Decision Analysis

Decision AnalysisDecision Analysis 11

Decision AnalysisDecision Analysis

• Ultimate objective of all engineering analysis

• Uncertainty always exist, hence satisfactory performance not guaranteed

• More conservative design reduces risk

• Same design SF for all?

• Component vs. System Risk

• Proper tradeoff between risk and investment


Solution by CalculusSolution by Calculus

• Set up objective function

where ’s are decision variables• From solution to

yields optimal values of decision variables

),......,,( 21 nxxxF

ix

n ..., 2, 1,i

;0ix

F


Contractor – submit a bidContractor – submit a bid (Example 2.1)(Example 2.1)

Bid Ratio, R

=

C cost, est.

B bid, his


To determine optimal bid and To determine optimal bid and optimal Roptimal R

• Establish the objective function

1.6)C-R(-R

R)-(1.6C1)-(R

pCC

C)-(B

C)p-(B

0p)(1- C)p-(B

profit expected X

2 6.2

3.1R

02.6)C(-2R R

X

opt

imummax

negative

2C- R

X2

2


• Case 2: Include Idling Cost

R)(-0.1C)1.6(1-

1.6)C-R62(-R

10p)(1- C)p-(B

profit expected X

2

.

).( C

25.1R

0 R

X

opt


Cofferdam for construction of Cofferdam for construction of Bridge Pier (2 yrs)Bridge Pier (2 yrs) (Example 2.2)(Example 2.2)

h?


InformationInformation

• Floods occur according to Poisson process with mean rate of 1.5/yr

• Elevation of each flood – exponential with mean 5 feet

• Each overtopping

loss due to pumping + delay = $25,000• Construction cost, hCCc 30000

h?


Expected damage cost, CExpected damage cost, C

!

3h)P(x 25000

yrs)2 in floods P(floods) |E(lossC

3

i

ei

ii

i

i

i

5

35

75000

3 25000

/

/

!h

i

i

h

e

i

eei

E (loss of flood)

5

h

5 5

1

ng)(overtoppi P

/

/

h

x

e

dxe


• Total Cost

= 8.05 ft

50 750003000 /h

C

T

ehC

CC

C

0

h

CTopth


Cost as Functions of cofferdam elevation above normal water level

8.05


Limitation of this ApproachLimitation of this Approach• Objective function may not be continuous

function of decision variables• Alternatives may be discrete

e.g. dam for flood control (height, location, other schemes)

• Consequences may be more than monetary costs

• Alternative may include acquiring new information before final decision

• Should we acquire or not?


Seepage under the Seepage under the Embankment Embankment (Example 2.4)(Example 2.4)

EmbankmentCooling Lake

Pump System

(100/120 gal/min)

Q = 95 or 120 gal/min?

Bentonite Seal


Decision tree for seepage Decision tree for seepage problemproblem

Pump System B (120)

Seal

Q1(0.9)Pump System A (100)

Q2(0.1)Add Pump System C

Q1(0.9)

Q2(0.1)

95

120

95

120


Decision Tree ModelDecision Tree Model

Decision Node

Chance Node

1a

2a

13 : ea

1a

2a

1a

2a

Alternatives)|P( , 111 a

)|P( , 121 a

)|P( , 211 a)|P( , 221 a

),,|P( , 11111 aez

),,|P( , 11111 aez),,|P( , 11111 aez

),,|P( , 11111 aez),,|P( , 11111 aez

),,|P( , 11111 aez),,|P( , 11111 aez

),,|P( , 11111 aez

Uncertainties

),

),

),

),

),

),

),

),

),

),

),

),

2

2221

1221

2121

1121

2211

1211

111

1111

22

12

21

11

a ,z ,(e

a ,z ,(e

a ,z ,(e

a ,z ,(e

a ,z ,(e

a ,z ,(e

a ,z ,(e

a ,z ,(e

(a

(a

(a

(a

u

u

u

u

u

u

u

u

u

u

u

u

Consequences


Click to enlarge

Example 2.17Example 2.17


Click to enlarge

Example 2.17Example 2.17


Decision CriteriaDecision Criteria

1.Pessimistic Minimize max loss Install

2.Optimistic Maximize max gain Not Install


3. Maximum EMV (Expected Monetary Value)

E(I) = 0.1x(-2000)+0.9x(-2000)

