Reliability Models & Applications (continued)

1. Systems Reliability, Supportability and Availability Analysis Reliability Models & Applications (continued)

2. The Normal or Gaussian Model Definition • A random variable T is said to have the Normal • (Gaussian) Distribution with parameters  and , • where  > 0, if the density function of T is: , for 0 < t < Definition If T ~ N(,) and if , then Z ~ N(0,1) the Standard Normal Distribution and Cumulative Probability is tabulated for various values of z.

3. Properties of the Normal Model - Failure Densities

4. Properties of the Normal Model Probability Distribution Function where (Z) is the Cumulative Probability Distribution Function of the Standard Normal Distribution. Reliability Function provided that P(T < 0)  0

5. Properties of the Normal Model MTBF (Mean Time Between Failure) Standard Deviation of Time to Failure = Failure Rate

6. The Normal Model - Example Examplem=1,000s=100

7. Table of Probabilities p Properties of the Normal Model - Standard Normal Distribution

8. Standard Normal Distribution Cumulative Probability Distribution Function F(x)

9. The Lognormal Model Definition A random variable T is said to have the Lognormal Distribution with parameters  and , where -  <  <  and  > 0, if the density function of T is: , for t > 0 , for t  0 Remark The Lognormal Model is often used as the failure distribution for mechanical items and for the distribution of repair times.

10. Properties of the Lognormal Model Failure Distribution where (z) is the cumulative distribution function Reliability Function If T ~ LN(,), then Y = lnT ~ N(,)

11. Properties of the Lognormal Model MTBF (Mean Time Between Failures) Standard Deviation of Time to Failure Failure Rate

12. The Lognormal Model • Failure rate • functions • for various • values of •  and 

13. The Lognormal Model

14. Example - Ductile Strength A theoretical justification based on a certain material failure mechanism underlies the assumption that ductile strength X of a material has a lognormal distribution. Suppose the parameters are  = 5 and  = 0.1 Compute E(X) and Var(X) Compute P(X > 120) (c) Compute P(110  X  130) (d) What is the value of median ductile strength? (e) If ten different samples of an alloy steel of this type were subjected to a strength test, how many would you expect to have strength at least 120? (f) If the smallest 5% of strength values were unacceptable, what would the minimum acceptable strength be?

15. Example - Solution The mean of x is and the variance of X is:

16. Example – Solution continued b)

17. Example – Solution continued c) d)

18. Example – Solution continued e) p=P(X>120)=0.983 from part (b) Y=number of tests with strength of at least 120 The Y~B(10,0.9834) and

19. Example – Solution continued f) We need to find the value of X, say x0.05, for which P(X < x0.05)=0.05 Since P(Z<-1.645)=0.05

20. The Binomial Model Definition If X is the number of successes in n trials, where: (1) The trials are identical and independent, (2) Each trial results in one of two possible outcomes success or failure, (3) The probability of success on a single trial is p, and is constant from trial to trial, then X has the Binomial Distribution with Probability Mass Function given by:

21. The Binomial Model Probability Mass Function , for x = 0, 1, 2, ... , n , otherwise where =

22. Binomial Distribution Mean or Expected Value  = np Standard Deviation , where q=1-p

23. The Binomial Model - Example Application 1 Four Engine Aircraft Engine Unreliability Q(t) = p = 0.1 Mission success: At least two engines survive Find RS(t)

24. The Binomial Model - Example Application 1- Solution X = number of engines failing in time t RS(t) = P(x  2) = b(0) + b(1) + b(2) = 0.6561 + 0.2916 + 0.0486 = 0.9963

25. Number of Failures Model Definition If T ~ E() and if X is the number of failures occurring in an interval of time, t, then X ~ P(t/ ), the Poisson Distribution with Probability Mass Function given by: for x = 0, 1, ... Where is the failure rate The expected number of failures in time t is

26. The Poisson Model X ~ P(2)

27. The Poisson Model p(x) Number of Failures ~ x

28. The Poisson Model - example continued 1.00 0 1 2 3 4 5 6 7 8 Number of Failures ~ x

29. The Poisson Model - Example Application An item has a failure rate of  = 0.002 failures per hour if the item is being put into service for a period of 1000 hours. What is the probability that 4 spares in stock will be sufficient?

30. The Poisson Model - Example Application - Solution Expected number of failures (spares required) = t = 2 P(enough spares) = P(X  4) = p(0) + p(1) + p(2) + p(3) + p(4) = 0.945 or about a 5% chance of not having enough spares!

31. Wheel Life Example The life of a given type of wheel for a specified operating environment has a Weibull distribution and its design life is t0.05 = 2000 hours. The median life is 3752 hours. What is the mean life and the median life? A system uses four wheels of this type. What is the probability of at least two of these wheels failing in 3000 hours?

32. Wheel Life Example We are given: t0.05 = 2000 hours and t0.50 = 3752 hours, but not  or θ. We have and Solve the first equation in terms of β

33. Wheel Life Example as follows

34. Wheel Life Example Substitute this for b in the other equation Solving this equation by Trial & Error we get Then

35. Wheel Life Example a) Calculate mean life: Standard Deviation

36. Wheel Life Example b) Probability of one wheel failing within 3000 hours

37. Wheel Life Example Probability of at least two wheels failing in 3000 hours Y = # of wheels failing in 3000 hours y = 0, 1, 2, 3, 4 Y has a Binomial Distribution with n = 4 and p = 0.2401 Y~B(4,0.2401)