Ch03 Practice Problems

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BA 360 Operations Management – Practice Problems Chapter 3: Project Management Problem #3-1: Rozales Manufacturing Co. is planning to install a new, flexible manufacturing system. The activities that must be performed, their immediate predecessors, and estimated activity times are shown below. Draw the project network and find the critical path computing early and late start days, early and late finish days, and activity slack. Estimated Immediate Activity Time Activity Description Predecessors (days) A Analyze current performance -- 3 B Identify goals A 1 C Conduct study of existing operation A 6 D Define new system capabilities B 7 E Study existing technologies -- 2 F Determine specifications D 9 G Conduct equipment analyses C, F 10 H Identify implementation activities C 3 I Determine organizational impacts H 4 J Prepare report E, G, I 2 K Establish audit procedure H 1 Problem #3-2: A computer-system installation project consists of eight activities. The immediate predecessors and activity times in weeks are shown below. Immediate Activity Activity Predecessor Time A -- 3 B -- 6 C A 2 D B, C 5 E D 4 F E 3 G B, C 9 H F, G 3 a. Draw the network for this project. b. What are the critical-path activities? c. What is the project completion time? d. Construct an early-start date Gantt chart. e. As a project manager, where would you focus your attention given your analysis? Problem #3-3: Page 1

BA 360 Operations Management – Practice Problems Chapter 3: Project Management Environment Recycling, Inc. must clean up a large automobile tire dump under a state environmental cleanup contract. The tasks, durations (weeks), costs, and predecessor relationships are as follows. Activit y Predecesso r(s) Normal Time Crash Time Normal Cost Crash Cost A --- 5 4 $400 $750 B A 12 9 1000 2200 C A 7 6 800 1100 D C 6 5 600 1000 E B,D 8 6 1200 2200 F D 3 2 800 1000 G D 3 2 500 650 H E 4 3 400 600 I F,G,H 6 5 900 1300 a. Draw the project network. b. Identify the critical path(s). c. What is the total project completion time and total cost? d. What is the total project completion time and lowest cost solution if the state wants to complete the project 3 weeks early? Problem #3-4: Two international banks are integrating two financial processing software systems as a result of their merger. Preliminary analysis and interviews with all parties involved resulted in the following project information. The “systems integration team” for this project plan to define and manage this project on two levels. The network defined below is the aggregate view and within each activity is a more detailed view with sub-tasks and project networks defined. All times are in weeks. Normal Crash Normal Crash Activity Predecessor Time Time Cost Cost A none 3 1 $1,000 $8,000 B A 1 1 4,000 4,000 C A 2 2 2,000 2,000 D B,C 7 5 3,000 6,000 E C 5 4 2,500 3,800 F C 3 2 1,500 3,000 G E 7 4 4,500 8,100 H E, F 5 4 3,000 3,600 I D, G, H 8 5 8,000 18,000 Page 2

BA 360 Operations Management – Practice Problems Chapter 3: Project Management a. Draw the project network. b. Identify the critical path. c. What is the total project completion time and total cost? d. What is the total project completion time and lowest cost solution if the bank wants to complete the project 2 weeks early? Problem #3-5: A competitor of Kozar International, Inc. has begun marketing a new instant-developing film project. Kozar has had a similar product under study in its R&D department but has not yet been able to begin production. Because of the competitor’s action, top managers have asked for a speedup of R&D activities so that Kozar can produce and market instant film at the earliest possible date. The predecessor information and activity time estimates in months are shown below. Most Immediate Optimistic Probable Pessimistic Activity Predecessors Time Time Time A --- 1 1.5 5 B A 3 4 5 C A 1 2 3 D B, C 3.5 5 6.5 E B 4 5 12 F C, D, E 6.5 7.5 11.5 G F 5 9 13 a. Draw the project network. b. Develop an activity schedule for this project using early and late start and finish times, compute activity slack time, and define the critical activities. c. What is the probability the project will be completed in time for Kozar to begin marketing the new product within 24 months? Problem #3-6: Suppose the estimates of activity times (weeks) for Kozar’s project (Previous problem) are as follows. Most Optimistic Probable Pessimistic Activity Time Time Time A 4 5 6 B 2.5 3 3.5 C 6 7 8 D 5 5.5 9 E 5 7 9 Page 3

