Encv 800503-Lecture-9 Mcs [25 Oct 2013] Dpk

8/13/2019 Encv 800503-Lecture-9 Mcs [25 Oct 2013] Dpk

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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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2

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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4

Where simulation fits in . . .

SimulationProgramming

Analysis

Modeling

Probability

&

Statistics


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5

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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11

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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14


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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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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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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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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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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42

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


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


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)


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)


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


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



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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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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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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



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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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Encv 800503-Lecture-9 Mcs [25 Oct 2013] Dpk