HOMEWORK SET 1Problem 1.5Problem 1.12Problem 2.1aProblem 2.1bProblem 2.1cProblem 2.14Case 2.4Case 2-4Slide 21Problem 3.2Problem 3.2Problem 3.14Problem 4.8Problem 4-14Problem 5-4Problem 5-5Problem 5-14 – Not assignedStitching ProblemStitching ProblemReview for Exam 1Exam StructureChapters coveredProblemsMore problemsChapter 1Activities in Operations Management include all but….Operations at a BankOperations at a retail storeOperations at hospitalOperations at a cable TV companyEvolution of OperationsProductivity is what exactly?Name some trends in OMName some barriers to entry of new enterprisesMore Chapter 1Slide 78How important is a country’s competitiveness?Operations StrategySlide 81Service OperationsCompeting On CostCompeting On QualityCompeting On FlexibilityCompeting On SpeedProduct-Process MatrixService-Process MatrixChapter 2--Quality ManagementChapter 2-Quality GurusName the first five of Deming’s 14 pointsTQMMeasures of QualitySlide 94Strategic Implications Of TQMDMAIC – what does it stand for?Cost Of QualityCompanies underestimate their failure costsSeven Quality Control ToolsChapter 3--Statistical Process ControlVariationTypes Of DataControl ChartsThe x-bar chart is based on which probability distribution?Typically control limits of control charts represent ____ sigma limitsSix sigma means what in terms of defectsChapter 4—Product DesignComputing ReliabilityA technique for improving the design of a product vis-à-visHouse Of QualityMeasures of Design Quality includeConcurrent DesignDFM GuidelinesChapter 6—Processes, Technologies & CapacityProcess Flowchart SymbolsSlide 116Slide 117The focus of operations today is not on functions, but _____Which of the following refers to dismantling a competitor’s product?BPR VS PROCESS IMPROVEMENTDisadvantages of handoffsWhat are byproducts of shortened cycle times?Process PlansWhat is this?Slide 125Slide 126Symbols for what kind of chart?HOMEWORK SET 1HOMEWORK SET 1Probs. 1-5, 1-12Probs. 1-5, 1-12Question 2-8Question 2-8Probs. 2-1, 2-14Probs. 2-1, 2-14Case Prob. 2.4Case Prob. 2.4Probs. 3-2, 3-14Probs. 3-2, 3-14Probs. 4-8, 4-14Probs. 4-8, 4-14Probs. 5-4 and 5-5Probs. 5-4 and 5-5Problem 1.5Problem 1.5%Labor Hours Units of OutputProductivityU.S. 89.5 136 1.52Germany 83.6 100 1.20Japan 72.7 102 1.40The U.S. is the most productive.Problem 1.12Problem 1.12%Alaskan Seal Brr Frost Cold CaseDeep Freeze Purchase cost$3,270 $4,000 $4,452 $5,450 Daily energy consumption (kwh)3.61 3.88 6.68 29.07Cost per kwh$0.10 $0.10 $0.10 $0.10Daily energy cost$0.36 $0.39 $0.67 $2.91Daily purchase cost$2.99 $3.65 $4.07 $4.98Total cost$3.35 $4.04 $4.73 $7.88Volume (cu ft)25 35 49 79Productivity (cu ft/$)7.47 8.66 10.35 10.02Cost/cu ft$0.13 $0.12 $0.10 $0.10Choose Cold Case.Problem 2.1aProblem 2.1aProblem 2.1bProblem 2.1bProblem 2.1cProblem 2.1cThese index values do not provide much information regarding the effectiveness of the quality assurance program. They are, however, useful in making