monte carlo simulation for system reliability and availability … · 2018-12-05 · 11 the state...

Monte Carlo Simulation for System Reliability and Availability Analysis

The experimental view

3

Simulation of system stochastic process

Simulation for reliability/availability analysis of a component

Examples

Contents

Sampling Random Numbers

4

Buffon

Gosset (Student)

Fermi, von Neumann, Ulam

Neutron transport

System Transport (RAMS)

Kelvin

The History of Monte Carlo Simulation

1707-88

1908

1824-1907

1950’s

1930-40’s

1990’s

SAMPLING RANDOM NUMBERS

6

Example: Exponential Distribution

Probability density function:

Expected value and variance:

00

0)(

=

= −

t

tetf t

T

2

1][

1][

=

=

TVar

TE

t

fT(t)

( ) t

Tf t e −=

7

Sampling Random Numbers from FX(x)

( ) x

X exF −−=1

Sample R from UR(r) and find X:

( )RFX X

1−=

( ) x

X exFR −−== 1

( ) ( )RRFX X −−== − 1ln11

Example: Exponential distribution

8

sample an

Graphically:

0 1, ,..., ,...kx x x =

0

k

k k i

i

F P X x P X x=

= = =

~ [0,1)R U

Sampling from discrete distributions

9

Arc number i Failure

probability Pi

1 0.050

2 0.025

3 0.050

4 0.020

5 0.075

Failure probability estimation: example

• 1- Calculate the analytic solution for the failure probability of the network, i.e., the probability of no connection between nodes S and T

• 2- Repeat the calculation with Monte Carlo simulation

10

Analytic solution

10

43

1

[ 1] [ ] 3.07 10T j

j

P X P M −

=

= = 3

1 1 4

3

2 2 5

4

3 1 3 5

5

4 2 3 4

[ ] 0.05 0.02 1 10

[ ] 0.025 0.075 1.875 10

[ ] 0.05 0.05 0.075 1.875 10

[ ] 0.025 0.05 0.02 7.5 10

P M p p

P M p p

P M p p p

P M p p p

−

−

−

−

= = =

= = =

= = =

= = =

1) Minimal cut sets

M2={2,5}

M3={1,3,5}

M4={2,3,4}

2) Network failure probability (rare event approximation):

11

The state of arc i can be sampled by applying the inverse transform method to the set of discrete probabilities of the mutually exclusive and exhaustive states

•We calculate the arrival state for all the arcs of the newtork

•As a result of these transitions, if the system falls in anyconfiguration corresponding to a minimal cut set, we add 1 to Nf

•The trial simulation then proceeds until we collect N trials. We obtain the estimation of the failure probability dividing Nf by N

0 1

R

S

B

B

1

31→

B

B

1

21→

SRU[0,1)

,1i ip p−

ip 1 ip−

Monte Carlo simulation

12

• N number of trials

• NF number of trials corresponding to realization of a minimal cut set

• The failure probability is equal to NF / N

clear all;

%arc failure probabilities

p=[0.05,0.025,0.05,0.02,0.075]; nf=0;

n=1e6; %number of MC simulations

for i=1:n

%sampling of arcs fault events

r=rand(1,5); s=zeros(1,5); rpm=r-p;

for j=1:5

if (rpm(1,j)<=0)

s(1,j)=1;

end

end

%cut set check

fault=s(1,1)*s(1,4)+s(1,2)*s(1,5)+s(1,1)*s(1,3)*s(1,5)+s(1,2)*s(1,3)*s(1,4);

if fault >= 1

nf=nf+1;

end

end

pf=nf/n; %system failure probability

For n=106, we obtain3[ 1] 3.04 10TP X −=


SIMULATION OF SYSTEM STOCHASTIC PROCESS

14Monte Carlo simulation of system

stochastic process

• PLANT = system of Nc suitably connected components.

• COMPONENT = a subsystem of the plant (pump, valve,...) which maystay in different exclusive (multi)states (nominal, failed, stand-by,... ).Stochastic transitions from state-to-state occur at stochastic times.

