Probability and Statistics with Reliability, Queuing and Computer Science Applications: Chapter 1 Introduction

1. Probability and Statistics with Reliability, Queuing and Computer Science Applications: Chapter 1 Introduction Dept. of Electrical & Computer engineering Duke University Email: bbm@ee.duke.edu, kst@ee.duke.edu

2. Sample Space • Probability implies random experiments. • A random experiment can have many possible outcomes; each outcome known as a sample point (a.k.a. elementary event) has some probability assigned. This assignment may be based on measured data or guestimates (“equally likely” is a convenient and often made assumption). • Sample Space S : a set of all possible outcomes (elementary events) of a random experiment. • Finite (e.g., if statement execution; two outcomes) • Countable (e.g., number of times a while statement is executed; countable number of outcomes) • Continuous (e.g., time to failure of a component)

3. Events • An event E is a collection of zero or more sample points from S • S is the universal event and the empty set • S and E are sets  use of set operations.

4. Algebra of events • Sample space is a set and events are the subsets of this (universal) set. • Useset algebra and its laws on p. 9 of the text. • Mutually exclusive (disjoint) events

5. Probability axioms (see pp. 15-16 of text for additional relations)

6. Combinatorial problems • Deals with the counting of the number of sample points in the event of interest. Assume equally likely sample points: P(E)= number of sample points in E / number in S • Example: Two successive execution of an if statement • S = {(T,T), (T,E), (E,T), (E,E)} {s1, s2, s3, s4} • P(s1) = 0.25= P(s2) = P(s3) = P(s4) (equally likely assumption) • E1: at least one execution of the then clause{s1,s2,s3} • E2: exactly one execution of the else clause{s2, s3} • P(E1) = 3/4; P(E2) = 1/2

7. Conditional probability • In some experiment, some prior information may be available, e.g., • What is the probability that Blue Devils will win the opening game, given that they were the 2000 national champs. • P(A|B): prob. that A occurs, given that ‘B’ has occurred. • In general,

8. Mutual Independence • A and B are said to be mutually independent, iff, • Also, then,

9. Independence Vs. Exclusive • Note that the probability of the union of mutually exclusive events is the sum of their probabilities • While the probability of the intersection of two mutually independent events is the product of their probabilities

10. Independent set of events • Set of n events, {A1, A2,..,An} are mutually independent iff, for each • Complements of such events also satisfy, • Pair wise independence (not mutually independent)

11. Reliability Block Diagrams

12. Reliability Block Diagrams (RBDs) • Schematic representation or model • Shows reliability structure (logic) of a system • Can be used to determine • If the system is operating or failed • Given the information whether each block is in operating or failed state • A block can be viewed as a “switch” that is “closed” when the block is operating and “open” when the block is failed • System is operational if a path of “closed switches” is found from the input to the output of the diagram

13. Reliability Block Diagrams: RBDs • Combinatorial (non-state space) model type • Each component of the system is represented as a block • System behavior is represented by connecting the blocks • Blocks that are all required are connected in series • Blocks among which only one is required are connected in parallel • When at least k out of n are required, use k-of-n structure • Failures of individual components are assumed to be independent for easy solution • For series-parallel RBD with independent components use series-parallel reductions to obtain the final answer

14. Series-ParallelReliability Block Diagrams (RBDs)

15. Series system • Series system: n statistically independent components. • Let, Ri = P(Ei), then series system reliability: • For now reliability is simply a probability, later it will be a function of time

16. Series system(Continued) This simple PRODUCT LAW OF RELIABILITIES, is applicable to series systems of independent components. R1 R2 Rn

17. Series system(Continued) • Assuming independent repair, we have product law of availabilities

18. Parallel system • System consisting of n independent parallel components. • System fails to function iff all n components fail. • Ei= "component i is functioning properly" • Ep= "parallel system of n components is functioning properly." • Rp = P(Ep).

19. Parallel system(Continued) Therefore:

20. Parallel system(Continued) R1 . . . • Parallel systems of independent components follow the PRODUCT LAW OF UNRELIABILITIES . . . Rn

21. Parallel system(Continued) • Assuming independent repair, we have product law of unavailabilities:

22. Series-Parallel System • Series-parallel system: n-series stages, each with ni parallel components. • Reliability of series parallel system

23. Series-Parallel system(example) Example: 2 Control and 3 Voice Channels voice control voice control voice

24. Series-Parallel system(Continued) • Each control channel has a reliability Rc • Each voice channel has a reliability Rv • System is up if at least one control channel and at least 1 voice channel are up. • Reliability:

25. For the following system, write down the expression for system reliability: Assuming that block i failure probability qi C D C B E A D C Homework :

26. Non-Series-Parallel Systems

27. Methods for non-series-parallel RBDs • State enumeration (Boolean truth table) • Factoring or conditioning (implemented in SHARPE) • First find minpaths • inclusion/exclusion (Relation Rd on p.15 of text) • SDP (Sum of Disjoint Products; Relation Re on p. 16 of text) (implemented in SHARPE) • BDD (Binary Decision Diagram) (implemented in SHARPE)