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BA 360 Operations Management – Practice Problems Chapter 3: Project Management F 2 3 4 G 8 10 12 H 6 7 14 Suppose that the critical path is A-D-F-H. What is the probability that the project will be completed within a. 20 weeks? b. 22 weeks? c. 24 weeks? Problem #3-7: The critical path for the activities noted below is C, F, H. Given this information, what is the probability that the project will be completed on time (28 weeks)? Activity Preceding Optimistic Most Likely Pessimistic Expected Variance A -- 5 11 14 10.50 2.25 B -- 3 3 9 4.00 1.00 C -- 6 10 14 10.00 1.78 D A, B 3 5 7 5.00 0.44 E B 4 6 11 6.50 1.36 F C 8 10 14 10.33 1.00 G D, E 2 4 6 4.00 0.44 H F 3 3 9 4.00 1.00 Problem #3-8: If an activity has a normal time of 15 days can be shortened to 10 days for $3,000 what is the crash cost per period? Problem #3-9: Given the following information, what is the Critical Path of this project and the Normal Cost to complete it? Activity Immediate Predecessor Normal Time Crash Time Normal Cost Crash Cost A None 3 2 $ 200 $ 400 Page 4

BA 360 Operations Management – Practice Problems Chapter 3: Project Management B A 4 3 $ 300 $ 600 C A 1 1 $ 200 $ 200 D B and C 3 2 $ 500 $ 550 E D 2 1 $ 500 $ 900 Problem #3-10: A project being analyzed by PERT has 60 activities, 13 of which are on the critical path. If the estimated time along the critical path is 204 days with a project variance of 100, what is the probability that the project will take 224 days or more to complete? Problem #3-11: What is the expected time for activity D? Activity Precedes Optimistic Most Likely Pessimistic A D 38 50 62 B E 90 100 108 C End 85 100 115 D F 85 90 140 E End 91 98 115 F End 62 65 68 Problem #3-12: Which activity should be crashed and what would the cost be to crash it on time? ACTIVITY SLACK NORMAL TIME CRASH TIME NORMAL COST CRASH COST J 1 7 6 $ 80,000 $ 120,000 K 1 6 4 $ 200,000 $ 320,000 L 3 3 2 $ 90,000 $ 100,000 M 3 2 1 $ 180,000 $ 250,000 N 0 4 2 $ 410,000 $ 450,000 P 0 3 2 $ 100,000 $ 150,000 Q 1 4 2 $ 310,000 $ 500,000 R 0 2 2 $ 400,000 $ 410,000 Page 5

BA 360 Operations Management – Practice Problems Chapter 3: Project Management * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * Activity Immediate Predecessor Time (days) Earliest Start Earliest Finish Latest Start Latest Finish Slack A --- 3 0 3 0 3 0 B A 1 3 4 3 4 0 C A 6 3 9 14 20 11 D B 7 4 11 4 11 0 E --- 2 0 2 28 30 28 F D 9 11 20 11 20 0 G C,F 10 20 30 20 30 0 H C 3 9 12 23 26 14 I H 4 12 16 26 30 14 J E,G,I 2 30 32 30 32 0 K H 1 12 13 31 32 19 L J,K 0 32 32 32 32 0 Critical path is A-B-D-F-G-J with a final project completion time of 32 days. Solution #3-2: Page 6 Solution #3-1: Here, L (or “ End ” ) is the ending activity. A D E C B F G H I J K L

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BA 360 Operations Management – Practice Problems Chapter 3: Project Management Activity Immediate Predecessor Time (days) Earliest Start Earliest Finish Latest Start Latest Finish Slack A --- 3 0 3 1 4 1 B --- 6 0 6 0 6 0 C A 2 3 5 4 6 1 D B, C 5 6 11 6 11 0 E D 4 11 15 11 15 0 F E 3 15 18 15 18 0 G B,C 9 6 15 9 18 3 H F.G 3 18 21 18 21 0 Critical path is B-D-E-F-H. Project completion time is 21. Gantt Chart: There is very little slack in the schedule, so the project manager would want to ensure that the activities remain on schedule. Instructors should ensure that students "see" the connection between the critical path and Gantt Chart (i.e., trace out the critical path on the chart). Page 7 A B C D E F G H

BA 360 Operations Management – Practice Problems Chapter 3: Project Management Activity Immediate Predecessor Time (days) Earliest Start Earliest Finish Latest Start Latest Finish Slack A --- 5 0 5 0 5 0 B A 12 5 17 6 18 1 C A 7 5 12 5 12 0 D C 6 12 18 12 18 0 E B,D 8 18 26 18 26 0 F D 3 18 21 27 30 9 G D 3 18 21 27 30 9 H E 4 26 30 26 30 0 I F,G,H 6 30 36 30 36 0 Critical path is A-C-D-E-H-I. Completion time = 36. Total normal cost = $6600. Activit y Normal Time Crash Time Normal Cost Crash Cost Crash cost/week A 5 4 $400 $750 $350 B 12 9 1000 2200 400 C 7 6 800 1100 300 D 6 5 600 1000 400 E 8 6 1200 2200 500 F 3 2 800 1000 200 G 3 2 500 650 150 H 4 3 400 600 200 I 6 5 900 1300 400 Page 8 Solution #3-3: A B C F D E G H I

