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1 Quantitative Methods for Management
Emanuele Borgonovo Quantitative Methods for Management First Edition Decision Market Return Structural Quantitative Methods for Management

2 Chapter three: Models Quantitative Methods for Management

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 Quantitative Methods for Management

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 Quantitative Methods for Management

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? Quantitative Methods for Management

6 Chapter II Introductory Elements of Probability theory
Quantitative Methods for Management

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) Quantitative Methods for Management

8 Kolmogorov Axioms U B A Quantitative Methods for Management

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 Quantitative Methods for Management

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 Quantitative Methods for Management

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) Quantitative Methods for Management

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) Quantitative Methods for Management

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? Quantitative Methods for Management

14 In the subjectivist approach, Probability changes with information
Quantitative Methods for Management

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 Quantitative Methods for Management

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: Quantitative Methods for Management

17 Bayes’ theorem Proof Starting point Conditional Probability formula
thesis Quantitative Methods for Management

18 the Total Probability theorem
C 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 : Quantitative Methods for Management

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. Quantitative Methods for Management

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. Quantitative Methods for Management

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: Quantitative Methods for Management

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: Quantitative Methods for Management

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. Quantitative Methods for Management

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. Quantitative Methods for Management

25 Pdf and Cdf of the Exponential Distribution
5 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 Quantitative Methods for Management

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 Quantitative Methods for Management

27 the Normal Distribution
Is a symmetric distribution around the mean Pdf: Cdf: Quantitative Methods for Management

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 Quantitative Methods for Management

29 Lognormal Distribution
Pdf Cdf Quantitative Methods for Management

30 Lognormal Distribution
Quantitative Methods for Management

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? Quantitative Methods for Management

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) Quantitative Methods for Management

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 Quantitative Methods for Management

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: Quantitative Methods for Management

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 Quantitative Methods for Management

36 Chapter III: Introductory Decision theory
Quantitative Methods for Management

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 Quantitative Methods for Management

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) Quantitative Methods for Management

39 Decision-Making Process Steps
Problem identification Alternatives identification Model implementation Alternatives evaluation Sensitivity Analysis Further Analysis? Yes Best Alternatives implementation No Quantitative Methods for Management

40 Decision-Making Problem Elements
Values and Objectives Attributes Decision Alternatives Uncertain Events Consequences Quantitative Methods for Management

41 Decision Problem Elements
Objectives: Maximize profit Attributes: Money Alternatives: Risky Less Risky Safe Random events: the Market Consequences: Profit or Loss Quantitative Methods for Management

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 Quantitative Methods for Management

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) Quantitative Methods for Management

44 ID Elements = Decision = Random Event = utility NODES ARCS
Informational Arcs probabilistic Dependency Arcs Structural Arcs = Decision = Random Event = utility Quantitative Methods for Management

45 ID Elements Informational Arc Structural Decision Node Chance Node
Value Node Conditional Arc probabilistic Dependency Informational Arc Sequential Decisions Structural Quantitative Methods for Management

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 Quantitative Methods for Management

47 Case Study 2 - Leaking SG tube
Influence Diagram for Case Study 2 Quantitative Methods for Management

48 Market Decision Return Influence Diagram Structural
Quantitative Methods for Management

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. Quantitative Methods for Management

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? Quantitative Methods for Management

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) Quantitative Methods for Management

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? Quantitative Methods for Management

53 Problem Solution Using the previous formula:
Quantitative Methods for Management

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? Quantitative Methods for Management

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 Quantitative Methods for Management

56 Example of a simple ID Decision Profit Engine failure
Final Classification Quantitative Methods for Management

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 Quantitative Methods for Management

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% Quantitative Methods for Management

59 Influence Diagram Quantitative Methods for Management

60 Decision Tree Quantitative Methods for Management

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 Quantitative Methods for Management

62 Example: investing Quantitative Methods for Management

63 EVPI for the Example Quantitative Methods for Management

64 EVPI Result Quantitative Methods for Management

65 Problems Quantitative Methods for Management

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 Quantitative Methods for Management

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? Quantitative Methods for Management

68 Solution Influence Diagram the Quantitative Methods for Management

69 Solution Corresponding Decision Tree
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70 Solution Influence Diagram II Quantitative Methods for Management

71 Solution Decision Tree II: Quantitative Methods for Management

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 Quantitative Methods for Management

73 Solution Sales_Costs Quantitative Methods for Management

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? Quantitative Methods for Management

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) Quantitative Methods for Management

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? Quantitative Methods for Management

77 Influence Diagram Quantitative Methods for Management

78 Decision Tree Quantitative Methods for Management

79 Probability Values Three years Quantitative Methods for Management

80 2 and 4 years 2 years 4 years Quantitative Methods for Management

81 Chapter IV Elements of Sensitivity Analysis
Quantitative Methods for Management

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) Quantitative Methods for Management

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) Quantitative Methods for Management

84 Sensitivity Analysis (Run or withdraw)
Underline the critical dependencies of the outcome Quantitative Methods for Management

