Question

5. With the booming world economy, the 2x4 lumber consumed in the construction industry is considerably increasing. To obtain
0.61 10.8 1 | 0.89] → [06 1 0.92] | 0.95 | | 0.65 | - 0.95 ] → 0.67 ] - 0.69 |
b) A system for milling 2x6s has four sensitive components connected in series with reliabilities of 0.93, 0.94, 0.92 and 0.
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Answer #1

0.6 | 10.89] | 0.8 →[092] | 0.92 | —[095 LAL 06] -05] > 0.65 | > 0.67] - 0.69 上

For parallels, the reliability is 1 – (1-reliability)*(1-reliability)*…..

Here:

For

→ 0.65] > 0.67 | 0.69

Reliability = 1 – (1-0.65)*(1-0.67)*(1-0.69) = 1 – (0.35*0.33*0.31) = 1 – (0.0358) = 0.9641 = 0.97

For

0.6 i - 0.6 |

Reliability = 1 – (1-0.6)*(1-0.6) = 1 – (0.4*0.4) = 1 – 0.16 = 0.84

Now the configuration is

自HIV 0.95 0.95 0.97

For series, reliability = reliability_1*reliability_2*….

Hence for

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Reliability = 0.84*0.89*0.92 = 0.687792

For

0.95 0.97

Reliability = 0.95*0.97 = 0.9215

Hence, now the configuration is:

10.687792 10.95 0.9215

Hence Reliability = 0.95* (1 – (1-0.687792)*(1-0.9215))

= 0.95* (1 – (0.312208*0.0785)) = 0.95*(1 – 0.0245) = 0.95*0.9754 = 0.92671

Hence the reliability is = 0.93

b.

Old:

Series reliability of 0.93, 0.94, 0.92, 0.97

Hence Reliability = 0.93*0.94*0.92*0.97 = 0.78

New:

Series reliability of 0.93, 0.94, 0.92, 0.97 with backup of 0.94

Hence Reliability = (0.93+0.94*(1-0.93))*(0.94+0.94*(1-0.94))*(0.92+0.94*(1-0.92))*(0.97+0.94*(1-0.97))

= (0.93+0.0658)*(0.94+0.0564)*(0.92+0.0752)*(0.97+0.0282)

= 0.9856

= 0.99

Third option:

Series reliability of 0.93, 0.94, 0.92, 0.97 with backup of 0.97

Hence Reliability = (0.93+0.97*(1-0.93))*(0.94+0.97*(1-0.94))*(0.92+0.97*(1-0.92))*(0.97+0.97*(1-0.97))

= (0.93+0.0679)*(0.94+0.0582)*(0.92+0.0776)*(0.97+0.0291)

= 0.99281

= 0.99

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