Monte Carlo Simulation for Reliability and Availability … · 2 Monte Carlo Simulation: Why? • Computation of system reliability and availability for industrial systems characterized

Monte Carlo Simulation for Reliability and Availability analyses

2

Monte Carlo Simulation: Why?

• Computation of system reliability and availability for industrial systems characterized by:

• many components

• multistate components

• complex architectures and production logics

• complex maintenance policy

• The use of analytical methods is unfeasible

• Monte Carlo Simulation

A real example of Monte Carlo 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

4System description:

basic scheme

TC

TEG

50%

50%

Wells Separation

Gas Export

3.0 mSm3/d, 60b

Oil export

TC

Flare

Oil Trt°

Water Inj.Sea

Wat. Trt°

60 b


gas-lift

5

First loop

Gas-lift pressure Production of the Well

100 100%

60 80%

0 60%

60 b


fuel gas generation and distribution

Second loop


electricity power production and distribution

8

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

9

State 0 = as good as new

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

State 2 = critical (function is lost)

2

l12

l02

m20

0 1

l01

m10

TC TG

l01 0.89 · 10-3 h-1 0.67 · 10-3 h-1

l02 0.77 · 10-3 h-1 0.74 · 10-3 h-1

l12 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

10Component failures and repairs:

EC and TEG

0 2

l

m

State 0 = as good as new

State 2 = critical (function is lost)

EC TEG

l 0.17 · 10-3 h-

1

5.7 · 10-5 h-1

m 0.032 h-1 0.333 h-1

11

2.2 MSm3/d

2.2 MSm3/d

4.4 MSm3/d


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


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

12

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

13

2.2 MSm3/d

2.2 MSm3/d

4.4 MSm3/d


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


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

14

Maintenance policy: reparation

Only 1 repair team

Two or more components are failed at the same time

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)

15Maintenance 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

16

Problem Statement

• Compute the system istantaneous and average availability

• The use of analytical methods is unfeasible

• Monte Carlo simulation

Monte Carlo Simulation

18

Buffon

Gosset (coefficient of the t-Student)

Fermi, von Neumann, Ulam

Neutron transport

System Transport (RAM,…)

Kelvin (kinetic theory of gas

- multidimensional integrals)

The History of Monte Carlo Simulation

1707-88

1908

1824-1907

1950’s

1930-40’s

1990’s

Random

sampling of

numbers

(Manhattan project,

interaction neutron-matter)

(design of shielding in

nuclear devices -

6 dimension-integral)

19

Simulation of system transport

Simulation for reliability/availability analysis of a component

Example

Contents

Sampling Random Numbers

SAMPLING RANDOM NUMBERS

21

rrRPrUR :cdf

1 :pdf dr

rdUru R

R

0

Sampling (pseudo) Random Numbers

Uniform Distribution

0𝒖𝑹(𝒓)

𝑼𝑹(𝒓)

22Sampling Random Numbers from an


1 million of random numbers

In the past: The Rand Book (1955)

23Sampling Random Numbers from an


Today:

𝑹~𝑼[𝟎, 𝟏)

24Sampling Random Numbers from a

Generic Distribution

𝑥

𝐹𝑋(𝑥)

1

25Sampling Random Numbers from a


𝑥

𝐹𝑋(𝑥)

1

𝑅

𝑋 = 𝐹−1(𝑅)

• Sample R from UR(r)

• RFX X

1

26

Sample R from UR(r) and find X:

RFX X

1

Question: which distribution does X obey?

xRFPxXP X 1

1

1 r 0

rUrR RPr xFX

R X x

Sampling Random Numbers from a


F

R

27


RFX X

1


xRFPxXP X 1

1

1 r 0

rUrR RPr xFX

R X x



F

xFRP X

R

28


RFX X

1


xRFPxXP X 1

1

1 r 0

rUrR RPr xFX

R X x



F

xFxFRP XX

R

29


RFX X

1


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 RPr xFX

R X x



R

30

Sampling from a Bernoulli distribution

• 𝑋 = ൝1 ′ℎ𝑒𝑎𝑙𝑡ℎ𝑦′

0 ′𝑓𝑎𝑖𝑙𝑒𝑑 ′with P 𝑋 = 0 = 𝑝0 = 0.1

𝑥

𝑃𝑋(𝑥)

0 1

0.9

𝑝𝑜 =0.1

𝑥

𝐹𝑋(𝑥)