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 Quantitative Methods for Management

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 Quantitative Methods for Management

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 Quantitative Methods for Management

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 Quantitative Methods for Management

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. Quantitative Methods for Management

90 Example of a Tornado Diagram
Quantitative Methods for Management

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 Quantitative Methods for Management

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 Quantitative Methods for Management

93 Process Identify how sensitivity analysis techniques work through analysis of several examples Formulate a definition Classify sensitivity analysis techniques accordingly Quantitative Methods for Management

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 Quantitative Methods for Management

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 Quantitative Methods for Management

96 Global Sensitivity Indices
Uncertainty in U and parameters is considered Sobol’’s decomposition theorem: Sobol’Indices Quantitative Methods for Management

97 Formal Definition of Sensitivity Analysis (SA) Techniques
SA technique are Operators on U: x1 x2 xn I(x1) I(xn) I(x2)  or  Quantitative Methods for Management

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 Quantitative Methods for Management

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 Quantitative Methods for Management

100 Techniques that fall under the definition of Local SA techniques
Quantitative Methods for Management

101 Global Importance Measures
Quantitative Methods for Management

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 Quantitative Methods for Management

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 Quantitative Methods for Management

104 Chapter V Uncertainty Analysis
Quantitative Methods for Management

105 Uncertainty Analysis Quantitative Methods for Management

106 Summary Distinction between Aleatory Uncertainty ed Epistemic Uncertainty Epistemic Uncertainty and Bayes‘ theorem Monte Carlo Method for uncertainty propagation Quantitative Methods for Management

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. Quantitative Methods for Management

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). Quantitative Methods for Management

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 Quantitative Methods for Management

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( ): Quantitative Methods for Management

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: Quantitative Methods for Management

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! Quantitative Methods for Management

113 In General the MOW will depend on m parameters , ,…:
the event Probability (P(t)) will be: Quantitative Methods for Management

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? Quantitative Methods for Management

115 Solution E[t]= Quantitative Methods for Management

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…. Quantitative Methods for Management

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 Quantitative Methods for Management

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!!! Quantitative Methods for Management

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 Quantitative Methods for Management

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 Quantitative Methods for Management

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) Quantitative Methods for Management

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 Quantitative Methods for Management

123 Conjugate Distributions
Likelihood Poisson Posterior: gamma Prior distribution Gamma with: Quantitative Methods for Management

124 Conjugate Distributions
Likelihood Normal Posterior: Normal Prior distribution: Normal with: Quantitative Methods for Management

125 Conjugate Distributions
Likelihood Binomial Posterior, Beta: Prior: Beta with: Quantitative Methods for Management

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. Quantitative Methods for Management

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 Quantitative Methods for Management

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: Quantitative Methods for Management

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 Quantitative Methods for Management

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 Quantitative Methods for Management

131 Inversion theorem 1 Inversion theorem:
Inversion theorem: the values of  sampled in this way have the Probability distribution from which we have inverted Quantitative Methods for Management

132 Example Let us evaluate the volume of the yellow solid through the Monte Carlo method. V0 V Quantitative Methods for Management

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. Quantitative Methods for Management

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? Quantitative Methods for Management

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. Quantitative Methods for Management

137 Prob. 5-2 Substituting: E[URisky]=50, E[USafe ]= 20, E[ULess Risky ]= 20 Quantitative Methods for Management

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 Quantitative Methods for Management

139 Problems Apply the one way, two way and Tornado Diagrams SA to the IDs and DTs of the previous chapters: Discuss your results Quantitative Methods for Management

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? Quantitative Methods for Management

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 Quantitative Methods for Management

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 Quantitative Methods for Management

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. Quantitative Methods for Management

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 Quantitative Methods for Management

146 Structuring Preferences
Indifference Curves: Points on the same curve leave you indifference x1 x2 Quantitative Methods for Management

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 Quantitative Methods for Management

148 V(x) V(x) is a value function if it satisfies the following properties: a) b) Quantitative Methods for Management

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? Quantitative Methods for Management

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 Quantitative Methods for Management

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: Quantitative Methods for Management

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 Quantitative Methods for Management

153 Stochastic Dominance Distribution 1 is dominated by
1 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 Quantitative Methods for Management

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 Quantitative Methods for Management

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 Quantitative Methods for Management

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: Quantitative Methods for Management

158 Mailmatical Definition
the Risk Aversion function is defined as: Or, equivalently: Supposing a constant risk aversion one gets an exponential utility function: Quantitative Methods for Management

159 Risk Preferences Constant Risk Aversion
Compute constant  through Certainty Equivalent (CE): Quantitative Methods for Management

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 Quantitative Methods for Management

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 Quantitative Methods for Management

163 Problems Quantitative Methods for Management

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 Quantitative Methods for Management

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? Quantitative Methods for Management

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? Quantitative Methods for Management

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. Quantitative Methods for Management

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 Quantitative Methods for Management

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 Quantitative Methods for Management

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 Quantitative Methods for Management

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. Quantitative Methods for Management

