PowerPoint presentation to accompany Heizer/Render – Principles of Operations Management, 5e, and Operations Management, 7e © 2004 by Prentice Hall, Inc.,

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For these two jobs rate each of Hackman and Oldham ´s core job characteristics on a scale from 1 to 10. What is your total score? What about the job could.
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PowerPoint presentation to accompany Heizer/Render – Principles of Operations Management, 5e, and Operations Management, 7e © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J Practice Problems 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).

PowerPoint presentation to accompany Heizer/Render – Principles of Operations Management, 5e, and Operations Management, 7e © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J Practice Problems 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). 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 Therefore: FR(N) = 6/148,000 = failures/hour MTBF = 1/FR(N) = 24,691 hours

PowerPoint presentation to accompany Heizer/Render – Principles of Operations Management, 5e, and Operations Management, 7e © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J Practice Problems 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? FR(N) = failures/hour

PowerPoint presentation to accompany Heizer/Render – Principles of Operations Management, 5e, and Operations Management, 7e © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J Practice Problems 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? FR(N) = failures/hour Converting the units of FR(N) to months: FR( N ) = * 24 hours/day * 30 days/month = failures/month FR(N) for the 300 units: FR( N ) = 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.

PowerPoint presentation to accompany Heizer/Render – Principles of Operations Management, 5e, and Operations Management, 7e © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J Practice Problems Problem 3: Find the reliability of this system:

PowerPoint presentation to accompany Heizer/Render – Principles of Operations Management, 5e, and Operations Management, 7e © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J Practice Problems Problem 3: Find the reliability of this system: R = [ (1 – 0.95)] * 0.98 * [ (1 – 0.90)] = * 0.98 * 0.99 = 96.6%

PowerPoint presentation to accompany Heizer/Render – Principles of Operations Management, 5e, and Operations Management, 7e © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J Practice Problems Problem 4: Given the probabilities below, calculate the expected breakdown cost. Assume a cost of $10 per breakdown. Number ofDaily BreakdownsFrequency

PowerPoint presentation to accompany Heizer/Render – Principles of Operations Management, 5e, and Operations Management, 7e © 2004 by Prentice Hall, Inc., Upper Saddle River, N.J Practice Problems Problem 4: Given the probabilities below, calculate the expected breakdown cost. Assume a cost of $10 per breakdown. Number ofDaily BreakdownsFrequency Probability Expected number of = (0)(0.3) + (1)(0.2) + (2)(0.2) + (3)(0.3) breakdowns = = 1.5 breakdowns/day ExpectedExpectedCost per breakdown=number of*breakdown costbreakdowns =1.5 * $10 =$15/day