1. Quantitative Methods for Management
Emanuele Borgonovo
Quantitative Methods for Management
First Edition
Decision
Market
Return
Structural
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2. Chapter three: Models
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3. 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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4. 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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5. 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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6. Chapter II Introductory Elements of Probability theory
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7. 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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8. Kolmogorov Axioms U B A
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9. Areas and rectangles? U C A B D and
Suppose one jumps into the area U randomly. Let P(A) be the Probability to jump into A. What is its value?
It will be the area of A divided by the area of U: P(A)=A/U
Note that in this case: P(U)=P(A)+ P(B)+ P(C)+ P(D)+ P(E), since there are no overlaps
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10. Conditional Probability
Consider events A and B. the conditional Probability of A given B, is the Probability of A given the B has happened. One writes: P(A|B) U B A AB
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11. Conditional Probability
Suppose now that B has happened, i.e., you jumped into area B (and you cannot jump back!). B AB A
You cannot but agree that:
P(A|B)=P(AB)/P(B)
Hence: P(AB)=P(A|B) *P(B)
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12. Thus, for independent events:
Independence
Two events, A and B, are independent if given that A happens does not influence the fact that B happens and vice versa. B B A A AB
Thus, for independent events:
P(AB)=P(A)*P(B)
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13. 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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14. In the subjectivist approach, Probability changes with information
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15. Bayes’ theorem Hypothesis: A and B are two events. A has happened.
Thesis: P(B) changes as follows:
P(B) before A
New value of the Probability of B
Probability of A
Probability of A given B
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16. 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:
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17. Bayes’ theorem Proof Starting point Conditional Probability formula
thesis
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18. the Total Probability theoremC D U
and
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 :
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19. 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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20. 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.
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21. Cumulative Distribution Function
Given a continuous random variable X, the Probability that X<x is given by:
If f(x) is continuous, then:
Note:
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22. 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:
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23. 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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24. 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.
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25. Pdf and Cdf of the Exponential Distribution5 10 15
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9 1
P(t<T)
f(t)
T/t
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26. Expected Value, Variance and Percentiles
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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27. the Normal Distribution
Is a symmetric distribution around the mean
Pdf:
Cdf:
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28. Graphs of the Normal Distribution
Cumulative Gaussian Distribution
10000
9000
8000
7000
6000
5000
4000
3000
2000
1000 -5 -4 -3 -2 -1 1 2 3 4 x
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29. Lognormal Distribution
Pdf
Cdf
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30. Lognormal Distribution
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31. 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?
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32. 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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33. 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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34. Problem II-3 For the example of the two rings, determine: P(B|o)
Solution: there are only two possible events, A or B. Thus, P(Bor)=1-P(Aor)=1/3
P(Ba)
P(Ba)=1, since B is the only event that has a silver ring. One can also show it using Bayes’ theorem:
P(Ba)=P(aB)*P(B)/[P(aB)* P(B)+P(aA)*P(A)]. Since P(aA)=0, one gets 1 at once.
the Probability of being in A given that the picked ring is golden in two consecutive extractions, having put the ring back in the box after the first extraction
Using Bayes’ theorem:
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35. Problem II-3
where, in the formula, subscript 1 indicates the probabilities after the information of the first extraction has been taken into account:
P1(B)=P(Bor)=1/3 and P1(A)=P(Aor)=2/3.
One can note that P(2oA)=1, and P(2oB)=1/2. P(2oB) is the Probability to pick a golden ring at the second run, given that one is in state B.
Thus, we have all the numbers to be substituted back in the theorem:
It is the same problem as in the example, but with adjourned probabilities.
the Probability of being in B given that the picked ring is golden in two consecutive extractions, having put the ring back in the box after the first extraction
Solution: 1-P(A2o)=0.2
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36. Chapter III: Introductory Decision theory
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37. 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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38. 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 Beaumont-en-Auge - March Paris)
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39. 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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40. Decision-Making Problem Elements
Values and Objectives
Attributes
Decision Alternatives
Uncertain Events
Consequences
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41. 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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42. Decision Analysis Tools
Influence Diagrams
Decision Trees
Market up
prob_up
Market down
1-prob_up
Less Risky
Risky
Safe
How should the
invest $1000?
