Monte Carlo Simulation for Reliability and Availability analyses

Monte Carlo Simulation: Why? 2 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 INSTALLATION Zio, 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

System description: basic scheme 4 TC TC 50% 50% 60 b TEG Flare Gas Export 3.0 msm3/d, 60b Oil export Wells Separation Oil Trt Wat. Trt Sea Water Inj.

System description: gas-lift 5 60 b First loop Gas-lift pressure Production of the Well 100 100% 60 80% 0 60% 5

System description: fuel gas generation and distribution 6 Second loop

System description: electricity power production and distribution 7

The offshore production plant 8 TC TC 0.1 MSm 3 0.1 MSm 3 2.2 MSm3/d 50% 50% 2.2 MSm 3 /d Fuel Gas 25 b Gas Lift 60b 4.4 MSm 3 /d TEG 6 MW 1 MSm 3 /d Flare Gas Export 3.0 MSm 3 /d, 60b Wells Production Gas : 5 MSm 3 /d Oil : 26500 Sm 3 / d Water : 8000 Sm 3 /d 4.4 MSm 3 /d Separation 7000 Sm 3 /d 23300 Sm 3 /d 0.1 MSm 3 TG 13 MW 50% Gas Lift 100b Oil Trt EC 6 MW 7 MW Oil export Wat. Trt 0.1 MSm 3 TG 50% 13 MW Sea 7 MW Water Inj.

Component failures and repairs: TCs and TGs 9 l 02 TC TG l 01 0 1 m 10 m 20 l 12 2 l 01 0.89 10-3 h -1 0.67 10-3 h -1 l 02 0.77 10-3 h -1 0.74 10-3 h -1 l 12 1.86 10-3 h -1 2.12 10-3 h -1 m 10 0.033 h -1 0.032 h -1 m 20 0.048 h -1 0.038 h -1 State 0 = as good as new State 1 = degraded (no function lost, greater failure rate value) State 2 = critical (function is lost)

Component failures and repairs: EC and TEG 10 l EC TEG 0 2 m l 0.17 10-3 h - 1 5.7 10-5 h -1 m 0.032 h -1 0.333 h -1 State 0 = as good as new State 2 = critical (function is lost)

Production priority: example 11 TC TC 0.1 MSm 3 0.1 MSm 3 2.2 MSm3/d 50% 50% 2.2 MSm 3 /d Fuel Gas 25 b Gas Lift 60b 4.4 MSm 3 /d TEG 6 MW 1 MSm 3 /d Flare Gas Export 3.0 MSm 3 /d, 60b 4.4 MSm 3 /d Gas Lift 100b EC Oil export Wells Production Gas : 5 MSm 3 /d Oil : 26500 Sm 3 / d Water : 8000 Sm 3 /d Separation 7000 Sm 3 /d 23300 Sm 3 /d 0.1 MSm 3 TG 13 MW 50% Oil Trt 7 MW Wat. Trt 0.1 MSm 3 TG 50% 13 MW Sea Water Inj.

Production priority 12 When a failure occurs, the system is reconfigured to minimise (in order): the impact on the export oil production the impact on export gas production The impact on water injection does not matter

Production priority: example 13 TC TC 0.1 MSm 3 0.1 MSm 3 2.2 MSm3/d 50% 50% 2.2 MSm 3 /d Fuel Gas 25 b Gas Lift 60b 4.4 MSm 3 /d TEG 6 MW 1 MSm 3 /d Flare Gas Export 3.0 MSm 3 /d, 60b 4.4 MSm 3 /d Gas Lift 100b EC Oil export Wells Production Gas : 5 MSm 3 /d Oil : 26500 Sm 3 / d Water : 8000 Sm 3 /d Separation 7000 Sm 3 /d 23300 Sm 3 /d 0.1 MSm 3 TG 13 MW 50% Oil Trt 7 MW Wat. Trt 0.1 MSm 3 TG 50% 13 MW Sea Water Inj.

