SCM 301 (Solo) Exam 1 Practice Exam Solutions

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1. D $118,000 www.liontutors.com SCM 301 (Solo) Exam 1 Practice Exam Solutions The first thing we need to do here is use the information given in the table to create a network diagram.

1 1. D $118,000 SCM 301 (Solo) Exam 1 Practice Exam Solutions The first thing we need to do here is use the information given in the table to create a network diagram. Once we have a network diagram, we can find the critical path through the network. Remember, a project s critical path is equal to the longest network path through the project. We will only crash activities that are on the critical path. Crashing activities that are not on the critical path will never actually shorten the duration of a project. This project s network diagram would look as follows: We see that there are 3 paths through the network, with the following lengths: A C F = 11 B D F = 14 B E G = 16 1

2 B E G is the critical path, so we should crash whichever activity has the cheapest crash cost per period on this path first. Therefore, we must calculate the crash cost period for all of the activities within the project. Crash cost per period = Crash cost Normal cost Normal time Crash time A (15,000 10,000) / (2 1) = 5,000 B (21,000 15,000) / (8 5) = 2,000 C (24,000 20,000) / (4 3) = 4,000 D (7,000 7,000) / (1 1) = 0 E (15,000 8,000) / (2 1) = 7,000 F (16,000 10,000) / (5 3) = 3,000 G (36,000 12,000) / (6 2) = 6,000 Since the crash cost for activity B ($2,000 per period) is lower than the crash cost for activity E ($7,000 per period) and activity G ($6,000 per period), we will crash activity B first. We can crash activity B up to a maximum of 3 periods (8 5). Doing so will cost $6,000 ($2,000 per period * 3 periods). After crashing B for 3 periods, the network paths will have the following lengths: A C F = 11 B D F = 11 B E G = 13 Since B E G is still the critical path, we should crash the next cheapest activity left on the path. Since we ve already crashed activity the maximum of 3 days, we cannot crash activity B any further. The crash cost per period for activity G ($6,000) is less than the crash cost per period for activity E ($7,000), so we should focus our efforts on activity G. It is possible to crash activity G up to a maximum of 4 days (6 2). However, doing so would cause the length of network path B E G to fall below 11 days, which is the length of the other network paths. If this were to happen, network path B E G would no longer be the critical path, and we never want to crash an activity that is not on the critical path. 2

3 Therefore, we should only crash activity G for 2 of the 4 possible periods. This means that we can still crash G for an additional 2 days at some point in the future, if necessary. Crashing G by 2 days will cost $12,000 ($6,000 per period * 2 periods). After crashing G for 2 days, the network paths will have the following lengths: A C F = 11 B D F = 11 B E G = 11 We now see that all three of the network paths have a length of 11 periods, which means that all three paths are now the critical paths. Comparing the activities that can still be crashed on the new critical paths (A, C, E, F, G), we see that activity F has the cheapest crash cost per period ($3,000). F can be crashed up to a maximum of 2 days. If we were to crash only activity F for 2 days, the network paths A C F and B D F would have a new length of 9 periods. However, network path B E G would still have a length of 11 days. This would not benefit the project. However, if we crash activity G by 2 periods at the same time as F for 2 periods, the lengths of all network paths will be reduced to 9 days. Crashing F for 2 days and G for 2 days costs a total of $18,000 ($3,000 * 2 periods + $6,000 * 2 periods). This will result in network paths of the following lengths. A C F = 9 B D F = 9 B E G = 9 We will now see that this project cannot be crashed any further, which means that the minimum completion time is 9 periods. We cannot crash any further because it is not possible to crash activity D. Therefore; it is not possible to bring the lengths of all network paths below 9 periods. Finally, we can find the total cost by adding the project s normal cost with the project s total crash costs. The project s normal cost is $82,000, and can be found by summing the normal costs of all of the project s activities (10, , , , , , ,000). The project s crash cost are $36,000, and can be found by adding up the crash costs from each step of the crashing process (6, , ,000). This gives us a total cost of $118,000 (88, ,000). 3

4 2. B Operations management 3. E Layout 4. C Operational 5. B Grey box design 6. D Transformation process 7. C Network-based approaches allow users to see predecessors, which makes it possible to find the critical path 8. B Marketing 9. A Organizational strengths that provide a foundation and a focus for a company s strategy 10. C Planning phase 11. A True 12. A $18,500 The first thing we need to do here is use the information given in the table to create a network diagram. Once we have a network diagram, we can find the critical path through the network. Remember, a project s critical path is equal to the longest network path through the project. We will only crash activities that are on the critical path. Crashing activities that are not on the critical path will never actually shorten the duration of a project. This project s network diagram would look as follows: 4

