One of the industrial robots designed by a leading producer of servomechanisms has four major components. Components’ reliabilities are 0.98, 0.95, 0.94, and 0.90. All of the components must function in order for the robot to operate effectively. a. Compute the reliability of the robot. (Round your answer to 4 decimal places.) Reliability b1. Designers want to improve the reliability by adding a backup component. Due to space limitations, only one backup can be added. The backup for any component will have the same reliability as the unit for which it is the backup. Compute the reliability of the robot. (Round your answers to 4 decimal places.) Reliability Component 1 Component 2 Component 3 Component 4 b2. Which component should get the backup in order to achieve the highest reliability? Component 1 Component 2 Component 3 Component 4 c. If one backup with a reliability of 0.92 can be added to any one of the main components, which component should get it to obtain the highest overall reliability? Component 1 Component 2 Component 3 Component 4

a) Reliability of the Robot = 0.7876

b1) Component 1: 0.8034

    Component 2: 0.8270

    Component 3: 0.8349

    Component 4: 0.8664

b2) Component 4 should get the backup in order to achieve the highest reliability.

c) Component 4 should get the backup with a reliability of 0.92, to obtain the highest overall reliability i.e. 0.8681.

Step-by-step explanation:

Component Reliabilities:

Component 1 (R1) : 0.98

Component 2 (R2) : 0.95

Component 3 (R3) : 0.94

Component 4 (R4) : 0.90

a) Reliability of the robot can be calculated by considering the reliabilities of all the components which are used to design the robot.

Reliability of the Robot = R1 x R2 x R3 x R4

                                      = 0.98 x 0.95 x 0.94 x 0.90

Reliability of the Robot = 0.787626 ≅ 0.7876

b1) Since only one backup can be added at a time and the reliability of that backup component is the same as the original one, we will consider the backups of each of the components one by one:

Reliability of the Robot with backup of component 1 can be computed by first finding out the chance of failure of the component along with its backup:

Chance of failure = 1 - reliability of component 1

                             = 1 - 0.98

                             = 0.02

Chance of failure of component 1 along with its backup = 0.02 x 0.02 = 0.0004

So, the reliability of component 1 and its backup (R1B) = 1 - 0.0004 = 0.9996

Reliability of the Robot = R1B x R2 x R3 x R4

                                         = 0.9996 x 0.95 x 0.94 x 0.90

Reliability of the Robot = 0.8034

Similarly, to find out the reliability of component 2:

Chance of failure of component 2 = 1 - 0.95 = 0.05

Chance of failure of component 2 and its backup = 0.05 x 0.05 = 0.0025

Reliability of component 2 and its backup (R2B) = 1 - 0.0025 = 0.9975

Reliability of the Robot = R1 x R2B x R3 x R4

                = 0.98 x 0.9975 x 0.94 x 0.90

Reliability of the Robot = 0.8270

Reliability of the Robot with backup of component 3 can be computed as:

Chance of failure of component 3 = 1 - 0.94 = 0.06

Chance of failure of component 3 and its backup = 0.06 x 0.06 = 0.0036

Reliability of component 3 and its backup (R3B) = 1 - 0.0036 = 0.9964

Reliability of the Robot = R1 x R2 x R3B x R4  

                = 0.98 x 0.95 x 0.9964 x 0.90

Reliability of the Robot = 0.8349

Reliability of the Robot with backup of component 4 can be computed as:

Chance of failure of component 4 = 1 - 0.90 = 0.10

Chance of failure of component 4 and its backup = 0.10 x 0.10 = 0.01

Reliability of component 4 and its backup (R4B) = 1 - 0.01 = 0.99

Reliability of the Robot = R1 x R2 x R3 x R4B

                                      = 0.98 x 0.95 x 0.94 x 0.99

Reliability of the Robot = 0.8664

b2) According to the calculated values, the highest reliability can be achieved by adding a backup of component 4 with a value of 0.8664. So, Component 4 should get the backup in order to achieve the highest reliability.

c) 0.92 reliability means the chance of failure = 1 - 0.92 = 0.08

We know the chances of failure of each of the individual components. The chances of failure of the components along with the backup can be computed as:

Component 1 = 0.02 x 0.08 = 0.0016

Component 2 = 0.05 x 0.08 = 0.0040

Component 3 = 0.06 x 0.08 = 0.0048

Component 4 =  0.10 x 0.08 = 0.0080

So, the reliability for each of the component & its backup is:

Component 1 (R1BB) = 1 - 0.0016 = 0.9984

Component 2 (R2BB) = 1 - 0.0040 = 0.9960

Component 3 (R3BB) = 1 - 0.0048 = 0.9952

Component 4 (R4BB) = 1 - 0.0080 = 0.9920

The reliability of the robot with backups for each of the components can be computed as:

Reliability with Component 1 Backup = R1BB x R2 x R3 x R4

                                                              = 0.9984 x 0.95 x 0.94 x 0.90

Reliability with Component 1 Backup = 0.8024

Reliability with Component 2 Backup = R1 x R2BB x R3 x R4

                                                              = 0.98 x 0.9960 x 0.94 x 0.90

Reliability with Component 2 Backup = 0.8258

Reliability with Component 3 Backup = R1 x R2 x R3BB x R4

                                                               = 0.98 x 0.95 x 0.9952 x 0.90

Reliability with Component 3 Backup = 0.8339

Reliability with Component 4 Backup = R1 x R2 x R3 x R4BB

                                                              = 0.98 x 0.95 x 0.94 x 0.9920

Reliability with Component 4 Backup = 0.8681

Component 4 should get the backup with a reliability of 0.92, to obtain the highest overall reliability i.e. 0.8681.

true

step-by-step explanation:

answer: both chords and diameters have two endpoints on a circle. diameters must intersect the center of a circle.