Maintenance policy: reparation 14 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)

Maintenance policy: preventive maintenance 15 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] Turbo-Generator and Turbo-Compressors Type 1 2160 (90 days) 4 Type 2 8760 (1 year) 120 (5 days) Type 3 43800 (5 years) 672 (4 weeks) Electro Compressor Type 4 2666 113

Problem Statement 16 Compute the system istantaneous and average availability The use of analytical methods is unfeasible Monte Carlo simulation

Monte Carlo Simulation

The History of Monte Carlo Simulation Buffon 18 1707-88 Kelvin (kinetic theory of gas - multidimensional integrals) 1824-1907 Random sampling of numbers Gosset (coefficient of the t-student) Fermi, von Neumann, Ulam (Manhattan project, interaction neutron-matter) Neutron transport (design of shielding in nuclear devices - 6 dimension-integral) System Transport (RAM, ) 1908 1930-40 s 1950 s 1990 s

Contents 19 Sampling Random Numbers Simulation of system transport Simulation for reliability/availability analysis of a component Example

SAMPLING RANDOM NUMBERS

Sampling (pseudo) Random Numbers Uniform Distribution 21 u R (r) 0 pdf : u R r dur dr r 1 U R (r) cdf : U R r P R r r 0

Sampling Random Numbers from an Uniform Distribution 22 In the past: The Rand Book (1955) 1 million of random numbers

Sampling Random Numbers from an Uniform Distribution 23 Today: R~U[0, 1)

Sampling Random Numbers from a Generic Distribution 24 F X (x) 1 x

Sampling Random Numbers from a Generic Distribution 25 F X (x) Sample R from U R (r) 1 X F R X 1 R X = F 1 (R) x

Sampling Random Numbers from a Generic Distribution 26 Pr R r U r R F X x R 1 Sample R from U R (r) and find X: X F 1 X R r 1 R 0 X x Question: which distribution does X obey? 1 X x P F R P X x F

Sampling (pseudo) Random Numbers Generic Distribution 27 Pr R r U r R F X x R 1 Sample R from U R (r) and find X: X F 1 X R r 1 R 0 X x Question: which distribution does X obey? 1 X x P F R P X x F P R F X x

Sampling (pseudo) Random Numbers Generic Distribution 28 Pr R r U r R F X x R 1 Sample R from U R (r) and find X: X F 1 X R r 1 R 0 X x Question: which distribution does X obey? 1 X x P F R P X x F P R F x F x X X

Sampling (pseudo) Random Numbers Generic Distribution 29 Pr R r U r R F X x R 1 Sample R from U R (r) and find X: X F 1 X R r 1 R 0 X x Question: which distribution does X obey? 1 X x P F R P X Application of the operator F x to the argument of P above yields P x X x P R F x F x Summary: From an R U R (r) we obtain an X F X (x) X X

Sampling from a Bernoulli distribution 30 X = 1 healthy 0 failed 0.9 P X (x) with P X = 0 = p 0 = 0.1 1 F X (x) p o =0.1 p o =0.1 0 1 x 0 1 x

Sampling from a Bernoulli distribution 31 X = 1 healthy 0 failed 0.9 P X (x) with P X = 0 = p 0 = 0.1 F X (x) R 1 p o =0.1 p o =0.1 0 1 x 0 1 x Sample R from U R (r) If R p 0 X = 0 If R > p 0 X = 1 Does it work? = R 0 p 0 1 P X = 0 = P R p 0 = p 0 P X = 1 = P R > p 0 = 1 p 0

Sampling from discrete distributions 32 Ω = x 0, x 1,, x k P X = x i = f i F i = P X x i i = j=0 f j

Sampling from discrete distributions 33 Ω = x 0, x 1,, x k, P X = x k = f k F k = P X x k k = i=0 f k Sampling procedure: Sample an R~ U[0,1) Find i such that F i 1 < R F i Graphically: Why does it work? P X = x i = P F i 1 < R F i = F i F i 1 = f i

Exercise 1: Sampling from the Exponential Distribution 34 Probability density function: f T ( t) le lt t 0 0 t 0 f T (t) Expected value and variance: 1 E[ T ] l 1 Var[ T ] 2 l t f () t le l T t 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)

Failure probability estimation: example 35 Arc number i Failure probability P i 1 0.050 2 0.025 3 0.050 4 0.020 5 0.075 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

Analytic solution 36 1) Minimal Failure Configuration (minimal cut set): M 1 ={1,4} M 2 ={2,5} M 3 ={1,3,5} M 4 ={2,3,4} 2) Network failure probability: P[ M ] p p 0.05 0.02 1 10 1 1 4 P[ M ] p p 0.025 0.075 1.875 10 2 2 5 P[ M ] p p p 0.05 0.05 0.075 1.875 10 3 1 3 5 P[ M ] p p p 0.025 0.05 0.02 7.5 10 4 2 3 4 3 3 5 4 4 P[ X 1] P[ M ] 3.07 10 T j 1 j 3 rare event approximation 36

