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Quantitative Methods for Management
Emanuele Borgonovo1
Emanuele Borgonovo
Quantitative Methods for Management
First Edition
Decision
Market
Return
Structural
Quantitative Methods for Management
Emanuele Borgonovo2
Chapter three:
Models
Quantitative Methods for Management
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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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Emanuele Borgonovo4
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
)B(P)A(P)BA(P
,eventsesclusivemutuallyBeAIf
0)A(P
1)U(P
U BA
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• 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
Areas and rectangles?
U
EDCBAU CA B D and
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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)
UB
AAB
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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
AAB
•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.
BAAB
BA
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:
)A(P
)BA(P)B(P)AB(P
P(B) before A
New value of the Probability of B Probability of A
Probability of A given 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)=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:
)o(P
)Ao(P)A(P)oA(P
3/24/3
12/1)oA(P
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Bayes’ theorem Proof
)B(P
)A(P)AB(P)BA(P
)A(P)AB(P)B(P)BA(P
)AB(P)AB(P
Starting point
Conditional Probability formula
thesis
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U
the Total Probability theorem
• 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:
• Bayes theorem in the presence of N events becomes :
)D(P
)C(P)B(P)A(P1
B A C
D
and
)A(P)AE(P...)A(P)AE(P)A(P)AE(P)E(P NN2211
N
1iii
111
)A(P)AE(P
)A(P)AE(P)EA(P
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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.
• Note: f(x0)dx is the Probability that x lies in an interval dx around x0.
1dx)x(f
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Cumulative Distribution Function
• Given a continuous random variable X, the Probability that X<x is given by:
• If f(x) is continuous, then:
• Note:
x
dt)t(f)xX(P
dx
dP)x(f
)Xx(P)Xx(Pdx)x(f)XxX(P 12
X
X
21
2
1
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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:
• and by the density function:
tλe1)tT(P te)t(f
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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. • Hence:
• Which solved leads:
• 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.
T)0(N/)T(Nlndt)t(N
)t(dNdt
)t(N
)t(dN T
0
T
0
Te)0(N
)T(N
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0 5 10 150
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
T/t
Pdf and Cdf of the Exponential Distribution
P(t<T)
f(t)
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Expected Value, Variance and Percentiles
xV:DeviationdardtanS
xExEdx)x(f)xEx(
)xEx(ExV:Variance
dx)x(xfxE:ValueExpected
222
2
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:
• Cdf:
Xe
2
1)x(f
2)x
(2
1
G
dxe
2
1)Xx(P
2)x
(2
1
G
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Graphs of the Normal Distribution
-4 -3 -2 -1 0 1 2 3 40
500
1000
1500
2000
2500
3000Distribuzione Normale Standard
x
f(x)
-5 -4 -3 -2 -1 0 1 2 3 40
1000
2000
3000
4000
5000
6000
7000
8000
9000
10000Cumulative Gaussian Distribution
x
)x(fG
)Xx(PG
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Lognormal Distribution
• Cdf
x0e2x
1)x(f
2)xln
(2
1
L
X
0
)xln
(2
1
L
2
e2x
1)Xx(P
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Lognormal Distribution
0 20 400
0.1
.20
0
f x( )
500.07 x
0 20 400
0.5
11
0
f2 x( )
500.07 x
)x(fL
)Xx(PL
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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?
• What is the Probability of the gear being integer after 9 years?
5/1dtet t
%5.16e
)e1(1)9t(P1)9t(P9)5/1(
T
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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(Bor)=1-
P(Aor)=1/3
– P(Ba)• P(Ba)=1, since B is the only event that has a silver ring. One can also
show it using Bayes’ theorem:• P(Ba)=P(aB)*P(B)/[P(aB)* P(B)+P(aA)*P(A)]. Since P(aA)=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:
)B(P)Bo2(P)A(P)Ao2(P
)A(P)Ao2(P)o2A(P
11
1
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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(Bor)=1/3 and P1(A)=P(Aor)=2/3.
– One can note that P(2oA)=1, and P(2oB)=1/2. P(2oB) 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:
– 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(A2o)=0.2
8.03/1*2/13/2*1
3/2*1
)B(P)Bo2(P)A(P)Ao2(P
)A(P)Ao2(P)o2A(P
11
1
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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 Further Analysis?
Yes
Best Alternatives implementation
No
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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
Decision
Market
Return
Structural
Market up
prob_upMarket down
1-prob_up
Less Risky
Market upprob_up
Market down
Risky
Safe
How should theinvest $1000?
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
= Decision
= Random Event
= utility
ARCS
• Informational Arcs
• probabilistic Dependency Arcs
• Structural Arcs
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ID Elements
Decision Node
Chance Node Value NodeChance Node
Decision Node
Conditional Arc
probabilistic Dependency
Informational Arc
Sequential Decisions
Structural
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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
DecisionsI - Normal Makeup
II - Shutdown
III - Reduce Power
IV - Isolate SG
Chemical VolumeControl System
SecondaryCooling
Primary
Cooling
Value
LeakageRate
days_to_shutdown
shutdown_cost
core_damage_cost
time_to_repair
Leakage from primaryto secondary, maximumrate of 20 l/hr
DeterministicInformation
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Influence Diagram
Decision
Market
Return
Structural
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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
Market down
1-prob_up
Less Risky
Market up
Market down
Risky
Safe
How should theinvest $1000?
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Decision Tree Solution
• Alternative Payoff or utility:
• 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)
j
Cjii jU)C(P]U[E
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example
Market up
P.upC1
Market down
1-P.upC2
Blue Chip Stock
Market up
P.upC3
Market down
1-P.upC4
Risky investment
CD paying 5%C5
How should theinvest $1000?
