Ops Mgmt 8e ch17pp

Ops Mgmt 8e ch17pp - Therefore FR(N = 6/148,000 = 0.0000405...

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Practice Problems: Chapter 17, Maintenance and Reliability Problem 1: California Instruments, Inc., produces 3,000 computer chips per day. Three hundred are tested for a period of 500 operating hours each. During the test, six failed: two after 50 hours, two at 100 hours, one at 300 hours, and one at 400 hours. Find FR(%) and FR(N). Problem 2: If 300 of these chips are used in building a mainframe computer, how many failures of the computer can be expected per month? Problem 3: Find the reliability of this system: 1

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Problem 4: Given the probabilities below, calculate the expected breakdown cost. Number of Breakdowns Daily Frequency 0 3 1 2 2 2 3 3 Assume a cost of \$10 per breakdown. 2
ANSWERS Problem 1: FR(%) = failures per number tested = 6/300 = 0.02 = 2% FR( N ) = failures per operating time: Total time = 300 * 500 = 150,000 hours Downtime = 2(450) + 2(400) + 1(200) + 1(100) = 2,000 hours Operating time = Total time – Downtime = 150,000 – 2,000 = 148,000

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Unformatted text preview: Therefore: FR(N) = 6/148,000 = 0.0000405 failures/hour MTBF = 1/FR(N) = 24,691 hours Problem 2: Converting the units of FR(N) to months: FR( N ) = 0.0000405 * 24 hours/day * 30 days/month = 0.029 failures/month FR(N) for the 300 units: FR( N ) = 0.029 failures/month * 300 units = 8.75 failures/month MTBF for the mainframe: MTBF = 1/FR( N ) = 1/8.75 = 0.11 month = 0.11 * 30 = 3.4 days Calculation for MTBF assumes that failure of any one chip brings down entire system. Problem 3: [0.95 0.92(1 0.95)] * [0.98] * [0.90 0.90(1 0.90)] R = +-+-= 0.996 * 0.98 * 0.99 = 96.6% Problem 4: 3 Number of Breakdowns Daily Frequency Probability 3 0.3 1 2 0.2 2 2 0.2 3 3 0.3 Expected number of breakdowns = (0)(0.3) + (1)(0.2) + (2)(0.2) + (3)(0.3) = 0 + 0.2 + 0.4 + 0.9 = 1.5 breakdowns/day Expected breakdown cost = Expected number of breakdowns * Cost per breakdown = 1.5 * \$10 = \$15/day 4...
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Ops Mgmt 8e ch17pp - Therefore FR(N = 6/148,000 = 0.0000405...

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