Monte Carlo simulation 37 The state of arc i can be sampled by applying the inverse transform method to the set of discrete probabilities pi,1 pi of the mutually exclusive and exhaustive states l B 1 2 B 1 p i l B l1 3 1 B p l1 i S R S R U[0,1) 0 1 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 N f The trial simulation then proceeds until we collect M trials. P MC = N f M

Monte Carlo simulation 38 N number of trials N F number of trials corresponding to realization of a minimal cut set The failure probability is equal to N F / 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=10 6, we obtain PX [ T 1] 3.04 10 3

SIMULATION OF SYSTEM TRANSPORT

Monte Carlo simulation for system reliability 40 PLANT = system of N c suitably connected components. COMPONENT = a subsystem of the plant (pump, valve,...) which may stay 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 N c components stay at t. The states of the plant are labeled by a scalar which enumerates all the 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 which the plant performs the n-th transition is called t n and the plant state thereby entered is called k n. PLANT LIFE = stochastic process.

Plant life: random walk 41 Random walk = realization of the system life generated by the underlying state-transition stochastic process. Phase space

Phase Space 42

Example: System Reliability Estimation 43 Divide the mission time, T M, in bins and associate a counter (of the system failure) to each bin: C F 1, C F 2,, C F τ, Initialize each counter to 0: C F 1 = 0, C F 2, = 0,, C F τ = 0, bin C F 1 = 0 C F 2 = 0 C F τ = 0 0 T M

Example: System Reliability Estimation 44 Divide the mission time, T M, in bins and associate a counter (of the system failure) to each bin: C F 1, C F 2,, C F τ, Initialize each counter to 0: C F 1 = 0, C F 2, = 0,, C F τ = 0, C F 1 = 0 C F 2 = 0 C F τ = 0 0 MC simulation of the first system life T M C F 2 = 0 C F 1 = 0 C F τ = 0 0 T M Events at component level, which do not entail system failure No system failure during first simulation C F τ = C F τ for τ [0, T m ]

Example: System Reliability Estimation 45 MC simulation of the first system life C F 2 = 0 C F 1 = 0 C F τ = 0 0 MC simulation of the second system life T M C F 1 = 0 CF 2 = 0 C F τ 2 = C F τ 2 +1=1 +1 +1 +1 +1 τ 2 T M Event at component level, which do not entail system failure Event at component level, which causes the system failure system failure at time τ 2 C F τ = C F τ + 1 for τ > τ 2

Example: System Reliability Estimation 46 MC simulation of the first system life C F 2 = 0 C F 1 = 0 C F τ = 0 0 MC simulation of the second system life T M +1 +1 +1 +1 MC simulation of the M-th system life T M +1 +1 τ M Event at component level, which do not entail system failure Event at component level, which causes the system failure T M system failure at time τ M C F τ = C F τ + 1 for τ > τ M

Example: System Reliability Estimation 47 MC simulation of the first system life C F 2 = 0 C F 1 = 0 C F τ = 0 0 MC simulation of the second system life T M +1 +1 +1 +1 MC simulation of the M-th system life T M +1 +1 τ M Event at component level, which do not entail system failure Event at component level, which causes the system failure T M MC estimation of the system reliability F T τ CF τ M R τ = 1 CF τ M

Example: System Reliability Estimation 48 4 F(t)=1-R(t) 1/4 1/2 t

SIMULATION FOR COMPONENT AVAILABILITY ESTIMATION 49

One component with exponential distribution of the failure time 50 l 0 m 1 values l 3 10-3 h -1 m 25 10-3 h -1 State X=0 (WORKING) State X=1 (FAILED) Limit unavailability: q =? MTTR MTTR + MTTF

One component with exponential distribution of the failure time 51 0 m l 1 values l 3 10-3 h -1 m 25 10-3 h -1 State X=0 (WORKING) State X=1 (FAILED) Limit unavailability: q = MTTR MTTR + MTTF = 1 μ 1 μ + 1 λ = 0.1071

One component with exponential distribution of the failure and repair time 52 l X(t) 0 1 m State 1 State 0 State X=0 (WORKING) State X=1 (FAILED)

One component with exponential distribution of the failure and repair time 53 l X(t) 0 1 m State 1 State 0 State X=0 (WORKING) State X=1 (FAILED)

One component with exponential distribution of the failure and repair time 54 l 0 1 m State 1 State 0 State X=0 (WORKING) State X=1 (FAILED)

