I am falling behind in my homework and need some help please. There are three sh

Question

I am falling behind in my homework and need some help please. There are three short questions: 1) A computer has three main modules that have individual reliabilities of 0.77, 0.87, and 0.82. Because of recent failures, management is now considering adding redundancy. Determine the reliability of the system if each module has a backup with a reliability equal to its own and a backup switch with a reliability of 1.00. 2) A computer has three main modules that have individual reliabilities of 0.76, 0.85, and 0.81. Because of recent failures, management is now considering adding redundancy. Determine the reliability of the system if the backup consists of an identical computer that operates as a whole rather than backing up individual sections. The single switch for the backup computer has a reliability of .96. 3) A DVD Burner has a mean time between failures of 83 weeks. Find the probability that the unit will not last 65 weeks.

Explanation / Answer

1. Relabilty of 1st module with the back up = 0.77 + (1-0.77)*0.77 = 0.9471 ( reliabilty of the module + failure of module*reliability of backup)

Relabilty of 2nd module with the back up = 0.87 + (1-0.87)*0.87= 0.9831

Relabilty of 3rd module with the back up = 0.82 + (1-0.82)*0.82 = 0.9676

Reliabilty of the system assuming the three modules act in series is = 0.9471*0.9831*0.9676 = 0.900

2. Reliability of the system = 0.76*0.85*0.81 = 0.52326

If the whole system is backed up as a whole rather than individuals then

reliability of the system = 0.52326 + (1-0.52326)*0.52326 = 0.7726

3. mean time between failures = 83 weeks

Reliability = e^(-time/mean time between failures) =  e^(-65/83)= 0.4570

Probability that it will not last = 1- reliability = 0.5430

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