Monte Carlo (MC) simulation for system reliability
By Fares Innal
RAMS Group Department of Production and Quality
Engineering NTNU
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Lecture plan
1. Motivation: MC simulation is a necessary method forcomplexe systems reliability analysis.
2. MC simulation principle
3. Random number generation (RNG)
4. Probability distributions simulation
5. Simulation realization (History)
6. Output analysis and data accuracy
7. Example
8. MC simulation and uncertainty propagation2
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1. Motivation: MC simulation is a necessity
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Complex systems
β’ High system states numberβ’ Non constant failure (component
aging) and repair ratesβ’ Components dependency: standby,
...β’ Complex maintenance strategies:
spare parts, priority, limitedresources,β¦
β’ Multi-phase systems (MPS)β’ Multi-state systems (MSS)β’ Reconfigurable systemsβ’ β¦
Analytical approachesbecome impractical:
β’ Approximatedrepresentation that doesnot fit to the real system
β’ Extensive time for thedevelopment of theanalytical model
β’ Infeasible
Simulation is required: a commonparadigm and a powerful tool foranalyzing complex systems, due toits capability of achieving a closeradherence to reality. It provides asimplified representation of thesystem under study.
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1. Motivation: example
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Analytical approaches
Fault tree
Markov chains
C1 C2
SOnly one repair man is available with FIFO repair strategy C1*
C2*
Ξ»C1
Ξ»C2
Ξ»C1
Ξ»C2
C1C2*
C1*C2
C1 (1)C2
C1C2 (1)
ΞΌC1
ΞΌC2
ΞΌC1
ΞΌC2
Non constant repairrates (e.g. lognormal)
β’ Markov chainsbecome impractical.
β’ Strong complicationsfor other analyticalapproaches
MC simulation
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2. MC simulation principle : definitions
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MC simulation may be defined as a process for obtainingestimates (numeric values) for a given system (guided by aprescribed set of goals) by means of random numbers.
Monte Carlo is one of Monacoβs four quarters. The nameβMonte Carloβ was given by Nicholas Metropolis after theMonte Carlo Casinos.
Random numbers: a set of numbers that have nothing to dowith the other numbers in the sequence. Thus, for any nrandom numbers, each appears with equal probability(independent, identically distributed).
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2. MC simulation principle : history
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The needle experiment of Comte de Buffon (Buffon's needles, 1733)(french biologist, 1707-1788) :
D
L β€ D
What is the probability p, that a needle (length L),which randomly falls on a wooden floor, crosses oneof the lines (Lathes) (distance D)?
ππ =2 πΏπΏπ·π· Ξ βΉ Ξ =
2 πΏπΏπ·π· ππ β
2 πΏπΏπ·π· οΏ½
ππππ
N: number of needlesn: number of needles that are crossing lines
The real development of the Monte Carlo methods is performed, under theleadership of Nicholas Metropolis, Stanislaw Marcin Ulam, Enrico Fermiand John von Neumann during World War II and research towards thedevelopment of nuclear weapons (Manhattan Project). In particular, theyused these probabilistic methods for solving certain types of differentialequation.
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2. MC simulation principle : Steps
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1. Problem analysis and information collection:β’ Objective of the study (system oriented problem): reliability, availability, production,
design options,β¦β’ System structure: its elementsβ’ Input parameters: failure and repair rates, processing time, β¦β’ Relationships between input parameters and performance measuresβ’ Rules governing the operation of systems components
2. Model construction. It consists in constructing the model (conceptual) andimplementing it as a computer program:β’ General-purpose language: C++, FORTRAN,β¦β’ Special-purpose language: simulation tools.
3. Model verification and validation: make sure that the model conforms toits specifications and does what it is supposed to do.
4. Simulation experiments. This step is achieved thanks to random numbers.The evolution of the simulated system within a large number of Historiesallows to statistically assess the established performance measure(average, variance, confidence interval, ...).