Monte carlo simulation for estimating the system availability at time t 55 Divide the mission time, T M, in bins and associate a counter (of the system failure) to each bin: C q 1, C q 2,, C q τ, Initialize each counter to 0: C q 1 = 0, C q 2, = 0,, C q τ = 0, If the component is failed in (t j, t j + t), the corrresponding counter, c q (t j ), increases c q (t j ) = c q (t j ) +1; otherwise, c q (t j ) = c q (t j ) Trial 1

Monte carlo simulation for estimating the system availability at time t 56 another trial Trial 1 Trial i-th

Monte carlo simulation for estimating the system availability at time t 57 another trial Trial 1 Trial i-th Trial M-th

Monte carlo simulation for estimating the system availability at time t 58 The counter c q (t j ) adds 1 until M trials have been sampled q t j = P X t j = 1 Cq t j M Trial 1 Trial i-th Trial M-th q t j Cq t j M

Unavailability 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 Falta de disponibilidad 59 1 0.8 único Single ensayo trial 0.6 Simulacion Monte Carlo Monte simulation Carlo Falta Limit de unavailability disponibilidad en estado estacionario 0.4 0.2 0 0 100 200 300 400 500 600 700 800 900 1000 Time Tiempo

SIMULATION FOR SYSTEM AVAILABILITY / RELIABILITY ESTIMATION 60

Stochastic Transitions: Governing Probabilities 61 T(t t ; 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)

Phase Space 62

Initial Indirect Monte Carlo: Example (1) 63 A C B Components times of transition between states are exponentially distributed i ( l = rate of transition of component i going from its state j i to the state m i ) j i m 1 1 - Arrival 1 2 3 2 AB ( ) - l 2 1 A( B) A 3 l 3 1 l ( B ) 3 2 - l A( B) 1 2 l l A( B) 1 3 A( B) 2 3

Initial Indirect Monte Carlo: Example (2) 64 1 - Arrival 1 2 3 4 l1 C 2 l1 C 3 l1 C 4 2 l C 1-2 l C 2 3 l C 2 4 3 - l C 3 1 l C 2 3 l C 3 4 4 - l C 4 1 l C 2 4 4 l C 3 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,*)).

Monte Carlo Trial: Sampling the time of transition 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: l1 A( B) A l1 ( B) 2 A l1 The rate of transition of component C out of its nominal state 1 is: C C l l1 C l1 The rate of transition of the system out of its current configuration (1, 1, 1) is: 1,1,1 l We are now in the position of sampling the first system transition time t 1, by applying the inverse transform method: ( B) 3 C l1 1 2 3 4 A B C l1 l1 l1 T(t t = 0, k = 1,1,1 = λ 1,1,1 e λ 1,1,1 t where R t ~ U[0,1) t t 1 l ln(1 ) 1,1,1 t 1 0 R

Monte Carlo Trial: Sampling the kind of Transition 66 How to build T(k t = t 1, k = (1,1,1))? Assuming that t 1 < T M (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 arrival state. The probabilities of components A, B, C undergoing a transition out of their initial nominal states 1, given that a transition occurs at time t 1, are: l l A 1, l l B 1 1,1,1 1,1,1 1,1,1 Thus, we can apply the inverse transform method to the discrete distribution, l l C 1 P(A ) P(B) P(C ) l = R l1 A C l1 B 1 1,1,1 1,1,1 1,1,1 l = = l l C 0 1

Sampling the Kind of Transition (2) 67 Given that at t 1 component B undergoes a transition, its arrival state can be sampled by applying the inverse transform method to the set of discrete probabilities B B l1 2 l1 3, B B l1 l1 of the mutually exclusive and exhaustive arrival states 0 l B 1 2 B 1 l As a result of this first transition, at t 1 the system is operating in configuration (1,3,1). The simulation now proceeds to sampling the next transition time t 2 with the updated transition rate l 1,3,1 l B 1 3 B 1 l A B C l1 l3 l1 S R S 1 R S U[0,1)

Sampling the Next Transition 68 The next transition, then, occurs at where R t ~ U[0,1). t t 1 l ln(1 ) 1,3,1 t 2 1 R Assuming again that t 2 < T M, 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 T M.