comparisons from one period to the next and in showing trends in product quality over time.Problem 2.14Problem 2.14With an 8% defect rate:Sales 151200Variable costs 61200Fixed costs 31000Profit 59000With zero defects:Sales 151200Variable costs 56305Fixed costs 31000Profit 63895% decr in var costs 7.998366% inc in profits 8.29661Return on inv = 19.58%Case 2.4Case 2.4Stage 1 yield: Y1 = (I)(%G) + (I)(1-%G)(%R)Stage 1 yield: Y1 = (I)(%G) + (I)(1-%G)(%R)–= 500 (.94) + 500(.06)(.23)= 500 (.94) + 500(.06)(.23)–= 470 + 6.9 = 476.9= 470 + 6.9 = 476.9Stg 2 yield: Y2=476.9(.96)+476.9*(.04)(.91)Stg 2 yield: Y2=476.9(.96)+476.9*(.04)(.91)–= 457.82 + 17.36 = 475.2= 457.82 + 17.36 = 475.2Stg 3 yield: Y3= 475.2(.95)+475.2*(.05)(.67)Stg 3 yield: Y3= 475.2(.95)+475.2*(.05)(.67)–= 451.44 + 15.92 = 467.6= 451.44 + 15.92 = 467.6Stg 4 yield: Y4=467.6*(.97)+467.6*(.03)(.89)Stg 4 yield: Y4=467.6*(.97)+467.6*(.03)(.89)–= 453.57 + 12.48 = 466.1= 453.57 + 12.48 = 466.1Stg 5 yield: Y5 = 466.1*(.98)+466.1*(.02)Stg 5 yield: Y5 = 466.1*(.98)+466.1*(.02)(.72)(.72)–= 456.78 + 6.71 = = 456.78 + 6.71 = 463.5463.5Case 2-4Case 2-4Average Weekly Yield        1Stage 1 yield : % 1 % %Y I G I G R        500 0.94 500 0.06 0.23  470 6.9  1476.9Y         2Stage 2 yield : 476.9 0.96 476.9 0.04 0.91Y   457.82 17.36  2475.2Y         3Stage 3 yield : 475.2 .95 475.2 .05 .67Y   451.44 15.92  467.6        4Stage 4 yield : 467.6 .97 467.6 .03 .89Y   453.57 12.48  466.1      5Stage 5 yield : 466.1 .98 466.1 .02 .72Y   456.78 6.71  463.5Case 2-4Case 2-4Increasing Good Quality Yield by 1% at Each Stage      1Stage 1 yield : 500 0.95 500 0.05 0.23Y   1480.75Y       2Stage 2 yield : 480.75 0.97 480.75 0.03 0.91Y   2479.45Y       3Stage 3 yield : 479.45 0.96 479.45 0.04 0.67Y   3473.12Y       4Stage 4 yield : 473.11 0.98 473.11 0.02 0.89Y   4472.07Y       5Stage 5 yield : 472.07 0.99 472.07 0.01 0.72Y   5470.75Y  Difference in yields 470.75 463.5 7.25 units   7.25Percentage increase 1.56%436.2 Problem 3.2 Problem 3.2 SampleProportion Defective P-Bar UCL LCLNumber Defective1 0.14 0.1535 0.2256 0.0814 142 0.12 0.1535 0.2256 0.0814 123 0.09 0.1535 0.2256 0.0814 94 0.10 0.1535 0.2256 0.0814 105 0.11 0.1535 0.2256 0.0814 116 0.07 0.1535 0.2256 0.0814 77 0.08 0.1535 0.2256 0.0814 88 0.14 0.1535 0.2256 0.0814 149 0.16 0.1535 0.2256 0.0814 1610 0.17 0.1535 0.2256 0.0814 1711 0.18 0.1535 0.2256 0.0814 1812 0.10 0.1535 0.2256 0.0814 1013 0.19 0.1535 0.2256 0.0814 1914 0.20 0.1535 0.2256 0.0814 2015 0.17 0.1535 0.2256 0.0814 1716 0.18 0.1535 0.2256 0.0814 1817 0.18 0.1535 0.2256 0.0814 1818 0.22 0.1535 0.2256 0.0814 2219 0.24 0.1535 0.2256 0.0814 2420 0.23 0.1535 0.2256 0.0814 23% % % % % 307P-Bar 0.154UCL 0.226LCL 0.081Problem 3.2Problem 3.2Problem 3.14Problem 3.14Problem 4.8Problem 4.8Problem 4-14Problem 4-14Provider MTBF MTTR SAProvider MTBF MTTR SAJCN (8*40)/50=6.4 3+2=5 6.4/JCN (8*40)/50=6.4 3+2=5 6.4/(6.4+5)=.56(6.4+5)=.56Bell (8*40)/100=3.2 2+1=3 3.2/Bell (8*40)/100=3.2 2+1=3