Midterm-review-examples-F23-solns

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Example 1 Labor productivity = 5000 / 200 = 25 units per hour Example 2 Multifactor productivity = 90×10 6 / (50×40×30×500 + 35×10 6 + 3×10 6 ) = 1.32 Example 3 Factor rating method. Alternative 1 score: 70.6 Alternative 2 score: 82.7 --- better alternative Example 4 Seminar Room Remodeling Project Path Total Expected Time A – C 8 + 6 = 14 A – D – E 8 + 4 + 7 = 19 B – E 8 + 7 = 15 The critical path is A – D – E. The expected completion time is 19 days. Project variance is 2.44 (1.00 + 0.44 + 1.00). Standard deviation of project completion time is then 1.56. 𝑃(? < 21) = 𝑃 ( ?−19 1.56 < 21−19 1.56 ) = 𝑃(𝑍 < 1.28) ≅ 90% 𝑃(? > 21 ≅ 10%) 𝑃(? < 17) = 𝑃 ( ?−19 1.56 < 17−19 1.56 ) = 𝑃(𝑍 < −1.28) ≅ 10% Consider path B-E. The expected completion time is 15 days. Variance of completion time for this path is 2.78. 𝑃(? > 17) = 𝑃 ( ?−15 1.67 > 17−15 1.67 ) = 𝑃(𝑍 > 1.20) = 1 − 𝑃(𝑍 < 1.20) = 11.5% Example 5 Activity Earliest Start Earliest Finish Latest Start Latest Finish A 0 3 0 3 B 3 8.5 3.5 9 C 3 9 3 9 D 9 15 9 15 E 8.5 12 9 12.5 F 9 14 9 14 G 15 19.5 15 19.5 H 14 19.5 14 19.5 I 12 19 12.5 19.5

A-C-D-G is a critical path with expected completion time of 19.5 weeks and standard deviation of completion time of 1.38 weeks. A-C-F-H is also a critical path with expected completion time of 19.5 weeks and standard deviation of completion time of 1.16 weeks. Since A-C-D-G has higher variation of completion time, we will use its standard deviation (1.38) to compute the probabilities we need. In doing so, we are considering the more volatile path as a worst-case scenario. ▪ What is the probability of completing the project within 21 weeks? 𝑃(? < 21) = 𝑃(𝑍 < 1.09) = 0.8621 ▪ What is the probability of completing the project within 18 weeks? 𝑃(? < 18) = 𝑃(𝑍 < −1.09) = 0.1379 ▪ What project deadline would give 70% probability of on-time completion? The z-value corresponding to 0.70 is 0.52. Then, the due date is 19.5+(0.52)(1.38)=20.2 weeks. Example 6 The critical path is C-D-E-F with a length of 20 days. To complete the project in 19 days, we would have to shorten a critical activity by one day. Comparing the daily crashing costs of C, D, E, and F, we decide to shorten activity C by one day. This would mean an additional cost of $300. The critical path is still C-D-E-F, but with the reduced length of 19 days. To complete the project in 18 days, we would have to shorten a critical activity by one day. Since activity C cannot be shortened to less than four days, the alternatives are D, E, F. Comparing the daily crashing costs of D, E, and F, we decide to shorten activity E by one day. This would mean an additional cost of $600. We now have two critical paths, C-D-E-F and A-B-F, both with the length of 18 days. To complete the project in 17 days, we would have to shorten both critical paths by one day. The following table summarizes which activities are available for crashing and by how many days as well as the daily cost of crashing. Activity Available days to shorten Crash cost per day A - - B 2 500 C - - D 3 700 E 1 600 F 1 800 Activity F is common to both critical paths. Shortening F by one day reduces both critical path lengths to 17. The additional cost of this is $800. This is the minimum-cost option. Note that if we consider the two critical paths separately, the total cost of crashing will be higher. For A-B-F, the reduction would be in B (cost of $500) and for C-D-E-F, the reduction would be in E (cost of $600). The total cost would be $1100.