Decision
Market
Return
Structural
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43. Influence Diagrams Influence diagrams (IDs) are… ID formal definition:
“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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44. ID Elements = Decision = Random Event = utility NODES ARCS
Informational Arcs
probabilistic Dependency Arcs
Structural Arcs
= Decision
= Random Event
= utility
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45. ID Elements Informational Arc Structural Decision Node Chance Node
Value Node
Conditional Arc
probabilistic
Dependency
Informational Arc
Sequential
Decisions
Structural
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46. 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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47. Case Study 2 - Leaking SG tube
Influence Diagram for Case Study 2
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48. Market Decision Return Influence Diagram Structural
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49. 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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50. the Decision Tree (DT) Market up Less Risky Market down How should the
1-prob_up
Less Risky
Risky
Safe
How should the
invest $1000?
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51. 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)
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52. example How should the invest $1000? Market up C1 P.up Market down C2
Blue Chip Stock C3 C4
Risky investment
CD paying 5% C5
How should the
invest $1000?
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53. Problem Solution Using the previous formula:
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54. the Best Investment for a Risk Neutral Decision - Maker
Market up
0.600
$200
Market down
0.400
($160)
Blue Chip Stock
$56
$500; P = 0.600
($600); P = 0.400
Risky investment
$60
CD paying 5%
return = $50
How should the
invest $1000?
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55. What to do? 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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56. Example of a simple ID Decision Profit Engine failure
Final Classification
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57. From IDs to Decision Trees
Out of first five
1.000
$20,000; P = 0.500
failure
Engine failure
0.500
$20,000
Win
$110,000; P = 0.250
In first five
0.300
$105,000; P = 0.150
0.200
$50,000; P = 0.100
No failure
$94,500
Run
Decision
$57,250
Old sponsor
$50,000
Withdraw
Engine_failure=0
pfailure=0.5
pfive=0.30
pout=0.2
pwin=0.5
Run : $57,250
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58. 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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59. Influence Diagram
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60. Decision Tree
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61. 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
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62. Example: investing
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63. EVPI for the Example
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64. EVPI Result
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65. Problems
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66. 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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67. 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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68. Solution
Influence Diagram the
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69. Solution Corresponding Decision Tree
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70. Solution
Influence Diagram II
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71. Solution
Decision Tree II:
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72. Sales_Costs
Given the following Influence Diagram and Decision Tree, given P_High and P_High|High, P_high|low, find the value of the Alternatives as a function of the assigned probabilities. Supposing P_high=0.5 and P_high|high=P_high|low=0.3, find the preferred alternative.
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
High
Sales
P_Alte|high
Low
1- P_Alte|high
-10
high
Cost
P_high
P_ P_Alte|high 20
Basso
1-P_high
Invest
Decision
Do not Invest 5
high=0.5
P_Alte=0.3
P_high=0.5
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73. Solution Sales_Costs
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74. 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 EUR 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 EUR With Probability P_10g the line will be as before the breakdown.
According to you, the residual life of the system is important for the decision?
Suppose that there are still three years of life for the system.
Which strategy should you carry out?
Determine the decision problem elements. Draw the Influence Diagram and the corresponding Decision Tree. Find the value or values of the probabilities for which a complete repair is more convenient than a partial one.
What would you would advise to the director of the system to do based on the engineer estimates?
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75. EVPI Problems
Determine the EVPI for the random event nodes in the previous IDs and DTs of the following problems:
Sales_Costs (lez. 2)
Production break-down (lez.2)
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76. 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 € 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 € With Probability P_10g the line will be as good as new.
In your opinion, the residual plant life is relevant to this decision?
Suppose that there are still three years of life for the plant. What should one decide?
Identify the decision making elements
Draw the Influence Diagram for the problem
Find the values of the probabilities for which one or the other intervention is more convenient
What would your suggestion to the plant director be?
What would happen if the plant life were 2 and 4 years instead of 3?
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77. Influence Diagram
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78. Decision Tree
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79. Probability Values
Three years
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80. 2 and 4 years
2 years
4 years
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81. Chapter IV Elements of Sensitivity Analysis
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82. Sensitivity Analysis Various Types of SA Uncertainty Analysis
One Way SA
Two Way SA
Tornado Diagrams
(Differential Importance Measure)
Uncertainty Analysis
Monte Carlo
(Global SA)
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83. How do we use SA? a) To check model correctness and robustness
b) To Further interrogate the model
Questions:
What is the most influential parameter with respect to changes?