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Problem Solution
• Using the previous formula:
21
2
1jjjRisky CP.up)1(CP.upC)C(P]U[E
43
2
1jjjLessRisky CP.up)1(CP.upC)C(P]U[E
5
2
1jjjLessRisky C1C)C(P]U[E
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the Best Investment for a Risk Neutral Decision - Maker
Market up
0.600$200
Market down
0.400($160)
Blue Chip Stock$56
Market up
0.600$500; P = 0.600
Market down
0.400($600); P = 0.400
Risky investment$60
CD paying 5%return = $50
How should the
invest $1000?
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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 Engine failure ProfitFinal Classification
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From IDs to Decision Trees
Out of first five
1.000$20,000; P = 0.500
failure
Engine failure
0.500$20,000
Win
0.500$110,000; P = 0.250
In first five
0.300$105,000; P = 0.150
Out of first five
0.200$50,000; P = 0.100
No failure
0.500$94,500
Run
Decision
$57,250
Old sponsor
1.000$50,000
Withdraw
Engine_failure=0$50,000
Decision
pfailure=0.5
pfive=0.30
pout=0.2
pwin=0.5
Run : $57,250
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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 Assicurazione Ruttura Costo
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Decision Tree
1
Assicurazione
-150-5%*(150) = (£158); P = 1.000
Sì
0.050-150-2%*150-150 = (£303)
No
0.950-150*(1+2%) = (£153)
2(£161)
A
Decision
1 : (£158)
1-(175+3%*(175)) = (£180)
Sì
0.030-175-2%*175-175 = (£354)
No
0.970-175-2%*175 = (£179)
2(£184)
B1 : (£180)
1-200-2%*200 = (£204)
Sì
0.020-200-1.5%*200-200 = (£403)
No
0.980-200-200*1.5% = (£203)
2(£207)
C1 : (£204)
pA=0.05pB=0.03pC=0.02
A : (£158)
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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:
• 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
]ingBeforeKnow[E]Knowing[EEVPI
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Example: investing
Decision Market Value
Up
Market
0.500£500; P = 0.500
Down
0.500(£400); P = 0.500
RISKY
Decision
£50
Up
0.500£200
Down
0.500(£160)
LESS RISKY£20
SAFE
Market=0£20
P_UP=0.5RISKY : £50
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EVPI for the Example
RISKY
Decision
£500; P = 0.500
LESS RISKY£200
SAFE£20
Up
Market
0.500RISKY : £500
RISKY(£400)
LESS RISKY(£160)
SAFE£20; P = 0.500
Down
0.500SAFE : £20
P_UP=0.5£260
DecisionMarket Value
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EVPI Result
210
50260
]r[E]Knowing[EEVPI
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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
D ec is io n
A
B
CC o n s eq u en c es
Skip Arc
• Influence Diagram the
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good
C
(No Payoff)
bad(No Payoff)
high
B
good(No Payoff)
bad(No Payoff)
low
up
A
down
B=0C=0
(No Payoff)
1
Decision
good(No Payoff)
bad(No Payoff)
high
good(No Payoff)
bad(No Payoff)
low
up
down
B=0C=0
(No Payoff)
2
Solution
• Corresponding Decision Tree
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Solution
• Influence Diagram II
Decision
A
B
CConsequences
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Solution
• Decision Tree II:good
C
(No Payoff)
bad(No Payoff)
high
B
good(No Payoff)
bad(No Payoff)
low
up
A
good(No Payoff)
bad(No Payoff)
down
B=0
1
Decision
good(No Payoff)
bad(No Payoff)
high
good(No Payoff)
bad(No Payoff)
low
up
good(No Payoff)
bad(No Payoff)
down
B=0
2
Quantitative Methods for Management
Emanuele Borgonovo72
• 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.
• 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
Sales_Costs
Decisione
Vendite
Costo
Payoff
High
Sales
P_Alte|high0
Low1- P_Alte|high
-10
high
Cost
P_high
HighP_ P_Alte|high
20
Low
1- P_Alte|high0
Basso
1-P_high
Invest
Decision
Do not Invest5
high=0.5P_Alte=0.3P_high=0.5
Quantitative Methods for Management
Emanuele Borgonovo73
Solution Sales_Costs
Alte
Vendite
0.300£0
Basse
0.700(£10)
Alto
Costo
0.500(£7)
Alte
0.300£20
Basse
0.700£0
Basso
0.500£6
Investo
Decisione
(£1)
Non-Investo£5; P = 1.000
alto=0.5P_Alte=0.3P_alto=0.5
Non-Investo : £5
Quantitative Methods for Management
Emanuele Borgonovo74
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?
Quantitative Methods for Management
Emanuele Borgonovo75
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)
Quantitative Methods for Management
Emanuele Borgonovo76
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?