• STATE of the PLANT at t = the set of the states in which the Nc

components stay at t. The states of the plant are labeled by a scalarwhich enumerates all the possible combinations of all the componentstates.

• PLANT TRANSITION = when any one of the plant componentsperforms a state transition we say that the plant has performed atransition. The time at which the plant performs the n-th transition iscalled tn and the plant state thereby entered is called kn.

• PLANT LIFE = stochastic process.

15

Random walk = realization of the system life generated by the underlying

state-transition stochastic process.

Realization of system, stochastic process:

random walk

16Phase Space of System Stochastic

Process

17

( ) ( ) 0,R R

MC t C t t T=

( ) ( ) 1 ,R R

MC t C t t T= +

( ) ( ) 1 ,R R

MC t C t t T= +

( ) ( ) 0,R R

MC t C t t T=

( )ˆ ( )R

T

C tF t

M=

Example: System Reliability Estimation

Events at components

level, which do not entail

system failure

18

• T(tt’; k’)dt = conditional probability of a transition at tdt, given that thepreceding transition occurred at t’ and that the state thereby enteredwas k’.

• C(k k’; t) = conditional probability that the plant enters state k, giventhat a transition occurred at time t when the system was in state k’. Boththese probabilities form the ”trasport kernel”:

K(t; k t’; k’)dt = T(t t’; k’)dt C(k k’; t)

• (t; k) = ingoing transition density or probability density function (pdf) ofa system transition at t, resulting in the entrance in state k

Stochastic Transitions: Governing

Probabilities

19

SIMULATION OF COMPONENT STOCHASTIC PROCESS FOR AVAILABILITY / RELIABILITY

ESTIMATION

20

12

m

State 2

State 1

State X=1 → ON

State X=2 → OFF

Single component with exponential

distribution of the failure time

21

12

m

State 2

State 1

State X=1 → ON

State X=2 → OFF



22

12

m

State 2

State 1

State X=1 → ON

State X=2 → OFF



23

12

m

State X=1 → ON

State X=2 → OFF



values

3· 10-3 h-1

m 25· 10-3 h-1

Limit unavailability:

1 /0.1071

1 / 1 /U

m

m = =

+

24Monte Carlo simulation for estimating

component availability at time t

• Nt time intervals Δt

• If the component fails in t+Δt, the counter increases cA(tj)= cA(tj) +1; otherwise, cA(tj) = cA(tj)

Trial 1



• … another trial

Trial 1Trial 1

Trial i-th



• … another trial

Trial 1

Trial i-th

Trial 1

Trial i-th

Trial M-th



Trial 1

Trial i-th

Trial M-th

Trial 1

Trial i-th

Trial M-th

The counter cA(tj) adds 1 until M trials have been sampled

= unavailability at time tj

= mean value of cA(tj) at time tj

( ) { ( ) 1}A j

G t P X t =

( )( )

A

j

A j

c tG t

M=

28

Pseudo Code

%Initialize parameters

Tm=mission time; Nt=number of trials; Dt=bin length;

Time_axis=0:Dt:Tm;

FOR i=1:Nt

%parameter initialization for each trial

t=0; failure_time=0; repair_time=0; state=1;

while t<Tm

if state=1

t=t+exprnd(1/lambda); state=2; failure_time=t;

lower_b=minimum(find(time_axis>=failure_time));

counter(i,lower_b)=1;

Else

t=t+exprnd(1/mu); state=1; repair_time=t;

upper_b=find(time_axis<up_time,1,'last');

counter(i,lower_b+1:upper_b)=1;

