Fault-Tolerant Computing

Transcription

1 Fault-Tolerant Computing Motivation, Background, and Tools Slide 1

2 About This Presentation This presentation has been prepared for the graduate course ECE 257A (Fault-Tolerant Computing) by Behrooz Parhami, Professor of Electrical and Computer Engineering at University of California, Santa Barbara. The material contained herein can be used freely in classroom teaching or any other educational setting. Unauthorized uses are prohibited. Behrooz Parhami Edition Released Revised Revised First Slide 2

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5 What Is? With respect to availability of resources and computational capabilities, a system can be viewed as being in one of several possible states The number of states can be large, if we want to make fine distinctions, or it can be relatively small if we lump similar states together State transitions: System moves from one state to another as resource availability and computational power change due to various events Great So-so State-space modeling entails quantifying transition probabilities so as to determine the probability of the system being in each state; from this, we derive reliability, availability, safety, and other desired parameters Good Lousy Slide 5

6 Markov Chains Represented by a state diagram with transition probabilities Sum of all transition probabilities out of each state is 1 The state of the system is characterized by the vector (s 0, s 1, s 2, s 3 ) (1, 0, 0, 0) means that the system is in state 0 Must sum to 1 (0.5, 0.5, 0, 0) means that the system is in state 0 or 1 with equal prob s (0.25, 0.25, 0.25, 0.25) represents complete uncertainty Transition matrix: M = s(t + 1) = s(t) M s(t + h) = s(t) M h Markov matrix (rows sum to 1) Example: 0.1 (s 0,s 1,s 2,s 3 ) = (0.5, 0.5, 0, 0) M = (0.4, 0.4, 0.15, 0.05) (s 0,s 1,s 2,s 3 ) = (0.4,0.4,0.15,0.05) M = (0.34,0.365,0.225,0.07) Self loops not shown Slide 6

7 Merging States in a Markov Model There are three identical units 1 = Unit is up 0 = Unit is down All solid lines λ Dashed lines λ Simpler equivalent model for 3-unit fail-soft system 3 3λ Whether or not states are merged depends on the model s semantics 2 2λ 1 0 λ Failed state if TMR Slide 7

8 Two-State Nonrepairable Systems Rate of change for the probability of being in state 1 is λ p 1 = λp 1 p 0 = 1 e λt p 1 + p 0 = 1 p 1 = e λt Reliability as a function of time: R(t) = p 1 (t) = e λt 1 Time Start State Initial condition: p 1 (0) = 1 Good Failure Failed λ 1 0 The label λ on this transition means that over time dt, the transition will occur with probability λdt (we are dealing with a continuous-time Markov model) Slide 8

9 k-out-of-n Nonrepairable Systems nλ n n 1 (n 1)λ n F 2 kλ k k 1 0 p n = nλp n p n 1 = nλp n (n 1)λp n 1. p k = (k + 1)λp k+1 kλp k p n + p n p k + p F = 1 p n = e nλt p 1 = ne (n 1)λt (1 e λt ). n k Initial condition: p n (0) = 1 p k = ()e (n k)λt (1 e λt ) k p F = 1 Σ j=k to n p j In this case, we do not need to resort to more general method of solving linear differential equations (LaPlace transform, to be introduced later) The first equation is solvable directly, and each additional equation introduces only one new variable Slide 9

10 Two-State Repairable Systems In steady state (equilibrium), transitions into/out-of each state must balance out λp 1 + p 0 = 0 p 1 + p 0 = 1 p 1 = /(λ + ) p 0 = λ/(λ + ) Availability as a function of time: A(t) = p 1 (t) = /(λ + ) + λ/(λ + ) e (λ+)t Derived in the next slide 1 Steady-state availability Time Start State Up Repair Failure Down 1 0 λ The label on this transition means that over time dt, repair will occur with probability dt (constant repair rate as well as constant failure rate) Slide 10

11 Solving the State Differential Equations p 1 (t) = λp 1 (t)+ p 0 (t) p 0 (t) = p 0 (t)+ λp 1 (t) 1 0 λ To solve linear differential equations with constant coefficients: 1. Convert to algebraic equations using LaPlace transform 2. Solve the algebraic equations 3. Use inverse LaPlace transform to find original solutions 1 sp 1 (s) p 1 (0) = λp 1 (s)+ P 0 (s) LaPlace Transform Table sp 0 (s) p (0) = P 0 (s)+ λp 1 (s) h(t) H(s) 0 k k/s P 1 (s) = (s + ) / [s 2 + (λ + )s] e at 1/(s + a) P 0 (s) = λ / [s 2 + (λ + )s] t n 1 e at /(n 1)! 1/(s + a) n p 1 (t) = /(λ + ) + λ/(λ + ) e (λ+)t p 0 (t) = λ/(λ + ) λ/(λ + ) e (λ+)t kh(t) h(t)+g(t) h (t) kh(s) H(s)+G(s) sh(s) h(0) Slide 11