Quantitative Methods for Management
Emanuele Borgonovo77
Influence Diagram
Decisione
Riparazione_2g_perfetta
Riparazione_10g_Perfetta
Perdite
Quantitative Methods for Management
Emanuele Borgonovo78
Decision Tree
Riparazione_2g_perfetta
Riparazione_2g_perfetta
P_2g-50000-500000
Riparazione_2g_non_perfetta
1-P_2g-50000-500000-25000*0.15*365*years
Intervento_2g
Decisione
Riparazione_10g_Perfetta=0
Riparazione_10g_Perfetta
Riparazione_10g_Perfetta
P_10g
Riparazione_2g_perfetta=0-250000-1000000
Riparazione_10g_non_Perfetta
1-P_10g
Riparazione_2g_perfetta=0-1000000-250000-years*.05*25000*365
Intervento_10g
P_10g=0.9P_2g=0.3years=3
Quantitative Methods for Management
Emanuele Borgonovo79
Probability Values
• Three years
Quantitative Methods for Management
Emanuele Borgonovo80
2 and 4 years
• 2 years
• 4 years
Quantitative Methods for Management
Emanuele Borgonovo81
Chapter IV
Elements of Sensitivity Analysis
Quantitative Methods for Management
Emanuele Borgonovo82
Sensitivity Analysis
• Various Types of SA– One Way SA– Two Way SA– Tornado Diagrams– (Differential Importance Measure)
• Uncertainty Analysis– Monte Carlo– (Global SA)
Quantitative Methods for Management
Emanuele Borgonovo83
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)
Quantitative Methods for Management
Emanuele Borgonovo84
• Underline the critical dependencies of the outcome
Sensitivity Analysis (Run or withdraw)
Tornado Diagram atDecision
Expected Value
$49K $55K $61K $67K $73K
pfailure: 0.25 to 0.75
pwin: 0.3 to 0.7
pfive: 0.2 to 0.4
Sensitivity Analysis on pfailure
pfailureE
xpec
ted
Val
ue0.450 0.525 0.600 0.675 0.750
$62K
$59K
$56K
$53K
$50K
$47K
$44K
$41K
$38K
Run
Withdraw
Threshold Values:
pfailure = 0.597EV = $50K
Quantitative Methods for Management
Emanuele Borgonovo85
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
Quantitative Methods for Management
Emanuele Borgonovo86
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
Quantitative Methods for Management
Emanuele Borgonovo87
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
pfailure
Exp
ecte
d V
alue
0.450 0.525 0.600 0.675 0.750
$62K
$59K
$56K
$53K
$50K
$47K
$44K
$41K
$38K
Run
Withdraw
Threshold Values:
pfailure = 0.597EV = $50K
Quantitative Methods for Management
Emanuele Borgonovo88
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
Quantitative Methods for Management
Emanuele Borgonovo89
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.
Quantitative Methods for Management
Emanuele Borgonovo90
Example of a Tornado Diagram
Tornado Diagram atDecision
Expected Value
$49K $55K $61K $67K $73K
pfailure: 0.25 to 0.75
pwin: 0.3 to 0.7
pfive: 0.2 to 0.4
Quantitative Methods for Management
Emanuele Borgonovo91
Upsides and Downsides
• Upsides
– Easy numerical calculations
– Results immediately understandable
• 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
Quantitative Methods for Management
Emanuele Borgonovo92
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
Quantitative Methods for Management
Emanuele Borgonovo93
Process
• Identify how sensitivity analysis techniques work through analysis of several examples
• Formulate a definition
• Classify sensitivity analysis techniques accordingly
Quantitative Methods for Management
Emanuele Borgonovo94
Sensitivity Analysis Types
• Model Output:
• 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
)x,...,x,f(xU n21
Quantitative Methods for Management
Emanuele Borgonovo95
the Differential Importance Measure
• Nominal Model output:– No uncertainty in the model parameters– and/or parameters fixed at nominal value
• Local Decomposition:
• Local importance measured by fraction of the differential attributable to each parameter
ox
i xdU
dU)DIM(x i
nn
22
11
dxx
f...dx
x
fdx
x
fdU
Quantitative Methods for Management
Emanuele Borgonovo96
Global Sensitivity Indices
• Uncertainty in U and parameters is considered• Sobol’’s decomposition theorem:
• Sobol’Indices
20
Ω
2
x
x
ni...ii
...iii...ii
fxd)x(f
...dxdxf...
D
D)(xS
i1
i1
1n1
s1
s1
n
1i nji1n...12jiijii0 )x(f...)x,x(f)x(ff)xf(U
Quantitative Methods for Management
Emanuele Borgonovo97
Formal Definition of Sensitivity Analysis (SA) Techniques
• SA technique are Operators on U:
)]x,...,x,f(x[UI(x)^ n21
x1x2 xn
I(x1)I(xn) I(x2)
or
or
Quantitative Methods for Management
Emanuele Borgonovo98
Importance Relations
• Importance relations:– X the set of the model parameters; – Binary relation
xi xj iff I(xixj)
xi~xj iff I(xixj)
xi xj iff I(xixj)
xi xj iff I(xixj)
• Importance relations induced by importance measures are complete preorder
Quantitative Methods for Management
Emanuele Borgonovo99
Additivity Property
• In many situation decision-maker interested in joint importance:
• An Importance measure is additive if:
• DIM is additive always
• Si are additive iff f(x) additive and xj’s are uncorrelated
)xx(I)x,x(I jiji
)x(I)x(I)x,x(I jiji
Quantitative Methods for Management
Emanuele Borgonovo100
Techniques that fall under the definition of Local SA techniques
IMPORTANCEMEASURE
EQUATIONTYPEADDITIVE
DIM
dU
dUix
Local Yes
L0
i
i
xU
x
x
U Local No
Tornado Diagrams U Local No
One Way Sensitivity U Local No
Fussell-Vesely0
0
i0x
)x(U
x)x(U Local No
Risk Achievement Worth
0
i
U
xU Local No
Quantitative Methods for Management
Emanuele Borgonovo101
Global Importance Measures
IMPORTANCEMEASURE
EQUATION TYPE ADDITIVE
Sobol’ IndicesD
Ds1...ii Global No
Extended Fast
1j
2j
2
1p
2wp
2wp
i
BA2
BA2
S
j
iiGlobal No
Morris
)x,...,x,...,x(f
xd ni1i Global No
PearsonUi
ii
)x,Ucov(
Global No
Smirnov )X(Y)X(Ysup i2i1 Global No
Standardizedregression coefficients
kkb Global No
Quantitative Methods for Management
Emanuele Borgonovo102
Sensitivity Analysis in Risk-Informed Decision-Making and Regulation
• Risk Metric:
• xi is undesired event Probability
• Fussell-Vesely fractional Importance:
• Tells us on which events regulator has to focus attention
)x,...,x,f(xR n21
R
)x(R,)FV(x i
i
Quantitative Methods for Management
Emanuele Borgonovo103
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 Risk-
informed Decision-making and in the use of model information
Quantitative Methods for Management
Emanuele Borgonovo104
Chapter V
Uncertainty Analysis
Quantitative Methods for Management
Emanuele Borgonovo105
Uncertainty Analysis
Monte Carlo Simulation atDecision
Value
Pro
babi
lity
$10K $40K $70K $100K $130K
1.000
0.900
0.800
0.700
0.600
0.500
0.400
0.300
0.200
0.100
0.000
Quantitative Methods for Management
Emanuele Borgonovo106
Summary
• Distinction between Aleatory Uncertainty ed Epistemic Uncertainty
• Epistemic Uncertainty and Bayes‘ theorem
• Monte Carlo Method for uncertainty propagation
Quantitative Methods for Management
Emanuele Borgonovo107
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 “”, 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.