0 1

1

𝑝𝑜=0.1

31

Sampling from a Bernoulli distribution

• 𝑋 = ൝1 ′ℎ𝑒𝑎𝑙𝑡ℎ𝑦′

0 ′𝑓𝑎𝑖𝑙𝑒𝑑 ′with P 𝑋 = 0 = 𝑝0 = 0.1

𝑥

𝑃𝑋(𝑥)

0 1

0.9

𝑝𝑜 =0.1

𝑥

𝐹𝑋(𝑥)

0 1

1

𝑝𝑜=0.1

• Sample R from UR(r)

• If 𝑅 ≤ 𝑝0 → 𝑋 = 0• If 𝑅 > 𝑝0 → 𝑋 = 1

0 𝒑𝟎 1

𝑅

Does it work?

𝑃 𝑋 = 0 = 𝑃 𝑅 ≤ 𝑝0 = 𝑝0𝑃 𝑋 = 1 = 𝑃 𝑅 > 𝑝0 = 1 − 𝑝0

𝑅

=

32

Ω = 𝑥0, 𝑥1, … , 𝑥𝑘

𝑃 𝑋 = 𝑥𝑖 = 𝑓𝑖

𝐹𝑖 = 𝑃 𝑋 ≤ 𝑥𝑖 =

𝑗=0

𝑖

𝑓𝑗

Sampling from discrete distributions

33

Ω = 𝑥0, 𝑥1, … , 𝑥𝑘 , …

𝑃 𝑋 = 𝑥𝑘 = 𝑓𝑘

𝐹𝑘 = 𝑃 𝑋 ≤ 𝑥𝑘 =

𝑖=0

𝑘

𝑓𝑘

Sampling procedure:

• Sample an

• Find 𝑖 such that 𝐹𝑖−1 < 𝑅 ≤ 𝐹𝑖

Graphically:

Why does it work?

𝑃 𝑋 = 𝑥𝑖 = 𝑃 𝐹𝑖−1 < 𝑅 ≤ 𝐹𝑖 = 𝐹𝑖 − 𝐹𝑖−1 = 𝑓𝑖

~ [0,1)R U

Sampling from discrete distributions

34Exercise 1: Sampling from

the Exponential Distribution

Probability density function:

Expected value and variance:

00

0)(

t

tetf t

T

ll

2

1][

1][

l

l

TVar

TE

t

fT(t)

( ) t

Tf t e ll

• Find the procedure to sample numbers from an exponential distribution.

Assume to have available a generator of random numbers from an uniform

distribution in the interval [0,1)

35

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

36

Analytic solution

36

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 Failure Configuration (minimal cut set):

M1={1,4}

M2={2,5}

M3={1,3,5}

M4={2,3,4}

2) Network failure probability:

rare event approximation

37

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 fails we add 1 to Nf

•The trial simulation then proceeds until we collect M trials.

0 1

R

S l

lB

B

1

31

l

lB

B

1

21

SRU[0,1)

,1i ip p

ip 1 ip

Monte Carlo simulation

𝑷𝑴𝑪 =𝑵𝒇

𝑴

38

• 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;

M=1e6; %number of MC simulations

for i=1:M

%sampling of arcs fault events

for j=1:5 % simulation of arc states

r=rand

if (r<=p(j))

s(1,j)=1; % arc j is failed

otherwise

s(1,j)=0; % arc j is working

end

end

(is the network faulty?)

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

if fault >= 1

nf=nf+1;

end

end

pf=nf/M; %system failure probability

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


SIMULATION OF SYSTEM TRANSPORT

40Monte Carlo simulation for system

reliability

• 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.

41

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

state-transition stochastic process.

Plant life: random walk

Phase space

42

Phase Space

43

Example: System Reliability Estimation

• Divide the mission time, 𝑇𝑀, in bins and associate a counter (of the system failure) to each bin:

𝐶𝐹 1 , 𝐶𝐹 2 ,… , 𝐶𝐹 𝜏 , …

• Initialize each counter to 0:𝐶𝐹 1 = 0, 𝐶𝐹 2 ,= 0,… , 𝐶𝐹 𝜏 = 0,…

𝑇𝑀0

𝐶𝐹 1 = 0

𝐶𝐹 𝟐 = 0𝐶𝐹 𝝉 = 0

bin

44



𝐶𝐹 1 , 𝐶𝐹 2 ,… , 𝐶𝐹 𝜏 , …

• Initialize each counter to 0:𝐶𝐹 1 = 0, 𝐶𝐹 2 ,= 0,… , 𝐶𝐹 𝜏 = 0,…

• MC simulation of the first system life𝑇𝑀

0

𝐶𝐹 1 = 0

𝐶𝐹 𝟐 = 0𝐶𝐹 𝝉 = 0

0

Events at component level, which do

not entail system failure No system

failure during first simulation

𝐶𝐹 1 = 0

𝐶𝐹 𝟐 = 0𝐶𝐹 𝝉 = 0

𝐶𝐹 𝜏 = 𝐶𝐹 𝜏 𝑓𝑜𝑟 ∀𝜏 ∈ [0, 𝑇𝑚]