Endif

End while

End For

unav=sum_channel(counter)/n_simulations

0 100 200 300 400 500 600 700 800 900 1000

0

0.2

0.4

0.6

0.8

1

Tiempo

Falta d

e d

isponib

ilidad

único ensayo

Simulacion Monte Carlo

Falta de disponibilidad en estado estacionario

Single trial


Limit unavailability

Time

Una

va

ilab

ility

29

SIMULATION OF SYSTEM STOCHASTIC PROCESS FOR AVAILABILITY / RELIABILITY

ESTIMATION

30Phase Space of System Stochastic

Process

31

A

B

C

Components’ times of transition between states are exponentially distributed

( = rate of transition of component i going from its state ji to the state mi)i

mji 1→

Arrival

Init

ial

1 2 3

1 -

2 -

3 -

)(

21BA

→ )(

31BA

→

( )2 1A B

→

)(32

BA→

)(

13BA

→ )(

23BA

→

Indirect Monte Carlo: Example

32

▪ The components are initially (t=0) in their nominal states (1,1,1)

▪One minimal cut set of order 1 (C in state 4:(*,*,4)) and one minimal

cut set of order 2 (A and B in 3: (3,3,*)).

Arrival

1 2 3 4

Init

ial

1 -

2 -

3 -

4 -

C

21→ C

31→ C

41→

C

12→ C

32→ C

42→

C

13→ C

23→ C

43→

C

14→ C

24→ C

34→

Indirect Monte Carlo: Example

33

SAMPLING THE TIME OF TRANSITION

The rate of transition of component A(B) out of its nominal state 1 is:

• The rate of transition of component C out of its nominal state 1 is:

•The rate of transition of the system out of its current configuration (1,1, 1) is:

•We are now in the position of sampling the first system transition timet1, by applying the inverse transform method:

where Rt ~ U[0,1)

)(

31)(

21)(

1BABABA

→→ +=

CCCC

4131211 →→→ ++=

( ) CBA

1111,1,1 ++=

( ) )1ln(1

1,1,101 Rtt t−−=

Analog Monte Carlo Trial

34

• Assuming that t1 < TM (otherwise we would proceed to the successive trial), we now need to determine which transition has occurred, i.e. which component has undergone the transition and to which arrivalstate.

• The probabilities of components A, B, C undergoing a transition out oftheir initial nominal states 1, given that a transition occurs at time t1, are:

• Thus, we can apply the inverse transform method to the discrete distribution

( ) ( ) ( )1,1,1

1

1,1,1

1

1,1,1

1 , ,

CBA

0

R

C ( )1,1,1

1

A

( )1,1,1

1

B

( )1,1,1

1

C

C

10

Sampling the kind of Transition (1)

35

• Given that at t1 component B undergoes a transition, its arrival state

can be sampled by applying the inverse transform method to the set of

discrete probabilities

of the mutually exclusive and exhaustive arrival states

→→

B

B

B

B

1

31

1

21 ,

0 1

R

S

B

B

1

31→

B

B

1

21→

S

• As a result of this first transition, at t1 the system is operating in

configuration (1,3,1).

• The simulation now proceeds to sampling the next transition time t2 with

the updated transition rate

( ) CBA

1311,3,1 ++=

Sampling the Kind of Transition (2)

RSU[0,1)

36

• The next transition, then, occurs at

where Rt ~ U[0,1).

•Assuming again that t2 < TM, the component undergoing the transition and

its final state are sampled as before by application of the inverse trasform

method to the appropriate discrete probabilities.

•The trial simulation then proceeds through the various transitions from one

system configuration to another up to the mission time TM.

( ) )1ln(1

1,3,112 Rtt t−−=

Sampling the Next Transition

37

• When the system enters a failed configuration (*,*,4) or

(3,3,*), where the * denotes any state of the component,

tallies are appropriately collected for the unreliability and

instantaneous unavailability estimates (at discrete times tj [0, TM]);

• After performing a large number of trials M, we can

obtain estimates of the system unreliability and

instantaneous unavailability by simply dividing by M, the

accumulated contents of CR(tj) and CA(tj), tj[0,TM]

Unreliability and Unavailability Estimation

38

For any arbitrary trial, starting at t=0 with the system in nominal

configuration (1,1,1) we would sample all the transition times:

where

)1ln(1

1,

1

01 Rtti

mtim

im i

i

i →

→

→ −−=

==

==

=

Cm

BAim

CBAi

i

i

ifor 4,3,2

,for 3,2

,,

)1,0[~1, URi

mt i→

A

B

C

Direct Monte Carlo: Example

39

Direct Monte Carlo: Example

The theoretical view

41

Sampling

Evaluation of definite integrals

Simulation of system transport

Contents

Simulation for reliability/availability analysis

42

Sampling



Contents

Simulation for system reliability/availability analysis

44

( ) rrRPrUR == :cdf

( )( )

1 :pdf ==dr

rdUru R

R

0

Sampling (pseudo) Random Numbers

Uniform Distribution

46

Sample R from UR(r) and find X:

( )RFX X

1−=

Question: which distribution does X obey?