Quantitative Methods for Management
Emanuele Borgonovo108
Example: Model of the World• the Probability of Earthquakes is usually modeled through a Poisson
model:
• 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).
!n
)t(e)t,n(P
nt
Quantitative Methods for Management
Emanuele Borgonovo109
Some useful information on Poisson Distributions
• the Poisson Probability that n events happen on 0-t is:
• the sum on n=0... of P(n,t) is, obviously, equal to 1.
• the Probability of k>N is given by:
• E[n]=t
!n
)t(e)t,n(P
nt
0n
ttn
t
0n
nt
1ee!n
)t(e
!n
)t(e
N
0n
nt
1Nn
nt
!n
)t(e1
!n
)t(e
tt
0n
1nt
0n
nt
0n
nt
ete!1n
)t(te
!1n
)t(e
!n
)t(ne
Quantitative Methods for Management
Emanuele Borgonovo110
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( ):
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.20
1
2
3
4
5
6
7
8Epistemic distribution for the frequency of earthquakes
lambda
f(lam
bda)
5/150/150/15/1)(u
5/150/10)(u
Quantitative Methods for Management
Emanuele Borgonovo111
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:
d)(u!n
)t(ed),t,n(p
nt
Quantitative Methods for Management
Emanuele Borgonovo112
….
• This expression tells us that not necessarily all Poisson distributions weight the same. Thus:
• In our case: u()=c;
• Hence, there is an expected Probability!
d)(u!n
)t(ed),t,n(p)t,n(P
nt
λd!n
)tλ(ecλd)λ,t,n(p)]t,n(P[E
ntλ
Quantitative Methods for Management
Emanuele Borgonovo113
In General
• the MOW will depend on m parameters , ,…:
• the event Probability (P(t)) will be:
),....,t(MOW
.....dd,....),(f),....,t(MOW)t(P
Quantitative Methods for Management
Emanuele Borgonovo114
An problem
• the failure time of a series of components is characterized by the exponential Probability function :
• From the available data, it emerges that:
• What is the mean time to failure?
dtedf t
2.0p10/1
3.0p8/1
5.0p5/1
Quantitative Methods for Management
Emanuele Borgonovo115
Solution
• E[t]=
0
t3
0
t2
0
t1 1.0)dtet(5.0)dtet(5.0)dtet( 321
1.0/13.0/15.0/1 321
9.6
Quantitative Methods for Management
Emanuele Borgonovo116
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….
Quantitative Methods for Management
Emanuele Borgonovo117
Formula
• the Probability density of a parameter, after having obtained evidence and, changes as follows:
• 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
λd)λ(π)λE(L
)λ(π)λE(L)λ(π
0
0
Quantitative Methods for Management
Emanuele Borgonovo118
From discrete to continuous• Let us take Bayes‘ theorem for discrete events:
• 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(EAj) 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!!!
n
1iii
jj
j
)A(P)AE(P
)A(P)AE(P)EA(P
Quantitative Methods for Management
Emanuele Borgonovo119
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:
• Substituting the various terms, one finds Bayes‘ theorem for continuous random variables we have shown before
d)()E(L)A(P)AE(P 0
n
1iii
Quantitative Methods for Management
Emanuele Borgonovo120
Is it a fair coin?• What is the MOW?
• It is a binomial distribution with parameter p:
• 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:
• Let us get some evidence. • At the first tossing it is head• At the second tail • At the third head
knk )p1(pk
n)kn,k(P
otherwise0)p(π
1p0if1)p(π
0
0
Quantitative Methods for Management
Emanuele Borgonovo121
Result• First tossing
– Evidence: h.– MOW: L(hp)=p
– Prior: 0
• Second tossing: – Evidence: t– MOW: L(tp)=(1-p)
– Prior: 1
• Third tossing:– Evidence: h– MOW: L(hp)=p
– Prior: 2
• Equivalently:– Evidence: h, t, h– L(hthp)=p2(1-p)– Prior: 0
p2
pdp
1p
dp)p(π)ph(L
)p(π)ph(L)p(π 1
0
0
01
)pp(6
dp)pp(
)p1(p
dp)p(π)pt(L
)p(π)pt(L)p(π
2
1
0
21
12
)pp(12
dp)p1(p
)p1(p
dp)p(π)ph(L
)p(π)ph(L)p(π
32
1
0
2
2
2
23
)pp(12
dp1)p1(p
1)p1(p
dp)p(π)phth(L
)p(π)phth(L)p(π
32
1
0
2
2
2
03
Quantitative Methods for Management
Emanuele Borgonovo122
Graph
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
0
1
2
3
Quantitative Methods for Management
Emanuele Borgonovo123
Conjugate Distributions
• Likelihood– Poisson
• Posterior: gamma
• Prior distribution– Gamma
• with:
!n
)t(e)t,n(P
nt
βλ
1αα
0 e)α(Γ
λβ)β,α,λ(π
'1''
e)'(
')',',(
t'
r'
Quantitative Methods for Management
Emanuele Borgonovo124
Conjugate Distributions
• Likelihood– Normal
• Posterior: Normal
• Prior distribution:– Normal
• with:
2
x
x )σ
μx(
2
1
x
X eπ2σ
1)x(f
2
μ
x )σ
μm(
2
1
μ0 e
π2σ
1)m(π
20μ
2x
20μ
2x'
)σ(n)σ(
)σ(xn)σ(μμ
2
'x
)σ
'μx(
2
1
xG e
π2'σ
1)x(f
n/)σ()σ(
)σ()n/σ(σ
2x
2μ
2μ
2x'
x
Quantitative Methods for Management
Emanuele Borgonovo125
Conjugate Distributions
• Likelihood– Binomial
• Posterior, Beta:
• Prior:– Beta
• with:
knk )p1(pk
n
kq'q
knr'r
1r)1q(0 )p1(p)p(π
1'r)1'q(1 )p1(p)p(π
Quantitative Methods for Management
Emanuele Borgonovo126
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
Quantitative Methods for Management
Emanuele Borgonovo127
Epistemic Uncertainty in Decision-Making Problems
• Investment:
• Suppose that P.up is characterized by a uniform pdf between 0.3 and 0.7
• How does the decision changes?
