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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