Homework Set on Probabilistic Failure Models

Transcription

1 Homework Set on Probabilistic Failure Models List of problems:. Time to failure:, p.2 (p.5). 2. Time to failure: 2, p.2 (p.6). 3. Time to failure: 3, p.3 (p.7). 4. Scratches, p.4 (p.8). 5. Many identical components, p.5 (p.9). 6. Pressurized tube, p.6 (p.2). 7. Adjusted machine, p.7 (p.2). 8. Passing grade, p.8 (p.22). 9. Switch envelopes? p.9 (p.23). Gear-tooth breakage:, p. (p.24).. Gear-tooth breakage: 2, p. (p.). 2. Rockets, p. (p.). 3. Two failure mechanisms, p.2 (p.25). 4. Crack density, p.3 (p.26). 5. Weibull distribution and the failure rate function, p.4 (p.27). \lectures \reltest \pfm ps.tex

2 pfm ps.tex Problem Set on Probabilistic Failure Models 2. Time to failure:. (p.5) A component with time to failure T has constant failure rate: z(t) = λ = [hours] () (a) Determine the probability that the component survives a period of 2 months without failure. (b) Find the mean time to failure (MTTF) of the component. (c) Find the probability that the component survives its MTTF. 2. Time to failure: 2. (p.6) A machine with constant failure rate λ will survive a period of hours without failure, with probability.5. (a) Determine the failure rate λ. (b) Find the probability that the machine will survive 5 hours without failure. (c) Determine the probability that the machine will fail within hours, when you know that the machine was functioning at 5 hours.

3 pfm ps.tex Problem Set on Probabilistic Failure Models 3 3. Time to failure: 3. (p.7) A component with time to failure T has failure rate: z(t) = kt for t > and k = 2. 6 [hours] 2 (2) (a) Determine the probability that the component survives 2 hours. (b) Determine the mean time to failure, MTTF, of the component. (c) Determine the probability that a component, which is functioning after 2 hours, is still functioning after 4 hours.

4 pfm ps.tex Problem Set on Probabilistic Failure Models 4 4. Scratches. (p.8) (Poisson and exponential distributions). Scratches are introduced randomly and independently along magnetic tape as it is produced by a particular procedure. The average density of scratches on the tape is λ =.3 scratches per meter. (a) What is the probability that a 2 meter segment is free of scratches? (b) meter tapes are acceptable if the number of scratches is no more than 2. In a box of 5 tapes, what is the probability that all tapes are acceptable? (c) What is the mean scratch-free length?

5 pfm ps.tex Problem Set on Probabilistic Failure Models 5 5. Many identical components. (p.9) (Weibull distribution). A device has a large number of identical components. The device fails as soon as the first component fails. The failure rate function for this device increases linearly in time, and the mean time to failure is 25 hours. What is the probability that the device will operate at least 75 hours?

6 pfm ps.tex Problem Set on Probabilistic Failure Models 6 6. Pressurized tube. (p.2) (Normal distribution). The fluctuation of pressure in a pressurized tube is due to the superposition of a large number of factors. The mean and standard deviation of the pressure are found to be.2[atm] and.4[atm], respectively. The tube ruptures at a pressure of.75[atm]. (a) What is the probability of failure? (b) A pressure gauge attached to the tube reports a pressure of.6[atm]. Do you believe the measurement, or do you suspect that the gauge is faulty? (c) The tube has been changed and is now operating at a new pressure. A reliable pressure gauge is used to measure the pressure 7 times, with the results: 2., 2., 2.3, 2.4,.9, 2.2, 2.2 [atm]. The new tube fails at 2.6[atm]. What is the probability of failure?

7 pfm ps.tex Problem Set on Probabilistic Failure Models 7 7. Adjusted machine. (p.2) Suppose that when a machine is properly adjusted, 5% of the products are high quality and the rest are medium quality. However, when the machine is improperly adjusted, 25% of the products are high quality and the rest are medium quality. The machine is improperly adjusted % of the time. At a particular time, 5 items produced by this machine were inspected and 4 were of high quality and of medium quality. What is the probability that the machine was properly adjusted when it produced these items?

8 pfm ps.tex Problem Set on Probabilistic Failure Models 8 8. Passing grade. (p.22) Consider the American system of examination (multiple choice) with 3 options for each of 2 questions. For each question, assume that either the student knows the answer and chooses the correct option, or he does not know the answer and chooses randomly with equal probability for each option. What value of a passing grade should be adopted in order that only % of candidates who do not know the answer to any question will pass? (Suggestion: use the normal approximation to the binomial distribution.)

