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CEE_597_Final2004

CEE_597_Final2004 - CEE 597 Risk Analysis and Management...

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CEE 597 – Risk Analysis and Management Final Exam Wednesday --- May 12, 2004 --- 9 am and noon. Test is open book & open notes. You have 150 minutes for this 150 point exam. Write clearly and show important steps. 1. Consider the fault tree below. A D E B C a) (6 pts.) Elements A-B-C are 99.5% reliable, and D-E are 90% reliable. What is the reliability of this system? b) (6 pts.) Draw a network diagram for this system, if it can be done. (If not, why not?) 2. Consider the system described by the network below. a) (8 pts.) What are ALL the minimal cut sets with 1, 2 or 3 components? b) (5 pts ) If all elements are 95% reliable, approximately what is the failure probability for the entire system? c) (2 pts.) If you could upgrade ANY ONE components in the original network to a reliability of 99.9 %, what upgrade would most improve system reliability? A B C D E F G 3. Unwanted e-mail messages (SPAM) are a very serious problem. New software can block such messages. You conducted a test of one expensive SPAM rejection program: it accepted only 8 of 6,000 undesirable messages over a 5-day period. A friend suggested a super deluxe program, which you tested for 3 days: only 1 of 4,000 undesirable messages was accepted. Assume each arrival can be considered as an independent test of the software. Consider if this is sufficient evidence to show that the super deluxe program is better. (a) (10 pts.) What are the null & alternative hypotheses, your test statistic, distribution of the test statistics under each hypothesis, rejection region for a test with α = 5% (type I error) ? (b) (2 pts.) What is the p-value for the observed data? What do you conclude? (c) (3 pts.) What is probability of concluding that there was no difference between the two packages if indeed the super deluxe program has a quarter the failure rate (is four times better)?
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P[ X = x | n = 9, p ] for binomial distribution with various values of p X 0.1250 0.1429 0.2500 0.3750 0.4000 0.5000 0.6000 0.6250 0.7500 0.8571 0.8750 0 0.3007 0.2497 0.0751 0.0146 0.0101 0.0020 0.0003 0.0001 0.0000 0.0000 0.0000 1 0.3866 0.3746 0.2253 0.0786 0.0605 0.0176 0.0035 0.0022 0.0001 0.0000 0.0000 2 0.2209 0.2497 0.3003 0.1886 0.1612 0.0703 0.0212 0.0147 0.0012 0.0000 0.0000 3 0.0736 0.0971 0.2336 0.2640 0.2508 0.1641 0.0743 0.0570 0.0087 0.0004 0.0002 4 0.0158 0.0243 0.1168 0.2376 0.2508 0.2461 0.1672 0.1426 0.0389 0.0040 0.0023 5 0.0023 0.0040 0.0389 0.1426 0.1672 0.2461 0.2508 0.2376 0.1168 0.0243 0.0158 6 0.0002 0.0004 0.0087 0.0570 0.0743 0.1641 0.2508 0.2640 0.2336 0.0971 0.0736 7 0.0000 0.0000 0.0012 0.0147 0.0212 0.0703 0.1612 0.1886 0.3003 0.2497 0.2209 8 0.0000 0.0000 0.0001 0.0022 0.0035 0.0176 0.0605 0.0786 0.2253 0.3746 0.3866 9 0.0000 0.0000 0.0000 0.0001 0.0003 0.0020 0.0101 0.0146 0.0751 0.2497 0.3007 4. (16 points) Prof. Stedinger’s old car has a serious squeak in the rear end and will need to go to the auto shop for repairs. This might be expensive. Prof. Stedinger estimates there is a 30% chance that something is just loose and can be fixed for only $50. That leaves a 70% chance that serious repairs are required. In that case there is a 10% probability it is the differential, and 90% that the brakes are the problem. If the differential has gone bad, then there is a 20% probability the differential needs only minor repairs ($100), otherwise a replacement unit is required. There is a 60% probability a used differential can be obtained as a replacement for $220, otherwise one needs to by a new unit for $430. On the other hand, if the brakes are a problem, there is a 75% probability only one rear brake needs to be fixed ($90), leaving a 25% probability one needs to fix both the left and right rear brake mechanisms ($150). Compute and plot the risk profile.
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