Chapter 3: Introduction to Reliability Theory Claver Diallo OUTLINE 1. Part 1: Basic Reliability Models 1. 2. 3. 1. 2. 3. 4. 5. System Reliability function Probability distributions Reliability Block Diagram Serial and Parallel Structures Stand-by Structure k-out-of n Structure Complex structure 2. Part 2: Reliability of Structures 3. Part 3: Reliability Allocation 4. References 2 2 2 2 2 2 2 2 Chapter 3 - Part 1: Basic Reliability Models SYSTEM System: a collection of components
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and average repair days below. Machine Repair time, y (days) | Probability of Repair time | Cumulative Probability | Random Number Range, r2 | | 1 | 0.20 | 0.20 | 0.00-0.20 | 0.2 | 2 | 0.45 | 0.65 | 0.21-0.65 | 0.9 | 3 | 0.25 | 0.90 | 0.66-0.90 | 0.75 | 4 | 0.10 | 1.00 | 0.91-1.00 | 0.4 | Average repair days | 2.25 | To find the repair time I used VLOOKUP /y days =VLOOKUP to probabilities in the Repair Time probability/ tool generating random number /which r2=RAND()/. If 0 < r2
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Gains (or, Payoffs) by the Probability for each States of Nature (in this case, Market Reactions: Good, Moderate, and Poor). Then, you add the totals of each Market Reaction and that becomes the expected value for each decision alternative. The highest expected value from all the decisions is generally the most favorable decision. The following are the calculations for the expected values of each of the four decision alternatives: The formula is: EV= Predicted Gains(Probability) Decision alternative
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c) P(X > 3) = 0 d) P(0 X 2) P( X 0) P( X 1.5) 1/ 3 1/ 3 2 / 3 e) P(X = 0 or X = 2) = 1/3 + 1/6 = ½ --------------------------------------------------------------------------------------------------------------------3-15. All probabilities are greater than or equal to zero and sum to one. a) P(X 1) = 1/8 + 2/8 + 2/8 + 2/8 = 7/8 b) P(X > 2) = 2/8 + 2/8 + 2/8 + 1/8 = 7/8 c) P(1 X 1) = 2/8 + 2/8 + 2/8 =6/8 = 3/4 d) P(X 1 or X = 2) = 1/8 + 2/8 +1/8 = 4/8 = ½ -------------
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shares of X ltd of 10, of which the current market price is Rs. 132 and the exercise price is Rs. 150. You expect the price to range between Rs. 120 to 190. The expected share price of X ltd and related probability is given below. Expected price | 120 | 140 | 160 | 180 | 190 | probability | .05 | .20 | .50 | .10 | .15 | Compute following 1. Expected share price at the end of 4months 2. Value of the ex call at the end of 4 months, if the exercise price prevails 3. In case
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million in first year, but bears the highest volatility, which may lead to losses of up to $170.5 million (See Exhibit 1). At this stage, we exclude the RNCN2 plan from further considerations because even though is offers low expected costs, the probability of going over the limit of $37 million and incurring costs of $60 million is significant for the purposes of this analysis. We plan to confirm this preliminary conclusion during the sensitivity analysis. Stage 2: The next step in our analysis is
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maeSome Implications of Belief in the Afterlife and the Allocation of Time to Spirituality∗ Constantino Hevia† University of Chicago September 2004 Abstract An otherwise standard model of intertemporal consumer choice is extended to incorporate the allocation of time to spiritual activities along the lines of the human capital literature. Several testable implications are analyzed. We study exogenous and endogenous changes in life expectancy, and we argue that the traditional value of life or
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Paper Reference(s) 6683/01 Edexcel GCE Statistics S1 Advanced/Advanced Subsidiary Wednesday 24 May 2006 ( Afternoon Time: 1 hour 30 minutes Materials required for examination Items included with question papers Mathematical Formulae (Green) Nil Candidates may use any calculator EXCEPT those with the facility for symbolic algebra, differentiation and/or integration. Thus
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Company A PERT/CPM Analysis | | | | | Task Detail Table 1.1 | | | | | | | | | | | | Task | Preceding Activity | Optimistic Time to Complete (weeks) | Probable Time to Complete (weeks) | Pessimistic Time to Complete (weeks) | Expected Time to Complete (weeks) | Variance (weeks) | START | | | | | | | A | START | 2 | 3 | 4 | 3 | 0.11 | B | START | 5 | 6 | 13 | 7 | 1.78 | C | A | 3 | 4 | 8 | 4.5 | 0.69 | D |
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Asset 0 is a risk-free bond with B0 = 1, and B1 = 1 + r for some constant interest rate r > −1, while asset 1 is a stock with prices S u ? 1 >> >> >> 1−p > > p d for some constant rates of return d < u and a probability 0 < p < 1. Use the definition of arbitrage to show that there is no arbitrage if and only if d < 1 + r < u. When there is no arbitrage, find all equivalent martingale measures Q (relative to asset 0, as usual) and their densities dQ with respect
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