5 A-B-C-G = 19 A-D-G = 14 A-E-F-G = 17 Focus on the critical path or paths Choose the activities that will shorten all critical paths at the least cost We ll focus on A-B-C-G first since it is the critical path and we ll want to crash the activity (A, B, C, or G) with the lowest cost first. We re unable to crash A at all. The fixed cost/day for the project is $1,000, so we would never consider crashing an activity with a crash cost that is greater than or equal to the fixed cost of $1,000. This means we won t ever want to crash activity C. Choosing from activities B and G, we ll want to crash B first because the crash cost is $800/day compared to $900/day for G. We can crash activity B for a maximum of 3 days; however, that would bring the length of path A-B-C-G to 16 days, which would be less than path A-E-F-G. Since we never crash an activity if it isn t on the critical path, we will only crash B for 2 days. This means that we can still crash B for 1 additional day at some point in the future, if necessary. Crash B 2 $800/day = 2 * $800 = $1,600 Recalculate the lengths of all paths The new path lengths are as follows: A-B-C-G = 19 3 = 17 A-D-G = 14 A-E-F-G = 17 5

6 After crashing activity B for 3 days, we have two critical paths (A-B-C-G and A-E-F-G), both with a length of 17. This means we ll only move forward with crashing the project if we re able to crash an activity that on both paths, because we will never crash an activity that isn t on the critical path. Both paths have activity G and it s possible to crash activity G for 1 day for $900. Crash G 1 $900/day = $900 The new path lengths are as follows: A-B-C-G = 17 1 = 16 A-D-G = 14 1 = 13 A-E-F-G = 17 1 = 16 We still have A-B-C-G and A-E-F-G as our critical paths, now with lengths of 16 days. Although these paths do not share any of the same activities that can still be crashed, it is still possible to crash activity B for one more day on path A-B-C-G, and it is also possible to crash activity E for 1 day on path A-E-F-G. Crashing B for 1 day and E for 1 day simultaneously will decrease the lengths of both paths down to 15. However, it would cost $800 to crash B for 1 day and $400 to crash E for 1 day, for a total of $1,200. Remember, we re trying to find the minimum completion cost for the project in this problem. Since the total crash cost of $1,200 to crash activities B and E is more than the fixed cost per day of $1,000 we will choose not to crash these activities. Therefore, we can conclude that we will not crash any additional activities. The final length of the project will be 16 days. The question asked for the minimum completion cost for the project. Remember that we have two components that make up the total cost of this project, the fixed cost per day and the crash cost. Total crash cost = $1,600 + $900 = $2,500 Cost of project = $1,000/day * 16 days = $16,000 Minimum completion cost for project = $2,500 + $16,000 = $18, A Forecast 14. B Mission statement 15. B Project 16. A True 6

7 17. C Activity C Network Paths A-B-E-F = 75 A-B-D-F = 75 A-C-D-F = 66 The critical paths are A-B-E-F and A-B-D-F, so answer choice C activity C is the only activity that isn t on the critical path 18. D Order qualifiers 19. D Business strategy 20. B Policies, decisions, and computer systems are all infrastructural elements A computer system is a structural element, not an infrastructural element 21. D A core competency 22. E All of the above 23. E All of the above 7

8 24. D Delivery 25. A Conformance quality 26. D Serviceability 27. A The first D represents design A is false because the first D in DMADV stands for define, not design 28. A True Use the Following Work for the Next Three Problems Network Paths 1) A-D-F-G = 19 2) A-D-F-H = 15 3) B-D-F-G = 21 4) B-D-F-H = 17 5) C-E-F-G = 19 6) C-E-F-H = 15 8

29. C 6 30. C B, D, F, G 31. D 21 weeks 32. B The earliest start time of an activity is dependent on the earliest finish time of its immediate predecessors 33. B False 34. B Downstream 35. C Rules 36.

9 29. C C B, D, F, G 31. D 21 weeks 32. B The earliest start time of an activity is dependent on the earliest finish time of its immediate predecessors 33. B False 34. B Downstream 35. C Rules 36. D Commercial preparation 37. A Accounting 38. A 45 Network Paths A-B-D = 45 A-C-D = 39 The critical path is A-B-D because it is the longest network path. The duration of a project is always equal to the duration of the critical path. 9

10 39. B False 40. C Intangible needs 41. B Infrastructural elements 42. C Pre-sourcing 43. C In many cases, a company s core competencies are the drive for the company s strategy 44. C Crashing a project is an effort to shorten the overall duration of a project by reducing the amount of time that it takes to perform certain activities 45. E All of the above 46. C A and B 47. B Key points in a project, such as when a major phase should be completed, are referred to as the scope of the project They are referred to as the milestones of the project, not the scope. 48. B Satisfied customers 49. D A supply chain 50. C Concurrent engineering 51. B Many organizations operate without being connected to a larger supply chain 52. D When the supplier that you re interested in working with is recognized as an expert 53. C Roof 10

11 54. A Backward pass through the network 55. C After-sales support 56. B A critical path is composed entirely of critical activities 57. D Negotiate 58. C Time 59. C Gantt chart 60. B The supply chain is made up of only direct partners 61. B Reliability 62. D All of the above 63. A Scientists and engineers 64. C Changeover flexibility 65. D Disseminate 66. D Customer requirements 67. B Planning 68. B Purchasing 69. A Customer requirements are potentially not being met by the product 11

SCM 301 (Lutz) Exam 1 Practice Exam Solutions

1. D $118,000 www.liontutors.com SCM 301 (Lutz) Exam 1 Practice Exam Solutions The first thing we need to do here is use the information given in the table to create a network diagram. Once we have a network