What is the most influential parameter on the uncertainty (data collection)
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84. Sensitivity Analysis (Run or withdraw)
Underline the critical dependencies of the outcome
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85. Summary Sensitivity Analysis Uncertainty Analysis One way sensitivity
Two way sensitivity
Tornado Diagrams
Uncertainty Analysis
Aleatory Uncertainty
Epistemic Uncertainty
Bayes‘ theorem for continuous distributions
Monte Carlo Method
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86. Sensitivity Analysis
By sensitivity analysis one means the study of the change in results (output) due to a change in one of the model parameters (input)
the simplest Sensitivity Analysis types are:
One way sensitivity
Two way sensitivity
Tornado diagrams
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87. 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
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88. Two-way Sensitivity Analysis
In a Two-way Sensitivity Analysis two parameters are varied at the same time.
Instead of a line, one obtains a plane, in which each region identifies the preferred alternative that correspond to the combination of the two parameter values
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89. Tornado Diagrams the analysis is focused on the preferred decision
An interval of variation for each input parameter is chosen
the parameters are changed one at a time, while keeping the oilrs at their reference value
the change in output is registered
the output change is shown by means of a horizontal bar
the most important variable is the one that corresponds to the longest bar.
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90. Example of a Tornado Diagram
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91. Upsides and Downsides Upsides Downsides
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
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92. Sensitivity Analysis and Parameter Importance
Relevance of parameter in a model with respect to a certain criterion
Sensitivity Analysis used to Determine Parameter Importance
Concept of importance not formalized, but extensively used
Risk-Informed Decision Making
Resource allocation
Need for a formal definition
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93. Process
Identify how sensitivity analysis techniques work through analysis of several examples
Formulate a definition
Classify sensitivity analysis techniques accordingly
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94. 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
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95. 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
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96. Global Sensitivity Indices
Uncertainty in U and parameters is considered
Sobol’’s decomposition theorem:
Sobol’Indices
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97. Formal Definition of Sensitivity Analysis (SA) Techniques
SA technique are Operators on U: x1 x2 xn
I(x1)
I(xn)
I(x2)
 or 
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98. Importance Relations Importance relations: xi xj iff I(xi)>I(xj)
X the set of the model parameters;
Binary relation
xi xj iff I(xi)>I(xj)
xi~xj iff I(xi)=I(xj)
xi xj iff I(xi)<I(xj)
Importance relations induced by importance measures are complete preorder
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99. 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
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100. Techniques that fall under the definition of Local SA techniques
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101. Global Importance Measures
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102. 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
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103. 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
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104. Chapter V Uncertainty Analysis
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105. Uncertainty Analysis
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106. Summary
Distinction between Aleatory Uncertainty ed Epistemic Uncertainty
Epistemic Uncertainty and Bayes‘ theorem
Monte Carlo Method for uncertainty propagation
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107. 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 “Epist”, knowledge
it reflects our lack of knowledge in the value of the Aleatory Model input parameters. the aleatory model or model of the world is the model chosen to represent the random event.
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108. 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).
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109. 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
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110. 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( ):
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111. 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:
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112. ….
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!
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113. In General the MOW will depend on m parameters , ,…:
the event Probability (P(t)) will be:
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114. 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?
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115. Solution
E[t]=
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116. Continuous form of Bayes Theorem
Epistemic Uncertainty and Bayes’ theorem are connected, in that we know that we can use evidence to update probabilities.
For example, suppose to have a coin in your hands. will it be a fair with, i.e., will the Probability of tossing the coin lead to 50% head and tails?.
How can we determine whether it is a fair coin?
….let us toss it….
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117. 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
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118. 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(EAj) has the meaning of Probability that the evidence and is realized given that  equals * . One writes: L(E   ) and it is the likelihood function
Note: it is the MOW!!!
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119. 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
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120. Is it a fair coin? What is the MOW? What is the value of p?
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
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121. Result Equivalently: Evidence: h, t, h L(hthp)=p2(1-p) First tossing
MOW: L(hp)=p
Prior: 0
Second tossing:
Evidence: t
MOW: L(tp)=(1-p)
Prior: 1
Third tossing:
Evidence: h
MOW: L(hp)=p
Prior: 2
Equivalently:
Evidence: h, t, h
L(hthp)=p2(1-p)
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122. Graph 3 Quantitative Methods for Management 0.1 0.2 0.3 0.4 0.5 0.6
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9 1
1.2
1.4
1.6
1.8 2 3
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123. Conjugate Distributions
Likelihood
Poisson
Posterior: gamma
Prior distribution
Gamma
with:
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124. Conjugate Distributions
Likelihood
Normal
Posterior: Normal
Prior distribution:
Normal
with:
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125. Conjugate Distributions
Likelihood
Binomial
Posterior, Beta:
Prior:
Beta
with:
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126. Summary of Conjugate Distributions
MOW - Likelihood
Prior Distribution
Posterior Distribution
Binomiale
Beta
Poisson
Gamma
Normal
Negative binominal
the table represents the distribution of the observations, the prior distribution of the parameter w and the posterior distribution of w. For more details, see Chapter 9 of DeGroot.