Application of the operator Fx to the argument of P above yields

Summary: From an R UR(r) we obtain an X FX(x)

( ) xRFPxXP X = −1

( ) ( )xFxFRPxXP XX ==

1

1 r 0

( )rUrR R=Pr ( )xFX

R X x

Sampling (pseudo) Random Numbers

Generic Distribution

47

• Markovian system with two states (good, failed)

• hazard rate, = constant

• cdf

•pdf

•Sampling a failure time T

( ) t

T etTPtF −−== 1

( ) dtedttTtPdttf t

T =+= −

( ) ( ) t

TR etFrFR −−== 1

( ) ( )RRFT T −−== − 1ln11

Example: Exponential Distribution

48

• hazard rate not constant

•CDF

•pdf

• Sampling a failure time T

( ) 1 t

Tf t dt P t T t dt t e dt − − = + =

( ) ( ) 1 t

R TR F r F t e− = = −

( ) ( )

1

1 1ln 1TT F R R

− = = − −

Example: Weibull Distribution

( ) 1 t

TF t P T t e−= = −

49

sample an

•

•

Graphically:

0 1, ,..., ,...kx x x =

0

k

k k i

i

F P X x P X x=

= = =

~ [0,1)R U

1 1

1 1

( ) ( )

~ [0,1) and ( )

k k R k R k

R

k k k k k k

P F R F F F F F

R U F r r

P F R F F F f P X x

− −

− −

= −

=

= − = = =

Sampling by the Inverse Transform

Method: Discrete Distributions

50

Sampling


Simulation of system transport

Contents

Simulation for reliability/availability analysis

51

MC analog dart game: sample x from f(x)

• the probability that a shot hits x dx is f(x)dx

• the award is g(x)

Consider N trials with result {x1, x2, …,xn}: the average award is

( )=

=N

i

iN xgN

G1

1

( ) ( )

( ) ( ) ( ) 1 ; 0 pdf =→

=

dxxfxfxf

dxxfxgGb

a

MC Evaluation of Definite Integrals (1D)

Analog Case

52

( ) ( )

( )

1

0

1

2 2

0

22

2

2cos 0.6366198

2

By setting: 1, g cos2

( )

1 ( ) cos

2 2

1 1 ( ) ( )

1 1 2 1 9

2

N

N

G x dx

f x x x

G E g x

E g x x dx

Var G Var g x E g x E g xN N

Var GN N

= = =

= =

=

= =

= = −

= − =

( )

2

4

2 6

.47 10

for 10 histories, ~ [0,1) cos2

0.6342, 9.6 10N

i i i

N G

N x U g x x

G s

−

−

= =

= =


Example

53

MC biased dart game: sample x from f1(x)

• the probability that a shot hits x dx is f1(x)dx

• the award is

( )( )( )

( ) ( )=

==N

i

iN xgN

Gxgxf

xfxg

1

11

1

1

1

The expression for G may be written

( )( )

( ) ( ) ( ) ( )

=

DDdxxfxgdxxfxg

xf

xfG 111

1


Biased Case

54

The pdf f1*(x) is:

From the normalization condition:

For the minimum value

* 2

1 ( )f x a bx= −

( )1 1

* 2

1

0 0

* 2

1

( ) 1 1 3( 1)3

( ) 3( 1)

bf x dx a bx dx a b a

f x a a x

= − = − = = −

= − −

31.5

2a = =

1

1

1 1

2

2

0 0 ( )( )

cos 12cos (1.5 1.5 )