Example 7 x = 390 sec, n = 9, = 10 sec = 0.337, = 0.184, = 1.816 = = 1.816(10 sec) = 18.16 sec = = 0.184(10 sec) = 1.84 sec = 2 x A R + = 390 sec + 0.337(10 sec) = 393.37 sec 2 x LCL x A R = − = 390 sec – 0.337(10 sec) = 386.63 sec Example 8 Sample Range (R) Sample mean ( 𝑥̅ ) 1 0.4 10.0 2 0.6 10.1 3 0.4 9.9 4 0.6 10.2 5 0.6 10.0 𝑅 ̅ = 0.52 𝑥̿ = 10.04 Sample size = 4 ??𝐿 ? = ? 4 𝑅 ̅ = (2.282)(0.52) = 1.19 𝐿?𝐿 ? = ? 3 𝑅 ̅ = 0 All sample ranges are within control limits. Variability is under control. ??𝐿 𝑥̅ = 𝑥̿ + 𝐴 2 𝑅 ̅ = 10.04 + (0.729)(0.52) = 10.42 𝐿?𝐿 𝑥̅ = 𝑥̿ − 𝐴 2 𝑅 ̅ = 10.04 − (0.729)(0.52) = 9.66 All sample means are within control limits. Process mean is under control. Sample Range (R) Sample mean ( 𝑥̅ ) 1 0.9 10.13 2 1.2 9.85 Sample 2 range is higher than UCL R . That is, the process is out of control with a potential issue on variability. Corrective action is required. R 2 A 3 D 4 D R UCL 4 D R R LCL 3 D R x UCL

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Example 9 We are asked to design an ͞ x-chart that has a type I error of 5 percent. That corresponds to setting the control limits so that there is a 2.5 percent chance a sample result will fall below the LCL and a 2.5 percent chance that a sample result will fall above the UCL. We read from the standard normal cumulative probability table the z -value (-1.96) that corresponds to the probability of 0.025 (see shaded area 1) to the left of LCL. Similarly, we know the z -value (1.96) that corresponds to 0.025 (see shaded area 2) to the right of UCL. ??𝐿 𝑥̅ = 𝑥̿ + 𝑧𝜎 𝑥̅ = 𝑥̿ + 𝑧 𝜎 √𝑛 = 5 + 1.96 1.5 √6 = 6.2 minutes 𝐿?𝐿 𝑥̅ = 𝑥̿ − 𝑧𝜎 𝑥̅ = 𝑥̿ − 𝑧 𝜎 √𝑛 = 5 − 1.96 1.5 √6 = 3.8 minutes Example 10 Sample Sample Mean Sample Range 1 601 24 2 602 10 3 582 22 4 602 32 5 604 24 Average 598.2 22.4 = = 2.282(22.4) = 51.1 = = 0(22.4) = 0 All sample ranges are within control limits. Variability is under control. ??𝐿 𝑥̅ = 𝑥̿ + 𝐴 2 𝑅 ̅ = 598.2 + (0.729)(22.4) = 614.5 𝐿?𝐿 𝑥̅ = 𝑥̿ − 𝐴 2 𝑅 ̅ = 598.2 − (0.729)(22.4) = 581.9 All sample means are within control limits. Process mean is under control. New sample mean is 616. New sample range is 43. Since the sample range is within control limits, process variability is in control. Since the sample mean (616) is above ??𝐿 𝑥̅ , the process average is out of control. We must explore assignable causes of variation. R UCL 4 D R R LCL 3 D R z = – 1.96 Shaded area 1 = 0.025 (2.5%) z = 1.96 Shaded area 2 = 0.025 (2.5%)