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127. 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
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128. 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:
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129. the Monte Carlo Method Sampling a value of P.up
For all sampled P.up the Model is re-evaluated.
Information:
Frequency of the preferred alternative
Distribution of each individual Alternative
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130. 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: 1 u
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131. Inversion theorem 1 Inversion theorem:
Inversion theorem:
the values of  sampled in this way have the Probability distribution from which we have inverted
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132. Example
Let us evaluate the volume of the yellow solid through the Monte Carlo method. V0 V
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133. Application to ID and DT
For every Model parameter one creates the corresponding epistemic distribution
Run nr. 1:
One generates n random numbers between 0 and 1, as many as the uncertain variables are
One samples the value of each of the parameters inverting from the corresponding distribution
Using these values one evaluates the model
One keeps record of the preferred alternative and of the value of the decision
the procedure is repeated N times.
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134. Results Strategy Selection Frequency Decision Value Distribution
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135. 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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136. 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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137. Prob. 5-2
Substituting: E[URisky]=50, E[USafe ]= 20, E[ULess Risky ]= 20
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138. 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
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139. 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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140. 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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141. Influence Diagram Quantitative Methods for Management Clienti Service
More than 50
Clienti
P_50_up
10000
Less than 50
1-P_50_up
-5000
Improves
Service
Pmigl
Does not improve
1-Pmigl
Invest
Decision
Not Invest
Clienti=0
Servizio=0
P=0.1
Pmigl=0.5
P_500=0.5
P_500_down=0.5
P_500_up=0.7
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142. Chapter VI Introduction to Decision theory
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143. Summary Preferences under Certainty Preferences under Uncertainty
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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144. 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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145. Preferences under Certainty
Here is a diagram for the Choice 1
Opction X1 2 X2 3 X3 4 X4 5 X5
X1=0.0
X2=0.0
X3=0.0
X4=0.0
X5=0.0
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146. Structuring Preferences
Indifference Curves:
Points on the same curve leave you indifference x1 x2
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147. 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 xi one is willing to exchange for an increase or decrease in xk x1 x2
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148. V(x)
V(x) is a value function if it satisfies the following properties: a) b)
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149. 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?
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150. Preferences under Uncertainty
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))
P11 U1
P12 U2
P13 U3
P14 U4 1 2 3
P41
P42
P43
P44 4
Decision
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151. 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:
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152. Utility vs. Value
One attribute Problem. Suppose that alternative 1 produces x1 and the 2 x2, then 12 if x1>x2
Let us take two Alternatives 1 and 2, with x1>x2, given with certainty.
the value function will give us: v(x1)>v(x2)
Let us now consider the following problem:
To choose one need u(x1) and u(2) P1 X1
1-P1 X2 XI 1 2
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153. Stochastic Dominance Distribution 1 is dominated by1 2 3 4 5 6 7 8 9 10
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9 x
Probability distributions over x
Distributions over attribute x
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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154. One Attribute Utility Functions
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155. 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] P1 X1
1-P1 XN X3 1 2 P1 X1
1-P1 X2 X* 1 2
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156. 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
£20 1
£10 2
£10; P = 1.000
2 : £10
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157. 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:
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158. Mailmatical Definition
the Risk Aversion function is defined as:
Or, equivalently:
Supposing a constant risk aversion one gets an exponential utility function:
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159. Risk Preferences Constant Risk Aversion
Compute constant  through Certainty Equivalent (CE):
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160. Investment Results with Risk Aversion
Market up
Market
0.600
1-exp(-200/70) = 1
Market Down
0.400
1-exp(-(-160)/70) = -9
Blue Chip Stock
Decision -3
1-exp(-500/70) = 1
1-exp(-(-600/70)) = -5,278
-2,110
Bond=1
1-exp(-50/70) = 1; P = 1.000
TwoStock
prob_up=0.6
Risky Investment
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161. A quale value accetterei l’investimento rischioso
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162. Esempi of funzioni utility
Linear: u=ax
Risk Properties:
Risk Neutral
Exponential:
- sign: Constant Risk Aversion, + sign: Constant Risk Proneness
Logarithmic:
Decreasing Risk Aversion
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163. Problems
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164. 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
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165. 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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166. 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?
