The guidance system of a ship is controlled by a computer that has 3 major modules. In order for the computer to function properly, all 3 modules must function. Two of the modules have reliabilities of 0.97 and the other has a reliability of 0.99.

a) What is the reliability of the computer?

b) A backup computer identical to the one being used will be installed to improve overall reliability. Assuming the new computer automatically functions if the main one falls, determine the 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 0.98, 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.)

b) The resulting reliability is now the reliability of the first computer, added to the possibility of failure of the first computer multiplied by the reliability of the second computer:

[tex]R= R_1 +(1-R_1)*R_2\\R= 0.931491+(1-0.931491)*0.97*0.97*0.99\\R=0.995307[/tex]

c) If a switch with reliability of 0.98 must be activated to turn on the second computer, the switch's reliability must be taken into account as follows:

[tex]R= R_1 +(1-R_1)*R_2*R_S\\R= 0.931491+(1-0.931491)*0.97*0.97*0.99*0.98\\R=0.994030[/tex]

MrRoyal MrRoyal · Answer 2 · 2021-12-31T08:01:35+01:00

The reliability of the system is simply its probability of not failing

(a) The reliability of the computer

This is the product of the reliabilities of the three modules.

So, we have:

[tex]R = 0.97 \times 0.97 \times 0.99[/tex]

[tex]R = 0.931491[/tex]

Approximate

[tex]R = 0.9315[/tex]

Hence, the reliability of the computer is 0.9315

(b) The reliability when a backup is used

In (a), the reliability of the computer is 0.9315

When the computer fails, the reliabilities of the other two are 1 - 0.9315 and 1 - 0.9315.

So, the reliability when a backup is used is calculated using the following complement rule

[tex]R = 1 - [(1 - 0.9315) \times (1 - 0.9315)][/tex]

[tex]R = 0.99530775[/tex]

Approximate

[tex]R = 0.9953[/tex]

Hence, the reliability of the computer when a backup is 0.9953

(c) The overall reliability of the system

In (a), the reliability of the computer is 0.9315.

Also, the reliability of the switch is 0.98

So, the reliability of the backup is:

[tex]R = 0.9315 \times 0.98[/tex]

[tex]R = 0.9129[/tex]

So, the overall system has:

Main computer with reliability of 0.9315
Back up of the computer system with reliability of 0.9129

The reliability of the overall system is then calculated using the following complement rule

[tex]R = 1 - [(1 - 0.9315) \times (1 - 0.9129)][/tex]

[tex]R = 0.9940[/tex]

Hence, the reliability of the overall system is 0.9940

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