Emanuele Borgonovo
Market
Structural
Decision
Return
Quantitative Methods for Management
First Edition
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Chapter three:
Models
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Models
• A Model is a mailmatical-logical instrument that the
analyst, the manager, the scientist, the engineer
develops to:
– foretell the behaviour of a system
– foresee the course of a market
– evaluate an investment decision accounting for
uncertainty factors
• Common Elements to the Models:
– Uncertainty
– Assumptions
– Inputs
• Model Results
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Building a Model
• To build a reliable model requires deep
acquaintance of:
– the Problem
– Important Events regarding the problem
– Factors that influence the behavior of the
quantities of interest
– Data and Information Collection
– Uncertainty Analysis
– Verification of the coherence of the Model by
means of empiric analysis and , if possible,
analysis of Sensitivity Analysis
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Example: the law of gravity
• We want to describe the vertical fall of a body on
the surface of the earth. We adopt the Model:
F=mg
for the fall of the bodies
• Hypothesis (?):
–
–
–
–
Punctiform Body (no spins)
No frictions
No atmospheric currents
Does the model work for the fall of a body placed to great
distance from the land surface?
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Chapter II
Introductory Elements of Probability
theory
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Probability
• Is it Possible to Define Probability?
• Yes, but there are two schools
• the first considers Probability as a property of
events
• the second school asserts that Probability is a
subjective measure of event likelihood (De
Finetti)
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Kolmogorov Axioms
U
B
A
P(U)  1
P( A )  0
If A e B mutually esclusive events ,
P( A  B)  P( A )  P(B)
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Areas and rectangles?
U
A B
C
D
and
U  A B  C D E
• Suppose one jumps into the area U randomly. Let P(A)
be the Probability to jump into A. What is its value?
• It will be the area of A divided by the area of U: P(A)=A/U
• Note that in this case: P(U)=P(A)+ P(B)+ P(C)+ P(D)+
P(E), since there are no overlaps
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Conditional Probability
• Consider events A and B. the conditional Probability of A
given B, is the Probability of A given the B has happened.
One writes: P(A|B)
U
B
AB A
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Conditional Probability
• Suppose now that B has happened, i.e., you jumped into
area B (and you cannot jump back!).
B
AB
A
•You cannot but agree that:
•P(A|B)=P(AB)/P(B)
•Hence: P(AB)=P(A|B) *P(B)
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Independence
• Two events, A and B, are independent if given
that A happens does not influence the fact
that B happens and vice versa.
B
B
A
AB A
Thus, for independent events:
P(AB)=P(A)*P(B)
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Probability and Information
• Problem: you are given a box containing two rings.
the box content is such that with the same
Probability (1/2) the box contains two golden rings
(event A) or a golden ring and a silver one (event B).
To let you know the box content, you are allowed to
pick one ring from the box. Suppose it is a golden
one.
– In your opinion, did you gain information from the draw?
– the Probability that the oil one is golden is 50%?
– Would you pay anything to have the possibility to draw
from the box?
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In the subjectivist approach,
Probability changes with information
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Bayes’ theorem
• Hypothesis: A and B are two events. A has
happened.
• Thesis: P(B) changes as follows:
P(B) before A
P(B A ) 
Probability of A given B
P(B)  P( A B)
P( A )
New value of the Probability of B
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Let us come back to the ring problem
• Events:
• A: both rings are golden
• o: the picked up ring is golden
• the theorem states: P( A o) 
P( A )  P(o A )
P(o)
• P(A)=Probability of both rings being golden before the extraction
=1/2
• P(o)=Probability of a golden ring=3/4
• P(o|A)=Probability that the extracted ring is golden given A=1
(since both rings are golden)
• So:
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1/ 2  1
P( A o) 
 2/3
3/4
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Bayes’ theorem Proof
Starting point
P( AB)  P( AB)
Conditional Probability formula
P( A B)  P(B)  P(B A )  P( A )
thesis
P( A B) 
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P(B A )  P( A )
P(B)
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the Total Probability theorem
D
B
Uand
A
C
1  P( A )  P(B)  P(C)
 P(D)
• the total Probability theorem states: given N mutually exclusive
and exhaustive events A1, A2,…,AN, the Probability of an event
and in U can be decomposed in:
P(E)  P(E A1 )  P( A1 )  P(E A 2 )  P( A 2 )  ...  P(E AN )  P( AN )
• Bayes theorem in the presence of N events becomes :
P( A 1 E) 
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P(E A 1 )  P( A 1 )
N
 P(E A )  P( A )
i1
i
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i
Continuous Random Variables
• Till now we have discussed individual events. there
are problems in which the event space is
continuous. For example, think of the failure time of
a component or the time interval between two
earthquakes. the random variable time ranges from
0 to +.
• To characterize such events one resorts to
Probability distributions.
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Probability Density Function
• f(x) is a Probability density function (pdf) if:
– It is integrable
– And
– the integral of f(x) over -:+  is equal to 1.

 f ( x )dx  1

• Note: f(x0)dx is the Probability that x lies in an
interval dx around x0.
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Cumulative Distribution Function
• Given a continuous random variable X, the
Probability that X<x is given by:
x
P(X  x )   f ( t )dt

dP
• If f(x) is continuous, then: f ( x ) 
dx
X2
• Note: P( X1  x  X2 )   f ( x )dx  P( x  X2 )  P( x  X1 )
X1
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the exponential distribution
• Consider events that happen continuously in time,
and with continuous time T.
• If the events are:
– Independents
– With constant failure rates
• the random variable T is characterized by an
exponential distribution:
P(T  t )  1  e
• and by the density function:
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 λt
f ( t )  e
 t
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Meaning of the Exponential Distribution
• We are dealing with a reliability problem, and we must characterize the
failure time, T. T is a random variable: one does not know when a
component is going to break. All one can say is that for sure the component
will break between 0 and infinity. Thus, T is a continuous random variable.
• Let us consider that failures are independent. This is the case if the failure
of one component does not influence the failure of the other components.
• Let us also consider constant failure rates. This is the case when repair
brings the component as good as new and when the component does not
age during its life.
• Under these Hypothesis, the failure times are independent and
characterized by a constant failure rate  at every dt. What is the
Probability distribution of T?
• Let us consider a population of N(t) components at time t. If  is the failure
rate of a component, then N(t)dt is the number of failues in dt around time
t.
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the Exponential Distribution
• Thus the change in the population is:
• -N(t)dt=N(t+dt)-N(t)=dN(t)
• Where the minus sign indicates that the number of
working components has decreased.
T
T
• Hence: dN(t )
dN( t )
N( t )
 dt  
• Which solved leads:
0
N( t )
   dt  lnN(T) / N(0)  T
0
N(T )
 T
e
N(0)
• N(T) is the number of components surviving till T.
N(0) is the initial number of components. Set N(0)=1.
then N(T)/N(0) is the Probability that a component
survives till T.
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Pdf and Cdf of the Exponential Distribution
1
0.9
0.8
0.7
P(t<T)
0.6
0.5
0.4
0.3
f(t)
0.2
0.1
0
0
5
10
15
T/t
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Expected Value, Variance and Percentiles

Expected Value : Ex    xf ( x )dx



Variance : Vx   E ( x  Ex ) 2 

 
  ( x  Ex ) f ( x )dx  E x  Ex 
2
2
2

S tan dard Deviation : Vx 
Percentile p: is the value xp of X such that the
Probability of X being lower than xp is equal to p/100
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the Normal Distribution
• Is a symmetric distribution around the mean
• Pdf:
1
fG ( x ) 
e
 2
1 x  2
 (
)
2 
   X  
• Cdf:

1
PG ( x  X)  
e
   2
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1 x  2
 (
)
2 
dx
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Graphs of the Normal Distribution
Distribuzione Normale Standard
3000
2500
fG ( x)
f(x)
2000
1500
1000
500
0
-4
-3
-2
-1
0
x
1
2
3
4
Cumulative Gaussian Distribution
10000
9000
8000
PG ( x  X)
7000
6000
5000
4000
3000
2000
1000
0
-5
-4
-3
-2
-1
0
1
2
3
4
x
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Lognormal Distribution
• Pdf
1
fL ( x ) 
e
x 2
1 ln x  2
(
)
2

0  x  
• Cdf
X
1
e
0 x 2
PL ( x  X)  
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1 ln x   2
(
)
2

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Lognormal Distribution
.20
fL ( x)
f ( x)
0.1
0
0
0
20
0.07
1
PL ( x  X)
x
50
1
f2( x) 0.5
0
0
0
0.07
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20
40
x
50
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Problem II-1 and solution
• the failure rate of a car gear is 1/5 for year
(exponential events).
• What is the mean time to failure of the gear?
 t  e
 t
dt  1/   5
• What is the Probability of the gear being integer after
9 years?
P( t  9)  1  P( t  9)  1  (1  e
 T
)
e (1/ 5 )9  16.5%
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Problem II-2
• You are considering a University admission test for a
particularly selective course. the admission test, as all tests
test, is not perfect. Suppose that the true distribution of the
class is such that 10% of the applicants are really qualified
and 90% are not. then you perform the test. If a student is
qualified, then the test will admit him/her with 90%
Probability. If the student is not qualified he/her gets
admitted at 10%. Now, let us consider a student that got
admitted:
– What is the Probability that the student is really qualified?
– Is it a good test? How would you use it?
– (Hint: use the theorem of Total Probability)
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Problem II-3
• For the example of the two rings, determine:
– P(B|o)
– P(B|a)
– the Probability of being in A given that the picked
ring is golden in two consecutive extractions,
having put the ring back in the box after the first
extraction
– the Probability of being in B given that the picked
ring is golden in two consecutive extractions,
having put the ring back in the box after the first
extraction
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Problem II-3
•
For the example of the two rings, determine:
– P(B|o)
• Solution: there are only two possible events, A or B. Thus, P(Bor)=1P(Aor)=1/3
– P(Ba)
• P(Ba)=1, since B is the only event that has a silver ring. One can also
show it using Bayes’ theorem:
• P(Ba)=P(aB)*P(B)/[P(aB)* P(B)+P(aA)*P(A)]. Since P(aA)=0,
one gets 1 at once.
– the Probability of being in A given that the picked ring is golden in
two consecutive extractions, having put the ring back in the box
after the first extraction
• Using Bayes’ theorem:
P( A 2o) 
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P(2o A )  P1( A )
P(2o A )  P1( A )  P(2o B)  P1(B)
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Problem II-3
• where, in the formula, subscript 1 indicates the probabilities after the
information of the first extraction has been taken into account:
– P1(B)=P(Bor)=1/3 and P1(A)=P(Aor)=2/3.
– One can note that P(2oA)=1, and P(2oB)=1/2. P(2oB) is the Probability
to pick a golden ring at the second run, given that one is in state B.
– Thus, we have all the numbers to be substituted back in the theorem:
P( A 2o) 
P(2o A )  P1( A )
P(2o A )  P1( A )  P(2o B)  P1(B)