𝑇𝑀

45


• MC simulation of the first system life

• MC simulation of the second system life

0

𝐶𝐹 1 = 0

𝐶𝐹 𝟐 = 0𝐶𝐹 𝝉 = 0

𝐶𝐹 𝜏 = 𝐶𝐹 𝜏 + 1 𝑓𝑜𝑟 ∀𝜏 > 𝜏2

𝐶𝐹 1 = 0𝐶𝐹 𝟐 = 0 𝐶𝐹 𝝉𝟐 = 𝐶𝐹 𝝉𝟐 +1=1

Event at component level,

which do not entail system failure


which causes the system failure

𝑇𝑀

𝑇𝑀

+1 +1 +1 +1

system failure at time 𝝉𝟐

𝝉𝟐

46




• MC simulation of the M-th system life

0

𝐶𝐹 1 = 0

𝐶𝐹 𝟐 = 0𝐶𝐹 𝝉 = 0

𝑇𝑀

𝑇𝑀

+1 +1 +1 +1

𝑇𝑀

+1 +1


which do not entail system failureEvent at component level,


𝐶𝐹 𝜏 = 𝐶𝐹 𝜏 + 1 𝑓𝑜𝑟 ∀𝜏 > 𝜏𝑀system failure at time 𝝉𝑴

𝝉𝑴

47




• MC simulation of the M-th system life

• MC estimation of the system reliability

0

𝐶𝐹 1 = 0

𝐶𝐹 𝟐 = 0𝐶𝐹 𝝉 = 0

𝑇𝑀

𝑇𝑀

+1 +1 +1 +1

𝑇𝑀

+1 +1


which do not entail system failureEvent at component level,


𝝉𝑴

𝐹𝑇 𝜏 ≅𝐶𝐹 𝜏

𝑀→ 𝑅 𝜏 = 1 −

𝐶𝐹 𝜏

𝑀

48


t

F(t)=1-R(t)

4

1/4

1/2

49

SIMULATION FOR COMPONENT AVAILABILITY ESTIMATION

50

𝑞 =𝑀𝑇𝑇𝑅

𝑀𝑇𝑇𝑅 +𝑀𝑇𝑇𝐹

01

l

m

State X=0 (WORKING)

State X=1 (FAILED)

One component with exponential

distribution of the failure time

values

l 3· 10-3 h-1

m 25· 10-3 h-1

Limit unavailability:

?

51

01

l

m


distribution of the failure time

values

l 3· 10-3 h-1

m 25· 10-3 h-1

Limit unavailability:State X=0 (WORKING)

State X=1 (FAILED)

𝑞 =𝑀𝑇𝑇𝑅

𝑀𝑇𝑇𝑅 +𝑀𝑇𝑇𝐹=

1𝜇

1𝜇+1𝜆

= 0.1071

52

01

l

m

State 1

State 0


distribution of the failure and repair time

State X=0 (WORKING)

State X=1 (FAILED)

𝑿(𝒕)

53

01

l

m

State 1

State 0

One component with exponential distribution

of the failure and repair time

State X=0 (WORKING)

State X=1 (FAILED)

𝑿(𝒕)

54

01

l

m

State 1

State 0


distribution of the failure and repair time

State X=0 (WORKING)

State X=1 (FAILED)

55Monte carlo simulation for estimating the

system availability at time t

• Divide the mission time, 𝑇𝑀, in bins and associate a

counter (of the system failure) to each bin:

𝐶𝑞 1 , 𝐶𝑞 2 ,… , 𝐶𝑞 𝜏 ,…• Initialize each counter to 0:

𝐶𝑞 1 = 0, 𝐶𝑞 2 ,= 0,… , 𝐶𝑞 𝜏 = 0,…

• If the component is failed in (𝑡𝑗 , 𝑡𝑗 + ∆𝑡) , thecorrresponding counter, cq(tj), increases cq(tj) = cq(tj) +1;otherwise, cq(tj) = cq(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 cq(tj) adds 1 until M trials have been sampled

