encv 800503-lecture-9 mcs [25 oct 2013] dpk
TRANSCRIPT
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Manajemen Sistem Rekayasa & Nilai - 2013
Department of Civil EngineeringGraduate ProgramUniversity of Indonesia25 October 2 13
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1
Simulation Models
Iconic
Physical replicas of real system, reduced scale
Interrelationship of components not well understood or
too complex Analog
Real system is modeled through a completely differentphysical media
Analytical System component and structure defined as
mathematical model
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Introduction to Simulation
Real situation rarely meet assumptions of analyticalmodel. Market uncertainties vs. competitive may make predicting
unit profit very difficult
Rate at which resources are consume may vary Availability of resources from suppliers may not be assured
Demand almost always uncertain
The more elegant the mathematical formulation of aproblem is, the less it matches reality.
Situations which problem does not meet the assumptionsrequired by standard analytical modeling approaches,simulation can be a valuable approach to modelingand solving a problem.
Evans & Olson, 1998
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Simulation Definition
Is the process of building a mathematical or logical
model of a system or a decision problem, and
experimenting with the model to obtain insight into
the systems behavior or to assist in solving thedecision problem.
Key elements:
Model Experiment
Evans & Olson, 1998
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Where simulation fits in . . .
SimulationProgramming
Analysis
Modeling
Probability
&
Statistics
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Examples of Simulation
Is used to forecast weather
In business, is used to predict, to explain, to trainand to help identify optimal solution
In manufacturing, is used to model production andassembly operations, develop realistic productionschedules, study inventory policies, analyzereliability, quality, and equipment replacement
problems, and design material handling andlogistics system
Finds extensive application in both profit-seekingservice firms and nonprofit service organization
Evans & Olson, 1998
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Models and Simulation
Model is an abstraction or representation of a real
system, idea, or object
Model
Prescriptive
Descriptive
Deterministicor
Probabilistic
Discrete or
Continuous
Linear
Programming
Queuing
models
Linear programming
(deterministic)
Queuing models
(probabilistic)
Determine
optimal policy
Describe
relationships and
provide information
for evaluation
Deterministic: all data are
known or assumed to be
known with certainty
Probabilistic: some data
are described byprobability distributions
Refers to the types
of the variables in
the model
Integer programming
(discrete)
Linear programming(continuous)
Evans & Olson, 1998
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Monte Carlo Simulation (MCS)
Is basically a sampling experiment whose purpose is to
estimate the distribution of an outcome variable that
depends on several probabilistic input variables
Example in financial problems: sales, costs, and inflationare random variables
The term MCS was first used during the development of
the atom bomb as a code name for computer
simulations of nuclear fission.
Researchers coined this term because of the similarity to
random sampling in games of chance such as roulette in
the famous casinos of Monte CarloEvans & Olson, 1998
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Generating Probabilistic Outcomes
Random Numbers
One that is uniformly distributed between 0 and 1
In Excel, =RAND()
Random Numbers Seed
A value from which a stream of a random numbers at a
later time.
Desirable when we wish reproduce an identicalsequence of random events in a simulation to test theeffects of different policies or decision variables under
the same circumstances
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Random Numbers and MCS
A random number, Ri, is defined as an independent
random sample drawn from a continuous uniform
distribution whose probability density function (pdf)
is given by
The procedure consists of two steps:
We develop the cumulative probability distribution(cdf) for the given random variable, and
We use the cdf to allocate the integer random numbers
directly to the various values of the random variables.