1* 2 / 3
 0 .8
1 * 2 / 3  1/ 2 * 1 / 3
– It is the same problem as in the example, but with adjourned probabilities.
• the Probability of being in B given that the picked ring is golden in two
consecutive extractions, having put the ring back in the box after the first
extraction
– Solution: 1-P(A2o)=0.2
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Chapter III:
Introductory Decision theory
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An Investment Decision
• At time T, you have to decide whether, and
how, to invest $1000. You face three mutually
exclusive options:
– (1) A risky investment that gives you $500 PV in
one year if the market is up or a loss of $400 if the
market is down
– (2) A less risky investment that gives you $200 in
one year or a loss of $160
– (3) the safe investment: a bond that gives you $20
in one year independently of the market
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Decision theory According to Laplace
• “the theory leaves nothing arbitrary in choosing
options or in making decisions and we can
always select, with the help of the theory , the
most advantageous choice on our own. It is a
refreshing supplement to the ignorance and
feebleness of the human mind”.
• Pierre-Simon Laplace
• (March 28 1749 Beaumont-en-Auge - March 5 1827 Paris)
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Decision-Making Process Steps
Problem identification
Alternatives identification
Model implementation
Alternatives evaluation
Sensitivity Analysis
Yes
Further Analysis?
No
Best Alternatives implementation
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Decision-Making Problem Elements
• Values and Objectives
• Attributes
• Decision Alternatives
• Uncertain Events
• Consequences
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Decision Problem Elements
• Objectives:
– Maximize profit
• Attributes:
– Money
• Alternatives:
– Risky
– Less Risky
– Safe
• Random events:
– the Market
• Consequences:
– Profit or Loss
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Decision Analysis Tools
• Influence Diagrams
• Decision Trees
Market up
Less Risky
Market
How should the
invest $1000? Risky
Structural
Decision
Return
Safe
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prob_up
Market down
1-prob_up
Market up
prob_up
Market down
1-prob_up
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Influence Diagrams
• Influence diagrams (IDs) are…
“a graphical representation of decisions and
uncertain quantities that explicitly reveals
probabilistic dependence and the flow of
information”
• ID formal definition:
– ID = a network consisting of a directed graph
G=(N,A) and associated node sets and functions
(Schachter, 1986)
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ID Elements
NODES
ARCS
• Informational Arcs
= Decision
• probabilistic Dependency
Arcs
= Random Event
• Structural Arcs
= utility
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ID Elements
Informational Arc
Decision Node
Decision Node
Sequential
Decisions
Structural
Chance Node
Conditional Arc
Chance Node
Value Node
probabilistic
Dependency
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Influence Diagram Levels
1. Physical Phenomena and
Dependencies
2. “Function level”: node output states
probabilistic relations (models)
3. “Number level”: tables of node
probabilities
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Case Study 2 - Leaking SG tube
• Influence Diagram for Case Study 2
Leakage
Rate
shutdown_cost
Leakage from primary
to secondary, maximum
rate of 20 l/hr
time_to_repair
Primary
Decisions
I - Normal Makeup
II - Shutdown
III - Reduce Power
IV - Isolate SG
Cooling
Chemical Volume
Control System
Value
Secondary
Cooling
days_to_shutdown
Deterministic
Information
core_damage_cost
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Influence Diagram
Market
Structural
Decision
Return
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Decision Trees
• Decision Trees (DTs) are constituted by the
same type of arcs of Influence Diagrams, but
highlight all the possible event combinations.
• Instead of arks, one finds branches that
emanate from the nodes as many as the
Alternatives or Outcomes of each node.
• With respect to Influence Diagrams, Decision
Trees have the advantage of showing all
possible patterns, but their structure becomes
quite complicated at the growing of the problem
complexity.
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the Decision Tree (DT)
Market up
Less Risky
Market down
How should the
invest $1000?
1-prob_up
Risky
Market up
Market down
Safe
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Decision Tree Solution
• Alternative Payoff or utility:
E[Ui ]   Pi (C j )  UC j
j
• j=1…mi spans all the Consequences
associated to alternative
the
• Uj is the utility or the payoff of consequence j
• Pi(Cj) is the Probability that consequence Cj happens given that
one chose alternative the
• In general, we will get: P(Cj) =P(E1E2… EN), where E1E2… EN
are the events that have to happen so that consequence Cj is
realized. Using conditional probabilities:
• P(Cj) =P(E1E2… EN)=P(EN| E1E2… )*…*P(E2| E1)*P(E1)
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example
Market up
Blue Chip Stock
How should the
invest $1000?
Risky investment
P.up
Market down
C1
C2
1-P.up
Market up
P.up
Market down
1-P.up
C3
C4
CD paying 5%
C5
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Problem Solution
• Using the previous formula:
2
E[URisky ]   P(C j )  C j  P.up  C1  (1  P.up)  C2
j1
2
E[ULessRisky ]   P(C j )  C j  P.up  C3  (1  P.up)  C4
j1
2
E[ULessRisky ]   P(C j )  C j  1 C5
j1
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the Best Investment for a Risk Neutral Decision - Maker
Market up
Blue Chip Stock
0.600
$56
Market down
$200
($160)
0.400
Market up
How should the
invest $1000?
Risky investment
$500; P = 0.600
$60 0.600
Market down
($600); P = 0.400
0.400
CD paying 5%
return = $50
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Run or Withdraw?
You are the owner of a racing team. It is the last race of the season, and it has been a
very good season for you. Your old sponsor will remain with you for the next season
offering an amount of $50000, no matter what happens in the last race. However, the
race is important and transmitted on television. If you win or end the race in the first five
positions, you will gain a new sponsor who is offering you $100000, besides $10000 or
$5000 praise. However there are unfavorable running conditions and an engine failure
is likely, based on your previous data.
It would be very bad for the image of you racing team to have an engine failure in such a
public race. You estimate the damage to a total of -$30000.
What to do? Run or withdraw?
• A) Elements of the problem:
–
–
–
–
–
What are your objectives
What are the decision alternatives
What are the attributes of the decision
What are the uncertain events
What are the alternatives
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Example of a simple ID
Decision
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Engine failure
Final Classification
Profit
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From IDs to Decision Trees
Engine failure
failure
Decision
Out of first five
$20,000
0.500
Run
$20,000; P = 0.500
1.000
Win
$57,250
$110,000; P = 0.250
0.500
No failure
Decision
0.500
Run : $57,250
pfailure=0.5
In first five
$94,500
0.300
Out of first five
$50,000; P = 0.100
pfive=0.30
0.200
pout=0.2
pwin=0.5
$105,000; P = 0.150
Withdraw
Engine_failure=0
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Old sponsor
$50,000
$50,000
1.000
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Sequential Decisions
• Are decision making problems in which more than
one decisions are evaluated one after the other.
• You are evaluating the purchase of a production machine. Three
models are being judged, A B and C. the machine costs are 150,
175 and 200 respectively. If you buy model A, you can choose
insurance A1, that covers all possible failues of A, and costs 5%
of A cost, or you can choose insurance policy A2, that costs 3% of
A cost, but covers only transportation risk. If you buy model B,
insurance policy B1 costs 3% of B cost and covers all B failures.
Insurance B2 costs 2% of B and covers only transportation. For
model C, the most reliable, the insurance coverages cost 2% and
1.5% respectively. Based on this information and supposing that
the machines production is the same, what will you choose?
• (failure Probability of A in the period of interest=5%)
• (failure Probability of B in the period of interest=3%)
• (failure Probability of C in the period of interest=2%
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Influence Diagram
Decision
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Assicurazione
Ruttura
Costo
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Decision Tree
Assicurazione
Decision
1
-150-5%*(150) = (£158); P = 1.000
A
1 : (£158)
2
Sì
0.050
-150-2%*150-150 = (£303)
(£161)
No
0.950
-150*(1+2%) = (£153)
1
-(175+3%*(175)) = (£180)
B
A : (£158)
pA=0.05
pB=0.03
pC=0.02
1 : (£180)
Sì
2
0.030
-175-2%*175-175 = (£354)
(£184)
No
0.970
-175-2%*175 = (£179)
1
-200-2%*200 = (£204)
C
1 : (£204)
2
Sì
0.020
-200-1.5%*200-200 = (£403)
(£207)
No
0.980
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-200-200*1.5% = (£203)
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the Expected Value of Perfect Information
• Data and information collection is essential to make
decisions. Sometimes firms hire consultants or experts to get
such information. But, how much should one spend?
• One can value information, since it is capable of helping the
decision-maker in selecting among alternatives
• the value of information is the added value of the information.
• the expected value of perfect information (EVPI) assumed
that the source of information is perfect, and then:
EVPI  E[Knowing ]  E[BeforeKnow ing]
• the definition is read as follows: how much is the decision
worth with the new information and without
• N.B.: we will refer only to aleatory uncertainty
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Example: investing
Decision
Value
Market
Market
Decision
Up
RISKY
0.500
£500; P = 0.500
£50
Down
0.500
(£400); P = 0.500
Up
LESS: £50
RISKY
RISKY
P _UP =0.5
0.500
£200
£20
Down
0.500
(£160)
SAFE
£20
Market=0
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EVPI for the Example
Market
Value
Decision
Decision
Market
Up
0.500
RISKY
£500; P = 0.500
LESS RISKY
RISKY : £500
£200
SAFE
£20
£260
RISKY
P _UP =0.5
(£400)
Down
0.500
LESS RISKY
SAFE : £20
(£160)
SAFE
£20; P = 0.500
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EVPI Result
EVPI  E[Knowing ]  E[r ] 
 260  50 
 210
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Problems
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How much to bid?
• Bob works for an energy production company. Your company
is engaged in the decision of how much to bid to salvage the
wreckage of the SS.Kuniang, a carbon transportation boat. If
the firm wins, the boat could be repaired and could come
back to its transportation activity again. Pending on the
possible winning and on the decision is the result of a
judgment by Coast Guard, which will be revealed only after
the opening of the bids. That is, if the Coast Guard will
assign a low value to the ship, this would mean that the ship
is considered as recoverable. Otherwise, the boat will be
deemed unusable. If you do not win, you will be forced to
buy a new boat.
• Identify the decision elements
• Structure the corresponding ID and DT
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Influence Diagram with three events
• Given the following elements:
– Alternatives 1 and 2
– Events: A=(up, down); (B=high, low);(C=good, bad);
– Consequences Ci (one distinct consequence for each
event combination)
– If A=Down happens, then CAdown is directly realized
• Draw the ID corresponding to the problem
• Draw the corresponding Decision Tree
• If C now depends on both A and B outcomes, how does the
ID become?
• How does the DT change?
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Solution
• Influence Diagram the
Skip Arc
A
C
Decision
Consequences
B
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Solution
• Corresponding Decision Tree
C
B
A
good
(No Payoff)
high
bad
up
Decision
(No Payoff)
good
(No Payoff)
low
1
bad
(No Payoff)
down
(No Payoff)
B=0
C=0
good
high
(No Payoff)
bad
(No Payoff)
up
good
(No Payoff)
low
2
bad
(No Payoff)
down
(No Payoff)
B=0
C=0
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Solution
• Influence Diagram II
A
C
Decision
Consequences
B
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Solution
• Decision Tree II:
C
B
A
good
(No Payoff)
high
bad
up
(No Payoff)
good
Decision
(No Payoff)
low
1
bad
(No Payoff)
good
(No Payoff)
down
bad
B=0
(No Payoff)
good
(No Payoff)
high
bad
(No Payoff)
up
good
(No Payoff)
low
2
bad
(No Payoff)
good
(No Payoff)
down
B=0
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bad
(No Payoff)
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Sales_Costs
•
Given the following Influence Diagram and Decision Tree, given P_High and
P_High|High, P_high|low, find the value of the Alternatives as a function of the
assigned probabilities. Supposing P_high=0.5 and P_high|high=P_high|low=0.3,
find the preferred alternative.
Sales
Cost
Vendite
Payoff
Decisione
high
Decision
P_high
Costo
Invest
High
P_Alte|high
Low
1- P_Alte|high
High
Basso
high=0.5
P_Alte=0.3
P_high=0.5
•
•
1-P_high
-10
20
P_ P_Alte|high
Low
1- P_Alte|high
Do not Invest
0
0
5
What would be the preferred decision if to a higher investment cost there would
correspond a better sale result? Set:
P_high|high=0.6 and P_high|low=0.2
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Solution Sales_Costs
Vendite
Costo
Decisione
Alte
Alto
0.500
0.300
(£7)
Basse
Investo
0.700
(£1)
alto=0.5
P _Alte=0.3
P _alto=0.5
(£10)
Alte
Basso
Non-Investo : £5
£0
0.500
0.300
£6
£20
Basse
0.700
£0
Non-Investo
£5; P = 1.000
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Breakdown in Production
• An industrial system composed from two lines has experience a breakdown
in one line. Production, therefore, is reduced by 50%. the management
asks you collaboration on the following decision. It is explained to you that
there are two ways to proceed: 1) an intermediate repair, of the duration of
two days, with a repair cost of EUR500000. For every day of production
loss of EUR25000 for day is sustained (Full production amounts at
EUR50000). From the engineer estimates, the Probability of perfect repair
in two days is equal to P_2g. In the case in which the repair it is not perfect
(partial repair), the line will come back with a loss of 15% of the productive
ability; 2) a more incisive intervention, of the duration of 10 days, with a
cost of repair of EUR1000000. With Probability P_10g the line will be as
before the breakdown.
–
–
–
–
According to you, the residual life of the system is important for the decision?
Suppose that there are still three years of life for the system.
Which strategy should you carry out?
Determine the decision problem elements. Draw the Influence Diagram and the
corresponding Decision Tree. Find the value or values of the probabilities for
which a complete repair is more convenient than a partial one.
– What would you would advise to the director of the system to do based on the
engineer estimates?
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EVPI Problems
• Determine the EVPI for the random event
nodes in the previous IDs and DTs of the
following problems:
• Sales_Costs (lez. 2)
• Production break-down (lez.2)
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Troubles in Production
•
•
•
One of the two production lines of the plant you manage has broke down. the plant
production capacity is therefore halved. the management faces the following
decision and asks you a collaboration. Technically one can a: 1) perform an
temporary repair, lasting two days, and costing €500000. For every lost production
day one has a revenue loss of €25000 for day (the total daily production value is
€50000). Based on the Engineer estimates, the Probability of perfect repair in two
days is P_2g . In the case of an imperfect repair, the production capacity will be
lowered by 15%. 2) perform a more incisive repair, lasting 10 days, and costing
€1000000. With Probability P_10g the line will be as good as new.
In your opinion, the residual plant life is relevant to this decision?
Suppose that there are still three years of life for the plant. What should one
decide?
– Identify the decision making elements
– Draw the Influence Diagram for the problem
– Find the values of the probabilities for which one or the other intervention is
more convenient
– What would your suggestion to the plant director be?
– What would happen if the plant life were 2 and 4 years instead of 3?
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Influence Diagram
Riparazione_10g_Perfetta
Perdite
Decisione
Riparazione_2g_perfetta
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Decision Tree
Riparazione_2g_perfetta
Decisione
Riparazione_2g_perfetta
-50000-500000
Intervento_2g
P_2g
Riparazione_2g_non_perfetta
Riparazione_10g_Perfetta=0
-50000-500000-25000*0.15*365*years
1-P_2g
Riparazione_10g_Perfetta
P_10g=0.9
P_2g=0.3
years=3
Riparazione_10g_Perfetta
-250000-1000000
Intervento_10g
Riparazione_2g_perfetta=0
P_10g
Riparazione_10g_non_Perfetta
-1000000-250000-years*.05*25000*365
Riparazione_2g_perfetta=0
1-P_10g
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Probability Values
• Three years
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2 and 4 years
• 2 years
• 4 years
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Chapter IV
Elements of Sensitivity Analysis
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Sensitivity Analysis
• Various Types of SA
– One Way SA
– Two Way SA
– Tornado Diagrams
– (Differential Importance Measure)
• Uncertainty Analysis
– Monte Carlo
– (Global SA)
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How do we use SA?
• a) To check model correctness and
robustness
• b) To Further interrogate the model
– Questions:
• What is the most influential parameter with respect to
changes?
• What is the most influential parameter on the uncertainty
(data collection)
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Sensitivity Analysis (Run or withdraw)
• Underline the critical dependencies of the
outcome
Tornado Diagram at
Decision
Sensitivity Analysis on
pfailure
$62K
pwin: 0.3 to 0.7
$59K
pfive: 0.2 to 0.4
$56K
Expected Value
pfailure: 0.25 to 0.75
Run
Withdraw
Threshold Values:
pfailure = 0.597
EV = $50K
$53K
$50K
$47K
$44K
$41K
$49K
$55K
$61K
$67K
Expected Value
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$73K
$38K
0.450
0.525
0.600
0.675
0.750
pfailure
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Summary
• Sensitivity Analysis
– One way sensitivity
– Two way sensitivity
– Tornado Diagrams
• Uncertainty Analysis
– Aleatory Uncertainty
– Epistemic Uncertainty
– Bayes‘ theorem for continuous distributions
– Monte Carlo Method
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Sensitivity Analysis
• By sensitivity analysis one means the study of the
change in results (output) due to a change in one of
the model parameters (input)
• the simplest Sensitivity Analysis types are:
– One way sensitivity
– Two way sensitivity
– Tornado diagrams
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One-way Sensitivity Analysis
• A one way sensitivity is obtained changing the Model input
variables one at a time, and registering the change in the
decision value.
• It enables the analyst to study the change in value of each of
the alternatives with respect to the change in the input
parameter under consideration
Sensitivity Analysis on
pfailure
$62K
Run
Expected Value
$59K
Withdraw
$56K
Threshold Values:
pfailure = 0.597
EV = $50K
$53K
$50K
$47K
$44K
$41K
$38K
0.450
0.525
0.600
0.675
0.750
pfailure
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Two-way Sensitivity Analysis
• In a Two-way Sensitivity Analysis two parameters are varied at the
same time.
• Instead of a line, one obtains a plane, in which each region identifies
the preferred alternative that correspond to the combination of the two
parameter values
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Tornado Diagrams
• the analysis is focused on the preferred decision
• An interval of variation for each input parameter is
chosen
• the parameters are changed one at a time, while
keeping the oilrs at their reference value
• the change in output is registered
• the output change is shown by means of a horizontal
bar
• the most important variable is the one that
corresponds to the longest bar.
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Example of a Tornado Diagram
Tornado Diagram at
Decision
pfailure: 0.25 to 0.75
pwin: 0.3 to 0.7
pfive: 0.2 to 0.4
$49K
$55K
$61K
$67K
$73K
Expected Value
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Upsides and Downsides
• Upsides
– Easy numerical
calculations
– Results immediately
understandable
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• Downsides
– Input range of variation
not considered together
with the output range:
should not be used to
infer parameter
importance
– One or two parameters
can be varied at the
same time
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Sensitivity Analysis and Parameter Importance
• Parameter importance:
– Relevance of parameter in a model with respect to a certain
criterion
• Sensitivity Analysis used to Determine
Parameter Importance
• Concept of importance not formalized, but
extensively used
– Risk-Informed Decision Making
– Resource allocation
• Need for a formal definition
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Process
• Identify how sensitivity analysis techniques
work through analysis of several examples
• Formulate a definition
• Classify sensitivity analysis techniques
accordingly
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Sensitivity Analysis Types
• Model Output:
U  f(x1, x2,..., xn )
• Local Sensitivity Analysis:
– Determines model parameter (xi) relevance with all
the xi fixed at nominal value
• Global Sensitivity Analysis:
– Determines xi relevance of xi’s epistemic/uncertainty
distribution
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the Differential Importance Measure
• Nominal Model output:
– No uncertainty in the model parameters
– and/or parameters fixed at nominal value
• Local Decomposition:
f
f
f
dU 
dx1 
dx 2  ... 
dx n
x1
x 2
x n
• Local importance measured by fraction of the
differential attributable to each parameter
DIM(x i ) 
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dUxi
dU
xo
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Global Sensitivity Indices
• Uncertainty in U and parameters is considered
• Sobol’’s decomposition theorem:
n
U  f( x)  f0   fi ( xi )  
i1
• Sobol’Indices
 f ( x , x )  ...  f
ij
1i jn
i
j
12...n
x i1
Si1...is (x i ) 
Di1...is
D
 ...  f
i1...in

