ackup computer identical to the one being useu will U ability. Assuming the new
ID: 398748 • Letter: A
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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
6)
a) Reliability of the robot = Product of reliabilities of all components = 0.98*0.95*0.94*0.90 = 0.79
b) The component with the lowest reliability should have a back up unit. In this case it is component 4.
If there is a backup, the reliability of component 4 will be = 0.9 + (1-0.9)*0.9 = 0.99
Overall reliability of robot will be = 0.98*0.95*0.94*0.99 = 0.87
c) The 4th component should get the backup due to lowest reliability. The new reliability of component 4 will be = 0.9 + (1-0.9)*0.92 = 0.992
Overall reliability of robot will be = 0.98*0.95*0.94*0.992 =0.87
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