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2. MC simulation principle : determining Ξ (classic example)
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Ξ is a ratio of a circle's circumference to its diameter: Ξ =πΆπΆπ·π·
D = 2 ππ
ππD
β’ The area of circle = π·π· ππππ
β’ The area of square = ππ πππππ·π· = 4 οΏ½
Area of circleArea of square
π·π· β 4 οΏ½Number of dots inside the circle
ππππππππππ number of dots
Using the hit and miss method (with ππ = 1):β’ Generate two sequences of random numbers: Ri and Rj (i, j= 0,β¦, N)β’β’β’ If
β’
π₯π₯ππ = β1 + 2 π π ππ (distribution range for x)π¦π¦ππ = β1 + 2 π π ππ(distribution range for y)
π₯π₯ππ2 + π¦π¦ππ2 β€ 1 πππ‘π‘π‘ππ ππ = ππ + 1 (dot inside) π‘π‘πππππ‘π‘ ππ = ππ (dot outside)
π·π· β 4 οΏ½Sππ
Solution with Excel
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2. MC simulation principle : integral calculus
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Xminxxmax
ymax
I = οΏ½π₯π₯ππππππ
π₯π₯ππππππππ π₯π₯ πππ₯π₯
ππ π₯π₯
πΌπΌArea of the rectangle
βNumber of dots under the curve
Total number of dots
π₯π₯ππ = ππ + (ππ β ππ) π π ππ (distribution range for x)π¦π¦ππ = π¦π¦πππππ₯π₯ π π ππ (distribution range for y)
π¦π¦ππ β€ ππ(π₯π₯ππ) πππ‘π‘π‘ππ ππ = ππ + 1 π‘π‘πππππ‘π‘ ππ = ππ
π°π° β π¦π¦πππππ₯π₯ οΏ½ (ππ β ππ) οΏ½Sππ
Solution with Excel
Using the hit and miss method (with 0 β€ f(x) β€ Ymax ):β’ Generate two sequences of random numbers: Ri and Rj (i, j= 0,β¦, N)β’β’β’ If
β’
Example:
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3. Random Numbers Generator (RNG)
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The generation of random numbers R uniformly distributed in [0,1) can beobtained physically by throwing a coin or dice, spinning the roulette,β¦
Monte Carlo simulations is based on computer generation of pseudo randomnumbers: the closest random number generator that can be obtained bycomputer algorithm.
In 1940's, von Neumann proposed to have the computer directly generate therandom numbers by means of an appropriate function: find the next numberXi+1 from the preceding one Xi. He proposed the middle-square method: obtainXi+1 by taking the central digits of the square of Xi (Example: 57721566492 =33317792380594909291).
Linear congruential generators (LCG) : ππππ+1 = ππ οΏ½ ππππ + ππ mod ππ
π π ππ+1 =ππππ+1ππ
Algorithm generates integers between 0 and m, map to zero and one.
a, c >= 0 , m > X0 , a, c
)2mod()65539( 311 nn XX Γ=+
Example: RANDU generator (1960βs IBM)
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4. Probability distributions simulation: principle
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Any given distribution can be generated from uniform random numberson [0,1]: U(0, 1).
0 1
1
Z
P(Z β€ z) = U(z) = z
0
1
X
P(X β€ x) = F(x)= P(Z β€ z)
xX= F-1(Z)
Principle: generating samples (x1, x2, ..., xn) of a random variable X (time tofailure, to repair, β¦) obeying any distribution F(x) from a sample (z1, z2, ...,zn) of the variable Z equally distributed between 0 and 1 by performing thetransformation: xi = F-1 (zi).
If the inversion of the distribution function cannot be performed explicitly,a numeric solution of the equation F(xi) = zi can be exploited. In addition,other simulation methods exist (rejection method,β¦).
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4. Probability distributions simulation: some useful distributions
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Uniform distribution: U(a, b)
πΏπΏ = ππ β ππ οΏ½ ππ + ππSimulation:
Exponential distribution: Exp(Ξ») πΉπΉ π₯π₯ = 1 β π‘π‘βπππ₯π₯
Example. Determine the MTTF and MTTR for a component with Ξ» = 1E-4/h and ΞΌ=1E-1/h(solution with Excel).