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167. Chapter VII the Logic of Failures
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168. Elements of Reliability theory
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169. 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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170. 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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171. Series 1 2 n
Every component is critical w.r.t. the system being able to perform its mission.
the fault of just one component is sufficient to provoke system failure
Redundancy: 0
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172. Parallel Systems 1 2 n In
Out
Each of the components is capable of assuring that the system accomplish its tasks.
Thus, to provoke the failure of the system, all the components must be contemporarily failed
Redundancy: n-1
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173. Elements of System Logics
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174. 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.
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175. Series Systems Let Ei denote the event “the i-th component failed.”
Let XT denote the event: “the System failed”.
XT takes the name of Top Event.
For the system failure, by definition of series, it is enough that one single component failes. Thus it is enough that E1 or E2 or …. or En is true.
From a set point of view: E1E2  ...  En
From a logical p.o.v., we get the following expression: E1 E2 E3 E1 E2 E3
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176. Parallel Systems
Let Ei denote the event “the i-th component has failed.”
Let XT denote the event: “the system has failed”.
the condition for failure of the system is that all component fail. This happens if E1 and E2 and … En are true at the same time.
From a Set point of view: E1E2  ...  En
the logical expression is: E1 E2 E3
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177. 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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178. 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.
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179. n/N Logics
n/N logics are intermediate logics between series and parallel.
N represents the total number of components in the system and n the number of components that must contemporarily fail to break the system.
As an example, a system has a 2/3 logic if it has 3 components and when two components have failed the system failes.
2/3 1 2 3 In
Out
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180. 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 :
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181. 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
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182. Golden Rule
In practice: the System Failure Probability is computed from the solved Structure Function, substituting to indicator Xi the corresponding Event Probability.
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183. 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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184. Problems
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185. 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 2 1 4 3
2/3 1 3 4 2 1 2 3
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186. 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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187. Chapter VIII Elements of Reliability
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188. 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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189. Even Trees
Event Trees: represent the sequence of events that lead to the event top.
Initiating Event
Event 1
Event 2
Top Event Sì No
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190. 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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191. Gas Leakage Fault - TreeNo
High
Pressure
Pipe
Room
Yes
Building
Top Event
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192. Fault Trees And Or event Base
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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193. Example Aeration 1 Static Filter Engine
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
Static Filter
Engine
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194. 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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195. Level II Aeration 1 Engine Static filter Fault Install. Elettr. Mech.
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196. From Fault Trees to Structure Functions
Engine=A
Static filter =B
Electr.=1, Mech=2, Fault=3, Install.=4
What are the minimal cut sets?
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197. 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:
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198. Event Trees & Fault TreesNo
High pressure
Pipe
Aeration 1
yes
Aeration 2
Top Event
Aeraz. 1
engine
filter
Electr.
Mech.
Installaz.
Fault
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199. Probability of the Top Event
From the Event Tree:
Expanding:
the conditional probabilities are found solving the corresponding Fault Trees
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200. Definitions
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201. Failure Density Given a system, tet us denote with
the Probability that the system fails between t and t+dt
It must hold that:
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202. 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:
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203. General Fault rate Infant (t) Mortality Aging Useful Life t
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204. 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:
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205. Legami between R(t), f(t) and (t)
From the above definition, there follows:
Relationship R(t)- (t):
Relationship f(t)- (t):
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206. time medio of failure (MTTF)
The mean time to failure is defined as:
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207. System Reliability
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208. Reliability of systems in Series
if independence is assumed:
Thus:
Faulure rate:
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209. Reliability of systems in Parallelo
Failure Probability of the system:
if independent:
Thus:
Failure Rate:
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210. 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:
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211. 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:
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212. 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
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213. Problems
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214. problem VIII-1
Write the Fault Trees for the following systems and derive the structure function: 2 1 4 3
2/3 1 3 4 2 1 2 3
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215. 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 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’ per anno.
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216. 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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217. 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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218. Problema VIII-5: Distribuzione Weibull
Dato un componente con rateo di guasto:
con  e 0t calcolare:
R(t), f(t), il MTTF e la varianza del tempo medio di guasto
R(t) è detta distribuzione di Weibull
Disegnare (t),f(t) ed R(t) per =-1,1, 2. Dedurne che la Weibull può essere utilizzata per descrivere il tasso di guasto di componenti in tutta la vita del componente.
