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<table border="0" cellpadding="3" cellspacing="1" width="100%" class="forumline"> <tr> <th class="thCornerL" width="22%">Autor</th> <th class="thCornerR">Wiadomość</th> </tr> <tr> <td width="22%" align="left" valign="top" class="row1"><span class="name"><a name=""></a><b>defgf10aj</b></span></td> <td class="row1" valign="top"><table width="100%" border="0" cellspacing="0" cellpadding="0"> <tr> <td width="100%"><img src="http://picsrv.fora.pl/Elveron/images/icon_minipost.gif" width="12" height="9" alt="Post" title="Post" border="0" /><span class="postdetails">Wysłany: Śro 22:36, 08 Gru 2010<span class="gen"> </span> Temat postu: ugg italia A typical system reliability prediction</span></td> </tr> <tr> <td colspan="2"><hr /></td> </tr> <tr> <td colspan="2"><span class="postbody">A typical distribution system reliability prediction and <br /> <br /> <br />Or part 2) the reliability of the Rl, the reliability of the components 2 R2 = I a ((I-R) / (1-R1)) (8) If a parallel system reliability, the reliability index of 0.99, the system consists of 2 components, parts I known reliability of 0.90, test for system reliability: from (8) yields the reliability of Part 2 is R: = I a ((I a 0.99) / ( i A 0.90)) = I 10. i0 = 0.90. ■ The second situation: if the proportion of the known reliability of the components due to I, ie I = Rl, then R = IR. By the R = I a (i-IR2) (i-R:) = i A (I + IRi a (i + I) R] then a i - lq-IR: + dry = o! R.: L ' 1Rl = IR2 ±()-- RI One! (9) (10) where, R are two solutions to be parallel systems according to the nature of R> max {R}, to take a joint understanding. If a parallel system reliability from 2 components, the system reliability index is 0.99, the two components of known reliability of the scale factor of 0.90, test for system reliability. known to R = 0.99, I = 0.90, from ( 9) the availability of parts 2 Reliability R.: l +0.90 =---------~ 2 × 0.90 -! l + I). 2IRlI ± (Qiang).一 0 ~ 99 = 1.0556 ± 0.1l92 take R2 = 1.0556 - 0.1l92 ---- 0.936, to meet the R> max {Ri}. And then by type (i0) can get the reliability of components l Rl = IR2 = 0.90 × 0. 936 = 0.843, then, after the distribution system reliability t112R = i-If (1 a Ri) = I-n ( 1-Ri) i-Ii = l = 1 a (i a 0.843) (i a 0.936) = 0.99 can be seen, system reliability to meet the requirements. 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Existing carrier communication system,<a href="http://www.ugg-bootsguenstig.com" target="_blank" class="postlink">ugg boots deutschland</a>, both at home and abroad with the Lord, prepare jj_ {Automatic and manual conversion of a redundant system to ensure its reliability. If I plant set for 1800 way superior system is the application of a typical example of a redundant system, shown in Figure 5. Figure 5 contains superior reliability logic diagram for the system components were mainly 1,2 ... ... 14,<a href="http://www.eccoshoesoutlet.net" target="_blank" class="postlink">ecco shoes outlet</a>, with the main vibration system, crossover, amplifier ... ... ... ... harmonic generator; parts were redundant 1,2 ... ... The main vibration system, the amplifier divider ... ... ... ... harmonic generator. The components 15,16,17,18 are band-pass filter, amplifier, automatic conversion part 19 for electronic devices, with automatic inspection found invalid, auto alarm, auto conversion function, that is for part 1, a l4 and 20 play conversion protection function, ensure the normal work load for the system, components of the mixture of 20 to match the impedance of the coil. This effect is clearly the system reliability is good, but also realistic. 42 Because the level of failure rate of electronic components status quo, I plant the production of PCM digital current power system reliability communications equipment was not very ideal. Power system due to a failure of normal communications equipment, things have occurred. Therefore, I think: This high-tech products to improve the reliability of the factory, the factory should the system for failure analysis, feedback to the design work, to make it more comprehensive and reliable. 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