otherwise0
101)(
xxf
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Monte Carlo Simulation
Similarity of statistical simulation to game of chance
Physical System
Probability
Distribution
function
Equations:
algebraic or
differentialMonte Carlo
simulation
Result Random
samplingSolution
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MONTE CARLO SIMULATIONS (1)
Similar to Playing Coins or Dice Many Times
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MONTE CARLO SIMULATIONS (2a)
Generating
Random
Numbers
between 0and 1
Calculating x
Using CDF
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MONTE CARLO SIMULATIONS (2b)
RV = 01
x = A + RV*(B-A) 1
0A B
RV
x
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MONTE CARLO SIMULATIONS (3)
0.382
0.88461
0.863247
0.03238
0.285043
0.371838
0.42616
0.991241
0.705039
0.3002110.074343
0.487045
0.040712
0.100314
0.809107
0.756157
0.552324
0.970275
0.805658
0.1148410.986297
0.500778
0.037965
0.584521
0.200476
0.300211
0.90405
0.516648
0.789026
0.619404
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
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MONTE CARLO SIMULATIONS (4)
0.382
0.88461
0.863247
0.03238
0.285043
0.371838
0.42616
0.991241
0.705039
0.300211
0.074343
0.487045
0.040712
0.100314
0.809107
0.756157
0.552324
0.970275
0.805658
0.114841
0.986297
0.500778
0.037965
0.584521
0.200476
0.300211
0.90405
0.516648
0.7890260.619404
Normal Distribution
Mean = 5, SD = 1
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
0 2 4 6 8 10 12
x
CDF
0.382 4.699768
0.88461 6.19835
0.863247 6.095023
0.03238 3.153089
0.285043 4.432075
0.371838 4.673009
0.42616 4.813842
0.991241 7.375655
0.705039 5.538948
0.300211 4.476205
0.074343 3.555813
0.487045 4.967521
0.040712 3.257517
0.100314 3.720236
0.809107 5.874609
0.756157 5.693994
0.552324 5.131536
0.970275 6.884846
0.805658 5.862008
0.114841 3.798821
0.986297 7.205688
0.500778 5.001951
0.037965 3.225194
0.584521 5.213473
0.200476 4.160078
0.300211 4.476205
0.90405 6.304979
0.516648 5.041741
0.789026 5.8030450.619404 5.303914
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USING MS-EXCEL 2003
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USING MS-EXCEL (1)
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USING MS-EXCEL (2)
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USING MS-EXCEL (3)
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USING MS-EXCEL (4)
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USING MS-EXCEL (5)
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USING MS-EXCEL 2007
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Using MS-EXCEL (1)
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Showing Up Data Analysis
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Showing Up Data Analysis
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Showing Up Data Analysis
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Showing Up Data Analysis
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Using MS-EXCEL (2)
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Using MS-EXCEL (3)
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Using MS-EXCEL (4)
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Using MS-EXCEL (5)
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SIMPLE PROBABILISTIC
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SIMPLE EXAMPLE: DICE
2 Dice: A = Die #1 + Die #2
SIMULATIONS: 2 DICE
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A = (Die No. 1 + Die No.2)
A < 7 A = 12
SIMULATIONS No. Samples = 60 3
Total Samples = 100 100
p = 0.6 0.03
THEORETICAL CALCULATION p = 7/12 1/360.5833 0.0278
RV1 Die No. 1 RV2 Die No. 2 A A < 7 A = 12
0.136265 1 0.306009 2 3 1 0
0.19541 2 0.600391 4 6 1 0
0.85876 6 0.085177 1 7 1 00.637135 4 0.079897 1 5 1 0
0.578234 4 0.735679 5 9 0 0
0.409223 3 0.141881 1 4 1 0
0.611133 4 0.920103 6 10 0 0
0.188238 2 0.262886 2 4 1 0
0.508499 4 0.542375 4 8 0 0
0.091098 1 0.106479 1 2 1 00.737358 5 0.328867 2 7 1 0
0.056459 1 0.003571 1 2 1 0
0.363964 3 0.403638 3 6 1 0
0.243812 2 0.380444 3 5 1 0
0.895169 6 0.389203 3 9 0 0
0.85638 6 0.915036 6 12 0 10.934294 6 0.063875 1 7 1 0
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0
5
10
15
20
25
1 2 3 4 5 6
Value of Die
Freq
uency
Die No. 1
Die No. 2
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SIMPLE EXAMPLE: DICE
RV1 = 0.136265 0 < RV1 < 1/6 Value of
Die #1 = 1
RV2 = 0.306009 1/6 < RV2 < 2/6 Value of
Die #2 = 2 A = Value of Die #1 + Value of Die #2 = 1 + 2 =
3
A 7? True 1 A = 12? False 0
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SIMPLE EXAMPLE: DICE
Same process repeated for 100 times (see TotalSamples = 100)
From 100 samples 60 samples satisfy A 7?
