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Chapter 14

1. B5-life of 10 years is equivalent to R(10yrs) = 0.95MTBF = 150 years, assuming exponential distribution, is equivalent to:R(10yrs) = exp(−10/150) = 0.9355The first case therefore gives us higher reliability.

2. Using Eq. (14.5):

N = ln(1 − 0.5)

ln 0.975= 27.38

Therefore, after rounding it takes 27 units3. a. Based on Eq. (14.4) with N = 50 and C = 0.9

R = (1 − C)1N = (1 − 0.9)

150

= 0.955

b. Using Eq. (14.3) with k = 2 and Excel function C = 1-BINOMDIST(k N, (1-R), TRUE) runningGoal Seek to fit C = 0.9 produces R = 0.897.

Alternatively, using the chi-square approximation Eq. (14.6) and Excel function:

R = EXP(−1∗CHIINV(1−C, 2∗(k + 1))/(2∗N)) produces R = 0.899.

4. From Eq. (14.5), the required number of samples for R = 0.95 is 45, and for R = 0.97 it is 76 (in bothcases C = 0.9). Therefore the sample size increases by 31. Thirty-one samples at the cost of $ 1600each, totals $ 49 600.

5. a. Similar to the calculation in Question 4, demonstration of R = 95.0 % with the confidence C =90 % will require 45 samples.

b. Now we need to demonstrate the same reliability and confidence with only 30 samples. For thatpurpose we need to calculate the life test ratio, L to determine how much longer the test needs torun. Based on Eq. (14.9):

L =(

ln(1 − C)

N ln R

) 1�

=(

ln(1 − 0.9)

30 ln 0.95

) 12.5

= 1.1749

Therefore, the new test duration is 1000L = 1175 cycles.

Practical Reliability Engineering, Fifth Edition. Patrick D. T. O’Connor and Andre Kleyner.© 2012 John Wiley & Sons, Ltd. Published 2012 by John Wiley & Sons, Ltd.

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96 Chapter 14

6. A full solution of this problem requires the use of degradation analysis software, such as Weibull++ R©,however it can be solved using Microsoft Excel, though with a lesser degree of accuracy.a. The linear extrapolation of the inductance measurements for each of the units produces the graph,

Figure A14.1.

Figure A14.1 Question 6. Extrapolation of the degradation data.

Based on the linear extrapolation, Figure A14.1, the expected failure times are:Unit 1 (middle plot) – 1500 hUnit 2 (upper plot) – 1611 hUnit 3 (lower plot) – 1167 hPlease note that the results may vary based on the particular curve fitting method and the softwareused.

b. Processing the failure data from part a. using regression on X (RRX) option produces the followingWeibull parameters: � = 5.84 and � = 1528 h.

c. B10 can be calculated directly from the main Weibull equation:

0.9 = exp

[−

(B10

1528

)5.84]

Therefore, B10 = 1039.3 hd. Alternative degradation data extrapolation methods:

Based on the exponential extrapolation y = beax the failure times are:Unit 1 – 1550 hUnit 2 – 1714 hUnit 3 – 1190 h

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Chapter 14 97

Processing these failure times using regression on X (RRX) option produces the following Weibullparameters: � = 5.19 and � = 1603 h with B10 = 1039.6 h.

Power extrapolation, y = bxa

Since the power model cannot handle x = 0, a small value, such as 1 hour should be assigned insteadas an initial x-value.

Based on this extrapolation, the failure times are:Unit 1 – 559 599 hUnit 2 – 983 799 hUnit 3 – 20 225 hProcessing these failure times using regression on X (RRX) option produces the following Weibullparameters: � = 0.845 and � = 4.88 × 105 h with B10 = 33 995 h.

This example shows that not every extrapolation method is appropriate for all data sets. The powermodel here produces results distinctly different from the linear and exponential. Therefore the choice ofthe extrapolation models should be based on a careful analysis of the degradation data and understandingof the process that causes those failures (physical, chemical, etc.).