dx i1 ...dx n
x i1
f
2
( x )dx  f0
2
Ω
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( x)
Formal Definition of Sensitivity Analysis (SA) Techniques
• SA technique are Operators on U:
x1
x2
xn
 or 
I(x)^ [U  f(x1, x2,..., xn )]
 or 
I(xn)
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I(x1)
I(x2)
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Importance Relations
• Importance relations:
– X the set of the model parameters;
–
Binary relation

xi  xj iff I(xi)I(xj)
xi~xj iff I(xi)I(xj)
xi  xj iff I(xi)I(xj)
xi  xj iff I(xi)I(xj)
• Importance relations induced by importance
measures are complete preorder
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Additivity Property
• In many situation decision-maker interested
in joint importance:
I( x i , x j )  I( x i  x j )
• An Importance measure is additive if:
I( x i , x j )  I( x i )  I( x j )
• DIM is additive always
• Si are additive iff f(x) additive and xj’s are
uncorrelated
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Techniques that fall under the definition of Local
SA techniques
IMPORTANCE
MEASURE
EQUATION TYPE ADDITIVE
DIM
dU xi
Local
Yes
Local
No
Local
No
Local
No
Local
No
Local
No
dU
L
Tornado Diagrams
One Way Sensitivity
Fussell-Vesely
U x i
 x0
x i U
U
U
U( x 0 )  x i
U( x 0 )
Risk Achievement Worth
U xi
x0
U0
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Global Importance Measures
IMPORTANCE
MEASURE
Sobol’ Indices
EQUATION
TYPE ADDITIVE
Di1...is
Global
No
Global
No
Global
No
Global
No
Global
No
Global
No
D
Extended Fast

Si 
2 A p2w i  Bp2w i
p 1

2 A 2j  B 2j
j1
Morris
Pearson
Smirnov
Standardized
regression coefficients
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d(x i ) 
i 
f ( x1,..., x i  ,..., x n )

cov(U, x i )
 i U
sup Y1( Xi )  Y2 ( Xi )
bk k

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Sensitivity Analysis in Risk-Informed Decision-Making
and Regulation
• Risk Metric: R  f(x , x ,..., x )
1
2
n
• xi is undesired event Probability
• Fussell-Vesely fractional Importance:
(R, x )

FV(x ) 
i
i
R
• Tells us on which events regulator has to
focus attention
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Summary of the previous concepts
• Formal Definition of Sensitivity Analysis
Techniques
• Definition of Importance Relations
• Definition enables to:
– Formalize use of Sensitivity Analysis
– Understand role of Sensitivity Analysis in Riskinformed Decision-making and in the use of model
information
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Chapter V
Uncertainty Analysis
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Uncertainty Analysis
Monte Carlo Simulation at
Decision
1.000
0.900
Probability
0.800
0.700
0.600
0.500
0.400
0.300
0.200
0.100
0.000
$10K
$40K
$70K
$100K
$130K
Value
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Summary
• Distinction between Aleatory Uncertainty ed
Epistemic Uncertainty
• Epistemic Uncertainty and Bayes‘ theorem
• Monte Carlo Method for uncertainty
propagation
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Uncertainty
• Aleatory Uncertainty:
– From “Alea”, die: “Alea jacta est”
It refers to the realization of an event.
– Example: the happening of an earthquake
• Epistemic Uncertainty:
– From GreeK “Eit”, knowledge
it reflects our lack of knowledge in the value of
the Aleatory Model input parameters. the
aleatory model or model of the world is the
model chosen to represent the random event.
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Example: Model of the World
• the Probability of Earthquakes is usually modeled through a Poisson
model:
n
P(n, t )  e
 t
( t )
n!
• that rappresents the Probability that the number of earthquakes between
0 and t is equal to n.
• the Poisson Distribution holds for independent events, in which next
events (arrivals) are not influenced by previous events and the
Probability of an event in a given interval of time is the same
independently of the time where the interval is located
• the Model chosen to describe the arrivals of earthquakes is given the
non-humble name of "model of the world" (MOW).
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Some useful information on Poisson Distributions
• the Poisson Probability that n events happen on 0-t
is:
e  t (t )n
P(n, t ) 
n!
• the sum on n=0... of P(n,t) is, obviously, equal to 1.

et (t )n
(t )n
t
e 
 et  et  1

n!
n 0
n0 n!