9 pfm ps.tex Problem Set on Probabilistic Failure Models 9 9. Switch envelopes? (p.23) I have two envelopes, each with a positive amount of money, one with ten times more than the other. There are many such pairs of envelopes, and the probability that the pair I am holding contains $ n and $ n+ is /2 n where n =, 2,.... This is illustrated as follows: Probability Small $ $ 2 $ 3 Large $ 2 $ 3 $ 4 You pick an envelope, open it, and find that it contains $ n. Do you want to switch this envelope for the other one in the pair? (3)

10 pfm ps.tex Problem Set on Probabilistic Failure Models. Gear-tooth breakage:. (p.24) The stress applied to the tooth of a gear is distributed with an exponential pdf: p L (l) = λe λl, l (4) Likewise, the yield strength of the tooth is exponentially distributed: (a) What is the probability of failure of a single specific tooth? p y (s y ) = µe µsy, s y (5) (b) Now consider an entire gear, with N teeth. Assume the gear fails if any single tooth fails. What is the probability of failure of the gear? What have you assumed in order to solve this? (c) Now suppose that the gear fails only when n or more out of the N teeth fail. What is the probability of failure?

11 pfm ps.tex Problem Set on Probabilistic Failure Models. Gear-tooth breakage: 2. (p.) Consider a gear with N teeth operating at ω = 2 gear cycles per minute. The load on the teeth is constant during each gear cycle, except that occasionally a single tooth gets an over-load far in excess of its yield strength, causing breakage of that tooth. The rate of occurrence of over-loads is λ = 9 per cycle. The gear fails when 2 or more teeth are broken. (a) What is the probability of gear failure in T = 2 hours? (b) What is the MTTF of the gear? 2. Rockets. (p.) A rocket is shot to a distance of about 4[km]. The payload must be activated when the rocket is 5[m] above the ground near the target. The barometric pressure is measured and used to determine the rocket altitude. However, barometric measurements are not accurate for two reasons: () variation of barometric pressure from day to day and (2) variation from one instrument to another as expressed by a random offset in the barometric reading of the sensor. The first problem is solved by calibrating the barometric measurement by measuring the barometric pressure at the time of launch. The second problem is solved by a voting method: N barometric sensors are used and the payload is activated when M sensors indicate that the altitude is no greater than 5[m]. Define failure of the activation mechanism to be: activation of the payload above 6[m] or below 4[m]. Assume that the distribution of barometric measurement error is normal with zero mean and variance σ 2 = 25[m 2 ]. Consider the following problems: (a) What is the probability of failure when: N = 3 and M =? N = 3 and M = 2? N = 3 and M = 3? (b) What is the utility of the marginal sensor if the payload is activated when a majority of sensors indicate an altitude no greater than 5[m]? That is, how does the probability of failure vary with N? (c) Suppose two different types of barometric sensors are available, an inexpensive but inaccurate model costing c and with variance σ, and an expensive but accurate model costing c 2 and with variance σ 2. Assume both types of sensors have zero-mean normal distributions. Develop an expression for choosing the equivalent number of cheap or expensive sensors, given the requirement that the probability of failure be no greater than P f.

12 pfm ps.tex Problem Set on Probabilistic Failure Models 2 3. Two failure mechanisms. (p.25) A component may fail due to two different causes: excessive stress and aging. Data for the component show that the time to failure T due to excessive stress is exponentially distributed: The time to failure T 2 due to aging has a gamma distribution: f (t) = λ e λ t, t (6) f 2 (t) = λ 2 Γ(k) (λ 2t) k e λ 2t, t (7) (a) What is the rationale for the following probability density for the time to failure T of the component? What is the meaning of the parameter p? (b) What is the failure rate function? (c) What is the mean time to failure? f(t) = pf (t) + ( p)f 2 (t), t (8) (d) Suppose design choices can influence the value of p. What is the optimal value of p?

13 pfm ps.tex Problem Set on Probabilistic Failure Models 3 4. Crack density. (p.26) (Poisson and exponential distributions). Solid rocket fuel has tiny internal cracks which are distributed randomly and independently in the material. Let s suppose that the average density of cracks is ρ = 5 cracks per meter 3. (a) What is the probability that a cm 3 sample is free of cracks? (b) What is the probability that a cm 3 sample has no more than crack? Given samples, each cm 3, what is the probability that no sample has more than crack? (c) What is the mean distance between cracks?