Example 11 No quality standard is provided in the question. Let’s consider 3 -sigma capability, so the critical ratio is 1.0. Process 1 and 2 are centered. Hence, C p can be used directly. Process 1: ? 𝑝 = 8−7 6(0.1) = 1.67 Process 2: ? 𝑝 = 4.9−4.3 6(0.12) = 0.83 Therefore, process 1 is 3-sigma capable (in fact, it is 5-sigma capable), however, process 2 is not capable ( C p < 1). Process 3 is not centered. Let’s compute C pk . Process 3: ? 𝑝𝑘 = min ( 6−5.5 3(0.14) , 6.7−6 3(0.14) ) = min(1.19,1.67) = 1.19 Since C pk > 1, we conclude that process 3 is 3-sigma capable. Note: You may simply go ahead and compute both process capability ratio ( C p ) and process capability index ( C pk ) for all three processes and make the same determination based on these results. process capability index ( C pk ) process capability ratio ( C p ) Process 1 1.67 1.67 Process 2 0.83 0.83 Process 3 1.19 1.43 As you can see in the table, process 1 is 3-sigma capable, while process 2 is not. For both processes, the capability index and capability ratio give the same value (1.67 and 0.83, respectively) since they are both centered perfectly. Process 3 is not centered perfectly, nevertheless, it is 3-sigma capable (note that capability ratio ( C p ) which focuses purely on process variability, is significantly higher than 3-sigma level of 1.0 – it is in fact even higher than 4-sigma level of 1.33 – and this compensates for the lack of centering of the process.) Example 12 c-chart is needed here (note that the safety department cannot determine the number of accidents that did not occur, so it has no way to compute a proportion defective at the intersection) ??𝐿 𝑐 = 𝑐̅ + 𝑧√𝑐̅ = 3 + 3√3 = 8.2 𝐿?𝐿 𝑐 = 𝑐̅ − 𝑧√𝑐̅ = 3 − 3√3 < 0. Hence, 𝐿?𝐿 𝑐 = 0 The number of accidents last month (seven) falls within the UCL and LCL of the chart. We conclude that no assignable causes are present and that the increase in accidents was due to chance.

Example 13 𝑝̅ = 300 (30)(250) = 0.04 , that is 4% ??𝐿 𝑝 = 0.04 + 3 √ 0.04(1 − 0.04) 250 = 0.077, that is 7.7% 𝐿?𝐿 𝑝 = 0.04 − 3 √ 0.04(1 − 0.04) 250 = 0.003, that is 0. 3% Sample Number of defective records Proportion defective Tuesday 17 17/250 = 0.068 Wednesday 15 15/250 = 0.060 Thursday 22 22/250 = 0.088 Friday 21 21/250 = 0.084 Thursday and Friday numbers show that the process is out of control. We must investigate, identify the issue(s), and take corrective action. Example 14 Upper specification limit (USL) = 1125+210=1335 hours Lower specification limit (LSL) = 1125-210=915 hours Note that these are specification limits (coming from our design specifications); they are not control limits. To determine process capability (decide if a process is capable at the desired level, that it is producing products that are within design specifications), we must have a stable process (we must have a process in control). In other words, for this question, we know that we have a stable process (in control) and we are working on understanding the capability of our process and make changes to increase that capability, if necessary. Process capability index ( C pk ): ? 𝑝𝑘 = min ( 𝜇−𝐿?𝐿 3𝜎 , ??𝐿−𝜇 3𝜎 ) = min ( 1050−915 3(55) , 1335−1050 3(55) ) = min(0.82,1.73) = 0.82 Process capability ratio ( C p ): ? 𝑝 = ??𝐿−𝐿?𝐿 6𝜎 = 1335−915 6(55) = 1.27 Note that three-sigma performance is targeted. Since C p > 1, we conclude that process variability is OK. But since C pk < 1, we conclude that there is a centering issue (the process is off-center). Note that our current process average is 1050 hours while our target is 1125 hours. Hence, we need to work on centering the process. Suppose the process is centered at 1125 hours. To what extent would the variability of the process have to be improved in order to achieve six-sigma capability? To achieve 6-sigma capability, both ratios have to be at least 2.0. Since we assume the process is centered, we will only be concerned with C p . ? 𝑝 = ??𝐿−𝐿?𝐿 6𝜎 = 1335−915 6𝜎 = 2 Solving for the standard deviation, we obtain σ = 35. In other words, we must reduce variability (measured in standard deviations) from the current level of 55 to 35.

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