• the Probability of k>N is given by:
N
e t (t )n
et (t )n
 1 

n!
n!
nN1
n 0

• E[n]=t
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

ne t (t )n
(t )n
(t )n1
t
t
e 
 e  t 
 et  t  et

n!
n 0
n0 n  1!
n0 n  1!

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the Corresponding Epistemic Model
• Now,in spite of all the efforts and studies, it is unlikely that a scientist would
tell you: the rate ( ) of arrivals of earthquakes is exactly xxx. More likely, he
will indicate you a range where the “true value” of  lies. For example 
cuold be between 1/5 and 1/50 (1/years). Suppose that the scientist state of
knowledge on  can be expressed by a uniform distribution u( ):
Epistemic distribution for the frequency of earthquakes
8
u( )  0   1/ 50   1/ 5

u( )  1/ 5  1/ 50 1/ 50    1/ 5
7
6
f(lambda)
5
4
3
2
1
0
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0
0.02
0.04
0.06
0.08
0.1 0.12
lambda
0.14
0.16
0.18
0.2
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Combining the Epistemic Model and MOW
• We have been dealing with two Models:
• MOW: the events happen according to a Poisson Distribution
• Epistemic Model: Uniform Uncertainty Distribution
• then, what is the Probability of having 1 earthquake in the
next year?
• Answer: there is no unique Probability, but a p(n,t, ) for all
values of .
• Thus, we have to write:
e  t (t )n
p(n, t,  )d 
u( )d
n!
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….
• This expression tells us that not necessarily
all Poisson distributions weight the same.
Thus:

  t
e (t )n
P(n, t )   p(n, t,  )d  
u( )d
n!


• In our case: u()=c;

  λt


E[P(n, t )]   p(n, t , λ)dλ  c 
e
(λt ) n
dλ
n!
• Hence, there is an expected Probability!
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In General
• the MOW will depend on m parameters , ,…:
MOW (t ,,.... )
• the event Probability (P(t)) will be:

P( t )   MOW ( t , ,.... )f (, ,....) d d.....

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An problem
• the failure time of a series of components is
characterized by the exponential Probability
function :
 t
df  e dt
• From the available data, it emerges that:
  1/ 5 p  0.5

  1/ 8 p  0.3
  1/ 10 p  0.2

• What is the mean time to failure?
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Solution
• E[t]=



0
0
0
(  t  1e 1t dt )  0.5 (  t   2e 2t dt )  0.5 (  t   3e3 t dt )  0.1
 1/ 1  0.5  1/  2  0.3  1/ 3  0.1
 6.9
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Continuous form of Bayes Theorem
• Epistemic Uncertainty and Bayes’ theorem
are connected, in that we know that we can
use evidence to update probabilities.
• For example, suppose to have a coin in your
hands. will it be a fair with, i.e., will the
Probability of tossing the coin lead to 50%
head and tails?.
• How can we determine whether it is a fair
coin?
• ….let us toss it….
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Formula
• the Probability density of a parameter, after having obtained
evidence and, changes as follows:
π( λ ) 
L(E λ )  π 0 ( λ )

 L(E λ)  π
0
( λ )dλ

• L(E) = MOW likelihood
• 0() is the pdf of  before the evidence, called Prior
Distribution
• () is the pdf of  after the evidence, called Posterior
Distribution
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From discrete to continuous
• Let us take Bayes‘ theorem for discrete events:
P( A j E) 
P(E A j )  P( A j )
n
 P(E A )  P( A )
i
i
i1
• Let us go to continuous events:
our purpose is to know the
Probability that a parameter of the MOW distribution
assumes a certain value, given a certain evidence
• Thus, event Aj is:  takes on value *
• Hence: P(Aj)0()d 0()=prior density
• therefore: P(EAj) has the meaning of Probability that the
evidence and is realized given that  equals * . One writes:
L(E   ) and it is the likelihood function
• Note: it is the MOW!!!
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From discrete to continuous
• the denominator in Bayes’ theorem expresses the
sum of the probabilities of the evidence given all the
possible states (the total Probability theorem). In the
case of epistemic uncertainty these events are all
possible values of . Thus:

n
 P(E A )  P( A )   L(E ) ()d
i1
i
i
0

• Substituting the various terms, one finds Bayes‘
theorem for continuous random variables we have
shown before
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Is it a fair coin?
• What is the MOW?
• It is a binomial distribution with parameter p:
 n k
P(k,n  k )     p  (1  p)nk
k 
• What is the value of p?
• Suppse we do not know anything about p. Let us assume a uniform
prior distribution between 0 and 1:
π 0 (p)  1 if 0  p  1

π 0 (p)  0 otherwise
• Let us get some evidence.
• At the first tossing it is head
• At the second tail
• At the third head
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Result
•
•
•
First tossing
– Evidence: h.
– MOW: L(hp)=p
– Prior: 0
Second tossing:
– Evidence: t
– MOW: L(tp)=(1-p)
– Prior: 1
Third tossing:
– Evidence: h
– MOW: L(hp)=p
– Prior: 2
• Equivalently:
– Evidence: h, t, h
– L(hthp)=p2(1-p)
– Prior: 0
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π1 (p) 
L( h p)  π 0 ( p )

p 1

 2p
1
 L(h p)  π (p)dp  pdp
0

π 2 ( p) 
0
L( t p)  π1 (p)


p  (1  p)
1
 L(t p)  π (p)dp  (p  p
1

2

)dp
0
 6(p  p )
2
π 3 ( p) 
L( h p )  π 2 ( p )


 L(h p)  π 2 (p)dp

p 2  (1  p)
1

2
p
  (1  p)dp
0
 12(p  p )
2
π 3 ( p) 
3
L(hth p)  π 0 (p)


 L(hth p)  π 2 (p)dp

p 2  (1  p) 1
1
2
p
  (1  p) 1dp
0
 12(p 2  p3 )
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
Graph
2
1.8
3
1.6
2
1.4
1.2
1
1
0
0.8
0.6
0.4
0.2
0
0
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0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
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Conjugate Distributions
• Likelihood
• Prior distribution
– Poisson
– Gamma
 t
e ( t )
P(n, t ) 
n!
n
• Posterior: gamma
β α λα 1 βλ
π 0 (λ, α, β) 
e
Γ(α)
• with:
'' ' 1 '
(, ' ,' ) 
e
(' )
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'    r
'    t
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Conjugate Distributions
• Likelihood
• Prior distribution:
– Normal
1
fX ( x ) 
e
σ x 2π
– Normal
1 x μx 2
 (
)
2 σx
• Posterior: Normal
1
π 0 (m) 
e
σμ 2π
• with:
μ 
'
1
fG ( x ) 
e
σ' x 2π
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1 x μ'
 ( ' )2
2 σx
σ 'x 
1 mμx 2
 (
)
2 σμ
μ(σ x )2  nx(σμ0 )2
(σ x )2  n(σμ0 )2
(σ x / n)2 (σ μ )2
( σ μ ) 2  ( σ x )2 / n
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Conjugate Distributions
• Likelihood
– Binomial
n k
  p (1  p)nk
k 
• Posterior, Beta:
• Prior:
– Beta
π 0 (p)  p( q1) (1  p)r 1
• with:
q'  q  k
π1(p)  p( q' 1) (1  p)r ' 1
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r'  r  n  k
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Summary of Conjugate Distributions
MOW - Likelihood
Prior
Distribution
Posterior
Distribution
Binomiale
Beta
Beta
Poisson
Gamma
Gamma
Normal
Normal
Normal
Normal
Gamma
Gamma
Negative binominal
Beta
Beta
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Epistemic Uncertainty in Decision-Making Problems
• Investment:
2
E[URisky ]   P(C j )  C j  P.up  C1  (1  P.up)  C2
j1
2
E[ULessRisky ]   P(C j )  C j  1 C5
2 j1
E[ULessRisky ]   P(C j )  C j  P.up  C3  (1  P.up)  C4
j1
• Suppose that P.up is characterized by a uniform pdf between 0.3
and 0.7
2
E[URisky P.up ]   P(C j )  C j  P.up  U1  (1  P.up)  U2
j1
• How does the decision changes?
• It is necessary to propagate the uncertainty in the model
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Analytical Propagation of Uncertainty
• It is the same problem of the MOW …


E URisky   E[URisky P.up ]  f (P.up )  dP.up 
 (P.up  U
1
 (1  P.up)  U2 )f (P.up )dP.up
• Repeating for the other decisions and comparing
the resulting mean values, one gets the optimal
decision.

• Recall that: E[g( x1, x 2 ,..., x n )]   g( x )  f ( x )dx

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the Monte Carlo Method
• Sampling a value of P.up
• For all sampled P.up the Model is re-evaluated.
• Information:
– Frequency of the preferred alternative
– Distribution of each individual Alternative
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the core of Monte Carlo
• 1) Random Number Generator “u” between 0 and 1
0
1
u
• 2) Numbers u are generated with a uniform
Distribution
• 3) Suppose that parameter  is uncertain and
characterized by the cumulative distribution reported
below:
Distribuzione cumulativa esponenziale
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
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0
0.1
0.2
0.3
0.4
0.5
x
0.6
0.7
0.8
0.9
1
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Inversion theorem
1
Distribuzione cumulativa esponenziale
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
• Inversion theorem:
0
0
0.1
0.2
0.3
0.4
0.5
x
0.6
0.7
0.8
0.9
1
1
  F (u)
• the values of  sampled in this way have the
Probability distribution from which we have inverted
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Example
• Let us evaluate the volume of the yellow solid through the
Monte Carlo method.
V0
V
nin
V  lim
 V0
n N
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Application to ID and DT
• For every Model parameter one creates the
corresponding epistemic distribution
• Run nr. 1:
• One generates n random numbers between 0 and 1, as many as
the uncertain variables are
• One samples the value of each of the parameters inverting from
the corresponding distribution
• Using these values one evaluates the model
• One keeps record of the preferred alternative and of the value of
the decision
• the procedure is repeated N times.
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Results
• Strategy Selection
Frequency
• Decision Value
Distribution
Monte Carlo Simulation at
Decision
1.000
0.900
Probability
0.800
0.700
0.600
0.500
0.400
0.300
0.200
0.100
0.000
$10K
$40K
$70K
$100K
$130K
Value
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Problem V-1
• the mean time to failure of a set of components is
characterized by an exponential distribution with parameter
. Suppose that  is described by a uniform epistemic
distribution between 1/100 and 1/10.
– Which is the MOW? Which is the epistemic model?
– What is the mean time to failure?
• Suppose you registered the following failure times: t=15,
22, 25.
– Update the epistemic distribution based on the new data
– What is the new mean time to failure?
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Problem V-2: Investing
•
•
•
•
•
•
We are again thinking of how to invest. Actually, we were not aware of the bayesian
approach before. Thus we start using data about P_up in Bayesian way. After 15
working days we get the evidence: up,down,
down,down,down,up,down,up,down,up,down,up,up,up. Assuming that each day is
independent of the previous one:
a) Which are the MOW and the epistemic model?
b) What is the best decision without incorporating the evidence?
c) What is the distribution of P_up after the evidence?
d) What do you decide when the new information is incorporated in the model?
Solution:
–
–
•
the MOW is the model of the events that accompany the decision. It is our ID or DT. More in
specific, there is a second mode which is the one utilized for modeling the fact that the market can
be up or down. This is a binomial distribution with parameter P_up.
the epistemic model is the set of the uncertainty distributions used to characterize the lack of
knowledge in the model parameters. In this case, it is the distribution of P_up. We need to choose
a prior distribution for P_up. We choose a uniform distribution between 0 and 1.
b) We write the alternative payoffs as a function of P_up.
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Prob. 5-2
E[ULessRisky P _ up ]  C5
2
E[URisky P_up ]   P(C j )  C j  P_up  C1  (1  P_up)  C2
j1
2
E[ULessRisky P _ up ]   P(C j )  C j  P_up  C3  (1  P_up)  C4
j1
2
E[ U LessRisky P _ up ]   P(C j )  C j  1 C5
j1
•
Substituting: E[URisky]=50, E[USafe ]= 20, E[ULess Risky ]= 20
E[ULessRisky ]   E[ULessRisky P_up ]f (P _ up )dP _ up  EP_up (C3  C4 )  0.5(C1  C2 )
1
0
1
1
0
0
E[URisky ]   E[URisky P_up ]f (P _ up )dP _ up   P_up  C1  (1  P_up)  C2   f (P _ up ) dP _ up 
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 EP_up (C1  C2 )  0.5(C1  C2 )
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Investment
•
c) Let us use Bayes’s theorem to update the prior uniform distribution
–
–
–
•
evidence: up,down, down,down,down,up,down,up,down,up,down,up,up,up
L(E|P_up):
15!
L(E | P_up) 
(P _ up )7 (1  P _ up )8
7!8!
Prior: 0 uniform bewteen 0 and 1
Bayes’theorem:
L(E | P_up) 
15!
(P _ up )7 (1  P _ up )8  1
7!8!
1
15!
7
8
0 7!8! (P _ up ) (1  P _ up )  1 dP _ up