28. Non-series-parallel RBD-Bridge with Five Components 1 2 3 S T 4 5

29. Truth Table for the Bridge Component System Probability 5 1 2 3 4 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 1 1 1 1 1 1 0 1 0 1 0 0 0

30. Truth Table for the Bridge Component System Probability 5 1 2 3 4 } 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 0 0 0 1 0 0 0 1 0 0 0

31. Bridge Reliability • From the truth table:

32. Conditioning & The Theorem of Total Probability • Any event A: partitioned into two disjoint events,

33. Example • Binary communication channel: P(R0|T0) T0 R0 Given: P(R0|T0) = 0.92; P(R1|T1) = 0.95 P(T0) = 0.45; P(T1) = 0.55 P(R0|T1) P(R1|T0) T1 R1 P(R1|T1) P(R0) = P(R0|T0) P(T0) + P(R0|T1) P(T1) (TTP) = 0.92 x 0.45 + 0.08 x 0.55 = 0.4580 =P(R0|T1) P(T1) + P(R1|T0) P(T0)

34. BridgeReliability usingconditioning/factoring

35. Bridge: Conditioning 1 2 C3 down S T 1 2 4 5 3 S T C3 up 4 5 1 2 S T Factor (condition) on C3 4 5 Non-series-parallel block diagram

36. Bridge (Continued) • Component C3 is chosen to factor on (or condition on) • Upper resulting block diagram: C3 is down • Lower resulting block diagram: C3 is up • Series-parallel reliability formulas are applied to both the resulting block diagrams • Use the theorem of total probability to get the final result

37. Bridge(Continued) RC3down= 1 - (1 - R1R2) (1 - R4R5) RC3up = (1 - Q1Q4)(1 - Q2Q5) = [1 - (1-R1) (1-R4)] [1 - (1-R2) (1-R5)] Rbridge = RC3down . (1-R3 ) + RC3up R3

38. Fault Trees • Combinatorial (non-state-space) model type • Components are represented as nodes • Components or subsystems in series are connected to OR gates • Components or subsystems in parallel are connected to AND gates • Components or subsystems in kofn (RBD) are connected as (n-k+1)ofn gate

39. Fault Trees (Continued) • Failure of a component or subsystem causes the corresponding input to the gate to become TRUE • Whenever the output of the topmost gate becomes TRUE, the system is considered failed • Extensions to fault-trees include a variety of different gates NOT, EXOR, Priority AND, cold spare gate, functional dependency gate, sequence enforcing gate

40. Fault Tree • Without repeated events or with repeated events • Reliability of series-parallel or non-series-parallel systems may be modeled using a fault tree • State vector X={x1, x2, …, xn} and structure function

41. or and and c1 c2 v1 v2 v3 Fault Tree Without Repeated Events • Structure Function: Reliability of the system 2 Control and 3 Voice Channels Example

42. Another Fault tree (w/o repeated events) • Example: DS1 NIC1 CPU DS2 NIC2 DS3

43. 2 control and 3 voice channels example with Fault Tree • Change the problem so that a control channel can also function as a voice channel • We need to use a fault tree with repeated events to model the reliability of the system

44. Fault tree with repeated events

45. failure and or or and and p1 p2 m1 m3 m2 m3 2 Proc 3 Mem Fault Tree • specialized for dependability analysis • represent all sequences of individual component failures that cause system failure in a tree-like structure • top event: system failure • gates: AND, OR, (NOT), K-of-N • Input of a gate: -- component (1 for failure, 0 for operational) -- output of another gate • Basic component and repeated component A fault tree example

46. Fault Tree (Cont.) • For fault tree without repeated nodes • We can map a fault tree into a RBD • Use algorithm for RBD to compute reliability • For fault tree with repeated nodes • Factoring algorithm • SDP algorithm • BDD algorithm

47. failure and or or and and p1 p2 m1 m3 m2 m3 Factoring Algorithm for Fault Tree failure • Basic idea: and M3 has failed or or p2 p1 m2 m1 failure and M3 has not failed p1 p2

48. Fault tree (Continued) • Major characteristics: • Fault trees without repeated events can be solved in linear time • Fault trees with repeated events -Theoretical complexity: exponential in number of components. • Find all minimal cut-sets & then use sum of disjoint products to compute reliability. • Use Factoring (conditioning) • Use BDD approach • Can solve fault trees with 100’s of components

49. Bernoulli Trial(s) • Random experiment  1/0, T/F, Head/Tail etc. • Two outcomes on each trial • Successive trial independent • Probability of success does not change from trial to trial • Sequence of Bernoulli trials: n independent repetitions. • n consecutive executions of an if-then-elsestatement • Sn: sample space of n Bernoulli trials • For S1:

50. Bernoulli Trials (contd.) • Problem: assign probabilities to points in Sn • P(s): Prob. of successive k successes followed by (n-k) failures. What about any k failures out of n ?