• It is necessary to propagate the uncertainty in the model
21
2
1jjjRisky CP.up)1(CP.upC)C(P]U[E
21
2
1jjjRisky UP.up)1(UP.upC)C(P]up.PU[E
43
2
1jjjLessRisky CP.up)1(CP.upC)C(P]U[E
5
2
1jjjLessRisky C1C)C(P]U[E
Quantitative Methods for Management
Emanuele Borgonovo128
Analytical Propagation of Uncertainty
• It is the same problem of the MOW …
• Repeating for the other decisions and comparing the resulting mean values, one gets the optimal decision.
• Recall that: xd)x(f)x(g)]x,...,x,x(g[E n21
up.dP)up.P(fUP.up)1(UP.up
up.dP)up.P(f]up.PU[EUE
21
RiskyRisky
Quantitative Methods for Management
Emanuele Borgonovo129
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
Quantitative Methods for Management
Emanuele Borgonovo130
the core of Monte Carlo
• 1) Random Number Generator “u” between 0 and 1
• 2) Numbers u are generated with a uniform Distribution
• 3) Suppose that parameter is uncertain and characterized by the cumulative distribution reported below:
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
x
Distribuzione cumulativa esponenziale
0 1u
Quantitative Methods for Management
Emanuele Borgonovo131
Inversion theorem
• Inversion theorem:
• the values of sampled in this way have the Probability distribution from which we have inverted
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
x
Distribuzione cumulativa esponenziale1
0
)u(F 1
Quantitative Methods for Management
Emanuele Borgonovo132
Example
• Let us evaluate the volume of the yellow solid through the Monte Carlo method.
0in
nV
N
nlimV
VV0
Quantitative Methods for Management
Emanuele Borgonovo133
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.
Quantitative Methods for Management
Emanuele Borgonovo134
Results
• Strategy Selection Frequency
Monte Carlo Simulation atDecision
Value
Pro
babi
lity
$10K $40K $70K $100K $130K
1.000
0.900
0.800
0.700
0.600
0.500
0.400
0.300
0.200
0.100
0.000
• Decision Value Distribution
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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
• Substituting: E[URisky]=50, E[USafe ]= 20, E[ULess Risky ]= 20
21
2
1jjjRisky CP_up)1(CP_upC)C(P]P_upU[E
43
2
1jjjLessRisky CP_up)1(CP_upC)C(P]up_PU[E
5
2
1jjjLessRisky C1C)C(P]up_PU[E
2121
1
0
21
1
0
RiskyRisky
CC5.0CCP_upE
up_dP)up_P(fCP_up)1(CP_upup_dP)up_P(f]P_upU[E]U[E
2143
1
0RiskyLessRiskyLess CC5.0CCP_upEup_dP)up_P(f]P_upU[E]U[E
5LessRisky C]up_PU[E
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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):
– Prior: 0 uniform bewteen 0 and 1
• Bayes’theorem:
• Posterior Distribution
• E[p_up]=0.47
• d) Posterior Decision: E[URisky]=23, E[USafe ]= 20, E[ULess Risky ]= 9.2
1
0
87
87
1
0
87
87
up_dP)up_P1()up_P(
)up_P1()up_P(
up_dP1)up_P1()up_P(!8!7!15
1)up_P1()up_P(!8!7!15
P_up)|L(E
0
0.5
1
1.5
2
2.5
3
3.5
0
0.07
0.14
0.21
0.28
0.35
0.42
0.49
0.56
0.63 0.7
0.77
0.84
0.91
0.98
p0
p1
87 )up_P1()up_P(!8!7
!15P_up)|L(E
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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
More than 50
Clienti
P_50_up10000
Less than 50
1-P_50_up-5000
Improves
Service
Pmigl
-5000Does not improve
1-Pmigl
Invest
Decision
Not Invest
Clienti=0Servizio=0
0
P=0.1Pmigl=0.5P_500=0.5P_500_down=0.5P_500_up=0.7
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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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1
Opction
X1
2X2
3X3
4X4
5X5
X1=0.0X2=0.0X3=0.0X4=0.0X5=0.0
Opzione Valore
Preferences under Certainty
• Here is a diagram for the Choice
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Structuring Preferences
• Indifference Curves:
• Points on the same curve leave you indifference
x1
x2
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the Value Function
• You can associate a numerical value representing you preferences to each indifference curve:
• V(x) is the function that says how much of x i one is willing to exchange for an increase or decrease in xk
x1
x2
)x,...,x,x(v)x(V n21
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V(x)
• V(x) is a value function if it satisfies the following properties:
• a)
• b)
)''x(v)'x(v''x'x
)''x(v)'x(v''x'x
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Example• For the “first job choice”, suppose that you value
function is as follows:
• 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?