Simulation: ππ = 1 β π‘π‘βπποΏ½ππ
ln(1 β ππ) = βππ οΏ½ ππ
ππ = βln(1 β ππ)
ππ
Z and 1-Z have thesame distribution
πΏπΏ = βln(ππ)ππ
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4. Probability distributions simulation: some useful distributions
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Weibull distribution: W(Ξ», Ξ²)Failure rate
πΉπΉ π₯π₯ = 1 β π‘π‘βπ₯π₯ππ
π½π½
Simulation: ππ = 1 β π‘π‘βππππ
π½π½
ln(1 β ππ) = βππππ
π½π½
ππ = ππ βln(ππ)1π½π½
Scale parameter (Ξ·)
Shape parameter (Ξ²)
Log Normal distribution: LN(ΞΌ, Ο) or LN(m, EF)
ππ π₯π₯ =1
π₯π₯ππ 2Ξ π‘π‘β
ln (π₯π₯)βππ 2
2ππ2
ππ = ln (ππ) βππ2
2
ππ =ln (ππ5%)
1.64
)2cos()]ln(2[ 221
1 zz ΟΞ½ β=
ππ = π‘π‘ ππ ππ+ππ
ππ1 ππ2
Average
Errorfactor
Remark β Marvin book: πΉπΉ π₯π₯ = 1 β π‘π‘β πππ₯π₯ πΌπΌ
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5. Simulation History
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History 1
time
Work
Failure
Time To Failure (TTF)
Time To Repair (TTR)
History 2
time
Work
Failure
TTF and TTR aregeneratedrandomly accordingto their respectivedistributionprobabilities
t1 t2
Reliability: R(t1) = 1; R(t2) = 0Availaibility: A(t1) = 1; A(t2) = 0.5;Mean unavailability: Q(0, t2)=(TTR1/t2+ (TTR2+Ο)/t2)/2 Mean failure number: W(0, t2)=(1+ 2)/2= 1.5MTTF = (TTF1,1+TTF2,1)/2 β¦
TTR1,1 TTR1,2
TTR2,1 TTR2,2
Ο
TTF1,1 TTF1,2
TTF2,1 TTF2,2
The results accuracydepends on the numberof stories: to obtain anaccurate result, it isnecessary to perform alarge number of stories.
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6. Output analysis and accuracy
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Statistical analysis of the simulated data sample of size n (number of histories orreplications).
n histories
n independent observations: xi : i = 1,β¦,n.
Point estimation
οΏ½ππ =1ππ οΏ½
ππ=1
ππ
π₯π₯ππ ππ2 =1ππ οΏ½
ππ=1
ππ
π₯π₯ππ β οΏ½ππ 2
Sample mean Sample variance
Confidence interval estimation
Confidence interval quantifies theconfidence (probability) that the true(but unknown) statistical parameterfalls within an interval whoseboundaries are calculated using pointestimates.
π·π·ππ οΏ½πΏπΏ β ππ β€ ππ β€ οΏ½πΏπΏ + ππ = ππ β πΆπΆCentral limit theorem
π·π·ππ οΏ½πΏπΏ β ππππβπΆπΆππβ ππππβ€ ππ β€ οΏ½πΏπΏ + ππππβπΆπΆππ
β ππππ
= ππ β πΆπΆ οΏ½πΏπΏ Β± ππππβπΆπΆππβ ππππ
Quantiles of thestandardized normal
distribution Accuracy is proportional to ππ
ππ
Example: find the confidenceinterval for MTTF at aconfidence of 90 % (slide 12).
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7. Example
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Consider a periodically tested component: once occurred, the failures ofsuch component remain hidden until the next periodic test. Failures andrepairs follow exponential distributions with rates 2.5E-3/h and 0.1/h,respectively. The duration between two successive periodic tests is 1000h.
1. Establish 3 histories, each having a duration of 4500 hours, using the followingrandom numbers according to their order of appearance:
- History nΒ°1 : 0.0235177 ; 0.13534 ; 0.082085 ; 0.00674 ; 0.006738.- History nΒ°2 : 0.28650 ; 0.00674 ; 0.093014 ; 0.36788 ; 0.0024788 ; 0.000045.- History nΒ°3 : 0.0024788 ; 0.36788 ; 0.0235177.
2. Compute the mean availability of the component.3. Compute its availability and reliability at t = 3000 h. 4. Compute the MTTF. 5. Estabilsh the confidence interval for the MTTF (confidence = 90 %). What can
you deduce ? 6. Compute the MUT and MDT of the component.
W
F
States
Test 1(1000)
Test 2(2000)
Test 3(3000)
Test 4(4000)
Restoration completed(2020)
RT11= 20 hTTF1 = 1500 h
DT11= 520 h
UT11 = 1000 h
3020
RT12= 50 h
DT12= 1030 h
4050
U
TTF: Time To first Failure RT: Repair Time DT: Down Time UT: Up Time
History 1:
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A1: histories
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TTF1 = - Ln(0.0235177)/2.5E-3 = 1500 h
RT11 = - Ln(0.13534)/0.1 = 20 h
DT11 = time required for failure detection + repair time = 2000 -1500 + 20 = 520 h
UT11 = - Ln(0.082085)/2.5E-3 = 1000 h
RT12 = - Ln(0.00674)/0.1 = 50 h
DT12 = time required for failure detection + repair time = 4000-3020 + 50 = 1030 h
UT*12 = - Ln(0.006738)/2.5E-3 = 2000 h.