14 pfm ps.tex Problem Set on Probabilistic Failure Models 4 5. Weibull distribution and the failure rate function. (p.27) In table are listed the lifetimes of 5 out of a population of 2 mechanical switches. The term lifetime refers to the number of cycles performed before failure occurs. All other switches in the population survived more than 45 cycles. Is the probability of failure of these switches constant, increasing or decreasing in time? Suggestion: Evaluate an empirical probability distribution, F (t). Suppose the true distribution is Weibull and consider the slope of ln[ ln( F )] vs ln t. Number of cycles Failure number to failure (t) (i) Table : Failure data.

15 pfm ps.tex Problem Set on Probabilistic Failure Models 5 Solutions Solution to problem. (p.2) The failure rate is constant, so the pdf of T can be exponential: f(t) = λe λt, F (t) = e λt, t, λ = [hr ] (9) (a) So: (b) (c) P (T > 2 months) = F (2 months) () 2 months 3 days 24 hours = 44[hour] () month day P (T > 2 months) = e 44λ = e (2) P MT T F = E(T ) = = 4, [hour] (3) λ ( T > ) ( = F = e λ λ) λ( λ) = e.368 (4) Question: why isn t the answer /2? Answer: the pdf is asymmetric.

16 pfm ps.tex Problem Set on Probabilistic Failure Models 6 Solution to problem 2. (p.2) Constant failure rate implies an exponential distribution: f(t) = λe λt, F (t) = e λt, t (5) (a) Hence: (b) P (T > hours) = F ( hours) = e λ =.5 (6) λ = ln 2 = λ = ln (7) (c) We seek: P (T > 5 hours) = F (5 hours) = e 5λ e (8) Conditional probability: P (A B) = Q = P (T T }{{}} > {{ 5 } ) = A B P (A and B) P (B) P [(T ) and (T > 5)] P (T > 5) P [(T ) and (T > 5)] = P [5 T ] = F () F (5) = e λ + e λ5 (2) So: (9) (2) P (T > 5) = F (5) = e λ5 (22) Q = e λ + e λ5 e λ5 = e λ5 = F (5).969 (23)

17 pfm ps.tex Problem Set on Probabilistic Failure Models 7 Solution to problem 3. (p.3) (a) The reliability function is related to the failure rate function as: [ t ] [ t ] R(t) = exp z(s) ds = exp ks ds = e kt2 /2 (24) R(2) = e k 22 /2 = e (25) (b) (c) The MTTF is related to the reliability function as: MT T F = Q = R(t) dt = e kt2 /2 dt = 2 2π k = 2π (26) P (T > 4) P (T > 2) = R(4) R(2) = e k 42/2 e k 22 /2 = e k 2 (42 2 2) e (27)

18 pfm ps.tex Problem Set on Probabilistic Failure Models 8 Solution to problem 4. (p.4) The probability of exactly n scratches in a length x of tape is given by the Poisson distribution: P n (x) = e λx (λx) n, λ =.3[m] (28) n! (a) P (2) = e λ2 = e (29) (b) The probability of no more than 2 scratches in x meters of tape is: For x = this is: ( ) Q = Prob(n 2) = P (x) + P (x) + P 2 (x) = e λx + λx + (λx)2 2! The probability that all of m tapes in a box are acceptable is: (3) Q.423 (3) A = Q m (32) Too bad! Sounded like a good deal. (c) We need the pdf for the scratch-free distance, x. To find this, we use the concept of probability balance: f(x) dx = Probability of first scratch in the interval [x, x + dx] (33) = P (x)λ dx (34) = λe λx dx (35) which is the exponential distribution. The mean free path is the expectation of x: E(x) = λ = [m] (36).3