= -2000

E(II) = 0.1x(-10000)+0.9x(0)

=-1000

ia

}{max)(

)(

jijij

iopt

jijiji

dpad

dpaE


Ex. 2.9 Decision tree for Ex. 2.9 Decision tree for construction projectconstruction project


]|1.0[]|[ 2 NCXxENCLE

dxxfxx x )()1.0( 2

4.8

5]259[1.0

)|()|()|(1.0

)|()|(1.02

2

NCxENCxENCxVar

NCxENCxE

)(]|[)(]|[)(]|[]|[ BPBCLENPNCLEGPGCLECLE

36.19

]|[4.0]|[4.002.0

BCLENCLE


Spillway DecisionsSpillway DecisionsAlternatives Capital Cost Annual

OMR Cost

• No Change 0 0• Lengthening

spillway 1.04M 0 • Plus lowering

crest, installing 1.30M 0flashboard

• Plus considerablecrest lowering, 3.90M 0installing radial gates

• 50years service; Discount rate 6%


Spillway DecisionsSpillway DecisionsSummary of Annual Costs (in Dollars)

0a

2a

3a

1a

Total Annual Cost

=Capital Cost x crf (i,n)

+Annual DMR Cost

+Expected Risk Cost (annual)08024.0

20

..%5..

1)1(

)1(),(

crf

yearsn

apige

i

iinicrf

n

n


Discount factorsDiscount factors

Given A to find P:

Given P to find A:

Where i = int. rate per period

n= no. of periods

1)1(

)1(),(

n

n

i

iinicrf

08024.0

462.12

20

..%5..

crf

pwf

yearsn

apige

n

n

ii

inipwf

)1(

1)1(),(


E2.11 Spillway DesignE2.11 Spillway Design


E2.11 Spillway DesignE2.11 Spillway Design

0

)()()( dxxfxcCE x

Risk Cost


Ex. 2.6 Prior AnalysisEx. 2.6 Prior Analysis

A (small)

B (large)

EH 0.7

EL 0.3

EH 0.7

EL 0.3

0

-100

-50

-20

E(A) = 0.7 x 0 + 0.3 x (-100) = -30

E(B) = 0.7 x (-50) + 0.3 x (-20) = -41

Hence, A is the optimal alternative.


Lab. Model test on Efficiency (Cost $10,000) will indicate: HR (high rating)

MR (medium rating)

LR (Low rating)

HR 0.8 HR 0.1

If EH MR 0.15 If EL MR 0.2

LR 0.05 LR 0.7

e.g. If the process is actually high efficiency (EH), then

the probability that the test will indicate HR is 0.8.


Suppose the test indicate HRSuppose the test indicate HR

Test HR

A (small)

B (large)

EH 0.95

EL 0.05

EH 0.95

EL 0.05

-10

-110

-60

-30

)(

)()|()|(

HRP

EHPEHHRPHREHP

95.059.0

56.0

3.01.07.08.0

7.08.0

)()|(

)()|(

ELPELHRP

EHPEHHRP


Suppose the test indicate HRSuppose the test indicate HR

• Similarly, P(EL|HR) = 0.05

• E(A|HR) =0.95x(-10)+0.05x(-110)

= -15• E(B|HR) =0.95x(-10)+0.05x(-110)

= -58.5

> 30 good news


Suppose the test indicate MRSuppose the test indicate MR

Test MR

A (small)

B (large)

EH 0.637

EL 0.363

EH 0.637

EL 0.363

-10

-110

-60

-30

637.0)(

)()|()|(

MRP

EHPEHMRPMREHP

• E(A|MR) = -46.3• E(B|MR) = -49.1

< -30


Suppose the test indicate LRSuppose the test indicate LR

143.0)|( LREHP

• E(A|LR) = -95.7• E(B|LR) = -34.3 < -30

Only if the test showed HR, saved money;

otherwise, more money with test


Should test be performed? Should test be performed? PrepostPreposterior analysiserior analysis

E(Test)

=0.59x(-15)+ 0.165 x(-46.3)+0.245 x(-34.3)

= -24.86

Better than -30 (without test)


Procedure for Preposterior AnalysisProcedure for Preposterior Analysis

• Determine updated probabilities using Bayes Theorem;

• Sub-tree analysis –Identify optimal alternative and maximum utility;

• Determine the best alternative in the next decision node (to the left);

• If Experimental alternative is optimal, wait for experimental outcome and select corresponding optimal alternative.