UT12 : we only keep from UT*12 the part included in the history duration:
4500 β 4050 = 450 h
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W
F
States
Test 1(1000)
Test 2(2000)
Test 3(3000)
Test 4(4000)
4 ( s1050
RT21= 50 h
TTF2 = - Ln(0.2865)/2.5E-3 = 500 h
RT21 = - Ln(0.00674)/0.1 = 50 h
DT21 = 1000 -500 + 50 = 550 h
UT21 = - Ln(0.093014)/2.5E-3 = 950 h
RT22 = - Ln(0.36788)/0.1 = 10 h
TTF2 = 500 h
DT21= 550 h
UT21 = 950 h
2010 4410
UT22 = 2400 h
History 2:
500
RT22=DT22= 10 h DT23 =
DT22 = 0+ 10 = 10 h
UT22 = - Ln(0.0024788)/2.5E-3 = 2400 h
RT23 : - Ln(0.000045)/0.1 = 100 h
DT*23 = 590+ 100= 690 h
DT23 : 4500 β 4410 = 90 h
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W
F
States
Test 1(1000)
Test 2(2000)
Test 3(3000)
Test 4(4000)
RT31= 10 hTTF3 = 2400 h
DT31= 610 h
UT31 = 1490 h
3010
History 3:
UT*31 = 1500 h
TTF3 = - Ln(0.0024788)/2.5E-3 = 2400 h
RT31 = - Ln(0.36788)/0.1 = 10 h
DT31 = 3000 -2400 + 10 = 610 h
UT*31 = - Ln(0.0235177)/2.5E-3 = 1500 h
UT31 = 4500 β 3010 = 1490 h
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A2. Mean availability : Aavg [0, 4500 h]
= [(TTF1+UT11+UT12)/4500 + (TTF2+UT21+UT22)/4500 + (TTF3+UT31)/4500]/3
= [(1500 + 1000+ 450)/4500 + (500 + 950 + 2400)/4500+ (2400 + 1490)/4500]/3
= 0.792
Aavg [0, 4500 h]
A3: Availability at t = 3000 h : A (3000) = (1 + 1 + 0) /3 = 2/3 = 0,667 Reliability at t = 3000 h : R (3000) = (0 + 0 + 0) /3 = 0
A4: MTTF = (TTF1+TTF2+TTF3)/3 = (1500 + 500 + 2400)/3= 1466,667 h.
A5: confidence interval (CI) at 90 % for the MTTF :
[ ])3/(64.1),3/(64.1 sMTTFsMTTF β +β β
hMTTFTTFi
i 0298.7763/)(3
1
2 =β= β=
Οwith :
[ ]hh 454.2201,880.731
CI =
CI =
This interval is very wide. This is due toreduced number of stories (3) (non-representative sample): it is thereforenecessary to significantly increase thenumber of histories.
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A6: MUT et MDT :
MUT = [(UT11+UT12)/2 + (UT21+UT22)/2 + UT31]/3
MDT = [(DT11+DT12)/2 + (DT21+DT22+DT23)/3 + DT31]/3
= [(1000+ 450)/2+ (950 + 2400)/2+ 1490]/3= 1296.66667 h.
= [(520+ 1030)/2+ (550 + 10+90)/2+ 610]/3= 533.889 h.
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7. MC simulation and uncertainty propagation
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Monte Carlo method is generally used to perform uncertainty study. Thistechnique has become the industry standard for propagatinguncertainties. It provides an efficient and straightforward way for thispurpose.
Uncertainty propagation shows how the uncertainty of input parameters(failure rate, for instance) spreads onto the output of the model at hand.
1. Construct a probability density function (pdf) for each input parameter (pdfreflects state of knowledge about the value of the parameter).
2. Generate one set of input parameters by using random numbers according topdfs assigned to those parameters.
3. Quantify the output function using the above set of random values. The obtainedvalue is a realization of a random variable (X).
4. Repeat steps 2 to 3 n times (until a sufficient number, e.g. 1000) producing nindependent output values. These n output values represent a random samplefrom the probability distribution of the output function.
5. Generate statistics from the obtained sample for the output result: mean,standard deviation Ο, confidence interval, etc.