19 pfm ps.tex Problem Set on Probabilistic Failure Models 9 Solution to problem 5. (p.5) Since the device fails as soon as the first of its many components fails, it is failing by a weakest link mechanism. Thus its lifetime distribution is asymptotically Weibull. More formally, let the number of components be N and let T n denote the lifetime of the nth component. Thus the lifetime of the device is: T = min {T,,..., T N } (37) n N When N is large, T has a Weibull distribution. The failure rate function is: z(t) = kt (38) For a Weibull distribution, the failure rate function is: Hence we conclude that: Hence: z(t) = αλ(λt) α = kt (39) The MTTF is 25 hours. For a Weibull distribution the MTTF is: α = 2 (4) MTTF = λ Γ ( α + ) = 25 (4) λ = Γ(3/2) (42) For a Weibull distribution the cumulative distribution function is: F (t) = e (λt)α (43) So: Pr(T 75) = F (75) = e (λ75)α.25 (44)

20 pfm ps.tex Problem Set on Probabilistic Failure Models 2 Solution to problem 6. (p.6) By the central limit theorem we conclude that the pressure P is normally distributed: f(p ) N (µ, σ 2 ) (45) (a) We know the mean and standard deviation: The probability of failure is: Pr(P.75) = ( P µ Pr σ µ =.2, σ =.4 [atm] (46) >.75 µ ) σ (47) = Φ(.375) =.955 =.846 (48) since P µ N (, ) (49) σ (b) We ask: what is the probability of obtaining a true pressure less than.6? ( P µ Pr(P.6) = Pr σ This is possible, but not likely. Better check the gauge. (c) We estimate the new mean as the sample mean:.6 µ ) σ (5) = Φ(.5) (5) = Φ(.5) =.9332 =.668 (52) P = N N P i = 2.7 (53) i= We estimate the new variance as the sample variance: s 2 = N N (P i P ) 2 =.257 = s =.6 (54) i= So, the probability of failure is estimated as: ( P P Pr(P 2.6) = Pr s 2.6 P s ) (55) = Φ(2.68) =.9963 =.37 (56) We have assumed that: P P N (57) s which is not true. This statistic has a t distribution with N degrees of freedom, which is a bit broader than the normal distribution. Hence our estimate of the failure probability is a bit too low.

21 pfm ps.tex Problem Set on Probabilistic Failure Models 2 Solution to problem 7. (p.7) Define the following events: A = adjusted properly. N = not adjusted properly. H = high quality. M = medium quality. Basic data: P (H A) =.5, P (M A) =.5, P (H N) =.25, P (M N) =.75, P (A) =.9, P (N) =. (58) Our observation of 5 items: 4 are H, is M. Call this observation O. We must calculate: P (A and O) P (A O) = P (O) (59) First calculate the denominator from the theorem of complete probability: Now calculate the numerator: P (O) = P (O A)P (A) + P (O N)P (N) (6) [( ) ( 5 4 ( ) = 4 2) ] [( ) ( 5 4 ( ) 3 (.9) ) ] (.) (6) 4 = [ ] 5 (62) P (A and O) = P (O A)P (A) = (63) So: P (A O) (64)

22 pfm ps.tex Problem Set on Probabilistic Failure Models 22 Solution to problem 8. (p.8) N = 2 questions. If the candidate knows no answers, then: p = probability of correct answer = /3. P r = probability that an ignorant student answers exactly r questions correctly, which is given by the binomial distribution: ( ) N P r = p r ( p) N r, r =,,..., N (65) r The mean and variance of the binomial distribution are: E(r) = Np, var(r) = Np( p) (66) The ignorant student passes if he answers correctly c or more questions. The probability of this is: P =. = P (r c) = r c P r (67) Define a normal random variable whose moments are those of the binomial distribution: So: Hence we require that c satisfy: which implies: Choose c = 2. x N (µ, σ 2 ) = N [Np, Np( p)] (68) µ = , σ , σ 2.8 (69) ( x µ P =. P (x c) = P σ c µ ) Φ σ ( ) c (7) ( ) c Φ =.99 (7) 2.8 c = 2.33 or c.58 (72)

