ∫0𝜏𝜏 𝐹𝐹𝑇𝑇 𝑡𝑡 𝑑𝑑𝑡𝑡 ≤ 𝜏𝜏

• 𝜏𝜏𝑅𝑅 ≪ 𝜏𝜏

(0 ) 0 00 0 0 0 0

0 0

1 1( ) (1 ) (1 ) ( )M

M

D TT T R T

M M M R



− = = + ⋅ + ⋅ + + − ⋅ ⋅ + − ⋅ + ∫ ∫

MT T

DOWNtimeqM=

MT T

DOWNtimeqM=


5252Piero Baraldi

• Consider an exponential system with small, constant failure rate

λ ⇒• Since typically γ0<<1, Q0<<1, the average unavailability reads:

00 0 0 0

0

1(1 ) ( )M

RT T

Qq Q F t dtττ γ γ

τ τ −

≅ + + − ⋅ + ⋅

∫

0 0 012M

RTq Qτ γ λ τ

τ≅ + + + ⋅ ⋅

Error after test Switching failure on demand

Random, unrevealed failures

between testsMaintenance

( ) 1 tTF t e tλ λ− ⋅= − ≅ ⋅

MT T

DOWNtimeqM=

MT T

DOWNtimeqM=


5353Piero Baraldi

Where to study?

• Slides• Lecture 5: Reliability of simple systems

• Chapter 5 (Red Book): Theory• Chapter 5 (Green Book): Exercises

• Lecture 6: Availability and Maintenability• Chapter 6 (Red Book): Theory• Chapter 6 (Green Book): Exercises

53

Availability and Maintainability...• The Laplace transform of the restoration density is g (t)...

Documents

Transcript of Availability and Maintainability...• The Laplace transform of the restoration density is g (t)...