MBA 675 Week 11 HW

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Industrial Engineering

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Jan 9, 2024

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Chapter 7, Problems, #14 1600 1200 Table 7.7 Data for the Billing Process Activity A - 4 1 5,000 8,000 B - 5 3 8,000 10,000 C A 1 1 4,000 4,000 D B 6 3 6,000 12,000 E B,C 7 6 4,000 7,000 F D 7 6 4,000 7,000 A. What is the minimum-cost schedule for this project? Normal Cost Schedule- Crash Cost Schedule- Minimum Cost Schedule- Paths: Weeks Costs Paths: Weeks Costs Paths: Weeks Costs A 4 $ 5,000 A 1 $ 8,000 A 4 $ 5,000 $ 5,600 C 1 $ 4,000 C 1 $ 4,000 C 1 $ 4,000 E 7 $ 4,000 E 6 $ 7,000 E 7 $ 4,000 Total: 12 $ 13,000 Total: 8 $ 19,000 Total: 12 $ 13,000 Critical path Critical path Critical path B 5 $ 8,000 B 3 $ 10,000 B 3 $ 10,000 D 6 $ 6,000 D 3 $ 12,000 D 3 $ 12,000 E 7 $ 4,000 E 6 $ 7,000 E 6 $ 7,000 F 7 $ 4,000 F 6 $ 7,000 F 6 $ 7,000 Total: 25 $ 22,000 Total: 18 $ 36,000 Total: 18 $ 36,000 Longest path: B-D-E-F 25 Longest path:B-D-E-F 18 Longest path: B-D-E-F 18 Indirect Costs: 1600 $ 40,000 Indirect Costs 1600 $ 28,800 Indirect Costs: 1600 $ 28,800 Penalty Costs: 13 $ 15,600 Penalty Costs 6 $ 7,200 Penalty Costs: 6 $ 7,200 TOTAL COST: $ 90,600 TOTAL COST: $ 91,000 TOTAL COST: $ 85,000 You are the manager of a project to improve a billing process at your firm. Table 7.7 contains the data you will need to conduct a cost analysis of the pr Indirect costs are $1,600 per week, and penalty costs are $1,200 per week after week 12. Immediate Predecessor(s) Normal Time (wks) Crash Time (wks) Normal Costs ($) Crash Costs ($) B. What is "normal tim
roject. the difference in total project costs between the earliest completion time of the project using me" and the minimum-cost scheduled derived from part (a)?
Chapter 7, Problems, #16 Table 7.9 Project Activity & Cost Data Normal Crash Activity Costs ($) Costs ($) A 12 1,300 11 1,900 1 600 B 13 1,050 9 1,500 4 112.5 C 18 3,000 16 4,500 2 750 D 9 2,000 5 3,000 4 250 E 12 650 10 1,100 2 225 F 8 700 7 1,050 1 350 G 8 1,550 6 1,950 2 200 Paths: Days H 2 600 1 800 1 200 ADG* 29 I 4 2,200 2 4,000 2 900 AEH 26 $ 13,050 $19,800 BFH 23 *= Critical Path CI 22 ADG 27 AEH 26 New Schedule BFH 23 Activity Duration CI 22 A 12 Cost= $ 13,450 B 13 C 18 D 7 ADG 25 E 11 AEH 25 F 8 BFH 23 G 6 CI 22 H 2 Cost= $ 14,175 I 4 The diagram in Figure 7.12 was developed for the project launch of Kitty Condo, a new product in the luxury cat cage market. Suppose that you, as project manager, are interested in finding ways to speed up the project at minimal additional cost. Determine the schedule for completing the project in 25 days at minimum cost. Penalty and project-overhead costs are negligible. Time and cost data for each activity are shown in Table 7.9. Time (days) Time (days) Crash Days Crash Cost/day Paths & New Durations: (crashing G by 2 days) Project Duration of 25 days at a cost of $14,175 Paths & New Durations: (crashing D by 2 days and E by 1)
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Time Estimates (days) Activity Optimistic Most Likely Pessimistic A 5 8 11 B 4 8 11 C 5 6 7 D 2 4 6 E 4 7 10 What is the expected completion time of the project? 19 days Activity Optimistic Most Likely Pessimistic Variance A - 5 8 11 8 1.00 B - 4 8 11 8 1.36 C A 5 6 7 6 0.11 D A 2 4 6 4 0.44 E B,D 4 7 10 7 1.00 Paths: *=Critical Path AC 14 ADE* 19 Project Variance= 2.44 BE 15 Chapter 7, Problems, #21 – answer all parts, on part b) provide the probability of completing it in 21 days and also provide the probability of completing it in 21 days or less; on part c) provide the probability of completing it in 17 days and also provide the probability of completing it in 17 days or less; Recently, you were assigned to manage a project to remodel the seminar room for your company. You have constructed a network diagram depicting the various activities in the project (Figure 7.13). In addition, you have asked your team to estimate the amount of time that they would expect each of the activities to take. Their responses are shown in the following table. Immediate Predecessor Expected Time
z= (given duration - critical path duration) / sqrt (critical path variance) What is the probability of completing the project in 21 days or les 1.28 0.8996 What is the probability of completing the project in 17 days or les -1.28 0.1004
Element A: 0.92 Element B: 0.94 Element C: 0.91 A. Draw how these three elements are connected B. What is the reliability of this system? R= 0.787 Elements: A B C = 0.85 0.92 0.94 C = 0.85 0.8648 C = 0.85 C = 0.983 R= 0.982 A piece of electronic equipment used for aviation has three elements connected in series, or sequence. The reliability of each of the three elements is as follows: C. Suppose we wanted the reliability to be at least 0.85 by upgrading Element C with a higher reliability. What would the reliability of Element C need to be? D. Instead of increasing the reliability of Element C, each of the elements is supported by an identical element in parallel as a back-up. What is the reliability of this system? A B C