23 pfm ps.tex Problem Set on Probabilistic Failure Models 23 Solution to problem 9. (p.9) Suppose your envelope contains $. Clearly you would want to switch, since you know the other envelope contains $. But suppose your envelope contains $? Do you want to switch? You have either the large envelop from from $ $ pair, or the small envelope from the $ $, pair. Let s make a decision based on expected return upon switching. Suppose your envelope contains $. What is the conditional probability that you hold the small envelope, and what is the conditional probability that you hold the large envelope? P (S ) = P (L ) = P (S, ) P () P (L, ) P () The expected return if you switch enveloped is: = = = + 2 = + 2 = 3 = 2 3 (73) (74) Since: ER = P (S ) $ 4 + P (L ) $ 2 = 3 $ $2 (75) ER > $ 3 (76) we conclude that it is a good idea, based on expectations, to switch envelopes in this case. Is this true for all values? Suppose your envelope contains $ n. Should you switch? The generalization of eqs.(73) and (74) is: P (S n ) = P (S, n ) P ( n ) P (L n ) = P (L, n ) P ( n ) Now, the generalization of eq.(75) is: = = 2 n 2 2 n n 2 2 n 2 2 n n 2 2 = + 2 = + 2 = 3 = 2 3 (77) (78) Since: ER = P (S n ) $ n+ + P (L n ) $ n = 3 $n $n (79) ER > $ n (8) we conclude that it is a good idea, based on expectations, to switch envelopes in every case. Does this make sense? This riddle is related to the Petersburg Paradox. What is the expected return from this game, even if you don t switch? It is: ER = n= ( 2 n 2 $n + ) 2 $n+ = 2 n= $ n = $ (8) 2n

24 pfm ps.tex Problem Set on Probabilistic Failure Models 24 Solution to problem. (p.) (a) The probability of failure is: F = Probability of failure = Prob(s y L) (82) = l = λµ = l µ µ + λ p L (l)p y (s y ) ds y dl (83) e λl e µs y ds y dl (84) The probability of failure is large if the load is much more widely distributed than the yield strength. That is, F is large if µ λ. The reliability is: R = F = λ (86) µ + λ which is large if the load-distribution is much narrower than the strength-distribution (λ µ). (b) The probability that the ith tooth does not fail is the reliability from eq.(86): (85) P i = λ µ + λ (87) So, the probability that no tooth fails, assuming that the teeth-failures are independent, is: So, the probability that the gear will fail is: N N [ ] λ R = P i = µ + λ = λ N (88) µ + λ i= i= For example, suppose λ =., µ =.5 and N = 2. Then: [ ] λ N F S = R = (89) µ + λ λ µ + λ = = F S = =.623 (9) (c) The probability that any particular tooth does not fail is P = P i, eq.(87). Likewise, the probability that any particular tooth does fail is F = P i. So, the probability that m out of N teeth fail is: ( ) ( ) ( ) N F m,n = F m P N m N λ m ( ) ( ) µ N m N λ m µ N m = = m m µ + λ µ + λ m (µ + λ) N (9) So, the overall probability of failure is: N F S = F m,n (92) m=n

25 pfm ps.tex Problem Set on Probabilistic Failure Models 25 Solution to problem 3. (p.2) (a) We can understand p as the probability of failure by excessive stress. Thus eq.(8) becomes the total probability of failure at time t. This is explained as follows. f (t) and f 2 (t) are conditional probability densities: f (t) is the probability density of failure at time t, given failure by excessive stress. f 2 (t) is the probability density of failure at time t, given failure by aging. p and p are marginal probabilities: p is the probability of failure by stress. p is the probability of failure by aging. f(t) is the total probability density of failure. The general formula for total probability of A given conditioning events B i is: P (A) = P (A B i )P (B i ) (93) i where: Eq.(8) is a special case of eq.(93). (b) The failure rate function is: B i B j = and i B i = S (entire space) (94) z(t) = where the cumulative distribution function F (t) is: F (t) = t t f(t) F (t) (95) f(τ) dτ (96) t = p f (τ) dτ + ( p) f 2 (τ) dτ (97) = pf (t) + ( p)f 2 (t) (98) ( ) = p e λ t + ( p) (Assuming that k is an integer.) (c) The MTTF is the expectation of t: MTTF = = p k j= (λ 2 t) j j! e λ 2t (99) tf(t) dt () tf (t) dt + ( p) tf 2 (t) dt () = pe (t) + ( p)e 2 (t) (2) = p λ + ( p) k λ 2 (3) (d) There are several ways to approach this problem. One is to choose p so as to enlarge the MTTF. From eq.(3) we know: MTTF = p λ + ( p) k λ 2 = ( λ k λ2 ) p + k λ 2 (4) Thus: Choose p large if λ > k λ 2. Choose p small if λ < k λ 2. A different approach is to choose p to make the failure rate function z(t) small. If k = one finds that these two approaches are the same.

