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 .87, and the other has a reliability of .97.

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 .95, 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           

2

Item 2

Item 2 3.33 points

Determine the availability for each of these cases:

a. MTBF = 48 days, average repair time = 6 days. (Round your answer to 2 decimal places.)

Availability              

b. MTBF = 278 hours, average repair time = 8 hours. (Round your answer to 2 decimal places.)

Availability        

A manager must decide between two machines. The manager will take into account each machine’s operating costs and initial costs, and its breakdown and repair times. Machine A has a projected average operating time of 139 hours and a projected average repair time of 6 hours. Projected times for machine B are an average operating time of 69 hours and a repair time of 5 hours. What are the projected availabilities of each machine? (Round your answers to 3 decimal places.)

	Availability
Machine A
Machine B

Answer 1

According to the guidelines provided by HomeworkLib, abiding by the HomeworkLib's honor code, only first question will be answered. Please post the other one separately. Thank you.

(a)

Since all three modules must function for the computer to function properly, the reliability of the

computer is the product of the reliabilities of three modules:

Reliability (Reliability of first module)*(Reliability of second module)*

(Reliability of third module)

Reliability (0.87)*(0.87)*(0.97)

= 0.7342 (rounded to four decimal places)

(b)

The reliability of the system
Reliability of first computer + (1- Reliability of first computer) (Reliability of second computer)
(0.7342) + (1-0.7342) (0.7342)
0.9294
So the reliability of the system 0.9294

(c)

The addition of the switch changes the reliability of the second computer to (Reliability of switch)*(Reliability of first module)*(Reliability of second module)*(Reliability of third module)

(0.95)*(0.87)*(0.87)*(0.97)

.6975

Then, the reliability of the two computers (Reliability of first computer) + (1- Reliability of first computer) (Reliability of second computer)

(0.7342) + (1- 0.7342) (0.6975)

0.9196

Then the overall reliability of the system 0.9196