Procedure for Preposterior AnalysisProcedure for Preposterior Analysis

B

C

Subtree B

Subtree C

*Bu

*Cu


Value of Information (Value of Information (VIVI))

• VI = E(T) – E( )*a

EMV of test alternative excluding test cost

EMV of optimal alternative without Test

VI = (-24.86 + 10) – (- 30)

= 15.14

(max. paid for that specific Test)


Suppose someone comes up with a better Suppose someone comes up with a better test, say cost 25,000, but doesn’t know that test, say cost 25,000, but doesn’t know that exact reliability, should the test be exact reliability, should the test be performed?performed?


VPI = E(PT) - E( ) *a

P(EH0) = P(EH0|EH) P(EH) + P(EH0|EL) P(EL)

= 1 x 0.7 + 0 x 0.3 = 0.7

E(PT) = 0.7 x 0 + 0.3 x (-20) = -6

VPI = -6 – (-30) = 24

Max. that should be paid for any information


Sensitivity AnalysisSensitivity Analysis

• If the probability estimates are off by +10%, would the alternative previously chosen be still optimal?

Method 1: Repeat analysis with several

values of p

Method 2: Determine value of probability p

that decision is switched


A

B

EH p

EL 1-p

EH p

EL 1-p

0

-100

-50

-20

E(A) = p x 0 + (1-p) x (-100)

E(B) = p x (-50) + (1-p) x (-20)


Sensitivity of Decision to ProbabilitySensitivity of Decision to Probability

p<0.62E(B) >E(A)P>0.62E(B) <E(A)

E(PT)=px0+(1-p)(-20)= -20(1-p)

E(T)VPIVI


Levee Elevation DecisionLevee Elevation Decision

• Annual max. Flood Level: median 10, c.o.v. 20%

• Cost of construction: a1: $ 2 million

a2: $ 2.5 million

• Service Life: 20 years

• Average annual damage cost due to inadequate protection: $ 2 million


Levee Elevation DecisionLevee Elevation Decision• Annual max. Flood Level: median 10, c.o.v. 20%

H=10’

H=14’

H=16’

E(C)=10.594

2.731

2.641

2x10.594

pwf (20yrs, 7%)


Value of Perfect InformationValue of Perfect Information

• E(CPI)

= 0.5x2.699

+0.5x2.482

= 2.59

• VPI

= 2.614–2.59

= $ 0.024 M

Max. Amount to be paid for verifying type of distribution of annual flood level


Consider a GameConsider a Game

• E(A) = 0.5 x 0 + 0.5 x 10¢ = 5 ¢• E(B) = 1.0 x 5 ¢ = 5 ¢

A

B

0.5

0.5

1.0

0

10 ¢

5 ¢

0

$1

$0.5

0

$100

$ 50

0

$100M

$ 50M

EMV criteria may not be applicableWe need something else!


EUV criteriaEUV criteria

• Expected Utility Value

• Definition: EUV is the true value to a decision maker with which he/she can make a proper decision based on the relative utility value.


Utility function of monetary valueUtility function of monetary value

Risk Indifferent

dollars

u(d) Risk aversive


Maximum Expected Utility Criterion Maximum Expected Utility Criterion (EUV)(EUV)

If all consequences expressed in monetary terms:

jijij

iopt

jijiji

upUE

upUE

max)(

)(

j

ijiji dupUE )()(


ExampleExample

• E( )= 0.1u (-2000)+0.9 u(-2000) = = -1.49• E( )= 0.1u (-10000)+0.9 u(0) =0.1x( ) +0.9x( ) = -1.64

*I

I

0.1 F

0.9 F

0.1 F

0.1 F

-2000

-2000

-10000

0

0

dollarsu(d)

-1

0

)( 5000/

d

edu d

IU)5000/2000( e

5000/10000 e 0eIU

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Decision Analysis