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 Quantitative Methods for Management

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 Quantitative Methods for Management

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. Quantitative Methods for Management

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. Quantitative Methods for Management

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 Quantitative Methods for Management

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 : Quantitative Methods for Management

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 Quantitative Methods for Management

182 Golden Rule In practice: the System Failure Probability is computed from the solved Structure Function, substituting to indicator Xi the corresponding Event Probability. Quantitative Methods for Management

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) Quantitative Methods for Management

184 Problems Quantitative Methods for Management

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 Quantitative Methods for Management

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 Quantitative Methods for Management

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 Quantitative Methods for Management

189 Even Trees Event Trees: represent the sequence of events that lead to the event top. Initiating Event Event 1 Event 2 Top Event No Quantitative Methods for Management

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. Quantitative Methods for Management

191 Gas Leakage Fault - Tree
No High Pressure Pipe Room Yes Building Top Event Quantitative Methods for Management

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 Quantitative Methods for Management

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 Quantitative Methods for Management

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: Quantitative Methods for Management

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? Quantitative Methods for Management

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: Quantitative Methods for Management

198 Event Trees & Fault Trees
No High pressure Pipe Aeration 1 yes Aeration 2 Top Event Aeraz. 1 engine filter Electr. Mech. Installaz. Fault Quantitative Methods for Management

199 Probability of the Top Event
From the Event Tree: Expanding: the conditional probabilities are found solving the corresponding Fault Trees Quantitative Methods for Management

200 Definitions Quantitative Methods for Management

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: Quantitative Methods for Management

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: Quantitative Methods for Management

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: Quantitative Methods for Management

205 Legami between R(t), f(t) and (t)
From the above definition, there follows: Relationship R(t)- (t): Relationship f(t)- (t): Quantitative Methods for Management

206 time medio of failure (MTTF)
The mean time to failure is defined as: Quantitative Methods for Management

207 System Reliability Quantitative Methods for Management

208 Reliability of systems in Series
if independence is assumed: Thus: Faulure rate: Quantitative Methods for Management

209 Reliability of systems in Parallelo
Failure Probability of the system: if independent: Thus: Failure Rate: Quantitative Methods for Management

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: Quantitative Methods for Management

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: Quantitative Methods for Management

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 Quantitative Methods for Management

213 Problems Quantitative Methods for Management

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 Quantitative Methods for Management

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. Quantitative Methods for Management

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? Quantitative Methods for Management

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 Quantitative Methods for Management

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. Quantitative Methods for Management

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) Quantitative Methods for Management

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 Quantitative Methods for Management

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 Quantitative Methods for Management

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? Quantitative Methods for Management

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? Quantitative Methods for Management

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. Quantitative Methods for Management

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: Quantitative Methods for Management

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 Quantitative Methods for Management

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? Quantitative Methods for Management

228 Capitolo IX Decisioni Operative: Ottimizzazione delle Manutenzioni
Quantitative Methods for Management

229 Decisioni Operative Decisioni di Affidabilita’ o Reliability Design
Decisioni di Optimal Replacement Decisioni di ispezione ottimale Decisioni di riparazione ottimale Quantitative Methods for Management

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: Quantitative Methods for Management

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 Quantitative Methods for Management

232 Effetto delle manutenzioni
Quantitative Methods for Management

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 Quantitative Methods for Management

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<<  : Quantitative Methods for Management

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. Quantitative Methods for Management

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’: Quantitative Methods for Management

237 Indisponibilita’ media con piu’ modi di guasto
L’indisponibilita’ media sull’intervallo 0 + r e’: Due assunzioni: 1) Eventi rari 2) + r  Quantitative Methods for Management

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 Quantitative Methods for Management

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 Quantitative Methods for Management

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 Quantitative Methods for Management

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 Quantitative Methods for Management

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 Quantitative Methods for Management

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 Quantitative Methods for Management

244 Esempio 2 Introduciamo poi il costo della indisponibilita’:
definito come multiplo del costo singola riparazione. Funzione energia: Intervallo ottimale: Quantitative Methods for Management

245 Esempio 2 Quantitative Methods for Management

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 Quantitative Methods for Management

247 Riparazioni Quantitative Methods for Management

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. Quantitative Methods for Management

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 Quantitative Methods for Management

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 Quantitative Methods for Management

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 Quantitative Methods for Management

252 Esempio Applichiamo uno schema a stati per il sistema in standby. Otteniamo: 1 2 3 P23 P32 P21 P12 P13 Quantitative Methods for Management

253 Equazioni di Markov/Kolmogorov
Dove A e’ la matrice di transizione del sistema, P e’ il vettore delle probabilita’ degli stati del sistema. Quantitative Methods for Management

254 Costruzione della matrice di transizione
1 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: Quantitative Methods for Management

255 La matrice di transizione
Analogamente: Quindi: La matrice di transizione e’: Il sistema di equazioni differenziali diventa: Quantitative Methods for Management

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: Quantitative Methods for Management

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: Quantitative Methods for Management

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) Quantitative Methods for Management

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