7. a. Section 2.15.2 (superimposed processes), Section 13.8.2 (multisocket systems).

b. MTBF = 1053

2= 526.5 h

From Appendix 2, for v = 2(2 + 1) = 6, and � = 0.1, � 2 = 10.6 (or this could be derived usingExcel function =CHIINV(0.1,6). Therefore lower 90 % MTBF = (2 × 1053)/10.6 = 198.7 h.

c. If MTBFL = 500, T = (500 × 10.6)/2 = 2650. Therefore the additional test time would be L(2650 − 1053) = 1597 h. This assumes that the failure rate will be constant over the whole of thetest time.

8. a. i. � = 2/1053 = 0.0019 failures per hourii. MTBF = 1053/2 = 526.5 hours

iii. From Appendix 2, for � = 2(2 + 1) = 6, and � = 0.1, � 2 = 10.6 or using Excel =CHIINV(0.1,6), i.e. MTBFL = 2 × 1053/10.6 = 198.7.

b. Using Table 14.2 for MTBFL = 500, T = 500 × 10.6/2 = 2650 hours.So further testing required (without more failures) = 2650 – 1053 = 1597 hours.

9. Based on the equations in Table 14.2 for failure truncated test:

2T

� 2( �

2 ,2k)

≤ MTBF ≤ 2T

� 2(1− �

2 ,2k)

T = 628 + 3444 + 822 + 846 + 236 = 5976 hours, 2k = 10� 2

(0.05, 10) = CHIINV(0.05, 10) = 18.307� 2

(0.95, 10) = CHIINV(0.95, 10) = 3.94

652.9 ≤ MTBF ≤ 3033.3

10. Time terminated test:

� ≥ � 2(�,2k+2)

2T

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98 Chapter 14

T = 4 × 1000 + 825 = 4825k = 1 (one failure)� 2

(0.1, 4) = CHIINV(0.1, 4) = 7.779� ≥ 7.779/(2 × 4825) = 0.00080616 failures/hour

11. Section 14.12.12. Design ratio is the acceptable MTBF divided by the low limit MTBF. The higher the risks (i.e. the higher

the values of � and �) and the lower the design ratio, the longer will be the expected test duration andtherefore the expected cost.

13. Based on Eq. (14.15):

�i = �c

1 − �

= 40 000

1 − 0.4= 66 667 hours

14. The major contributor to reliability growth and continuous process improvement is early detection andelimination of failure causes. See also Section 14.3.

15.

Failures (F) Time (T) Cumulative MTBF �C(= T/F)

1 12 12.002 36 18.003 80 26.674 120 30.005 200 40.006 360 60.007 400 57.148 440 55.009 480 53.33

(Total 2128)

The Duane plot appears in Figure A14.2. As there is an upward slope, growth is present. The slope ofthe fitted straight line (measured directly from the plot) is approximately 0.45, indicative of an effectiveimprovement programme. Note that the plotted points, in contrast to a probability plot, are not bound tobe in increasing order, as evidenced by the last four points. Note also that the results from a Duane plotare heavily dependent on the assumed starting point. A plot of the last four points alone would show adeterioration (‘negative growth’).

As the test is time-terminated, Eq. (2.46) can be used direct.

U = (2128/9) − (500/2)

500√

(1/12 × 9)= −0.28

Reference to Appendix 1 shows that this value has a probability of (1 − 0.6103) ≈ 0.39 (or =NORMSDIST(–0.28) = 0.3897) of arising by chance, so that the trend is not statistically significant.(It is the last four points that are the problem. The logarithmic scale of the Duane plot masks their

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Chapter 14 99

effect – a linear trend plot gives a different picture. The moral is not to blindly pursue Duane plotswithout checking by other means that growth is really occurring.)

Figure A14.2 Question 15. Duane plot.