p = 60 / 100 = 0.6 (theoretical p = 0.5833) From 100 samples 3 samples satisfy A = 12?
p = 3/100 = 0.03 (theoretical p = 0.0278)
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HOW MANY SAMPLES (1)
0.01
0.1
1
1 10 100 1000 10000
No. Samples
ProbabilityA
=12
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1 10 100 1000 10000
No. Samples
Probability
A
10 (1 / Probability)
More Samples
Better Typically
0.01
0.1
1
1 10 100 1000 10000
No. Sample Meeting Criterion
Probabilit
y
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ENGINEERING ECONOMY
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Benefit / Cost Ratio
Design A Design BInitial Investment* (4,074,088) (426,209)
Annual Maintenance (1,500,000) (150,000)
Total (5,574,088) (576,209)
Annual Benefits 6,500,000 650,000
Benefit/Cost Ratio 1.17 1.13
Both Alternatives Economically Feasible
* Note:
Initial Investment = (40,000,000) (4,000,000)
i = 8% 4%
n = 20 12
Annual Worth = (4,074,088) (426,209)
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Initial Rate Annual Annual Annual Annual B/C Ratio
Investment Worth Maintenance Cost Benefit
Min = 36000000 6.00% 1350000 5850000
Max = 44000000 10.00% 1650000 7150000 Prob of Failure
0.06
Average = 39975895 8.02% 4086120 1510867 5596987 6571531 1.183707
Min = 36113773 6.02% 3187186 1351492 4642614 5860633 0.906326
Max = 43923582 9.90% 4978663 1647977 6504055 7148334 1.481975
0.382 0.101 0.596 0.899 39056001 6.40% 3517241 1528945 5046186 7018838 1.390919 0
0.014 0.407 0.863 0.139 36115970 7.63% 3577709 1608974 5186683 6030160 1.162623 0
0.032 0.164 0.22 0.017 36259041 6.66% 3331766 1415883 4747649 5872217 1.236868 00.554 0.357 0.372 0.356 40429090 7.43% 3944536 1461551 5406087 6312282 1.167625 0
0.426 0.304 0.976 0.807 39409284 7.22% 3782525 1642712 5425238 6898665 1.271588 0
0.952 0.053 0.705 0.817 43613514 6.21% 3868671 1561512 5430182 6911480 1.27279 0
0.3 0.75 0.351 0.776 38401685 9.00% 4207016 1455445 5662461 6858356 1.211197 0
0.064 0.358 0.487 0.511 36512467 7.43% 3563465 1496113 5059579 6514580 1.287574 0
0.041 0.231 0.005 0.926 36325694 6.92% 3408363 1351492 4759855 7053989 1.481975 0
0.776 0.68 0.809 0.724 42205512 8.72% 4531116 1592732 6123848 6791624 1.109045 0
0.756 0.627 0.174 0.405 42049257 8.51% 4445346 1402095 5847441 6376237 1.090432 00.555 0.181 0.97 0.687 40441298 6.72% 3736059 1641082 5377141 6743023 1.254016 0
0.806 0.262 0.178 0.867 42445265 7.05% 4021760 1403386 5425146 6976783 1.286008 0
0.762 0.738 0.986 0.926 42092471 8.95% 4595839 1645889 6241728 7053275 1.13002 0
0.501 0.675 0.49 0.146 40006226 8.70% 4289219 1496947 5786166 6039523 1.043787 0
0.672 0.732 0.585 0.152 41372478 8.93% 4508568 1525356 6033924 6047894 1.002315 0
/
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Benefit / Cost Ratio (line1)
RV1 = 0.382 Initial Investment = 36M + RV1*8M =39.1M
RV2 = 0.101 Rate = 6% + RV2*4% = 6.4%
Annual Worth = PMT(Rate,20,-Initial Investment) = 3.52M
RV3 = 0.596 Annual Maintenance = 1.35M +RV3*0.3M = 1.53M
Annual Cost = 3.52M + 1.53M = 5.05M
RV4 = 0.899 Annual Benefit = 5.85M + RV4*1.3M =
7.02M
B/C Ratio = 7.02M / 5.05M = 1.39
B/C Ratio > 1.0 Not fail = 0
/
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Benefit / Cost Ratio
Same process repeated for 100 times (see TotalSamples = 100)
Simulation: B/C Ratio = 1.18
Theoretical: B/C Ratio = 1.17 From 100 samples 6 samples do not satisfy
B/C Ratio 1.0 probability of failure = 6 /100 = 0.06
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RISK ASSESSMENT
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Series Networks