(P _ up )7 (1  P _ up )8
1
 (P _ up ) (1  P _ up )
7
8
 dP _ up
0
3.5
3
2.5
•
Posterior Distribution
2
p0
1.5
p1
1
0.5
0.98
0.91
0.84
0.7
0.77
0.63
0.56
0.49
0.42
0.35
0.28
0.21
0.14
0
•
•
0.07
0
E[p_up]=0.47
d) Posterior Decision: E[URisky]=23, E[USafe ]= 20, E[ULess Risky ]= 9.2
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Problems
• Apply the one way, two way and
Tornado Diagrams SA to the IDs and
DTs of the previous chapters:
• Discuss your results
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Bayesian Decision
•
•
•
•
•
You are the director of a library shop. To improve the sales, you are thinking of
hiring additional sale personnel. This should, in your opinion, improve the
service level in the shop. If this happens, you expect an increase in costumer
number, and correspondingly, an increase in revenue sales. Suppose that the
number of people entering the shop is, any day, distributed according to a
Poisson distribution with  uncertain. the prior distribution of  is a gamma with
mean equal to 55 and standard deviation equal to 15. the cost increase due to
the hiring is 5000EUR for month. If the service quality improves and the library
receives more than 50 customers per day, revenues increase would amount at
15000EUR (on the average). If less than 50 customers visit the shop, then
revenues would not increase (and you loose the 5000EUR). What to you
decide?
You decide to monitor the number of customers on the next 6 working days:
75,45,30,80,72,41.
You update the Probability. What do you decide now?
How much do you expect to gain now?
Perform a sensitivity analysis on the probabilities. What information do you
get?
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Influence Diagram
Decisione
Servizio
Clienti
Guadagno
Clienti
Service
Decision
Improves
Pmigl
Invest
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1-Pmigl
Not Invest
Clienti=0
Servizio=0
P_50_up
Less than 50
1-P_50_up
Does not improve
P=0.1
Pmigl=0.5
P_500=0.5
P_500_down=0.5
P_500_up=0.7
More than 50
10000
-5000
-5000
0
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Chapter VI
Introduction to Decision theory
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Summary
• Preferences under Certainty
– Indifference Curves
– the Value Function [V(x)]: properties
– Preferential independence
• Preferences under Uncertainty
– Axioms of rational choice
– utility Function [U(x)] in one dimension
– Risk Aversion
• Preferences with Multiple Objectives
– Multi-attribute utility Function
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Preferences Under Certainty
• Example: you are choosing your first job. You select your
attributes as: location (measured in distance from home),
starting salary and career perspectives. You denote the
attributes as x1, x2, x3. you have to select among five
offers a1, a2,…,a5. Every offer gives you certain values of
x1, x2, x3 for certain. How do you decide?
• It is a multi-attribute decision problem in the presence of
certainty, since once you decide you will receive x1,x2,x3
for certain.
• In this case you have to establish how much of one
attribute to forego to receive more of anoilr attribute.
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Preferences under Certainty
Opzione
Valore
• Here is a diagram for the Choice
Opction
1
2
3
X1=0.0
X2=0.0
X3=0.0
X4=0.0
X5=0.0
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4
5
X1
X2
X3
X4
X5
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Structuring Preferences
• Indifference Curves:
x2
x1
• Points on the same curve leave you
indifference
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the Value Function
• You can associate a numerical value representing
you preferences to each indifference curve:
x2
x1
V( x)  v( x1, x2,..., xn )
• V(x) is the function that says how much of xi one is
willing to exchange for an increase or decrease in xk
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V(x)
• V(x) is a value function if it satisfies the
following properties:
• a)
x'  x' '  v( x' )  v( x' ' )
• b)
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x'  x' '  v( x' )  v( x' ' )
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Example
• For the “first job choice”, suppose that you value
function is as follows:
v( x)  3 / x  4x2  x3
2
1
2
• where x1 measures the distance from home in
100km, x2 is the career perspective measured on a
scale from 0 a 10 and x3 the starting salary in
kEUR.
• Suppose to have received the following offers:
– (1, 5, 20), (5, 4, 10), (8,3,60), (10, 5, 20), (10,2,40)
• Which one would you pick?
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Preferences under Uncertainty
Opzione
Utilità
Evento Casuale
Suppose one has to choose between lotteries that offer a mix
the previous job offers: to choose one does not use the value
function, but must resort to the utility function (U(x))
1
2
P11
U1
P12
U2
P13
U3
P14
U4
P41
U1
P42
U2
P43
U3
P44
U4
Decision
3
4
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utility Function
• the utility function is the appropriate one to
express preferences over the distributions of the
Attributes.
• Given two distributions 1 and 2 on the
Consequences , Distribution 1 is more or as
much desirable than Distribution 2 if and only if:
EU( x1)  EU( x2)
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Utility vs. Value
– One attribute Problem. Suppose that alternative 1
produces x1 and the 2 x2, then 12 if x1>x2
– Let us take two Alternatives 1 and 2, with x1>x2, given with
certainty.
– the value function will give us: v(x1)>v(x2)
– Let us now consider the following problem:
1
P1
2
1-P1
X1
X2
XI
– To choose one need u(x1) and u(2)
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Stochastic Dominance
Distributions over attribute x
1
2
0.9
1
0.8
Probability distributions over x
0.7
Distribution 1 is dominated by
distribution 2, if obtaining more of x is
preferable.
Vice versa, if less of x
is preferable, then Distribution 2
is dominated by distribution 1
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
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2
3
4
5
x
6
7
8
9
10
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One Attribute Utility Functions
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Certainty Equivalent
• Given the lottery:
1
X1
P1
2
XN
1-P1
X3
• the value of x such that you are indifferent between
x* for certain and playing the lottery.
1
X1
P1
2
X2
1-P1
X*


• In equations: u( x*)  E u( x)
• N.B.: if you are risk neutral, then x*=E[x]
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definition of Risk Aversion
• a decision-maker is risk averse if preferisce sempre
the expected value of a lottery alla lottery
1
2 : £10
2
£10
0.500
0.500
£40
£(£20)
20
£10; P = 1.000
• Hp: increasing utility function. Th: You are risk
averse if the Certainty Equivalent of a lottery is
always lower than the expected value of the lottery
• You are risk averse if and only if your utility function
utility is concave
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Risk Premium and Insurance Premium
• the Risk Premium (“RP”) of a lottry is the difference
between the expected value of the lottery and your
Certainty Equivalent for the lottery:
RP  Ex  x *
• Intuitively, the Risk Premium is the quantity of attribute
you are willing to forego to avoid the risks connected with
the lottery.
• Suppose now that E[x]=0. the insurance premium is how
much one would pay to avoid a lottery:
IP :  x *
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Mailmatical Definition
• the Risk Aversion function is defined as:
r( x ) : 
u" ( x )
u' ( x )
d
r( x )   ln( u' ( x ))
dx
• Or, equivalently:
• Supposing a constant risk aversion one gets an exponential utility
function:
d
   ln( u' ( x ))  ln( u' ( x ))  x  u' ( x )  e x
dx
u( x )
x
1  x
  u' ( t )dt   e dt  u( x )  u( x 0 )  (e  e x 0 )