Steps:
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7. MC simulation and uncertainty propagation: IEC 61508 standard
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Route 2H: the IEC 61508 standard (clause 7.4.4.3.3) stipulates that βIfroute 2H is selected, then the reliability data uncertainties shall be takeninto account when calculating the target failure measure (i.e. PFDavg orPFH) and the system shall be improved until there is a confidence greaterthan 90 % that the target failure measure is achievedβ.
The confidence on the obtained SIL according to the value of PFDavg or PFHmay be established by checking that the upper limit of the confidenceinterval is encompassed in the corresponding required SIL zone. Also, adirect measure is the evaluation of the cumulated density function (cdf) atthe target performance measure (PFDmax, PFHmax): Pr πΏπΏ β€ π·π·π·π·π·π·ππππππ .
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7. MC simulation and uncertainty propagation: probabilitydistributions
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Probability distribution for the variable x
Probability density function (pdf) and its main properties
Comments and propositions to handle uncertainties
Uniform (a, b)
ππ π₯π₯ = οΏ½1
ππ β ππππππ ππ β€ π₯π₯ β€ ππ
0 πππππ‘π‘π‘πππππππππ‘π‘πππ‘π‘ππππ = πππ‘π‘ππππππππ = β(ππ + ππ) 2πππππππ‘π‘ = any value in [a, b]
ππππππ = β(ππ β ππ) 12
- This distribution expresses a big lack of knowledge about theparameter value (non-informative). The source of data provides aninterval [a, b].
- Simulation: π₯π₯ = ππ + ππππππππ οΏ½ (ππ β ππ) , where rand is a numberuniformly distributed between 0 and 1. MATLAB use directly thefunction unifinv (rand, a, b).
Triangular (a, b, c)
ππ π₯π₯ =
2 π₯π₯ β ππππ β ππ οΏ½ ππ β ππ
ππππ ππ β€ π₯π₯ β€ ππ
2 ππ β π₯π₯ππ β ππ οΏ½ ππ β ππ
ππππ ππ β€ π₯π₯ β€ ππ
0 πππππ‘π‘π‘πππππππππ‘π‘πππ‘π‘ππππ = β(ππ + ππ + ππ) 3
πππππππ‘π‘ = ππ
ππππππ = (ππ β ππ)2+(ππ β ππ)(ππ β ππ) /18
- The distribution is more precise than the Uniform distribution. Thesource of data provides, beside a and b, an estimation of the mostlikely value of the parameter: c. Using the Mean value, one maycompute c (if not available).
- Simulation: MATLAB does not include this law. So, one can use thefollowing:
ππ =ππ β ππππ β ππ
π π = ππππππππ
π₯π₯ = οΏ½ππ + ππ β ππ οΏ½ ππ β ππ οΏ½ π π , ππππ π π β€ ππ
ππ β ππ β ππ οΏ½ ππ β ππ οΏ½ 1 β π π , πππππ‘π‘π‘πππππππππ‘π‘
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7. MC simulation and uncertainty propagation: probabilitydistributions
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Lognormal (ΞΌ, Ο)
f x =1
xΟ 2ΟοΏ½ exp
β(ln x β Β΅)2
2Ο2
where: x > 0; ββ < Β΅ < β; Ο > 0
Mean = exp Β΅ +Ο2
2
Std = exp Β΅ +Ο2
2οΏ½ exp(Ο2) β 1
Median = x50 = exp Β΅
x95 = exp Β΅ + 1.645 Ο
x05 = exp Β΅ β 1.645 Ο
EF =x50x05
=x95x50
= exp 1.645 Ο
- The lognormal distribution is used frequently in safety and reliabilitystudies to model uncertainty. The distribution is skewed to the right,it allows therefore a pessimistic value of x.
- The source of data provides the average value of x or its pointestimate (οΏ½x) and an uncertainty range (EF: error factor): e.g., EF= 10.In that case, we may assume that οΏ½x = x50. By doing so, one canderivate ΞΌ and Ο using equations that give the Mean and EF.
- The source of data provides the average value of x or its pointestimate (οΏ½x) and an estimation of the standard deviation Ο (e.g. Ξ» =1Eβ4 Β± 1Eβ5). In that situation, one can derivate ΞΌ and Ο for thelognormal distribution using equations that give the Mean and Std.
- The source of data provides the average value of x or its pointestimate (οΏ½x), lower and upper bounds (xL, xU). One could assume thatthese bounds x05 and x95, respectively. The problem been over-determined, the use of two among the three available (the mostaccurate) value gives ΞΌ and Ο for the lognormal distribution.