If either one of A, B, or C fails, system failsR = 0.8521 0.9712 0.9357 = 0.7743
Pf = 10.7743 = 0.2257Reliability Probability of Failure
A 0.8521 0.1479
B 0.9712 0.0288
C 0.9357 0.0643
A B C Probability Sum
1 0.7743 0.7743
2
0.05323 0.0230
4 0.1344
5 0.0016 0.2257
6 0.0092
7 0.0040
8 0.0003
SIMULATIONS: SERIES SYSTEM
A S h d l d
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A B C
SIMULATIONS No. Samples = 88 99 92 79
Total Samples = 100 100 100 100
p = 0.88 0.99 0.92 0.79
THEORETICAL p = 0.8521 0.9712 0.9357 0.7743
RV1 RV2 RV3 A B C
0.136265 0.306009 0.382 1 1 1 3 1
0.19541 0.600391 0.100681 1 1 1 3 1
0.85876 0.085177 0.596484 0 1 1 2 0
0.637135 0.079897 0.899106 1 1 1 3 1
0.578234 0.735679 0.88461 1 1 1 3 1
0.409223 0.141881 0.958464 1 1 0 2 0
0.611133 0.920103 0.014496 1 1 1 3 1
0.188238 0.262886 0.407422 1 1 1 3 1
0.508499 0.542375 0.863247 1 1 1 3 1
0.091098 0.106479 0.138585 1 1 1 3 10.737358 0.328867 0.245033 1 1 1 3 1
0.056459 0.003571 0.045473 1 1 1 3 1
0.363964 0.403638 0.03238 1 1 1 3 1
0.243812 0.380444 0.164129 1 1 1 3 1
0.895169 0.389203 0.219611 0 1 1 2 0
0.85638 0.915036 0.01709 0 1 1 2 0
0.934294 0.063875 0.285043 0 1 1 2 0
As Scheduled
As Scheduled
System
System
S N ( )
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Series Networks (line1)
RV1 = 0.136265 RV1 < Reliability of A =0.8521 A is reliable A = 1
RV2 = 0.306009 RV2 < Reliability of B =0.9712 B is reliable B = 1
RV3 = 0.382 RV3 < Reliability of C =0.9357 C is reliable C = 1
Sum of A, B, C = 3 all subsystems reliable
series system network works
= 1
S N k (l 3)
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Series Networks (line3)
RV1 = 0.85876 RV1 > Reliability of A =0.8521 A is not reliable A = 0
RV2 = 0.085177 RV2 < Reliability of B =0.9712 B is reliable B = 1
RV3 = 0.596484 RV3 < Reliability of C =0.9357 C is reliable C = 1
Sum of A, B, C = 2 NOT all subsystems
reliable
series system network fails
= 0
S N k
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Series Networks
Same process repeated for 100 times (see TotalSamples = 100)
From 100 samples 79 samples satisfy series
system network requirement
p = 79 / 100 =0.79 (theoretical p = 0.7743)
P ll l N k
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Parallel Networks
If both A and B fail, system failsRcomponent = 0.7500
Rsystem = 0.9375
Reliability Probability of FailureA 0.7500 0.2500
B 0.7500 0.2500
A B Probability Sum
1 0.56252 0.1875 0.93753 0.18754 0.0625 0.0625
SIMULATIONS: PARALLEL SYSTEM
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A B
SIMULATIONS No. Samples = 79 72 8
Total Samples = 100 100 100
p = 0.79 0.72 0.080.9200
THEORETICAL p = 0.75 0.75 0.9375
RV1 RV2 A B
0.306009 0.382 1 1 2 0
0.600391 0.100681 1 1 2 00.085177 0.596484 1 1 2 0
0.079897 0.899106 1 0 1 0
0.735679 0.88461 1 0 1 0
0.141881 0.958464 1 0 1 0
0.920103 0.014496 0 1 1 0
0.262886 0.407422 1 1 2 00.542375 0.863247 1 0 1 0
0.106479 0.138585 1 1 2 0
System
System
0.870815 0.991241 0 0 0 1
P ll l N k (li 1)
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Parallel Networks (line1)
RV1 = 0.306009 RV1 < Reliability of A =0.7500 A is reliable A = 1
RV2 = 0.382 RV2 < Reliability of B = 0.7500
B is reliable
B = 1
Sum of A, B = 2 NOT all subsystems
unreliable parallel system network works =
0
P ll l N k (li 4)
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Parallel Networks (line4)
RV1 = 0.079897 RV1 < Reliability of A =0.7500 A is reliable A = 1
RV2 = 0.899106 RV2 < Reliability of B =
0.7500
B is not reliable
B = 0 Sum of A, B = 1 NOT all subsystems
unreliable parallel system network works =
0
P ll l N k (li ?)
