nce system of a ship is controlled by a computer that has three major modu ofunc
ID: 420575 • Letter: N
Question
nce system of a ship is controlled by a computer that has three major modu ofunction properly, all three modules must function. Two of the mo modules. In order laces, what is the largest component proba 5. The guidance system of a ship is controlled by a co for the computer to function properly, all three reliabilities of.97, and the other has a reliability of.99. a. What is the reliability of the computer? b. A backup computer identical to the one being used will be installed to improve overall reli ability bility, Assuming the new computer automatically functions if the main one fails, dete resulting reliability. c. If the backup computer must be activated by a switch in the event that the first computer fails, and the switch has a reliability of 98, what is the overall reliability of the system? (Both the switch and the backup computer must function in order for the backup to take over.) One of the industrial robots designed by a leading producer of servomechanisms has four major components. Components' reliabilities are 98, .95, .94, and .90. A tion in order for the robot to operate effectively Il of the components must func Compute the reliability of the robot. . Designers want to improve the reliability by adding a backup component. Due to 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? 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? roduction line has three machines A. B, and C, with reliabilities of .99,.96, and .93, respec ly. The machines are arranged so that if one breaks down, the others must shut down. Engineers weighing two alternative designs for increasing the line's reliability. Plan 1 involves adding dentical backup line, and plan 2 machines (A. B, and C) would be used with reliabilities e hich plan will provide the higher reliability? plain w hat other factors involves providing a backup for each machine. In either case. equal to the original three. hy the two reliabilities are not the same might enter into the decision of which plan to adopt?Explanation / Answer
Answer to question 5 :
Reliability of the 3 modules are 0.97, 0.97 and 0.99
Therefore reliability of the computer = 0.97 x 0.97 x 0.99 = 0.9315
In such case, reliability of the system
= 1 – ( 1 – 0.9315)^2
= 1 – 0.0685 x 0.0685
= 1 – 0.0047
= 0.9953
Therefore , reliability of the backup computer with switch = 0.98 x 0.9315 = 0.91287
Since the main computer with reliability 0.9315 and back up computer with switch with reliability 0.91287 are connected in parallel, the reliability of the overall system
= 1 – ( 1 – 0.9315) x ( 1 – 0.91287)
= 1 – 0.0685 x 0.08713
= 1 – 0.00596
= 0.9940
Answer to question 6 :
In such case , reliability of the robot = 0.98 x 0.95 x 0.94 x 0.90 = 0.7876
Reliability of the component with reliability 0.98 with identical backup = 1 – ( 1- 0.98)^2 = 1 – 0.02 x 0.02 = 1 – 0.0004 = 0.9996
Reliability of the component with reliability of 0.95 with identical back up = 1 – ( 1 – 0.95)^2 = 1 – 0.05 x 0.05 = 1 – 0.0025 = 0.9975
Reliability of the component with reliability 0.94 with identical back up = 1 – ( 1 – 0.94)^2 = 1 – 0.06 x 0.06 = 1 – 0.0036 = 0.9964
Reliability of the component with reliability of 0.90 with identical back up = 1 – ( 1 – 0.90)^2 = 1 – 0.1 x 0.1 = 1 – 0.01 = 0.9900
Since reliability of the back up system is highest( i.e. 0.9996 ) for the component with reliability 0.98 , this component should get the back up to achieve the highest reliability
Reliability of component with reliability 0.95 with back up 0.92 = 1 – ( 1 – 0.95 ) x ( 1 – 0.92) = 1 – 0.05 x 0.08 = 1 – 0.004 = 0.996
Reliability of the component with reliability 0.94 with back up of 0.92 = 1 – ( 1 – 0.94) x ( 1 – 0.92 ) = 1 – 0.06 x 0.08 = 1 – 0.0048 = 0.9952
Reliability of one component with reliability 0.90 with back up of 0.92 = 1 – ( 1 – 0.90)x ( 1 – 0.92) = 1 – 0.1 x 0.08 = 1 – 0.008 = 0.992
The highest reliability is achieved by adding back up to component with reliability of 0.98 . Therefore , component with reliability of 0.98 should get it
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