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PM= $ 1,200 Weeks between preventive maintenance Expected # of breakdowns between preventive maintenance Breakdown: 1 1.1 Week 1= 2200 2 2.52 Week 2= 2400 3 4.63 Week 3= 2600 4 8.27 Week 4= 2800 Total Cost per week: $ 3,620 Weekly preventative maintenance is recommended since it has the lowest total cost per week Total Costs every 2 weeks: $ 3,624 Total Cost every 3 weeks: $ 4,413 Total Cost every 4 weeks: $ 6,089 Let’s revisit the preventive maintenance problem discussed in the video. A hospital has seven large identical heat pumps. If one of the heat pumps malfunctions, it could seriously impact hospital operations. If preventive maintenance is performed weekly, it costs $1200 to perform preventive maintenance (PM) on all seven of them. If one of the heat pumps breaks down between PM inspections, the average cost to the company of a breakdown is $2200. The historical data for the heat pumps is as follows: Now suppose that for each week that preventive maintenance is delayed, the cost to the company of a breakdown increases by $200 (for example, if preventive maintenance is every week, the cost of a breakdown is $2200 and if preventive maintenance is every two weeks, the cost of a breakdown is $2400, etc.) Based on the above information, what Preventive Maintenance policy do you recommend? Why?
Discuss the pros and cons of McDonnell’s position as expressed by the VP. What strategy would you have recommended McDonnell Douglas follow to address the failure? Why? When a McDonnell Douglas DC-10 crashed over Iowa, a subsequent investigation suggested that the plane’s hydraulic systems did not provide enough protection. The DC-10 had three separate hydraulic systems, all of which failed when an engine exploded. The engine threw off shreds of metal that severed two of the lines, and the third line required power from the demolished engine that was no longer available. The DC-10, unlike other commercial jets, had no shutoff valves that might have stemmed the flow of hydraulic fluid. Lockheed’s similar L-1011 tri-jet had four hydraulic systems. A McDonnell Douglas VP said at the time. “You can always be extreme and not have a practical airplane. You can be perfectly safe and never get off the ground.” Discuss the pros and cons of McDonnell’s position as expressed by the VP. The viewpoint expressed by the McDonnell Douglas VP highlights the need to strike a balance between practicality and safety when designing aircraft. The VP argues that extreme safety measures can compromise the functionality and feasibility of an aircraft. This raises concerns about the practicality and cost-effectiveness of implementing additional safety features, such as hydraulic shutoff valves. However, the absence of these safety measures in the McDonnell Douglas DC-10, as revealed during the investigation of a crash in Iowa, exposed a critical design flaw. The incident highlighted the vulnerability of the DC-10's hydraulic system and prompts questions about whether McDonnell Douglas could have incorporated alternative designs with greater redundancy, as seen in Lockheed's L-1011 tri-jet. It is worth noting that the VP's perspective may downplay the importance of safety, potentially eroding public trust in the company and its products. Ultimately, while considering practicality and operational feasibility in aircraft design is important, it is crucial to recognize the significant safety implications and maintain public confidence in the aviation industry. McDonnell Douglas could have addressed the hydraulic system failure in the DC-10 by implementing a comprehensive strategy. Firstly, a thorough review and redesign of the system should have been conducted to identify weaknesses, explore alternative designs, and enhance reliability and redundancy. Lessons learned from the investigation should have guided the implementation of critical safety measures like shutoff valves or alternative prevention methods. Collaboration with industry experts and regulators would have provided valuable insights and ensured compliance with safety standards. Transparent communication about the implemented safety measures would have been crucial for rebuilding trust. Additionally, establishing systems for continuous monitoring, data analysis, and improvement would have ensured long-term reliability and safety. By following these strategies, McDonnell Douglas could have demonstrated a proactive commitment to safety and taken the necessary steps to regain confidence in their aircraft.