- Simulation: MATLAB use the function logninv (rand,ΞΌ,Ο), byimplementing a numeric approach.
Gamma (Ξ±, Ξ²)
ππ π₯π₯ =π₯π₯πΌπΌβ1
π½π½πΌπΌΞ(πΌπΌ)οΏ½ exp β
π₯π₯π½π½
where: 0 < π₯π₯ < β; πΌπΌ > 0; π½π½ > 0
πππ‘π‘ππππ = πΌπΌπ½π½
πππππππ‘π‘ = π½π½ πΌπΌ β 1 ππππππ πΌπΌ β₯ 1
ππππππ = π½π½ πΌπΌ
- The Gamma distribution can be a posterior distribution obtainedusing a Bayesian approach.
- All information provided above, especially for the lognormaldistribution, may be used to compute Ξ± and Ξ², once the Mean andStd are available or calculated.
- Simulation: MATLAB use the function gaminv(rand,Ξ±,Ξ²), byimplementing a numeric approach.
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7. MC simulation and uncertainty propagation: probabilitydistributions
27
Chi-square (k):ππππ2
f x =xk2β1
2(k2) Ξ(k2)οΏ½ exp β
x2
where: 0 < x < β
k β N (degree of freedom)
Mean = k
Mode = max(k β 2, 0)
Std = 2k
- The Ο2 distribution is a special case of the gamma distribution where Ξ² = 2and Ξ± =k/2.
- For n observed failures over a cumulated observation time T, it isestablished that the failure rate (Ξ») follows the distribution Ο2n2 /2T. Inthat condition:
Mean =k
2T=Ξ±T
=nT
the point estimate of Ξ»: οΏ½Ξ» .
Std = 2k2T
= 1T
k2
= 1T
Ξ± = 1T
n. Therefore, if the Mean and Std are
provided, one can easily derivate parameters for the gamma distribution (Ξ±=nand Ξ²=1/T) and vice versa.
Simulation: x = 1/(2T) . chi2inv (rand, 2n).
Beta (Ξ±, Ξ²)ππ π₯π₯ =
Ξ(πΌπΌ + π½π½)Ξ(πΌπΌ) β Ξ(π½π½)
οΏ½ π₯π₯πΌπΌβ1(1 β π₯π₯)π½π½β1
where: 0 β€ π₯π₯ β€ 1; πΌπΌ > 0; π½π½ > 0
πππ‘π‘ππππ =πΌπΌ
πΌπΌ + π½π½
πππππππ‘π‘ =πΌπΌ β 1
πΌπΌ + π½π½ β 2ππππππ πΌπΌ > 1; π½π½ > 1
ππππππ =πΌπΌπ½π½
(πΌπΌ + π½π½)2 β (πΌπΌ + π½π½ + 1)
- The Beta distribution can be a posterior distribution obtained using a Bayesian approach (only for 0 β€ π₯π₯ β€ 1).
- If the bounds of x are more accurate or out of that range [π₯π₯πΏπΏ, π₯π₯ππ] β [0, 1], the Mean and Std become:
πππ‘π‘ππππ = π₯π₯πΏπΏ + (π₯π₯ππ β π₯π₯πΏπΏ) β πΌπΌ
πΌπΌ + π½π½
ππππππ = (π₯π₯ππ β π₯π₯πΏπΏ) β πΌπΌπ½π½
(πΌπΌ + π½π½)2 β (πΌπΌ + π½π½ + 1)
- If the Mean and Std are provided, the solving of the above equations gives Ξ± and Ξ² for the Beta distribution.
- If Std is not available, one may use the approximation: Std =(π₯π₯ππ β π₯π₯πΏπΏ)/6).- Simulation: x = π₯π₯πΏπΏ+ (π₯π₯ππ β π₯π₯πΏπΏ)Β·betainv (rand, Ξ±,Ξ²).
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7. MC simulation and uncertainty propagation: Example
28
Paper: H. Jin, M. A. Lundteigen & M. Rausand
Uncertainty assessment of reliability estimates for safety instrumented systems
Excel solution
Log-normal :Median = 3E-7EF=2
Log-normal :Median = 0.1EF=2
Uniform :a = 1.5b=2
- Compute the PFDavg- Establish the confidence interval at 90 %- Compute the confidence to comply with the different SILs (1, 2, 3, 4)
2929