2 1.5 1.5f x

g x

x

G xdx x dxx

= = −−

1

0

2cos 0.6366198

2G x dx

= = =


Biased Case: Example

55

2

4

1 1

1 2 10.406275 9.9026 10NVar G

N N

−

= − =

Finally we obtain:

( )( )

1

4 *

1 1 2

2 8

1

cos2

for 10 histories, ~1.5 1.5

0.6366, 9.95 10N

i

i i

i

N G

x

N x f g xx

G s

−

= =−

= =


Biased Case: Example

56

Sampling



Contents


57Monte Carlo simulation for system

reliability

• PLANT = system of Nc suitably connected components.

• COMPONENT = a subsystem of the plant (pump, valve,...) which may stay in different exclusive (multi)states (nominal, failed, stand-by,... ). Stochastictransitions from state-to-state occur at stochastic times.

• STATE of the PLANT at t = the set of the states in which the Nc componentsstay at t. The states of the plant are labeled by a scalar which enumerates allthe possible combinations of all the component states.

• PLANT TRANSITION = when any one of the plant components performs a state transition we say that the plant has performed a transition. The time at whichthe plant performs the n-th transition is called tn and the plant state therebyentered is called kn.

• PLANT LIFE = stochastic process.

60

T(tt’; k’)dt = conditional probability of a transition at t dt, given that the preceding transition occurred at t’ and that the state thereby entered was k’.C(k k’; t) = conditional probability that the plant enters state k, given that a transition occurred at time t when the system was in state k’.Both these probabilities form the ”trasport kernel”:

K(t; k t’; k’)dt = T(t t’; k’)dt C(k k’; t)

(t; k) = ingoing transition density or probability density function (pdf) of a system transition at t, resulting in the entrance in state k

Stochastic Transitions: Governing

Probabilities

61The von Neumann’s Approach and

the Transport Equation

61

The transition density (t; k) is expanded in series of the partialtransition densities:n(t; k) = pdf that the system performs the n−th transition at t, entering the state k.Then,

Transport equation for the plant states

+=

==

=

'

0

0

0

)','|,()','('),(

),(),(

k

t

t

n

n

ktktKktdtkt

ktkt

62

)**,,(),,(),(),( 11001100

0

*011

1

0

1

ktktKktktKktdtktk

t

t==

),,()**,,(

),,(),(),(

112211*

1

112211

1

*122

2

1

2

1

2

ktktKktktKdt

ktktKktdtkt

k

t

t

k

t

t

=

==

……

),,(),,(*)*,,(...

...),(

11112211*

1

*2

,...,,*

1

2

1

121

−−

−−

=

−

−

nn

t

t

t

tn

kkk

t

tn

n

ktktKktktKktktKdt

dtdtktn

n

n

Initial Conditions: (t*, k*)

Formally rewrite the partial transition densities:

Monte Carlo Solution to the Transport

Equation

63

•MC analog dart game: sample x = (t1, k1; t2, k2; ...) from

f(x)=

•

•the probability that a shot hits x dx is f(x)dx

• the award is g(x)=1

Consider N trials with result {x1, x2, …,xn}: the average award is

( )=

=N

i

iN xgN

G1

1

( ) ( )

( ) ( ) ( ) 1 ; 0 pdf =→

=

dxxfxfxf

dxxfxgGb

a

),,(),,(*)*,,( 11112211 −− nn ktktKktktKktktK

MC Evaluation of Definite Integrals

64

Sampling



Contents


65

Monte Carlo Simulation in RAMS

=k

t

k dtRktG0

),(),()(

• = subset of all system failure states

• Rk(,t) = 1 G(t) = unreliability

• Rk(,t) = prob. system not exiting before t from the state k

entered at <t

G(t) = unavailability

Monte Carlo solution of a definite integral:

expected value sample mean

Expected value

A real example of Indirect Simulation

PRODUCTION AVAILABILITY EVALUATION OF AN

OFFSHORE INSTALLATIONZio, E., Baraldi, P., Patelli E. Assessment of the availability of an offshore installation by Monte Carlo simulation. International