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Parallel Networks (line?)
RV1 = 0.870815 RV1 < Reliability of A =0.7500 A is not reliable A = 0
RV2 = 0.991241 RV2 < Reliability of B =
0.7500
B is not reliable
B = 0 Sum of A, B = 0 all subsystems unreliable
parallel system network fails = 1
P ll l N t k
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Parallel Networks
Same process repeated for 100 times (see TotalSamples = 100)
From 100 samples 8 samples fails parallel
system network requirement
probability offailure = 8 / 100 = 0.08 reliability = 10.08 = 0.92 (theoretical reliability = 0.9375)
P b bilit f F il
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Probability of Failure
ProbabilityDe
nsityFunction
Capacity, Q
Load, F
mF mQ
sF
sQ
Safety Margin, M = Q - F
ProbabilityDensityFunction
mM
pf
sM
Reliability Index = mM/sM= (mQ-mF)/(sQ
2+sF2)0.5
Probabilit of Fail re (2)
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Probability of Failure (2)
C: C = 30 C = 6D:D = 20 D = 6
M = CDM = 10 M = 8.49
= M / M = 10 / 8.49= 1.18 probability of failure = 0.1193
Probability of Failure (3)
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Probability of Failure (3)C D M = C - D
mean 30 20 10
stdev 6 6 8.485281COV 0.2 0.3 0.848528
mean 30.09954 19.78092 10.31861 0.115
stdev 6.015097 6.007956 8.722103
COV 0.19984 0.303725 0.845279 1.20036
0.382 0.100681 28.19861 12.3339 15.86471 0
0.88461 0.958464 37.1901 30.3988 6.791303 0
0.863247 0.138585 36.57014 13.4798 23.09034 00.03238 0.164129 18.91853 14.13422 4.784312 0
0.285043 0.343089 26.59245 17.57571 9.016736 0
0.371838 0.355602 28.03806 17.77856 10.2595 0
0.037965 0.796258 19.35116 24.96999 -5.618825 1
P b bilit f F il (li 1)
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Probability of Failure (line1)
RV1 = 0.382 C = NORMINV(RV1,mean-C,stdev-C) = 28.19861
RV2 = 0.100681 D = NORMINV(RV2,mean-
D,stdev-D) = 12.3339 M = CD = 15.86471 > 0 M NOT fails
= 0
P b bilit f F il (li ?)
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Probability of Failure (line?)