69 Unreliability and Unavailability Estimation 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 t j [0, T M ]); After performing a large number of trials M, we can obtain estimates of the system unreliability at any time t j by F t j = CF (t j ) M system instantaneous unavailability at any time t j by q t j = Cq (t j ) M

Example: System Reliability Estimation 70 Divide the mission time, T M, in bins and associate a counter (of the system failure) to each bin: bin Initialize each counter to 0: C F 1, C F 2,, C F τ, C F 1 = 0 C F 2 = 0 C F 1 = 0, C F 2, = 0,, C F τ = 0, C F τ = 0 0 T M

Example: System Reliability Estimation 71 R R CC F τ ( t) = CC F ( tτ) t 0, T M τε[0, T M ] τ 1 τ r1 R R CC F ( tτ ) = CC( F t) τ 1 + t1, τε[τ T M 1, T M ] R R CC ( t) C ( t) 1 t, T τ F τ = C F τ + 1 τε[τ M 2, T M ] r2 τ 2 C F τ = C F τ τε[0, T M ] Events at components level, which do not entail system failure Repair Time F T τ CF τ M R τ = 1 CF τ M

Example: System Availability Estimation 72 Divide the mission time, T M, in bins and associate a counter (of the system failure) to each bin: bin Initialize each counter to 0: C q 1, C q 2,, C q τ, C q 1 = 0 C q 2 = 0 C q 1 = 0, C q 2, = 0,, C q τ = 0, C q τ = 0 0 T M

Example: System Availability Estimation 73 R R CC q τ ( t) = CC q ( tτ) t 0, T M τε[0, T M ] C q τ = C q τ + 1 τε[τ 1, τ r1 ] τ 1 τ r1 τ 2 τ r2 C q τ = C q τ + 1 τε[τ 1, τ r2 ] C q τ = C q τ τε[0, T M ] Events at components level, which do not entail system failure Repair Time q τ Cq τ M p τ = 1 Cq τ M

The offshore production plant 75 TC TC 0.1 MSm 3 0.1 MSm 3 2.2 MSm3/d 50% 50% 2.2 MSm 3 /d Fuel Gas 25 b Gas Lift 60b 4.4 MSm 3 /d TEG 6 MW 1 MSm 3 /d Flare Gas Export 3.0 MSm 3 /d, 60b Wells Production Gas : 5 MSm 3 /d Oil : 26500 Sm 3 / d Water : 8000 Sm 3 /d 4.4 MSm 3 /d Separation 7000 Sm 3 /d 23300 Sm 3 /d 0.1 MSm 3 TG 13 MW 50% Gas Lift 100b Oil Trt EC 6 MW 7 MW Oil export Wat. Trt 0.1 MSm 3 TG 50% 13 MW Sea 7 MW Water Inj.

MONTE CARLO APPROACH Plant state? Production levels oil gas water 76 TG 1 TG 2 TC 1 TC 2 TEG EC Associate a production level to each of the 453 plant states A systematic procedure

A systematic procedure 77 7 different production levels Plant state (Production Level) Gas [ksm 3 /d] Oil [k m 3 /d] Water [m 3 /d] 0=(100%) 3000 23.3 7000 1 900 23.3 7000 Failure of ony 1 TG 2 2700 21.2 0 3 1000 21.2 0 4 2600 21.2 6400 Failure of the EC 5 900 21.2 6400 6 0 0 0 Failure of TEG, Failure of both TCs Failure of both TGs

Real system with preventive maintenance 78 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

Real system with preventive maintenance 79 OIL Mean Std Oil [k m 3 /d] 22.60 0.42 Gas [k Sm 3 /d] 2687 194. 3 GAS Water [k m 3 /d] 6.04 0.76 WATER

Case A: perfect system and preventive maintenances 80 Type of maintenance Frequency [hours] Duration [hours] Turbo-Generator and Turbo-Compressors Type 1 2160 (90 days) 4 Type 2 8760 (1 year) 120 (5 days) Type 3 43800 (5 years) 672 (4 weeks) Electro Compressor Type 4 2666 113

Case A: perfect system and preventive maintenances 81 P.Maintenance Type 1 (TC,TG) P.Maintenance Type 2 (EC) OIL P.Maintenance Type 3 (TC,TG) Mean Std Oil [k m 3 /d] Gas [k Sm 3 /d] Water [k m 3 /d] 23.230 0.263 2929 194.0 6.811 0.883 GAS WATER

Case B: Component Failure and no preventive maintenances 82

Conclusions 83 Complex multi-state system with maintenance and operational loops MC simulation Systematic procedure to assign a production level to each configuration oil gas water Investigation of effects maintenance on production

Where to study? 84 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