Journal of Pressure Vessels and Piping 83 (2006) 312–320

68

2.2 MSm3/d

2.2 MSm3/d

4.4 MSm3/d

Production Gas : 5 MSm

3/d

Oil : 26500 Sm3/ d

Water : 8000 Sm3/d

4.4 MSm3/d

23300 Sm3/d

7000 Sm3/d

TC

TEG

Fuel Gas 25 b

50%

50%

Gas Lift 100b

Wells Separation

50%

50%

Gas Lift 60b

Gas Export 3.0 MSm

3/d, 60b

Oil export

TC

Flare

EC

Oil Trt °

TG

TG

Water Inj . Sea Wat . Trt °

13 MW

13 MW

7 MW

7 MW

6 MW

6 MW

0.1

MSm3

0.1

MSm3

0.1

MSm3

0.1

MSm3

1

MSm3/d

The offshore production plant

69

State 0 = as good as new

State 1 = degraded (no function lost, greater failure rate value)

State 2 = critical (function is lost)

2

12

02

m20

0 1

01

m10

TC TG

01 0.89 · 10-3 h-1 0.67 · 10-3 h-1

02 0.77 · 10-3 h-1 0.74 · 10-3 h-1

12 1.86 · 10-3 h-1 2.12 · 10-3 h-1

m10 0.033 h-1 0.032 h-1

m20 0.048 h-1 0.038 h-1

Component failures and repairs:

TCs and TGs

70Component failures and repairs:

EC and TEG

0 2

m

State 0 = as good as new

State 2 = critical (function is lost)

EC TEG

0.17 · 10-3 h-

1

5.7 · 10-5 h-1

m 0.032 h-1 0.333 h-1

71

Production priority

When a failure occurs, the system is reconfigured tominimise (in order):

• the impact on the export oil production

• the impact on export gas production

➢ The impact on water injection does not matter

72

2.2 MSm3/d

2.2 MSm3/d

4.4 MSm3/d

Production Gas : 5 MSm

3/d

Oil : 26500 Sm3/ d

Water : 8000 Sm3/d

4.4 MSm3/d

23300 Sm3/d

7000 Sm3/d

TC

TEG

Fuel Gas 25 b

50%

50%

Gas Lift 100b

Wells Separation

50%

50%

Gas Lift 60b

Gas Export 3.0 MSm

3/d, 60b

Oil export

TC

Flare

EC

Oil Trt °

TG

TG

Water Inj . Sea Wat . Trt °

13 MW

13 MW

7 MW

6 MW

0.1

MSm3

0.1

MSm3

0.1

MSm3

0.1

MSm3

1

MSm3/d

Production priority: example

73

Maintenance policy: reparation

Only 1 repair team

Priority levels of failures:1. Failures leading to total loss of export oil (both TG’s or both TC’s

or TEG)

2. Failures leading to partial loss of export oil (single TG or EC)

3. Failures leading to no loss of export oil (single TC failure)

74Maintenance policy: preventive

maintenance

➢ Only 1 preventive maintenance team

➢ The preventive maintenance takes place only if the system is in perfect state of operation

Type of maintenance Frequency [hours] Duration [hours]

Type 1 2160 (90 days) 4

Type 2 8760 (1 year) 120 (5 days) Turbo-Generator and

Turbo-Compressors Type 3 43800 (5 years) 672 (4 weeks)

Electro Compressor Type 4 2666 113

75

Number of components = 6

Number of states for component = 2 or 322·34 = 324 plant states

+1 repair team 129 new plant states

1 maintenance team Non homogeneous Markov chain

+

MONTE CARLO APPROACH

Markov approach too complex

MARKOV APPROACH

76

Mean Std

Oil [k

m3/d]23.230 0.263

Gas[k Sm3/d]

2929 194.0

Water[k m3/d]

6.811 0.883

P.Maintenance Type 1 (TC,TG)

P.Maintenance Type 2 (EC)

P.Maintenance Type 3(TC,TG)

Case A: perfect system and preventive

maintenances

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Questions & Answers

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