RV1 = 0.038965 C = NORMINV(RV1,mean-C,stdev-C) = 19.35116
RV2 = 0.796258 D = NORMINV(RV2,mean-
D,stdev-D) = 24.96999 M = CD = -5.618825 < 0 M fails = 1
P b bilit f F il (4)
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Probability of Failure (4)
Same process repeated for 1000 times C: C = 30 C = 6 (theoretical)
C: C = 30.1 C = 6.0 (simulation)
D: D = 20 D = 6 (theoretical)D: D = 19.8 D = 6.0 (simulation)
M: M = 10 M = 8.49 (theoretical)
M: M = 10.3 M = 8.72 (simulation)
Probabilit of Fail re (5)
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Probability of Failure (5)
Same process repeated for 1000 times Theoretical: 0.1193 = 1.18 Simulation: From 1000 samples 115 samples
fail probability of failure = 115 / 1000 =0.115 = 1.20036
Probability of Failure (6)
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Probability of Failure (6)
0
50
100
150
200
250
300
350
5 15 25 35 45 55 65
C
Frequency
0
50
100
150
200
250
300
350
5 15 25 35 45 55 65
D
Frequency
0
50
100
150
200
250
-20 -10 0 10 20 30 40 50M = D - C
Frequency
Risk Calculation
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Risk Calculation
Risk Calculation (B)
Pipeline: uniform distribution w. range between 22 and 34PDF Max value = 1 / (3422) = 0.0833Total area of PDF = total prob = 1
Prob. t > 28 = p(t > 28) = 1/2
18 20 22 24 26 28 30 32 34 36 38 t
PDF
Risk Calculation (B)
t > 28penalty
1,000,000per day
-
2,000,000
4,000,000
6,000,000
8,000,000
10,000,000
12,000,000
18 20 22 24 26 28 30 32 34 36 38
Duration (day)
Penalty
Risk Calculation
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Risk Calculation
f(t) = 0.0833
C(t) = 1,000,000 (t-28)
Risk = 28
34f(t) C(t) dt
Risk = 2834
(0.0833)[1,000,000 (t-28)] dt
Risk = (41,667 t22,333,333 t) |28
34
Risk = -31,166,667-32,666,667
Risk = 1,500,000
SIMULATIONS: RISK ASSESSMENT
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Fail? Penalty
SIMULATIONS No. Samples = 1482
Total Samples = 3000
p = 0.494
THEORETICAL p = 0.5
RISK Average = 1469726.9
Min = 22.01941
Max = 33.99854
Mean = 27.96697
RV1A Days Fail? Penalty
0.148473 23.78167 0 0.0
0.037507 22.45009 0 0.0
0.300363 25.60436 0 0.0
0.074923 22.89908 0 0.0
SIMULATIONS: RISK ASSESSMENT
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Fail? Penalty
SIMULATIONS No. Samples = 1559
Total Samples = 3000
p = 0.519667
THEORETICAL p = 0.5
RISK Average = 1522088.3
Min = 22.00183
Max = 33.99268
Mean = 28.11062
RV1B Days Fail? Penalty
0.352824 26.23389 0 0.0
0.974761 33.69713 1 5697134.3
0.949675 33.3961 1 5396099.7
0.908567 32.9028 1 4902798.5
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No. RV1A RV1B
Samples
50 1003427 1818366
100 1233486 1782784
250 1323745 1629570
500 1448791 1545574
1000 1440903 1508369
1500 1444841 1545439
2000 1468108 1555730
2500 1478144 1523507
3000 1469727 1522088
0
500000
1000000
1500000
2000000
0 1000 2000 3000
No. Samples
Risk
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EFFECT OF R.V. SETS
Small number of RVeffect of RV sets may besignificant
Large number of RVeffect of RV sets notsignificant
Number of RV shouldsufficient to ensure theeffect of RV setsinsignificant
0
500000
1000000
1500000
2000000
0 1000 2000 3000
No. Samples
Risk
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DECISION MAKING
DECISION MAKING (1)
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DECISION MAKING (1)
Tornado DiagramCost of CM1 = -75 -125 [-100]
Cost of CM2 = -50 -110 [-80]
