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Chap004S

Chap004S - Supplement to Chapter 04 Reliability SUPPLEMENT...

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Supplement to Chapter 04 - Reliability SUPPLEMENT TO CHAPTER 4 RELIABILITY Teaching Notes The main topics of this chapter are: 1. Quantifying Reliability 2. Availability 3. Improving Reliability Reliability is a measure of the ability of a product, part, or system to perform its intended function under a prescribed set of conditions. There are three important aspects of reliability: (1) reliability as a probability; (2) definition of failure; and (3) prescribed set of operating conditions. Quantitative methods include the use of probabilities (addition, products, complements) in determining point-in-time reliability and the use of exponential and normal distributions in determining the mean time between failures. Students seem to have some difficulty with exponential distribution, especially if they have not had it in their statistics courses. The coverage of exponential distribution can be omitted without loss of continuity. The normal distribution should be included because it paves the way for later use of inventory management and quality control sampling theory. Answers to Discussion and Review Questions 1. Reliability is a measure of the ability of a product or service to perform its intended function under a prescribed set of conditions. 2. If a product is composed of a large number of parts, it can conceivably have a low reliability because its reliability is a function of the products of the individual reliabilities. For example, if a product has 20 parts, each with a reliability of .99, and all must operate, the overall product reliability will only be about .99 20 = .818. 3. Redundancy refers to backup parts or systems built into a product (or service). Their purpose is to increase reliability by taking over in the event that a primary part or system fails. 4S-1

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Supplement to Chapter 04 - Reliability Solutions 1. a. P(operate) = .9 2 = .81 b. [.90 + .10(.90)] [.90 + .10(.90)] = .9801 c. [.90 + .99(.10)(.90)] 2 = .9783 2. .96 x .96 x .99 x .99 = .9033 3. X 3 = .92x = .9726 4. C = (10P) 2 per component 2 (10P) 2 = 173 100P 2 = 86.5 P 2 = .865 P = .93 5. a. 97 x .97 x .99 = .9315 b. .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. a. .98 x .95 x .94 x .90 = .7876 b. If 1 st : [.98 + (1 – .98) x .98] x .95 x .94 x .90 = .8034 If 2 nd : .98 x [.95 + (1 – .95) x .95] x .94 x .90 = .8270 If 3 rd : .98 x .95 x [.94 + (1 – .94) x .94] x .90 = .8349 If 4 th : .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 reliability of .90 since it poses the greatest risk of failure. The system reliability will then be .86814. 4S-2 .9 .9 .9 .9
Supplement to Chapter 04 - Reliability Solutions (continued) 7. a. #1: P line = .99 x .96 x .93 = .8839 P(line works) + P(line fails) x P(backup works) = .8839 + [(1 – .8839) x (.8839)] = .9865 #2: P: .99 = [(1 – .99) x .99] .96 + [(1 – .96) x .96] .93 + [(1 – .93) x .93] = .9999 = .9984 = .9951 Overall: .9999 x .9984 x .9951 = .9934 b. In #1 the system will fail if any one original and any one backup fails.

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