Number 7 ackup computer identical to the one being useu will U ability. Assuming
ID: 398784 • Letter: N
Question
Number 7
ackup computer identical to the one being useu will U ability. Assuming the new computer automatically functions if the main one fail resulting reliability. c. If the backup computer must be activated by a switch in the event that the first ter fails h the and the switch has a reliability of .98, what is the overall reliability of the syste switch and the backup computer must function in order for the backup to take 6. One of the industrial robots designed by a leading producer of servomechanism t func components. Components' reliabilities are .98,.95, .94, and tion in order for the robot to operate effectively. a. Compute the reliability of the robot. b. Designers want to improve the reliability by adding a backup component. Due t space limita- tions, only one backup can be added. The backup for any component will have the same reli- ability as the unit for which it is the backup. Which component should get the backup in order to achieve the highest reliability? c. If one backup with a reliability of .92 can be added to any one of the main components, which component should get it to obtain the highest overall reliability? A production line has three machines A, B, and C, with reliabilities of .99, .96, and .93, respec tively. The machines are arranged so that if one breaks down, the others must shut down. Engineers are weighing two alternative designs for increasing the line's reliability. Plan 1 involves adding an identical backup line, and plan 2 involves providing a backup for each machine. In either case three machines (A, B, and C) would be used with reliabilities equal to the original three. Which plan will provide the higher reliability? b. Explain why the two reliabilities are not the same What other factors might enter into the decision of which plan to adopt? C.Explanation / Answer
7
Reliability of production line with three components in series is product of reliability of each component = 0.99*0.96*0.93 = 0.884
Plan 1:
Identical back up line in parallel, So, system reliability will be = 0.884 + (1-0.884)*0.884 = 0.987
Plan 2:
Back up for each component:
Component 1 = 0.99 + (1-0.99)*0.99 = 1.000
Component 2 = 0.96 + (1-0.96)*0.96 =0.998
Component 3 = 0.93 + (1-0.93)*0.93 = 0.995
Overall system reliability = 1*0.998*0.995 = 0.993
a) So, Plan 2 of back up for each component is better
b) Two reliabilities are not same as in 1st case, even if any one component fails, entire load switches to other line. If some component fails in 2nd line also, then it will be failure of system. In Plan 2, there is back up for each individual component, so, the reliability is higher. The Plan 2 fails only if both the equipment of same component fails, whereas in plan 1 if any two components fail, then system fails
c) Factors to be considered for which plan to adopt are:
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