u( x 0 )
x0
 t
 u( x )  a  e x  b
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Risk Preferences
• Constant Risk Aversion
U  1 e
 r
• Compute constant  through Certainty
Equivalent (CE):
0.5  e
x1
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 0.5  e
 x1 / 2
e
CE
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Investment Results with Risk Aversion
Market
Decision
Market up
1-exp(-200/70) = 1
Blue Chip Stock 0.600
-3
Market Down
1-exp(-(-160)/70) = -9
0.400
Market up
TwoStock
1-exp(-500/70) = 1
Risky Investment 0.600
prob_up=0.6
-2,110
Market Down
1-exp(-(-600/70)) = -5,278
0.400
Bond=1
1-exp(-50/70) = 1; P = 1.000
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A quale value accetterei l’investimento rischioso
Sensitivity Analysis on
prob_up
400.0
Blue Chip Stock
0.0
Risky Investment
Expected Value
-400.0
Bond
-800.0
Threshold Values:
-1,200.0
prob_up = 0.96
EV = 0.5
-1,600.0
prob_up = 1.00
EV = 0.9
-2,000.0
-2,400.0
-2,800.0
-3,200.0
0.40
0.52
0.64
0.76
0.88
1.00
prob_up
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Esempi of funzioni utility
• Linear: u=ax
– Risk Properties:
• Risk Neutral
• Exponential:
u( x)  a  ecx  b
– Risk Properties:
• - sign: Constant Risk Aversion, + sign: Constant Risk Proneness
• Logarithmic:
– Risk Properties: u( x )  a ln( x )  b
• Decreasing Risk Aversion
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Problems
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problem VI-1
• For the following three utility functions,
u( x )  a  x
u( x )  a  e x  b
u( x )  a  ln( x )  b
• compute:
– the risk aversion function r(x)
– the risk premium for 50/50 lotteries
– the insurance premium
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Problem VI-2
• Consider a 50/50 lottery. Determine your
Risk Aversion constant, assuming an
exponential utility function.
• Reexamine some of the problems discussed
till now utilizing instead of the monetary
payoff the corresponding exponential utility
function with the constant determined above.
How do the decisions change?
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Problem VI-3
• You are analyzing some alternatives for your next vacations:
– A guided tour through Italian cultural cities (Rome, Florence, Venice,
Siena …an infinite list..), duration 10 days, cost 500EUR, for a total of
1500km by bus.
– A journey to the Caribbean, lasting 1 week, cost 2000EUR, by plane.
– 15 days in a wonderful mountain in Trentino, for a cost of 2000EUR,
with 500km of promenades.
• Do you need a utility or a value function to decide?
• Suppose that, after some thinking, you discover to have the following
three attribute utility function:
1
2
  x1
x
(
x
)
s( x )  1  e a  2  3
b
c
• where x1 is the vacation cost in kEUR, x2 is distance in km and x3 is a
merit coefficient regarding relax/amusement to be assigned between 1
and 10.
• What do you choose?
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Chapter VII
the Logic of Failures
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Elements of Reliability theory
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Safety and Reliability
• Safety and Reliability study the performance of systems.
• Reliability and safety study cover two wide areas:
– System Failures and Failure Modes
• Structure Function
– Failure Probability
• Failure Data Analysis
• the approach can be static or dynamic. Static approach is
analytically simpler and is more diffuse.
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Systems
• A system is a set of components connected
through some logical relations with respect to
operation and failure of the system
• More simple structures are:
– Series
– Parallel
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Series
1
2
n
• Every component is critical w.r.t. the system
being able to perform its mission.
• the fault of just one component is sufficient to
provoke system failure
• Redundancy: 0
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Parallel Systems
1
In
2
Out
n
• Each of the components is capable of assuring
that the system accomplish its tasks.
• Thus, to provoke the failure of the system, all the
components must be contemporarily failed
• Redundancy: n-1
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Elements of System Logics
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Boolean Logic
• An event (and) can be True or False
• State Variable or Indicator:
1
Xj  
0
if E has happened
if E has not happened
• Properties:
– (XJ)n=Xj
– X j  X j  0 where X j is the complementary of XJ
• This simple definition enables one to use
algebraic operations to describe the logical
behavior of systems.
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Series Systems
•
•
•
•
Let Ei denote the event “the i-th component failed.”
Let XT denote the event: “the System failed”.
XT takes the name of Top Event.
For the system failure, by definition of series, it is enough
that one single component failes. Thus it is enough that E1
or E2 or …. or En is true.
• From a set point of view: E1E2  ...  En
E1E3E2
• From a logical p.o.v., we get the following expression:
n
XT  1  (1  X1 )  (1  X2 )  ...  (1  Xn )  1   (1  Xi )
i1
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Parallel Systems
• Let Ei denote the event “the i-th component has failed.”
• Let XT denote the event: “the system has failed”.
• the condition for failure of the system is that all component
fail. This happens if E1 and E2 and … En are true at the
same time.
• From a Set point of view: E1E2  ...  En
E1E3E2
• the logical expression is:
n
X T  X1  X 2  ...  Xn   Xi
i1
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the Structure Function
• In general, a system will be formed by a
combination of series or parallel elements, or other
logics (as we will see next).
• One defines the Structure Function of a system the
logical expression that expresses the top event (XT)
as function of the individual failure events.
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the Logic of Performance
• Let Ai denote the event “the i-th component is working (=Not
failed).”
• Let YT denote the event: “the system is working”.
• For a series system: all the components must be working for
the system to work. Thus: A1 , A2 , … and An must be true at
the same time
YT
Series
n
  Yi
i 1
• In parallel: for the system to work it is sufficient that just one
component is working. Thus: A1 or A2 or An must be true.
YT
Parallel
n
 1   (1  Yi )
i 1
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n/N Logics
• n/N logics are intermediate logics between series and
parallel.
• N represents the total number of components in the
system and n the number of components that must
contemporarily fail to break the system.
• As an example, a system has a 2/3 logic if it has 3
components and when two components have failed the
system failes.
2/3
1
In
2
Out
3
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Example: 2/3 System Logics
• Let us find XT for a 2/3 system
• Events: E1, E2, E3, Indicators: X1, X2, X3
• Events that provoke a failure: E1E2 E3, E1 E2 , E1E3, E3 E2 .
• Let us denote E1E2 E3=Z1, E1 E2 =M1, E1 E3= M2, E3 E2 = M3.
• For XT to happen: Z1(M1  M2  M3).
• the structure function expression is:
XT  1  (1  Z1 )  (1  M1 )  (1 M2 )  (1  M3 )
• Let us go to a level below:
XT  1  (1  X1  X2  X3 )  (1  X1  X2 )  (1  X1  X3 )  (1  X2  X3 )
• Let us solve the calculations, noting that (Xi)n=Xi :
XT  X1X2  X1X3  X2 X3  2X1X2 X3
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Probability Sum Rules
• We recall that, for generic Events:
n
P( Ei )   P(Ei )   P(EiE j ) 
i1
j
ij,i j
 P(E E E )   P(E E E E )
ijk ,i jk
i
j
i
ijkh,i jhk
i
j
k
h
 ...  ( 1)n  P(E1E2 ...En )
• We recall that, if the events are independent:
n
P(EiE j ,..., En )   P(E s )
s 1
• Rare Event Approssimation: neglect all terms
corresponding to multiple events
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Golden Rule
• In practice: the System Failure Probability is
computed from the solved Structure Function,
substituting to indicator Xi the corresponding
Event Probability.
XT  X1X2  X1X3  X2 X3  2X1X2 X3
P( XT )  P(E1E2 )  P(E1E3 )  P(E2E3 )  2P(E1E2E3 )
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Proof
•
the System Failure Probability is:
•
P(XT)=P[(Z1(M1  M2  M3)]= P[(Z1 Z2]=P(Z1)+P(Z2)-P(Z1Z2)
•
where:
– P(Z1)=P(E1E2 E3)
– P(Z2)=P(M1  M2  M3)= P(M1)+ P(M2)+P(M3)-P(M1M2)- P(M1M3)P(M3 M2)+P(M1 M2 M3). Ma: M1= E1E2, M2= E3E1, M3= E3E2. Noting,
that: M1 M2= M1 M3= M2 M3= M1 M2 M3 = E1E2E3. Substituting:
P(Z2)=P(E1E2)+ P(E3E1)+P(E3E2)-P(E1E2E3)- P(E1E2E3)P(E1E2E3)+P(E1E2E3). Thus:
• P(Z2)=P(E1E2)+ P(E3E1)+P(E3E2)-2P(E1E2E3)
– P(Z1Z2)=P(E1E2E3 E1E2 E3E1  E3E2)=P(E1E2E3 )
•
Thus: P(XT)= P(E1 E2 E3)+P(E1E2)+ P(E3E1)+P(E3E2)-2P(E1E2E3)-P(E1E2E3)=
P(E1E2)+ P(E2E1)+P(E3E2)-2P(E1E2E3)
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Problems
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Problem VII-1
• For the following systems compute:
– the Structure Function for System Failure
– the Structure Function for System Operation
– the Failure Probability
– the Operation Probability
1
1
2/3
3
2
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2
3
1
2
3
4
4
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Problem VII-2
• for the following system:
1
2
3
4
• Compute the Failure Probability supposing
independent events and denoting the component
failure probability by p.
• Repeat the computation starting with the system
success function, YT. Verify that the two results
coincide.
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Chapter VIII
Elements of Reliability
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Cut and Path sets
•
•
•
•
Failure Logic
By cut set one means an event/set of events whose
happening causes system failure
By minimal cut set one means a cut set that does not
have other cut sets as subsets
Success Logic
By path set one means an event/set of events whose
happening causes system to work
By minimal path set one means a path set that does
not have other path sets as subsets
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Even Trees
• Event Trees: represent the sequence of events
that lead to the event top.
Initiating
Event
Event 1
Event 2
Top Event
Sì
No
Sì
No
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Example
• One has to establish the sequence of events that lead to
leakage of toxic chemicals from a production plant. High
pressure in one of the pipes can cause a breach in the pipe
itself, with leakage of toxic material in the room where the
machine works. the filtering of the air conditioning could
prevent the passage of the toxic gas to the outside of the
room. A fault on the air circulation system due to air filter fault
or maintenance error, would lead to the diffusion of the gas to
the entire firm building. At this point, public safety would be
protected only by the building air circulation system, last
barrier for the gas going to the outsides.
• Draft the event tree for this sequence.
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Gas Leakage Fault - Tree
High
Pressure
Pipe
Room
Building
Top Event
Yes
No
No
No
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Fault Trees
• Fault Trees: represent the logical connection among failures that lead to
the failure of a barrier
• they are characterized by a set of logic symbols that connect a series of
events
And
Or
event Base
• Basic Event: is the event that represents the base of the fault-tree.
From a physical point of view, it represents the failure of a component or
of part of it. From a modeling point of view, it represents the lowest level
of detail.
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Example
• Let us consider the failure of the aeration system. Suppose that the
system is composed by two main parts: an suction engine and a static filter.
the failure of the aeration system, thus, happens either due to engine
failure or for filter fault.
Aeration 1
Engine
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Example
• We could however realize that the level of detail
could be Further increased. In fact, we discover that
the engine can brake for a failure of its mechanical
components and, in particular, of the fan or for a fault
of the electric feeder. the filter can break because of
wrong installation after maintenance or for an
intrinsic fault.
• the fault tree becomes as follows:
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Level II
Aeration 1
Engine
Elettr.
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Mech.
Static filter
Fault
Install.
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From Fault Trees to Structure Functions
• Engine=A
• Static filter =B
• Electr.=1, Mech=2, Fault=3, Install.=4
XT  1  (1  A )  (1  B) 
1  1  [1  (1  X1 )  (1  X 2 )] 1  [1  (1  X3 )  (1  X 4 )] 
 1  1  ( X1  X2  X1X2 )]  [1  ( X3  X 4  X3 X 4 ) 
 X1  X2  X3  X 4  X1X3  X1X 4  X1X2  X3 X 4  X2 X3  X 2 X 4
 X1X3 X 4  X2 X3 X 4  X1X 2 X3  X1X2 X 4
 X1X2 X3 X 4
• What are the minimal cut sets?
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Rare Events Approximation
• If the event probabilities are low (rare events),
then lower the event intersection probabilities
will be.
• One neglects the probabilities of
intersections.
• the Failure Probability is computed as sum of
the minimal cut sets Probabilities:
P( XT )  P( X1 )  P( X2 )  P( X3 )  P( X4 )
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Event Trees & Fault Trees
High pressure
Pipe
Aeration 1
Aeration 2
Top Event
yes
No
No
No
Aeraz. 1
engine
Electr.
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filter
Mech.
Fault
Installaz.
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Probability of the Top Event
• From the Event Tree:
P(Top )  P(Condutt .  Aeraz.1 Aeraz.2 AP)
• Expanding:
P(Top )  P( Aeraz.1 Aeraz.2 Condutt .) * P(Condutt . AP)
 P( Aeraz.1Aeraz.2,Condutt ., AP)  P( Aeraz.2 Cond., AP)  P(Cond AP)
• the conditional probabilities are found solving
the corresponding Fault Trees
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Definitions
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Failure Density
• Given a system, tet us denote with
fs ( t )dt
the Probability that the system fails between t
and t+dt
• It must hold that:

 f (t)dt  1
s
0
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Reliability
• The Reliability of a system between 0 and t is the
Probability that the system fulfills its function between 0 and t
• The Unreliability of a system between 0 and t is the
Probability that the system breaks withint time T:
F( t )  Pr(T  t )   f ( t)dt
0
• Thus the Reliability [R(t)] is related to the failure time pdf as
follows:
t
R( t )  1  Pr(T  t )  1   f ( t)dt
• Note that, if f(t) is continuous:
0
R' ( t )  f ( t )
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General Fault rate
(t)
Infant
Mortality
Useful
Life
Aging
t
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Hazard/Failure Rate
• Failure rate, (t), is the Probability that the system si
rompa between t and t+dt, given that is sopravvissuto
fino a t.
• Dalla definition segue immediatamente the relaction
with the Reliability and the function densità:
( t )  R( t )dt  f ( t )dt
• Thus:
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f (t )
R' ( t )
( t ) 

R( t )
R( t )
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Legami between R(t), f(t) and (t)
• From the above definition, there follows:
f (t )
R' ( t )
( t ) 

R( t )
R( t )
• Relationship R(t)- (t):
R( t )  e
t

  ( t ) dt
0
• Relationship f(t)- (t):
t
f ( t )  ( t )  R( t )  ( t )  e
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
  ( t ) dt
0
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time medio of failure (MTTF)
• The mean time to failure is defined as:

MTTF   t  f ( t )dt
0
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System Reliability
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Reliability of systems in Series
• Series:
P(TS  t )  R S (t )  1  FS (t )  P(T1  T2  ...  TN  t )
• if independence is assumed:
P(TS  t )  P(T1  t )  P(T2  t )  P(TN  t )
n
• Thus:
R s ( t )   Ri ( t )
i 1
• Faulure rate:
n
S (t)   i (t)
i 1
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Reliability of systems in Parallelo
• Failure Probability of the system:
P(TS  t )  FS ( t )  1  R S (t )  P(T1  T2  ...  TN  t )
• if independent:
P(TS  t )  P(T1  t )  P(T2  t )  P(TN  t )
• Thus:
n
N
Fs ( t )   Fi ( t )  RS ( t )  1  Fs ( t )  1   1  Ri ( t )
i1
• Failure Rate:
i1
1
1
1
1
1
1