𝑞 𝑡𝑗 = 𝑃 𝑋 𝑡𝑗 = 1 ≅𝐶𝑞 𝑡𝑗

𝑀

q 𝑡𝑗 ≅𝐶𝑞 𝑡𝑗

𝑀

59

Pseudo Code

%Initialize parameters

Tm=mission time;

M=number of trials;

Dt=bin length;

Time_axis=0:Dt:Tm;

counter_q=zeros(length(1,Time_axis))

for i=1:M

%parameter initialization for each trial

t=0;

state=0; %state=working

while t<Tm

if state=0 %system is working sampling of failure time

t=t-[log(1-rand)]/lambda %failure time

state=1; % state=failed

failure_time=t;

lower_b=minimum(find(time_axis>=failure_time)); % first unavailability counter to be increased

else %system is failed sampling of repair time

t=t-[log(1-rand)]/mu;

state=1;

repair_time=t;

upper_b=minimum(find(time_axis > repair time)); %last unavailability counter to be increased

counter(lower_b:upper_b)= counter(lower_b:upper_b)+1; %increase all unavailability counter between lower_b and upper_b

end

end

end

Unav(:)=counter(:)/M

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

60

SIMULATION FOR SYSTEM AVAILABILITY / RELIABILITY

ESTIMATION

61

• 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)

Stochastic Transitions: Governing

Probabilities

62

Phase Space

63

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 1l

Arrival

Init

ial

1 2 3

1 -

2 -

3 -

l)(

21BA

l)(

31BA

( )2 1A B

l l

)(32

BA

l)(

13BA

l)(

23BA

Indirect Monte Carlo: Example (1)

64

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

▪ The failure configuration are (C in state 4:(*,*,4)) and (A and B in 3:

(3,3,*)).

Arrival

1 2 3 4

Init

ial

1 -

2 -

3 -

4 -

lC

21 lC

31 lC

41

lC

12 lC

32 lC

42

lC

13 lC

23 lC

43

lC

14 lC

24 lC

34

Indirect Monte Carlo: Example (2)

65

How to build 𝑇(𝑡|𝑡′ = 0, 𝑘′ = (1,1,1)?

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 time t1, byapplying the inverse transform method:

where Rt ~ U[0,1)

lll)(

31)(

21)(

1BABABA

llllCCCC

4131211

llll CBA

1111,1,1

)1ln(1

1,1,101 Rtt tl

Monte Carlo Trial:

Sampling the time of transition

𝑇(𝑡|𝑡′ = 0, 𝑘′ = 1,1,1 = 𝜆 1,1,1 𝑒−𝜆1,1,1 𝑡

66

How to build 𝑇(𝑘|𝑡 = 𝑡1, 𝑘′ = (1,1,1))?

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

• The probabilities of components A, B, C undergoing a transition out of theirinitial 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 , ,l

l

l

l

l

lCBA

0

R

C 1,1,1

1

l

lA

1,1,1

1

l

lB

1,1,1

1

l

lC

C

10

Monte Carlo Trial:

Sampling the kind of Transition

𝑷(𝑨 ) 𝑷(𝑩) 𝑷(𝑪 )

= = =

67

• 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

l

l

l

lB

B

B

B

1

31

1

21 ,

0 1

R

S l

lB

B

1

31

l

lB

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

llll CBA

1311,3,1

Sampling the Kind of Transition (2)

RSU[0,1)

68

• 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 tl

Sampling the Next Transition

69

• 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 at any time 𝑡𝑗 by

𝐹 𝑡𝑗 =𝐶𝐹(𝑡𝑗)

𝑀

• system instantaneous unavailability at any time 𝑡𝑗 by

q 𝑡𝑗 =𝐶𝑞(𝑡𝑗)

𝑀

Unreliability and Unavailability Estimation

70



𝐶𝐹 1 , 𝐶𝐹 2 ,… , 𝐶𝐹 𝜏 , …

• Initialize each counter to 0:

𝐶𝐹 1 = 0, 𝐶𝐹 2 ,= 0,… , 𝐶𝐹 𝜏 = 0,…

𝑇𝑀0

𝐶𝐹 1 = 0

𝐶𝐹 𝟐 = 0𝐶𝐹 𝝉 = 0

bin

71

( ) ( ) 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


Events at components

level, which do not entail

system failure

𝝉𝒓𝟏

𝝉𝒓𝟐

Repair Time

𝐶𝐹 𝜏 = 𝐶𝐹 𝜏 ∀𝜏𝜖[0, 𝑇𝑀]

