C again involves the client in the design process.D allows the PM to regain control of the meeting, and allows the client to respond to theissue so they don't think anything is being swept under the rug, but the PM is not makingany commitment to any particular course of action. This would be my second choice, and Iwould do this if the client pressed the issue. Still, by airing the issue in front of the client,you will increase the client's angst about the project.The PM needs to provide immediate feedback to the engineer after the meeting in asituation like this. When a client is depending on the expertise of the project team, it isimperative that the team provides a uniform face to the client, and that all conflict isresolved internally instead of in front of the client. Open conflict in front of the client willundermine the client's confidence in the project team and jeopardize the project. If theengineer has issues with the design, those should be discussed openly at internal projectmeetings, and resolution should be reached internally, and then everybody should abide bythe resolution.95. When components in a system are arranged in parallel, system reliability be:a. Better than the best componentb. As good as the best componentc. As good as the weakest componentd. The product of the reliability of individual componentsAnswer is b. (No, correct answer will be A)Regarding parallel and serial reliability: for components arranged in parallel, the reliability isbetter than the best componentand for components arranged in serial, the reliability of thesystem is worse than the worst component. When components are arranged in parallel,only one of the units need to be successful for the whole system to be a success (i.e. thesystem fails only when every unit fails). There is redundancy built into the system. In aseries arrangement, the failure of any one of the components leads to the failure of thewhole system.Here is an example:Suppose two components with reliability 0.3 and 0.6 are arranged in parallel; then thereliability of the system is given by:R = 1 - [(1-R1)(1-R2)] = 1 - (0.7)(0.4) = 0.72, which is better than both 0.7 and 0.4.If two components with reliability 0.3 and 0.6 are arranged in series, then the reliability ofthe system is given by:R = R1 * R2 = 0.3 * 0.6 = 0.18, which is lower than both 0.3 and 0.6.Also, a parallel configuration (because of its redundancy) gives you a higher reliability (.72)than a similar serial configuration (.18).[Another way to look into this is:If, 1stS -> 1stbeing successful, 1stF -> 1stbeing unsuccessful, and 2ndS -> 2ndbeingsuccessful, 2ndF -> 2ndbeing unsuccessful, thenProbability of 1stS + 2ndS = 0.3 * 0.6 = 0.18Probability of 1stS + 2ndF = 0.3 * 0.4 = 0.12
Probability of 1stF + 2ndS = 0.7 * 0.6 = 0.42Probability of 1stF + 2ndF = 0.7 * 0.4 = 0.28So, when they are in parallel, Probability of success = Probability of at least one beingsuccessful = 0.18 + 0.12 + 0.42 = 0.72And, when they are in series, Probability of success = probability of both being successful =0.18]So, the answer to this question is (a).
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