232
21 xx4x/3)x(v
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Preferences under Uncertainty
Opzione Evento Casuale Utilità
P11 U1
P12 U2
P13 U3
P14 U4
1
2
3
P41 U1
P42 U2
P43 U3
P44 U4
4
Decision
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))
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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:
)2x(UE)1x(UE
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Utility vs. Value
– One attribute Problem. Suppose that alternative 1 produces x1 and the 2 x2, then 12 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:
– To choose one need u(x1) and u(2)
P1X1
1-P1X2
XI
1
2
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Stochastic Dominance
0 1 2 3 4 5 6 7 8 9 100
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
x
Pro
babi
lity
dis
trib
utio
ns o
ver
x
Distributions over attribute x
1
2
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
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One Attribute Utility Functions
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Certainty Equivalent• Given the lottery:
• the value of x such that you are indifferent between x* for certain and playing the lottery.
• In equations:• N.B.: if you are risk neutral, then x*=E[x]
)x(uE*)x(u
P1X1
1-P1XN
X3
1
2
P1X1
1-P1X2
X*
1
2
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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
• 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
(£20)
0.500£40
0.500 £20
1£10
2£10; P = 1.000
2 : £10
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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:
• 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:
*xxERP
*x:IP
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Mailmatical Definition
• the Risk Aversion function is defined as:
• Or, equivalently:
• Supposing a constant risk aversion one gets an exponential utility function:
)x('u
)x("u:)x(r
))x('uln(dx
d)x(r
bea)x(u
)ee(1
)x(u)x(udtedt)t('u
e)x('ux))x('uln())x('uln(dx
d
x
xx0
x
x
t)x(u
)x(u
x
0
00
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Risk Preferences
• Constant Risk Aversion
• Compute constant through Certainty Equivalent (CE):
re1U
CE2/xx ee5.0e5.0 11
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Investment Results with Risk Aversion
Market up
Market
0.6001-exp(-200/70) = 1
Market Down
0.4001-exp(-(-160)/70) = -9
Blue Chip Stock
Decision
-3
Market up
0.6001-exp(-500/70) = 1
Market Down
0.4001-exp(-(-600/70)) = -5,278
-2,110
Bond=11-exp(-50/70) = 1; P = 1.000
TwoStock
prob_up=0.6 Risky Investment
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A quale value accetterei l’investimento rischioso
Sensitivity Analysis on prob_up
prob_up
Exp
ecte
d V
alue
0.40 0.52 0.64 0.76 0.88 1.00
400.0
0.0
-400.0
-800.0
-1,200.0
-1,600.0
-2,000.0
-2,400.0
-2,800.0
-3,200.0
Blue Chip Stock
Risky Investment
Bond
Threshold Values:
prob_up = 0.96EV = 0.5
prob_up = 1.00EV = 0.9
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Esempi of funzioni utility
• Linear: u=ax– Risk Properties:
• Risk Neutral
• Exponential:– Risk Properties:
• - sign: Constant Risk Aversion, + sign: Constant Risk Proneness
• Logarithmic: – Risk Properties:
• Decreasing Risk Aversion
bea)x(u cx
b)xln(a)x(u
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Problems
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problem VI-1
• For the following three utility functions,
• compute: – the risk aversion function r(x)– the risk premium for 50/50 lotteries– the insurance premium
xa)x(u
bea)x(u x
b)xln(a)x(u
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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:
• 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?
c
x
b
)x(e1)x(s 3
22
xa
11
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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
• 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
1 2 n
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Parallel Systems
• 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
1
2
n
In Out
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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:
• Properties:
– (XJ)n=Xj
– where is the complementary of XJ
• This simple definition enables one to use algebraic operations to describe the logical behavior of systems.
happenednothasEif0
happenedhasEif1X j
0XX jj jX
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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: E1E2 ... En
• From a logical p.o.v., we get the following expression:
n
1iin21T X11)X1(...)X1()X1(1X
E1 E2E3E1 E2E3
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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: E1E2 ... En
• the logical expression is:
n
1iin21T XX...XXX
E1 E2E3
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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
• 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.
n
ii
SeriesT YY
1
n
ii
ParallelT YY
1
)1(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.
1
2
3
In Out
2/3
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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:
• Let us go to a level below:
• Let us solve the calculations, noting that (Xi)n=Xi :
)M1()M1()M1()Z1(1X 3211T
)XX1()XX1()XX1()XXX1(1X 323121321T
321323121T XXX2XXXXXXX
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Probability Sum Rules
• We recall that, for generic Events:
• We recall that, if the events are independent:
• Rare Event Approssimation: neglect all terms corresponding to multiple events
)E...EE(P)1(...
)EEEE(P)EEE(P)EE(P)E(P)E(P
n21n
khji,ijkhhkji
kji,ijkiji
j ji,ijjii
n
1ii
)E(P)E,...,EE(P s
n
1snji
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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.
)EEE(P2)EE(P)EE(P)EE(P)X(P 321323121T
321323121T XXX2XXXXXXX
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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
23 2
1
4
3
2/3 1
3 4
2
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Problem VII-2
• for the following system:
• 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.
1
3 4
2
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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
Sì
No
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
No
HighPressure Pipe Room
Yes
No
No
Building Top Event
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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
• 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.