𝐶𝐹 𝜏 = 𝐶𝐹 𝜏 + 1 ∀𝜏𝜖[𝜏1, 𝑇𝑀]

𝝉𝟏

𝝉𝟐

𝐶𝐹 𝜏 = 𝐶𝐹 𝜏 + 1 ∀𝜏𝜖[𝜏2, 𝑇𝑀]

𝐶𝐹 𝜏 = 𝐶𝐹 𝜏 ∀𝜏𝜖[0, 𝑇𝑀]

𝐹𝑇 𝜏 ≅𝐶𝐹 𝜏

𝑀→ 𝑅 𝜏 = 1 −

𝐶𝐹 𝜏

𝑀

72

Example: System Availability Estimation


𝐶𝑞 1 , 𝐶𝑞 2 ,… , 𝐶𝑞 𝜏 , …

• Initialize each counter to 0:

𝐶𝑞 1 = 0, 𝐶𝑞 2 ,= 0,… , 𝐶𝑞 𝜏 = 0,…

𝑇𝑀0

𝐶𝒒 1 = 0

𝐶𝒒 𝟐 = 0𝐶𝒒 𝝉 = 0

bin

73

( ) ( ) 0,R R

MC t C t t T

Example: System Availability Estimation

Events at components

level, which do not entail

system failure

𝝉𝒓𝟏

𝝉𝒓𝟐

Repair Time

𝐶𝑞 𝜏 = 𝐶𝑞 𝜏 ∀𝜏𝜖[0, 𝑇𝑀]

𝐶𝑞 𝜏 = 𝐶𝑞 𝜏 + 1 ∀𝜏𝜖[𝜏1, 𝜏𝑟1]

𝝉𝟏

𝝉𝟐

𝐶𝑞 𝜏 = 𝐶𝑞 𝜏 ∀𝜏𝜖[0, 𝑇𝑀]

q 𝜏 ≅𝐶𝑞 𝜏

𝑀→ 𝑝 𝜏 = 1 −

𝐶𝑞 𝜏

𝑀

𝐶𝑞 𝜏 = 𝐶𝑞 𝜏 + 1 ∀𝜏𝜖[𝜏1, 𝜏𝑟2]

A real example of Monte Carlo 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

75

2.2 MSm3/d

2.2 MSm3/d

4.4 MSm3/d


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


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

76

MONTE CARLO APPROACH

Associate a production level to

each of the 453 plant states A systematic

procedure

Production levels

…

oil gas water

Plant state ?TG1 TG2 TC1 TC2 TEG EC

77

A systematic procedure

7 different production

levels

Plant state

(Production

Level)

Gas

[kSm3/d]

Oil [k

m3/d]

Water

[m3/d]

0=(100%) 3000 23.3 7000

1 900 23.3 7000

2 2700 21.2 0

3 1000 21.2 0

4 2600 21.2 6400

5 900 21.2 6400

6 0 0 0Failure of TEG,

Failure of both TCs

Failure of both TGs

Failure of ony 1 TG

Failure of the EC

78Real system with preventive

maintenance

Production

level

Average

availability

0 8.13E-1

1 5.68E-2

2 6.58E-2

3 1.19E-2

4 3.55E-2

5 2.34E-3

6 1.50E-2

79

Mean Std

Oil [k m3/d] 22.60 0.42

Gas[k Sm3/d]

2687194.

3

Water[k m3/d]

6.04 0.76

Real system with preventive

maintenance

OIL

GAS

WATER

80Case A: perfect system and preventive

maintenances

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

81

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

OIL

GAS

WATER

82Case B: Component Failure and no

preventive maintenances

83

Conclusions

• Complex multi-state system with maintenance and operational loops

MC simulation

• Systematic procedure to assign a production level to each configuration

• Investigation of effects maintenance on production

oil gas water

84

Where to study?

• Slides

• Book:

• Chapter 1

• Sections 3.1,3.3.1,3.3.2 (no examples of the multivariate normal distribution)

• Sections 4.1,4.2,4.3,4.4,4.4.1

• Section 5.1

Documents

Monte Carlo Simulation for Reliability and Availability … · 2 Monte Carlo Simulation: Why? • Computation of system reliability and availability for industrial systems characterized