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Stochastic Control Part 15

Tham khảo tài liệu 'stochastic control part 15', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 552 Stochastic Control Goodness-of-fit testing. It is assumed that Xi f x a b S x51exp - x5-p x p s 0 i 1 1 15 93 where the parameters p and are unknown 5 0.87 . Thus for this example r n 15 k 3 m 5 1-a 0.95 X5 6.1 and S 170.8. It can be shown that the Uj 1 1 j 1 j g n - i 1 X5- X5-1 4 2-------------------- j 2 n - i 1 X5- X5-1 I i 2 J j 1 1 n - 2 94 are i.i.d. U 0 1 rv s Nechval et al. 1998 . We assess the statistical significance of departures from the left-truncated Weibull model by performing the Kolmogorov-Smirnov goodness-of-fit test. We use the .K statistic Muller et al. 1979 . The rejection region for the a level of significance is .K .Kn a . The percentage points for .Kn a were given by Muller et al. 1979 . For this example .K 0.220 .Kn 13 a 0.05 0.361. 95 Thus there is not evidence to rule out the left-truncated Weibull model. It follows from 92 for a 0.05 km 0.5 96 n km that h X8 - h x1 n 1 -if8 km i n-1 _ . a n km J - J -.X 1 0.87 z xX 15 i14_ 1 0.05 15 15 J 1 -J 5. 97 1 6.1 -1708 Thus the manufacturer has 95 assurance that before h 5 month intervals. no failures will occur in each shipment 5. Examples 5.1 Example 1 An electronic component is required to pass a performance test of 500 hours. The specification is that 20 randomly selected items shall be placed on test simultaneously and 5 failures or less shall occur during 500 hours. The cost of performing the test is 105 per hour. The cost of redesign is 5000. Assume that the failure distribution follows the one-parameter 553 Stochastic Decision Support Models and Optimal Stopping Rules in a New Product Lifetime Testing exponential model 15 . Three failures are observed at 80 220 and 310 hours. Should the test be continued We have from 19 and 20 80 220 310 17 X 310 0 -------------------------1960 hours 98 3 ppas - f 17 - 12 14 500 X6 Since 310 p J- p 1960 -12 310 117 1 expl 1960 I -115 6 - 0.79665 99 expl 1960 T0 Jxsf xs I xr Ô dxs - 430.05 hours xk ppas xr c 5000 -2 - 310 0.79665 - - 347.94 hours.