And Or event Base
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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 Static Filter
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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 Static filter
Fault Install.Elettr. Mech.
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• Engine=A• Static filter =B• Electr.=1, Mech=2, Fault=3, Install.=4
• What are the minimal cut sets?
4321
421321432431
4232432141314321
43432121
4321
T
XXXX
XXXXXXXXXXXX
XXXXXXXXXXXXXXXX
)XXXX(1[)]XXXX(11
)]X1()X1(1[1)]X1()X1(1[11
)B1()A1(1X
From Fault Trees to Structure Functions
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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:
)X(P)X(P)X(P)X(P)X(P 4321T
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Event Trees & Fault Trees
No
High pressurePipe Aeration 1
yes
NoNo
Aeration 2 Top Event
Aeraz. 1
engine filter
Fault Installaz.Electr. Mech.
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Probability of the Top Event
• From the Event Tree:
• Expanding:
• the conditional probabilities are found solving the corresponding Fault Trees
)AP2.Aeraz1.Aeraz.Condutt(P)Top(P
)AP.Condutt(P*).Condutt2.Aeraz1.Aeraz(P)Top(P
)APCond(P)AP,.Cond2.Aeraz(P)AP.,Condutt,2.Aeraz1.Aeraz(P
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Definitions
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Failure Density
• Given a system, tet us denote with
the Probability that the system fails between t and t+dt
• It must hold that:
dt)t(fs
1dt)t(f0
s
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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 within time T:
• Thus the Reliability [R(t)] is related to the failure time pdf as follows:
• Note that, if f(t) is continuous:
t
0
d)(f)tTPr()t(F
t
0
d)(f1)tTPr(1)t(R
)t(f)t('R
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General Fault rate
t
(t) InfantMortality
Useful Life
Aging
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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:
• Thus:
dt)t(fdt)t(R)t(
)t(R
)t('R
)t(R
)t(f)t(
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Legami between R(t), f(t) and (t)
• From the above definition, there follows:
• Relationship R(t)- (t):
• Relationship f(t)- (t):
)t(R
)t('R
)t(R
)t(f)t(
t
0
d)(
e)t(R
t
0
d)(
e)t()t(R)t()t(f
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time medio of failure (MTTF)
• The mean time to failure is defined as:
0
dt)t(ftMTTF
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System Reliability
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Reliability of systems in Series
• Series:
• if independence is assumed:
• Thus:
• Faulure rate:
n
1iis )t(R)t(R
)tT...TT(P)t(F1)t(R)tT(P N21SSS
)tT(P)tT(P)tT(P)tT(P N21S
n
1iiS )t()t(
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Reliability of systems in Parallelo
• Failure Probability of the system:
• if independent:
• Thus:
• Failure Rate:
N
1iisS
n
1iis )t(R11)t(F1)t(R)t(F)t(F
)tT...TT(P)t(R1)t(F)tT(P N21SSS
)tT(P)tT(P)tT(P)tT(P N21S
)t(...)t()t()t(
1)1(
....)t()t()t(
1
)t()t()t(
1
)t()t(
1
)t()t(
1
)t(
1...
)t(
1
)t(
1
)t(
1
n321
n
321321
3121n21s
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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:– Thus:
– where 2 indicates that there are two components in standby, while the subscript denotes the second component
– 2) Three components:
– Thus:
)TtTtT(P)tT(P)tT(P 12112
S
t
0
1121112
S dt)tt(R)t(f)t(R)t(R
)TTtTTtTtT(P
)TtTtT(P)tT(P)tT(P
213121
12113
S
21
t
0
tt
0
21322112
S3
S dtdt)ttt(R)t(f)t(f)t(R)t(R1
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Standby Systems with const. failure rates
• For a standby system, it holds that:
• then P(t<T) is given by the convolution of the fi(t).
• If these distributions are exponential and the failure rates identical:
N21 T...TTT
1n
0i
itn
s !i
)t(eR
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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 :
• and
t
0
1121112
S dt)tt(R)t(f)q1()t(R)t('R
21
t
0
tt
0
213221122
S3
S dtdt)ttt(R)t(f)t(f)q1()t('R)t(R1
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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
23 2
1
4
3
2/31
3 4
2
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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:
• con e 0t 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.
1t
)t(
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Problema VIII-6
• Dato un componente con il tasso di guasto (t) seguente:
• calcolare:• R(t), f(t), e il MTTF del componente
10tset100
1
10t1se1
1t0se1t
1
)t(λ
2
0 5 10 150
1
2
3
4
5
6
7
8
9
10
t
l(t)
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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 SoluzioneRottura Primario
Spegnimento
No
Si’
Raffreddamento Top Event
Spegnimento
Bar.Sof. El.
Si’Raffreddamento
Pompa 1. Pompa 2
On Demand
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Prob. VIII-2 Soluzione
• Assumiamo eventi rari.• La frequenza si calcola dalla combinazione degli eventi:
• dove: frottura=.000001 per anno
• P(Spegn|rottura prim.)=P(Sof|rottura prim.)+P(E|rottura prim.)+P(Bar|rottura prim.)=0.00111
• P(Raff| rottura prim.)= =.0010995
• Quindi:
).Pr()Pr( imRotturaSpegnPimRotturaRaffPfF rottura
2
)10010000
1(
e1q
9102.20011.00109.1000000/1F
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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).– – Ore in un anno: 8760. – Sostituendo i numeri:
• Confrontatela con quella di un sistema in cui i componenti sono messi in:
– Parallelo: – ¾ supponendo I ratei di guasto =1/8000.
• Risultato: 0.11
– 2/4 supponendo I ratei di guasto =1/8000• Risultato: 0.41
tttt 4321 eeee)t(R
006.0eeee)t(R 5000
8760
10000
8760
8000
8760
6000
8760
75.0)e1()e1()e1()e1(1)t(R 5000
8760
10000
8760
8000
8760
6000
8760
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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).
