# Arithmetic Mean and Reliability

**Topics:**Arithmetic mean, Reliability engineering, Battery

**Pages:**6 (2190 words)

**Published:**April 29, 2012

1. Consider the following system: Determine the probability that the system will operate under each of these conditions: a. The system as shown. b. Each system component has a backup with a probability of .90 and a switch that is 100 percent reliable. c. Backups with .90 probability and a switch that is 99 percent reliable.Solutions1.a.P(operate) = .92 = .81.9 .9

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b.[.90 + .10(.90)] [.90 + .10(.90)] = .9801c.[.90 + .99(.10)(.90)]2 = .9783 2. A product is composed of four parts. In order for the product to function properly in a given situation, each of the parts must function. Two of the parts have a .96 probability of functioning, and two have a probability of .99. What is the overall probability that the product will function properly? Solution: .96 x .96 x .99 x .99 = .9033 3. A system consists of three identical components. In order for the system to perform as intended, all of the components must perform. Each has the same probability of performance. If the system is to have a .92 probability of performing, what is the minimum probability of performing needed by each of the individual components?3.X3 = .92x = .9726 4. A product engineer has developed the following equation for the cost of a system component: C = (10P)2 where C is the cost in dollars and P is the probability that the component will operate as expected. The system is composed of two identical components, both of which must operate for the system to operate. The engineer can spend $173 for the two components. To the nearest two decimal places, what is the largest component probability that can be achieved?4.C = (10P) 2 per component2 (10P) 2 = 173100P2 = 86.5P2 = .865P = .93 5. 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. d. What is the reliability of the computer? e. 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. f. 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 .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.)5.a.97 x .97 x .99 = .9315b..9315 + (1 – .9315) x .9315 = .9953[i.e., P(work) + P(not work) x P(backup works)]c..9315 + [(1 – .9315) x .98 x .9315] = .994[i.e., P(work) + [P(not work) x P(switch works) x P(backup works)] 6. One of the industrial robots designed by a leading producer of servomechanisms has four major components. Components’ reliabilities are .98, .95, .94, and .90. All of the components must function in order for the robot to operate effectively. g. Compute the reliability of the robot. h. 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. Which component should get the backup in order to achieve the highest reliability? i. If one backup with a reliability of .92 can be added to any one of the main components, which component should get it to obtain the highest overall reliability?6.a..98 x .95 x .94 x .90 = .7876b.If 1st: [.98 + (1 – .98) x .98] x .95 x .94 x .90 = .8034If 2nd: .98 x [.95 + (1 – .95) x .95] x .94 x .90 = .8270If 3rd: .98 x .95 x [.94 + (1 – .94) x .94] x .90 = .8349If 4th: .98 x .95 x .94 x [.90 + (1 – .90) x .90] = .8664[i.e., for any case, P(all other work) x P(that one fails) x P(backup works)]c.The one with a...

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