 ... 



 s ( t )  1( t )  2 ( t )
 n ( t )  1( t )   2 ( t )  1 ( t )   3 ( t )
1
1

 .... 
 1( t )   2 ( t )   3 ( t )  1 ( t )   2 ( t )   3 ( t )
 ( 1)n
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1( t )   2 ( t )   3 ( t )  ...   n ( t )
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Reliability of Standby Systems
• A standby system is a system where a subsystem is
operational and the other subsystems become operational
only after the failure of the system which is operating at the
time of failure.
• An example is the fifth wheel of a car.
• In this case the System Reliability is given by:
– 1) Two components: P(TS2  t)  P(T1  t)  P(T1  t  T2  t  T1)
t
– Thus:
2
RS ( t )  R1( t )   f1( t1 )  R2 ( t  t1 )dt1
0
– where 2 indicates that there are two components in standby, while the
subscript denotes the second component
P(TS  t )  P(T1  t )  P(T1  t  T2  t  T1 )
3
– 2) Three components:
– Thus:
 P(T1  t  T2  t  T1  T3  t  T1  T2 )
t
t  t1
0
0
RS ( t )  RS ( t )   f1( t1 )  f2 ( t 2 )  R3 ( t  t1  t 2 ) dt1  dt 2
3
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Standby Systems with const. failure rates
• For a standby system, it holds that:
T  T1  T2  ...  TN
• then P(t<T) is given by the convolution of the fi(t).
• If these distributions are exponential and the failure rates
identical:
n1
R  e
n
s
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i 0
 t
( t )

i!
i
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Failure on Demand
• If a system is called in function and does not respond
(i.e. does not begin to work), one talks about a “failure
on demand”.
• For a standby system, one denotes with q the failure on
demand probability :
t
RS ' ( t )  R1( t )  (1  q)   f1( t1 )  R2 ( t  t1 )dt1
2
0
• and
t
t  t1
0
0
RS ( t )  RS ' ( t )  (1  q)2  f1( t1 )  f2 ( t 2 )  R3 ( t  t1  t 2 ) dt1  dt 2
3
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Problems
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problem VIII-1
• Write the Fault Trees for the following
systems and derive the structure function:
1
1
2/3
3
2
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2
3
1
2
3
4
4
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Problema VIII-2
• Una delle sequenze incidentali di un piccolo reattore di ricerca prevede la
spaccatura della conduttura principale del circuito idraulico primario. Se
la conduttura si rompe, si ha perdita immediata di raffreddamento del
nocciolo - la zona del reattore dove avviene la reazione nucleare.
L’incidente si può evitare se il sistema di raffreddamento ausiliario
interviene per tempo e se il sistema di spegnimento del reattore
interviene con successo. L’insuccesso dello spegnimento può avvenire
se uno dei seguenti avvenimenti si realizza: mancata lettura del segnale
per un guasto al software [P(Sof|alta press.)=10-4], mancato arrivo del
segnale per un guasto del sistema elettrico [P(E|alta press.)= 10-5],
mancato sganciamento delle barre per un guasto meccanico [P(Bar|alta
press.)= 10-3]. Il sistema di raffreddamento ausiliario è costituito da due
pompe in parallelo, con rateo di guasto 1/10000 [1/h] e probabilita’ di
guasto on demand di 10-3. Le pompe devono funzionare per 100 ore
affinche’ l’impianto sia fuori pericolo. Determinare:
– L’albero degli eventi
– Gli alberi dei guasti
– La probabilità di fondere il reattore dato che si è verificato l’incidente in un
anno dato che la frequenza di eventi di alta pressione e’ 0.0001 per anno.
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Problema VIII-3
•
•
•
•
Un test di polizia per la determinazione del grado di alcool nei guidatori,
ha probabilità 0.8 di essere corretto, cioè di dare risposta positiva
quando il contenuto di alcool nel sangue è elevato o negativa quando il
contenuto è basso. Coloro che risultano positivi al test, vengono
sottoposti ad un esame da parte di un dottore. Il test del dottore non fa
mai errori con un guidatore sobrio, ma ha un 10% di errore con
guidatori ebbri. I due test si possono supporre indipendenti.
1) Determinare la frazione di guidatori che, fermati dalla polizia
subiranno un secondo test che non rivela alto contenuto di alcool
2) Qual è la probabilità a posteriori che tale persona abbia un alto
contenuto di alcool nel sangue?
3) Quale frazione di guidatori non avrà un secondo test?
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Problema VIII-4
•
•
Un impianto elettrico ha due generatori (1 e 2). A causa di manutenzioni e occasionali
guasti, le probabilità che in una settimana le unità 1 e 2 siano fuori serivizio (eventi che
chiamiamo E1 ed E2 rispettivamente) sono 0.2 e 0.3 rispettivamente. C’è una probabiltà di
0.1 che il tempo sia molto caldo (Temperatura>30 gradi) durante l’estate (chiamiamo H
questo evento). In tal caso, la domanda di elettricità potrebbe aumentare a causa del
funzionamento dei condizionatori. La prestazione del sistema e la potenzialità di soddisfare
la domanda può essere classificata come:
– Soddisfacente (S): se tutte e due le unità sono funzionanti e la temperatura è inferiore
a 30 gradi
– Marginale (M) : se una delle due unità è funzionante e la temperatura è maggiore di 30
gradi
– Insoddisfacente (U): se tutte e due le unità sono non funzionanti
1) Qual è la probabilità che esattamente una unità sia fuori servizio in una settimana?
•
2) Definire gli eventi S, M e U in termini di H, E1 ed E2
•
3) Scrivere le probabilità: P(S), P(U), P(M)
•
Suggerimenti: Utilizzate alberi degli eventi e dei guasti per determinare la funzione di
struttura e poi passate alle probabilità
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Problema VIII-5: Distribuzione Weibull
• Dato un componente con rateo di guasto:
  t  
( t )  

  
 1
• con  e 0t calcolare:
• R(t), f(t), il MTTF e la varianza del tempo medio di
guasto
• R(t) è detta distribuzione di Weibull
• Disegnare (t),f(t) ed R(t) per =-1,1, 2. Dedurne che
la Weibull può essere utilizzata per descrivere il tasso
di guasto di componenti in tutta la vita del
componente.
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Problema VIII-6
• Dato un componente con il tasso di guasto
(t) seguente:
10
9
8
7
6
l(t)
 1
se 0  t  1

 t  1
λ ( t )  1
se 1  t  10
 1

t 2 se t  10
100
5
4
3
2
1
0
0
5
• calcolare:
• R(t), f(t), e il MTTF del componente
10
t
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15
Problema VIII-7
• Calcolare l’espressione dell’affidabilità [R(t)]
di un sistema k su n con rateo di guasto
generico.
• Calcolare la stessa espressione con
distribuzioni esponenziali
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Problema VIII-8
• Calcolare l’affidabilità annuale di un sistema
con 4 componenti in serie con ratei di guasto
[1/h]: (1/6000, 1/8000, 1/10000, 1/5000).
• Confrontatela con quella di un sistema in cui i
componenti sono messi in:
– Parallelo
– In logica 3/4
– In logica 2/4
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Problema VIII-9
• Due componenti identici, con tasso di guasto
=3x10-7 [1/h] devono essere messi in parallelo o
standby. Determinate la configurazione migliore e il
guadagno in affidabilità (in percentuale).
• Supponete ora che il sistema di switch sia difettoso,
con probabilità q=0.01. Quale delle due
configurazioni è più conveniente?
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Problema VIII-10
• Considerate un sistema in standby di due
componenti diversi, con densità di guasto
esponenziali. Il MTTF del primo componente è 2
anni, quello del secondo è 3 anni. Calcolate:
• La densità di guasto del sistema
• Il MTTF
• Cosa succede se i due componenti sono identici
con MTTF di 2.5 anni?
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Prob. VIII-2 Soluzione
Rottura
Primario
Spegnimento
Raffreddamento
Top Event
No
Si’
Si’
Raffreddamento
Spegnimento
On Demand
Sof.
El.
Bar.
Pompa 1.
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Pompa 2
Prob. VIII-2 Soluzione
• Assumiamo eventi rari.
• La frequenza si calcola dalla combinazione degli eventi:
F  frottura  P(Raff Rottura Pr im)  P(Spegn Rottura Pr im.)
• dove: frottura=.000001 per anno
• P(Spegn|rottura prim.)=P(Sof|rottura prim.)+P(E|rottura
prim.)+P(Bar|rottura prim.)=0.00111
2
1

(
100 )  
• P(Raff| rottura prim.)= q  1  e 10000  

 

• Quindi:
=.0010995
F  1/ 1000000  .00109  .0011  2.2  109
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Problema VIII-8 Soluzione
• Calcolare l’affidabilità annuale di un sistema con 4
componenti in serie con ratei di guasto [1/h]: (1/6000,
1/8000, 1/10000, 1/5000).
– R( t )  e   t  e   t  e   t  e   t
– Ore in un anno: 8760.
8760
8760
8760
8760




– Sostituendo i numeri: R(t )  e 6000  e 8000  e 10000  e 5000  0.006
1
2
3
4
• Confrontatela con quella di un sistema in cui i
componenti sono messi in:

8760
6000

8760
8000
)  (1  e
)  (1  e
– Parallelo: R( t )  1  (1  e
– ¾ supponendo I ratei di guasto =1/8000.

8760
10000
)  (1  e

8760
5000
)  0.75
• Risultato: 0.11
– 2/4 supponendo I ratei di guasto =1/8000
• Risultato: 0.41
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Problema VIII-9 Soluzione
• Due componenti identici, con tasso di guasto =3x10-7 [1/h] devono
essere messi in parallelo o standby. Determinate la configurazione
migliore e il guadagno in affidabilità (in percentuale) per t=7 anni
(61320hs).
Rp (t )  1  (1  e

61320
30000000
)  (1  e

61320
30000000
)  0.99967
t
RS ( t )  R1( t )   f1( t1 )  R2 ( t  t1 )dt1  e t (1  t )  0.99983
2
0
– Il guadagno di affidabilita’ e’ dell’ordine del 10^-2% (0.0002), quindi
trascurabile
• Supponete ora che il sistema di switch sia difettoso, con probabilità
q=0.01. Quale delle due configurazioni è più conveniente?
Rp (t )  1  (1  e

61320
30000000
)  (1  e

61320
30000000
)  0.99967
t
RS ( t )  R1( t )  (1  q) f1( t1 )  R2 ( t  t1 )dt1  et (1  (1  q)t )  0.99965
2
0
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Capitolo IX
Decisioni Operative:
Ottimizzazione delle Manutenzioni
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Decisioni Operative
• Decisioni di Affidabilita’ o Reliability Design
• Decisioni di Optimal Replacement
• Decisioni di ispezione ottimale
• Decisioni di riparazione ottimale
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Indisponibilita’
• Sistemi riparabili o manutenibili: il sistema
puo’ ritornare a funzionare dopo la rottura
• Indisponibilta’ istantanea:
– q(t):= P(sistema indisponibile per T=t)
• Indisponibilita’ limite:
lim q( t )
t  
T
• Indisponiblita’ media in T: q 
 q(t )dt
0
T
T
• Indisponibilta’ media limite:
qlim  lim
T  
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 q(t )dt
0
T
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Disponibilita’
• La disponibilita’ istantanea e’ il complementare
della indisponibilita’. Le altre definizioni seguono
immediatamente
• Note: la disponibilita’/indisponibilita’ non e’ una
densita’ di probabilita’ e l’indisponibilita’ media non
e’ una probabilita’.
• Interpretazione: la disponibilita’/indisponibilita’
media e’ la frazione media di tempo in cui il
sistema e’ disponibile in [0 T].
• Le riparazioni/manutenzioni introducono
periodicita’ nel problema
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Effetto delle manutenzioni
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Calcolo della Indisponibilita’: un unico componente, una
sola modalita’ di guasto
• Evoluzione temporale:
t
tr
t
tr
t
• A t=0 il sistema entra in funzione dopo la manutenzione.
Dopo un tempo t= t torna di nuovo in manutenzione. La
manutenzione dura tr. t e’ il tempo in cui il componente e’
soggetto a rotture casuali con (t).
• Si nota che il problema e’ periodico, di periodo T= tr+t
• Durante t il sistema ha una indisponibilita’ istantanea pari
alla sua probabilita’ di rottura, se, come da ipotesi, non ci
sono riparazioni:
 ( t ')dt '
t
q( t )  FS ( t )  1  R( t )  1  e 0
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Calcolo della Indisponibilita’: un unico
componente, una sola modalita’ di guasto