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219. Problema VIII-6
Dato un componente con il tasso di guasto (t) seguente:
calcolare:
R(t), f(t), e il MTTF del componente 5 10 15 1 2 3 4 6 7 8 9 t
l(t)
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220. 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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221. 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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222. 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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223. 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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224. Prob. VIII-2 Soluzione Rottura Primario Spegnimento Raffreddamento
Top Event No
Si’
Si’
Raffreddamento
Pompa 1.
Pompa 2
On Demand
Spegnimento
Bar.
Sof.
El.
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225. Prob. VIII-2 Soluzione Assumiamo eventi rari.
La frequenza si calcola dalla combinazione degli eventi:
dove: frottura= per anno
P(Spegn|rottura prim.)=P(Sof|rottura prim.)+P(E|rottura prim.)+P(Bar|rottura prim.)=
P(Raff| rottura prim.)= =
Quindi:
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226. 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
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227. 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?
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228. Capitolo IX Decisioni Operative: Ottimizzazione delle Manutenzioni
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229. Decisioni Operative Decisioni di Affidabilita’ o Reliability Design
Decisioni di Optimal Replacement
Decisioni di ispezione ottimale
Decisioni di riparazione ottimale
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230. 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:
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231. 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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232. Effetto delle manutenzioni
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233. 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
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234. 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<<  :
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235. 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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236. Indisponibilita’ istantanea con piu’ modi di guasto
Consideriamo per un componente i modi di guasto indicati in precedenza.
A t=0 il componente puo’ essere gia’ guasto se disabilitato dall’erronea manutenzione. Questo evento ha probabilita’ 0. Con probabilita’ (1- 0) il componente invece potra’ invece aver superato con successo la manutenzione. In questo caso il componente potra’ rompersi “on demand” (E1) o con tasso di guasto (t) (E2). Si ha: P(E1E2)=P(E1)+P(E2)-P(E1E2)=Q0+F(t)-Q0F(t).
Riassumendo, tra 0 e  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’:
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237. Indisponibilita’ media con piu’ modi di guasto
L’indisponibilita’ media sull’intervallo 0 + r e’:
Due assunzioni: 1) Eventi rari 2) + r 
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238. Rappresentazione equivalente
Componente 0 
t Q0
La funzione struttura e’:
XC=1- (1-Xt) (1-XQ0)(1-X0)(1-X)=
= Xt +XQ0+X0+X-termini di ordine superiore….
Approssimazione eventi rari:
XC= Xt +XQ0+X0+X
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239. Il caso di due componenti
Sostituzioni distanziate
Sostituzioni successive
Periodo: +2r 
r1
r2 
r1
r2 r r r r r
+2r
+2r
Indisponibilita’ media e’ la somma di piu’ termini:
“R”: random, “C” common cause, “D” demand e
“M” maintenance
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240. 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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241. 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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242. 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/ (1/h) ott=2200hr
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243. 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
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244. Esempio 2 Introduciamo poi il costo della indisponibilita’:
definito come multiplo del costo singola riparazione.
Funzione energia:
Intervallo ottimale:
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245. Esempio 2
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246. 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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247. Riparazioni
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248. 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.
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249. 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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250. 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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251. 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
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252. Esempio
Applichiamo uno schema a stati per il sistema in standby. Otteniamo: 1 2 3
P23
P32
P21
P12
P13
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253. Equazioni di Markov/Kolmogorov
Dove A e’ la matrice di transizione del sistema, P e’ il vettore delle probabilita’ degli stati del sistema.
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254. Costruzione della matrice di transizione1 2
P21
P12 1 2  
Esempio: componente soggetto a rottura e riparazione. 2 stati: in funzione o in riparazione, con tassi di guasto  e riparazione.
Chi sono P12 e P21? Sono le probabilita’ di transizione in dt. Quindi:
P12= e P21= 
La matrice di transizione e’ costruita con le seguenti regole:
(+) se il salto e’ in entrata allo stato, (-) se il salto e’ in uscita
Prendiamo lo stato 1: si entra in 1 da due con tasso  (+), si esce con tasso  (-).
Quindi:
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255. La matrice di transizione
Analogamente:
Quindi:
La matrice di transizione e’:
Il sistema di equazioni differenziali diventa:
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256. 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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257. Risultati per un componente
Disponibilita’ istantanea:
Disponibilita’ asintotica:
Interpretazione: tempo che occorre in media alla riparazione diviso il tempo totale
Disponibilita’ media su T:
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258. 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)
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