Probability of Successful CM2 = 0.5 0.7 [0.6]Cost of Unsuccessful CM2 = -20 -50 [-35]
Decision to Make:
Constr. Method 1 or
Constr. Method 2
Method 1; -100m
Method 2;-80m
Successful; 0
p= 0.6
Unsuccessful; -35m
p= 0.4
-100m
-80m[= (-80m)
+ 0]
-115m[= (-80m)
+ (-35m)]
A= Constr.Method 1
= Constr.Method 2
E= Successful;
= UnsuccessfulL= Cost
DECISION MAKING (2)
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DECISION MAKING (2)
Choose Branch with Highest Value at a Decision Node:CM1: E(LA) = -100m CM2: E(L) = -94mE(LA) < E(L) Do Constr. Method 2
Decision to Make:
Constr. Method 1 or
Constr. Method 2
CM1: E(LA) = -100m
CM2: E(L) = -94m
A= Constr.Method 1
= Constr.Method 2
E= Successful; = Unsuccessful
L= Cost
DECISION MAKING (3)
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DECISION MAKING (3) Tornado Diagram
Decision to Make:
Constr. Method 1 or
Constr. Method 2
Method 1; -100m
Method 2;
-80m
Successful; 0p= 0.6
Unsuccessful; -35m
p= 0.4
-100m
-80m[= (-80m)
+ 0]
-115m[= (-80m)
+ (-35m)]
A= Constr.Method 1
= Constr.Method 2
E= Successful;= Unsuccessful
L= Cost
DECISION MAKING (4)
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DECISION MAKING (4) Simulations
UPPER BRANCH Cost of CM2 Prob of Successful MINIMUMCost of CM1 Losses When Unsucc. LOWER BRANCH
Min = 75 50 20 0.5
Max = 125 110 50 0.7
Average = 100.95 81.657 35.546 0.6032 95.725 89.05 Prob of "CM2" =
Min = 76.506 51.54 20.149 0.5006 64.038 64.038 0.62
Max = 124.89 109.41 49.713 0.6972 127.67 123.96
0.382 0.101 0.596 0.899 94.1 56.041 37.895 0.6798 68.174 68.174 CM2 1
0.885 0.958 0.014 0.407 119.23 107.51 20.435 0.5815 116.06 116.06 CM2 1
0.863 0.139 0.245 0.045 118.16 58.315 27.351 0.5091 71.742 71.742 CM2 1
0.032 0.164 0.22 0.017 76.619 59.848 26.588 0.5034 73.051 73.051 CM2 1
0.285 0.343 0.554 0.357 89.252 70.585 36.609 0.5715 86.273 86.273 CM2 1
0.372 0.356 0.91 0.466 93.592 71.336 47.309 0.5932 90.581 90.581 CM2 1
0.426 0.304 0.976 0.807 96.308 68.234 49.271 0.6613 84.921 84.921 CM2 10.991 0.256 0.952 0.053 124.56 65.376 48.551 0.5107 89.132 89.132 CM2 1
0.705 0.817 0.973 0.466 110.25 98.991 49.175 0.5933 118.99 110.25 CM1 0
0.3 0.75 0.351 0.776 90.011 95.012 30.544 0.6551 105.55 90.011 CM1 0
0.074 0.198 0.064 0.358 78.717 61.906 21.922 0.5717 71.296 71.296 CM2 1
0.487 0.511 0.373 0.986 99.352 80.673 31.204 0.6972 90.122 90.122 CM2 1
0.041 0.231 0.005 0.926 77.036 63.843 20.149 0.6852 70.186 70.186 CM2 1
0.1 0.257 0.776 0.68 80.016 65.401 43.271 0.6359 81.155 80.016 CM1 0
DECISION MAKING (line1)
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DECISION MAKING (line1)
RV1 = 0.382 Cost CM1 = 75 + 50*RV1 =94.1 A = 94.1
RV2 = 0.101 Cost CM2 = 50 + 60*RV2 =56.0
RV3 = 0.596 Cost Unscful.CM2 = 20 + 30*RV3= 37.9
RV4 = 0.899 Prob.Scful.CM2 = 0.5 + 0.2*RV4= 0.68
= Cost CM2*Prob.Scful.CM2 + (1-Prob.Scful.CM2)*(Cost CM2+ Cost Unscful.CM2) =68.2
A greater than choose CM2
DECISION MAKING (5)
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DECISION MAKING (5)
0
2
4
6
8
10
12
14
16
18
65 70 75 80 85 90 95 100
105
110
115
120
125
Minimum Cost
No.
Samples
DECISION MAKING (6)
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DECISION MAKING (6)
Same process repeated for 100 times From 100 samples 62 samples result in CM2
probability of CM2 is better option = 62 /
100 = 0.62
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ETC.
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