  ( t ')dt '

0
0tt
• L’indisponibilta’ istantanea risulta quindi: q( t )  1  e
1
t  t  t  tr

t
t
t
• Da cui l’ind. Media: q 
  ( t ')dt '
0
1

e

0
t  tr
dt
tr

t  tr
• Supponiamo  cost e 1. Quindi utilizziamo approssimaz. Taylor:
t
  ( t ')dt '
1 e 0
 1  e t  1  (1  t  ...)  t
• Sostituiamo nella ind. Media, e assumiamo tr<< t :
q
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1
t
T  r
2
T
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Modi di Guasto
• Guasto in funzionamento: f(t) [1/T]
• Guasto in hot standby: s(t) [1/T]
• Guasto a seguito di manutenzione errata: 0,
1, 2…. Dove:
– 0=incondizionale,
– 1=dato che 1 manutenzione errata,
– 2= dato che 2 manutenzioni errate
• Guasto on demand: Q0,Q1 etc.
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Indisponibilita’ istantanea con piu’ modi di guasto
• Consideriamo per un componente i modi di guasto indicati in precedenza.
• A t=0 il componente puo’ essere gia’ guasto se disabilitato dall’erronea
manutenzione. Questo evento ha probabilita’ 0. Con probabilita’ (1- 0) il
componente invece potra’ invece aver superato con successo la
manutenzione. In questo caso il componente potra’ rompersi “on
demand” (E1) o con tasso di guasto (t) (E2). Si ha:
P(E1E2)=P(E1)+P(E2)-P(E1E2)=Q0+F(t)-Q0F(t).
• Riassumendo, tra 0 e t si ha:
q(t)= 0+(1- 0)*[Q0+F(t)-Q0F(t)].
• Introduciamo ora una probabilita’ esponenziale per le rotture. Utilizziamo
la approssimazione di Taylor. Abbiamo:
q(t)= 0+(1- 0)*[Q0+(1-Q0)t].
• Quindi l’indisponibilita’ istantanea e’:
  0  (1   0 )Q0  t  Q0t  0  t  t
q( t )  
t  t  t  tr
1
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Indisponibilita’ media con piu’ modi di guasto
• L’indisponibilita’ media sull’intervallo 0 t+ tr e’:

t2
t2 
 0 t  (1   0 )Q0 t    Q0    tr
2
2

q
t  tr
• Due assunzioni: 1) Eventi rari 2) t+ tr t
t tr
q   0  Q0   
2 t
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Rappresentazione equivalente
Componente
t
Q0
0
t
• La funzione struttura e’:
• XC=1- (1-Xt) (1-XQ0)(1-X0)(1-Xt)=
• = Xt +XQ0+X0+Xt-termini di ordine superiore….
• Approssimazione eventi rari:
• XC= Xt +XQ0+X0+Xt
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Il caso di due componenti
• Sostituzioni successive
t tr1tr2
t tr1 tr2
• Sostituzioni distanziate
tr
t t+2t trtr
r
t t+2tr
r
r
• Periodo: t+2tr
Indisponibilita’ media e’ la somma di piu’ termini:
q  qR  qC  qD  qM
“R”: random, “C” common cause, “D” demand e
“M” maintenance
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Modi di guasto
• Causa comune: sono quei guasti che colpiscono
il sistema come uno e rendono inutili le
ridondanze e/o annullano indipendenza
condizionale dei guasti.
• Es.: difetto di fabbrica in parallelo di componenti
identici
• Errori in manutenzione: human errors
• Human Reliability
• CC e HR sono due importanti rami dello studio
dell’affidabilita’ dei sistemi
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Modelli decisionali corrispondenti
• Come stabilire una politica di replacement
ottimale?
• Costruzione della funzione obiettivo
– i) Individuazione del Criterio
– ii)Costruzione della funzione obiettivo o utilita’
– iii) Ottimizzazione
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Esempio 1
• 1 componente soggetto replacement
periodico e manutenzione periodica
• Criterio = disponibilita’ media
• Funzione obiettivo: q(t)
• t ottimale:
t
2  tr

– tr=24 h, =1/10000 (1/h)  tott=700hr
• Con =1/100000 (1/h) tott=2200hr
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Esempio 2
• Ottimizzazione in considerazione del costo di
sostituzione e della disponibilita’
• Funzione obiettivo: B( t)  Cq  q( t)  C( t)
• t ottimale:
 dB( t)
 dt  0
 2
 d B( t)  0
 d2 t
• Occorre introdurre vita del’impianto L.
L
• Si ha: C( t)   c 0 dove c0 e’ il costo unitario di
t
riparazione
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Esempio 2
• Introduciamo poi il costo della indisponibilita’:
Cq  a c0
• definito come multiplo del costo singola riparazione.
• Funzione energia:
E( t)  Cq q( t  0 Q0 tr  )  C( t)
1
tr 
L

E( t)  a c0   

 c0


2
2
2
dt
t  t

d
1
• Intervallo ottimale:
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 tr
L 1 

t  2    
    a 
2
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Esempio 2
c0  5
34000
3.51210
L  70000
a  10
tr  24
1
 
10000
E( t ) 2000
59.363
0
0
0
5 10
t
4
1 10
100000
5
0  0.001
Q0  0.001
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t  1.185  10
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Applicazione del modello
• Il modello si applica al meglio a componenti
in standby o sistemi di sicurezza passivi.
• Infatti si ipotizza che il componente sia
rimpiazzato secondo un intervallo di tempo
prestabilito t.
• Si valuta percio’ la convenienza rispetto alla
minimizzazione del costo di replacement e/o
alla massimizzazione della disponibilita’
• Per sistemi in funzionamento occorre
considerare invece la possibilita’ di riparare il
sistema
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Riparazioni
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Il tasso di riparazione (t)
•
Analogamente alla rottura, anche il processo di riparazione di un componente ha
delle caratteristiche di casualita’. Per esempio, non si sa il tempo necessario
alla individuazione del guasto, cosi’ come puo’ essere non noto a priori il tempo
necessario all’arrivo delle parti di ricambio o il tempo richiesto dall’esecuzione
della riparazione. Tutto cio’ viene condensato in una quantita’ analoga al rateo
di guasto, e, precisamente, il tasso di riparazione (t) . E’ uso comune
assumere un tasso di riparazione costante - e spesso questa assunzione non e’
peggiore di quella di assumere (t) costante.- Ne seguono:

rip( t )    e t


 t
Rip( t )  1  e


MTTR  t    e t  1
0


•
Dove rip(t) e’ la densita’ di riparazione, ovvero la probabilita’ che la riparazione
avvenga tra t e t+dt e Rip(t) e’ la probabilita’ che la riparazione avvenga entro t.
Notiamo che (t) e’ la probabilita’ che il componente sia riparato tra t e t+dt dato
che non e’ stato ancora riparato a t.
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Esempio
• Consideriamo un sistema composto da due componenti, di cui uno in
standby.
• Per modellizzare questo problema occorre un approccio diverso sia dai
due casi precedenti.
• Occorre introdurre gli stati del sistema
• Nell’esempio. Il sistema puo’ essere: in funzione con il componente 1
funzionante (stato 1), in funzione ma con il componente 2 funzionante e il
componente 1 in riparazione (stato 2), (stato 3) con entrambe i
componenti rotti. Da 3 puo’ tornare a 2 e da 2 ad 1. Puo’ passare da 1 a
3 se c’e’ failure on demand.
1
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2
3
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Assunzioni
• Stato del sistema al tempo t e’ indipendente dalla
storia del sistema.
• Questa assunzione e’ alla base dei processi
stocastici di Markov.
• In particolare, supponiamo che il sistema possa
avere M stati e denotiamo con Xt lo stato del
sistema al tempo t. Allora Xt potra’ assumere
valori 1,2,….M.
• Cosa accade in dt?
• Il sistema puo’ transitare in un altro stato
(eventualmente con dei vincoli): i  j
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Matrice di transizione
• Indichiamo con Pij la probabilita’ che il sistema
passi dallo stato i allo stato j
P11 P12
P
M  Pij   21


PM1
 
P1M 




PMM 
• Proprieta’:
M
• 1)  Pij  1
i1
• 2)Se
Pii  1
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allora stato i e’ detto assorbente
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Esempio
• Applichiamo uno schema a stati per il sistema in
standby. Otteniamo:
P13
1
P12
P21
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2
P23
3
P32
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Equazioni di Markov/Kolmogorov
dP
 A P
dt
• Dove A e’ la matrice di transizione del sistema, P e’
il vettore delle probabilita’ degli stati del sistema.
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Costruzione della matrice di transizione

P12
1
2
P21
1

2
• Esempio: componente soggetto a rottura e riparazione. 2 stati: in
funzione o in riparazione, con tassi di guasto  e riparazione.
• Chi sono P12 e P21? Sono le probabilita’ di transizione in dt. Quindi:
P12= e P21= 
• La matrice di transizione e’ costruita con le seguenti regole:
• (+) se il salto e’ in entrata allo stato, (-) se il salto e’ in uscita
• Prendiamo lo stato 1: si entra in 1 da due con tasso  (+), si esce con
tasso  (-).
dP1
• Quindi:
 P1  P2
dt
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La matrice di transizione
•
Analogamente:
1
2
• Quindi:
1  
2  
dP2
 P1  P2
dt
   
A






•
La matrice di transizione e’:
•
Il sistema di equazioni differenziali
dP
 A  P diventa:
dt
dP1
 P1  P2
dt
dP2
 P1  P2
dt
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Disponibilita’ asintotica e media
• E’ la probabilita’ che a t il componente sia nello
stato 1. Occorre risolvere il sistema di equazioni
differenziali lineari precedente. Modo piu’ usato in
affidabilita’ e’ mediante trasformata di Laplace.
• Con trasf. Laplace, le equazioni da differenziali
diventano algebriche. Dopo aver lavorato con
equazioni algebriche, occorre poi antitrasformare.
• Si ottiene dunque la disponibilita’ come funzione
del tempo. A questo punto due disponibilita’
interessano: quella asintotica e quella media. Il
risultato per un componente singolo soggetto a
riparazioni e rotture e’ il seguente:
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Risultati per un componente
• Disponibilita’ istantanea:
μ
λ
P1 ( t ) =
+
e
μ+λ μ+λ
( μ+ λ ) t
• Disponibilita’ asintotica:
μ
MTTF
lim P1 ( t ) =
=
t →∞
μ + λ MTTF + MTTR
•
Interpretazione: tempo che occorre in media alla riparazione diviso il tempo
totale
• Disponibilita’ media su T:
q( T) 
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
(   ) 2
 expT (   ) 

(   ) 2



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Problema IX-1
• Calcolare l’ indisponibilita’ media di un componente
in standby soggetto a sostituzione periodica con le
seguenti probabilita’ di guasto per t=5000:
0  0.002
Q0  0.002
tr  24
1
 
15000
• (Soluzione: q=.175)
• Calcolare l’intervallo di sostituzione ottimale e
l’indisponibilita’corrispondente, con L=70000, a=10
e a=.
• (Soluzione: t=14500, q=0.5; t=849, q=0.06)
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