12 Systems with Multiple Failure States In steady state (equilibrium), transitions into/out-of each state must balance out λp 2 + p 1 + p 0 =0 p 1 + λ 1 p 2 = 0 p 2 + p 1 + p 0 = 1 p 2 = /(λ + ) p 1 = λ 1 /(λ + ) p 0 = λ 0 /(λ + ) Start State Good Failure Failure Safe Failed Unsafe Failed Safety evaluation: Total risk of system is Σ failure states c j p j 1 1 p 2 (t) p 1 (t) p 0 (t) Time 2 λ 1 λ 0 λ 1 + λ 0 = λ 0 Failure state j has a cost (penalty) c j associated with it Slide 12

13 Systems with Multiple Operational States λ 2 p p 1 = 0 λ 1 p 1 1 p 0 = 0 p 2 + p 1 + p 0 = 1 Start State Repair Up 2 Up 1 Partial failure Partial repair Down Failure Let δ = 1/[1 + λ 2 / 2 + λ 1 λ 2 /( 1 2 )] p 2 = δ p 1 = δλ 2 / 2 p 0 = δλ 1 λ 2 /( 1 2 ) Performability evaluation: Performability = Σ operational states b j p j 2 λ λ 1 Operational state j has a benefit b j associated with it Example: λ 2 = 2λ, λ 1 = λ, 1 = 2 = (single repairperson or facility), b 2 = 2, b 1 = 1, b 0 = 0 P = 2p 2 + p 1 = 2δ + 2δλ/ = 2(1 + λ/)/(1 +2 λ/+ 2λ 2 / 2 ) Slide 13

14 3λp 3 + p 2 = 0 ( + 2λ)p 2 + 3λp 3 = 0 p 3 + p 2 + p F = 1 Steady-state analysis of no use p 3 = p 2 = 0, p F = 1 TMR System with Repair Mean time to failure evaluation: See [Siew92], pp , for derivation MTTF = 5/(6λ) + /(6λ 2 ) = [5/(6λ)]( /λ) MTTF for TMR Improvement due to repair 3 2 2λ F 3λ Improvement factor Assume the voter is perfect Upon first module malfunction, we switch to duplex operation with comparison MTTF Comparisons (λ = 10 6 /hr, = 0.1/hr) Nonredundant 1/λ 1M hr TMR 5/(6λ) M hr TMR with repair [5/(6λ)]( /λ) 16,668 M hr Slide 14

15 Fail-Soft System with Imperfect Coverage λ 2 p p 1 = 0 λ 2 (1 c)p 2 + λ 1 p 1 1 p 0 = 0 p 2 + p 1 + p 0 = 1 Start State Repair Up 2 Up 1 Partial failure Partial repair Down Failure We solve this in the special case of λ 2 = 2λ, λ 1 = λ, 2 = 1 = Let ρ = λ/ and θ = 1 / (1 + 4ρ 2cρ + 2ρ 2 ) p 2 = θ p 1 = 2ρθ p 0 = 2ρ(1 c + ρ)θ λ 2 c λ 1 λ 2 (1 c) If malfunction of one unit goes undetected, the system fails We can also consider coverage for the repair direction Slide 15

16 Birth-and-Death Processes Special case of Markov model with states appearing in a chain and transitions allowed only between adjacent states λ 3λ This model is used in queuing theory, where the customers arrival rate and provider s service rate determine the queue size and waiting time Closed-form expression for state probabilities can be found, assuming n + 1 states and s service providers (repair persons): M/M/s/n/n queue 2λ λ p j = (n j + 1) (λ/) p j 1 / j p j = (n j + 1) (λ/) p j 1 / r for j = 1, 2,..., r for j = r + 1, r + 2,..., n Equation for p 0 [Siew92], p. 347 Slide 16

17 The Dependability Modeling Process Choose modeling approach Combinational State-space Construct model Derive model parameters Solve model Iterate until results are satisfactory Interpret results Validate model and results Slide 17

18 Software Aids for Reliability Modeling Relex (company specializing in reliability engineering) Reliability block diagram: Markov: University of Virginia Galileo: Iowa State University HIMAP: See Appendix D, pp , of [Shoo02] for more programs Dept. of Mechanical Engineering, Univ. of Maryland: List of reliability engineering tools: Slide 18