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MT2Sol2010Sec15W

MT2Sol2010Sec15W - C p =(US-LS/6 σ =(100.5-99.5(6x0.414 =...

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DS 412 Sec 15 Midterm 2 Solution (White) Fall 2009 1. Let V = annual volume P,Q: (36-24)V – 300,000 = (32-22)V – 220,000, so 2V = 80,000 or V = 40,000 P: (36-24)V – 300,000 = 0, so 12V = 300,000 or V = 25,000 Q: (32-22)V – 220,000 = 0, so 10V = 220,000 or V = 22,000 Location Volume P ≥ 40,000 Q 22,000 – 40,000 2. Warehouse at P: Cost/year = (100x140x0.12) + (70x110x0.14) = $2,758 Warehouse at Q: Cost/year = (80x140x0.12) + (70x120x0.11) = $2,268 Warehouse at R: Cost/year = (80x110x0.14) + (100x120x0.11) = $2,552 Choose Q 3a. = 99.765 gms., = 1.077 gms. n = 5, so A 2 = 0.58, D 3 = 0, D 4 = 2.11 Mean chart : UCL = + A 2 = 99.765 + (0.58x1.077) = 100.39 gms. LCL = - A 2 = 99.765 - (0.58x1.077) = 99.14 gms. Range chart : UCL = D 4 = 2.11x1.077 = 2.272 gms. LCL = D 3 = 0x1.077 = 0 gms. b. Sample mean = (99.6+98.8+99.4+100.6+99.9)/5 = 99.66 gms.. Sample range = 100.6 – 98.8 = 1.8 gms. Yes, the sample satisfies control limits for both charts. c. = 99.765 gms., σ = 0.414 gms., US = 100.5 gms., LS = 99.5 gms.
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Unformatted text preview: C p = (US-LS)/6 σ = (100.5-99.5)/(6x0.414) = 0.40; Not capable because C p < 1. C pk = Min{(US-)/3 σ, (-LS)/3 σ} = Min{(100.5-99.765)/(3x0.414), (99.765-99.5)/(3x0.414)} = 0.725; Not capable because C pk < 1. Since C p < 1, the variability (std. dev.) must be reduced. Further, the mean must be increased, since C pk < 1. 4a. A p-chart should be used because control limits are required for the proportion of wrong reports. Also, the sample size and # defective are given. b. = 30/1500 = 0.02, n = 150, z = 2 = = 0.0114 UCL = + z σ = 0.02 + (2x0.0114) = 0.0428 LCL = - z σ = 0.02 – (2x0.0114) = -0.0028 → 0 c. Sample p 1 2/150 = 0.0133 2 0/150 = 0 3 6/150 = 0.04 4 3/150 = 0.02 5 1/150 = 0.0067 All 5 samples (100%) lie between control limits. We need at least 95.5% with 2-sigma limits, so the process is in control....
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