– 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?
99967.0)e1()e1(1)t(R 30000000
61320
30000000
61320
p
99983.0t1edt)tt(R)t(f)t(R)t(R tt
0
1121112
S
99967.0)e1()e1(1)t(R 30000000
61320
30000000
61320
p
99965.0t)q1(1edt)tt(R)t(f)q1()t(R)t(R tt
0
1121112
S
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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:
• Indisponiblita’ media in T:
• Indisponibilta’ media limite:
)t(qlimt
T
dt)t(q
q
T
0
T
dt)t(q
limq
T
0
Tlim
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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:
• A t=0 il sistema entra in funzione dopo la manutenzione. Dopo un tempo t= torna di nuovo in manutenzione. La manutenzione dura r. e’ il tempo in cui il componente e’ soggetto a rotture casuali con (t).
• Si nota che il problema e’ periodico, di periodo T= r+
• Durante il sistema ha una indisponibilita’ istantanea pari alla sua probabilita’ di rottura, se, come da ipotesi, non ci sono riparazioni:
r t r
'dt)'t(
S
t
0e1)t(R1)t(F)t(q
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Calcolo della Indisponibilita’: un unico componente, una sola modalita’ di guasto
• L’indisponibilta’ istantanea risulta quindi:
• Da cui l’ind. Media:
• Supponiamo cost e 1. Quindi utilizziamo approssimaz. Taylor:
• Sostituiamo nella ind. Media, e assumiamo r<< :
r
'dt)'t(
t1
t0e1)t(q
t
0
r
r
r
0
'dt)'t(
dte1
q
t
0
t...)t1(1e1e1 t'dt)'t(
t
0
TT
2
1q r
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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(E1E2)=P(E1)+P(E2)-P(E1E2)=Q0+F(t)-Q0F(t).
• Riassumendo, tra 0 e 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’:
r
0000
t1
t0tQtQ)1( )t(q
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Indisponibilita’ media con piu’ modi di guasto
• L’indisponibilita’ media sull’intervallo 0 + r e’:
• Due assunzioni: 1) Eventi rari 2) + r
r
r
2
0
2
000 2Q
2Q)1(
q
r00 2
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Rappresentazione equivalente
• La funzione struttura e’: • XC=1- (1-Xt) (1-XQ0)(1-X0)(1-X)=
• = Xt +XQ0+X0+X-termini di ordine superiore….• Approssimazione eventi rari:
• XC= Xt +XQ0+X0+X
Componente
0 t Q0
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Il caso di due componenti
• Sostituzioni successive
• Periodo: +2r
• Sostituzioni distanziate
r1r2 r2 r1rr r r r +2r+2r
Indisponibilita’ media e’ la somma di piu’ termini:
“R”: random, “C” common cause, “D” demand e “M” maintenance
MDCR qqqqq
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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()
ottimale: r=24 h, =1/10000 (1/h) ott=700hr
• Con =1/100000 (1/h) ott=2200hr
r2
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Esempio 2
• Ottimizzazione in considerazione del costo di sostituzione e della disponibilita’
• Funzione obiettivo:
ottimale:
• Occorre introdurre vita del’impianto L.
• Si ha: dove c0 e’ il costo unitario di riparazione
)(C)(qC)(B q
0d
)(Bd
0d
)(dB
2
2
0cL
)(C
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Esempio 2
• Introduciamo poi il costo della indisponibilita’:
• definito come multiplo del costo singola riparazione.• Funzione energia:
• Intervallo ottimale:
Cq a c0 c0
E Cq q 0 Q0 r C C
E d
da c0
1
2
r
2
L
2
c0
2r
L
1
a
1
2
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Esempio 2
c0 5
L 70000
a 10
r 24
1
10000
0 0.001
Q0 0.001 1.185 10
4
0 5 104
1 105
0
2000
40003.512 10
3
59.363
E t( )
1000000 t
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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 .
• 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:
• 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.
1etMTTR
e1)t(Rip
e)t(rip
0
t
t
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 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
• Proprieta’: • 1)
• 2)Se allora stato i e’ detto assorbente
MM1M
21
M11211
ij
PP
P
PPP
PM
1PM
1iij
1Pii
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Esempio
• Applichiamo uno schema a stati per il sistema in standby. Otteniamo:
1 2 3P23
P32P21
P12
P13
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Equazioni di Markov/Kolmogorov
• Dove A e’ la matrice di transizione del sistema, P e’ il vettore delle probabilita’ degli stati del sistema.
PAdt
dP
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Costruzione della matrice di transizione
• 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 (-). • Quindi:
1 2P21
P12
1 2
211 PP
dt
dP
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La matrice di transizione• Analogamente:
• Quindi:
• La matrice di transizione e’:
• Il sistema di equazioni differenziali diventa:
2
1
21
A
211 PP
dt
dP
212 PP
dt
dP
212 PP
dt
dP
PAdt
dP
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Emanuele Borgonovo256
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:
• Disponibilita’ asintotica:
• Interpretazione: tempo che occorre in media alla riparazione diviso il tempo totale
• Disponibilita’ media su T:
t)λμ(1 e
λμ
λ
λμ
μ)t(P +
++
+=
MTTRMTTF
MTTF
λμ
μ)t(Plim 1
t +=
+=
∞→
q T( )
2exp T
2
Quantitative Methods for Management
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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 =5000:
• (Soluzione: q=.175)• Calcolare l’intervallo di sostituzione ottimale e
l’indisponibilita’corrispondente, con L=70000, a=10 e a=.
• (Soluzione: =14500, q=0.5; =849, q=0.06)
0 0.002
Q0 0.002
r 24
1
15000