26 pfm ps.tex Problem Set on Probabilistic Failure Models 26 Solution to problem 4, Crack density (p.3). (4a) The probability of exactly n cracks in a volume x is given by the Poisson distribution: Thus: P n (x) = e ρx (ρx) n, ρ = 5 [m] 3 (5) n! P ( 6 ) = e 5 6 = e..948 (6) (4b) The probability of no more than 2 cracks in x meters 3 of fuel is: For x = 6 m 3 this is: Q = Prob(n ) = P (x) + P (x) = e ρx ( + ρx) (7) Q = P ( 6 ) + P ( 6 ) = e. ( +.).9953 (8) The probability that all of the samples have no more than crack is: A = Q (9) (4c) The crack density is ρ cracks/m 3. Thus, on average, each crack owns a cubic volume of ( ) /3 ρ m3. This cube has a side-length of l = ρ m. For ρ = 5 m 3 this is l =.25 m. The distance between midpoints of adjacent cubes is l, which is an approximation to the mean distance between cracks. To evaluate the mean distance between cracks more precisely, consider an infinitesimal crack at some point, and ask what is the average radius, r, around this crack, of the sphere that contains the first additional crack. The probability density of r is: f(r)dr = probability of first crack in shell at (r, r + dr) () ( ) 4 = P 3 πr3 4πr }{{ 2 dr} ρ vol of shell }{{} ( = exp 4 3 πr3 ρ One can verify by integration that this pdf is normalized. The mean distance between cracks is: E(r) = This integral is not easy to evaluate analytically. () prob of crack in shell ) 4ρπr 2 dr (2) rf(r) dr (3)

27 pfm ps.tex Problem Set on Probabilistic Failure Models 27 Solution to problem 5, Weibull distribution and the failure rate function (p.4). The solution is based on sections.2 and.3 of Lecture Notes on Censoring and Estimation in Statistical Sampling. Weibull Analysis We have a population of M items of the same sort. N < M items have failed at times t t 2,..., t N. We wish to determine the best coefficients of a Weibull distribution: F T (t) = Prob(T t) = { e (λt) α, t > t (4) Why? These coefficients tell us what stage of the bathtub curve the population is in: burn-in (phase ), central (phase 2) or burn-out (phase 3). The general approach is to use this data to estimate F (t) empirically, and then fit the parameters. In this solution we use only the data on the failed items, and ignore the fact that M N items are still functional because we do not know when these items will fail. The lecture on censoring is devoted to learning how to incorporate this additional data. To establish the empirical distribution function, note that the times are ranked in increasing order. Thus the fraction /N of the sample failed in time t, a fraction 2/N failed in time t 2 and so on. We could thus estimate F (t) as: F (t i ) = i, i =,..., N (5) N This is not a good approximation to the cdf because the data are not at random times but rather at failure times at which the empirical function makes a step increase. A better approximation, which is strictly ad hoc, is: F (t i ) = i.3, i =,..., N (6) N +.4 We now have an estimate of the distribution at N instants: F (t i ), i =,..., N. We can use either graphical methods or algebraic methods to find the best estimates for α and λ. The graphical method for estimating the Weibull parameters could be based on noting that: ln ( ln[ F (t)]) = α ln t + α ln λ (7) Thus, plotting ln( ln[ F (t)]) versus ln t would result in a straight line whose slope is α and whose intercept is α ln λ. An algebraic method could be based on choosing α and λ to minimize the deviation of the data from the fitted function. This least-squares approach is to choose α and λ to minimize: S 2 = N i= [ ] 2 F (t i ) F (t i ) (8) where F (t i ) is the estimate of the cpf, eq.(5) or eq.(6), and F (t i ) is the Weibull distribution in eq.(4) evaluated at time t i. We choose α and λ to minimize S 2. Weibull Distribution and Failure Rate Function The failure rate function of the Weibull distribution is: z(t) = f(t) F (t) = αλ(λt)α, t > (9) From the 5 data points of table on p.4 we calculate the empirical cdf F (t i ) according to eq.(6). Then we plot ln[ ln( F (t))] versus ln t as shown in fig.. An approximate eyeball linear fit shows the slope to be α = 2.. So the failure rate function is increasing in time, since α >. This population is thus in the burn out end-of-life phase. \lectures\reltest\censor.tex.

28 pfm ps.tex Problem Set on Probabilistic Failure Models 28 ( ln ln[ F ) (t)] ln t Figure : Circles: estimated cdf versus failure time, eqs.(6) and (7). Line: approximate linear fit with slope = 2.