Question

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 .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 reliability. Assuming the new computer automatically functions if the
main one fails, determine the resulting reliability.

Answer 1

Answer: (a) The reliability of the computer is approximately 0.9419.

(b) The resulting reliability of the system with the backup computer installed is approximately 0.9966.

Explanation:

To find the reliability of the computer, we need to find the probability that all three modules function. Since all three modules must function for the computer to work properly, we can multiply their reliabilities together.

Given:

- Reliability of Module 1 = 0.97

- Reliability of Module 2 = 0.97

- Reliability of Module 3 = 0.99

(a) Reliability of the computer:

$\[ \text{Reliability} = \text{Reliability of Module 1} \times \text{Reliability of Module 2} \times \text{Reliability of Module 3} \]$

$\[ \text{Reliability} = 0.97 \times 0.97 \times 0.99 \]$

$\[ \text{Reliability} = 0.941883 \]$

So, the reliability of the computer is approximately 0.9419.

(b) Now, if a backup computer is installed, the reliability of the system would improve as either the main computer or the backup computer needs to function for the system to work properly.

The reliability of the system with the backup computer functioning as a failover can be calculated using the formula for complementary events. That is, the probability of the system working is equal to 1 minus the probability of both computers failing.

$\[ \text{Reliability of system with backup} = 1 - \text{(Probability of both computers failing)} \]$

Since the backup computer automatically functions if the main one fails, the probability of both computers failing is the probability that both main and backup computers fail simultaneously.

$\[ \text{Probability of both computers failing} = (1 - \text{Reliability of main computer}) \times (1 - \text{Reliability of backup computer}) \]$

Given that the reliability of the main computer is approximately 0.9419, the reliability of the backup computer is also the same.

$\[ \text{Probability of both computers failing} = (1 - 0.9419) \times (1 - 0.9419) \]$

$\[ \text{Probability of both computers failing} = (0.0581) \times (0.0581) \]$

$\[ \text{Probability of both computers failing} = 0.00337961 \]$

Now, we can find the reliability of the system with the backup computer functioning:

$\[ \text{Reliability of system with backup} = 1 - 0.00337961 \]$

$\[ \text{Reliability of system with backup} \approx 0.9966 \]$

So, the resulting reliability of the system with the backup computer installed is approximately 0.9966.