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The guidance system of a ship is controlled by a computer that has three major modules....

The guidance system of a ship is controlled by a computer that has three major modules. In order for the computer to function properly, all three modules must function. Two of the modules have reliabilities of .95, and the other has a reliability of .99.

a.What is the reliability of the computer? (Round your answer to 4 decimal places.)

Reliability             

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 fails, determine the resulting reliability. (Round your intermediate calculations and final answers to 4 decimal places.)

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 .99, 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.) (Round your intermediate calculations and final answers to 4 decimal places.)

Reliability            

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Answer #1

a)The computer has 3 modules which are connected in series. The reliability of modules are 0.95, 0.95 and 0.99

Thus, reliability of the computer = 0.95 x 0.95 x 0.99 = 0.8934

b)The backup computer with same reliability of 0.8934 is connected in parallel to the main computer with probability of 0.8934

Thus, probability of the system with 2 computers in parallel each with reliability of 0.8934

= 1 – ( 1 – 0.8934) x ( 1 – 0.8934)

= 1 – 0.1066 x 0.1066

= 1 – 0.0113

= 0.9887

RESULTING RELAIBILITY = 0.9887
  1. The back up system now consists of following items in series :

Computer with reliability of 0.8934

Switch with reliability of 0.99

Reliability of the back up system therefore will be = 0.8934 x 0.99 = 0.8844

The overall system thus consists of :

Main computer with reliability of 0.8934

Back up computer system with reliability of 0.8844

The reliability of overall system

= 1 – ( 1 – 0.8934)x ( 1 – 0.8844)

= 1 – 0.1066 x 0.1156

= 1 – 0.0123

= 0.9877

OVERALL RELIABILITY OF THE SYSTEM = 0.9877
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