BA 360 Operations Management – Practice Problems Chapter 3: Project Management Crashing options Crash A by 1 week: $350 Crash C by 1 week: $300 Crash D by 1 week: $400 Crash E by 2 weeks: $1000 Crash H by 1 week: $200 Crash I by 1 week: $400 To reduce completion time to 33 weeks, the least cost combination is to crash A, C and H by 1 week each for a total cost of $850. The critical path remains the same. Total project costs are now $6600 + $850 = $7450. Activity Immediate Predecessor Time (days) Earliest Start Earliest Finish Latest Start Latest Finish Slack A --- 3 0 3 0 3 0 B A 1 3 4 9 10 6 C A 2 3 5 3 5 0 D B,C 7 5 12 10 17 5 E C 5 5 10 5 10 0 F C 3 5 8 9 12 4 G E 7 10 17 10 17 0 H E,F 5 10 15 12 17 2 Page 9 Solution #3-4: A C B F D E G H I

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BA 360 Operations Management – Practice Problems Chapter 3: Project Management I D,G,H 8 17 25 17 25 0 Critical path is A-C-E-G-I. Completion time is 25 weeks. Total normal cost = $29,500 Activit y Normal Time Crash Time Normal Cost Crash Cost Crash cost/wee k A 3 1 1000 8000 3500 B 1 1 4000 4000 -- C 2 2 2000 2000 -- D 7 5 3000 6000 1500 E 5 4 2500 3800 1300 F 3 2 1500 3000 1500 G 7 4 4500 8100 1200 H 5 4 3000 3600 600 I 8 5 8000 18000 3333 The solution can be easily seen as crashing G by 2 weeks for an additional cost of $2400. This is the optimal solution that can also be found using linear programming. Problem #3-5: Using the 3-point estimates of times, we obtain expected times and variances: Activity Expected Time Variance A 2 0.44 B 4 0.11 C 2 0.11 D 5 0.25 E 6 1.78 F 8 0.69 Page 10 A B C E D F G

BA 360 Operations Management – Practice Problems Chapter 3: Project Management G 9 1.78 Activity Immediate Predecessor Time (days) Earliest Start Earliest Finish Latest Start Latest Finish Slack A --- 2 0 2 0 2 0 B A 4 2 6 2 6 0 C A 2 2 4 5 7 3 D B,C 5 6 11 7 12 1 E B 6 6 12 6 12 0 F C,D.E 8 12 20 12 20 0 G F 9 20 29 20 29 0 Critical path is A-B-E-F-G; completion time is 29 months. Variance of the critical path is 4.81. Assuming a normal distribution, the probability of completion in 24 months is calculated as: z = (24 – 29)/sqrt(4.81) = –2.28 → Probability = 0.0113, or 1.13% There is little chance of meeting this deadline. Solution #3-6: Activity Expected Time Variance A 5 0.111 B 3 0.028 C 7 0.111 D 6 0.444 E 7 0.444 F 3 0.111 G 10 0.444 H 8 1.778 The expected completion time for the critical path is 22; the variance along the critical path is 2.44, so the standard deviation is 1.56. a. z = (20 – 22)/1.56 = –1.28, → Probability = 0.1003 = 10.03% b. z = (22 – 22)/1.56 = 0, Probability = 0.5 = 50% c. z = (24 – 22)/1.56 = 1.28, → Probability = 0.89973 = 89.97% Page 11

BA 360 Operations Management – Practice Problems Chapter 3: Project Management Solution #3-7: The critical path is C, F, H. Given this information, what is the probability to project will be completed on time (28 weeks)? Sum of Expected times for C, F, H = 10.00 + 10.33 + 4.00 = 24.33 Sum of Variances for C, F, H = 1.78 + 1.00 + 1.00 = 3.78 Std Deviation = Sq rt of Variances = sqrt(3.78) = 1.944 Z = Specified Time 28 – Path Duration 24.33 / Path Std Dev 1.944 = 1.886 Z = 1.89, look at the Normal Curve Areas Chart = 97.06% probability Solution #3-8: $3,000 - $0 / (15 -10 = 5) = $600 crash cost per period Solution #3-9: Path: A-B-D-E = 12 weeks (critical path) Path: A-C-D-E = 11 weeks Normal cost to complete the project = $200 + $300 + $200 + $500 + $500 = $1,700 Solution #3-10: On Time or Less Than = Z probability. Or More = 1- probability. Specified Time 224 – Path Duration 204 = 20. Project Variance = 100 Std. Dev = Sqrt (100) = 10 Z = 20 / 10 = 2.00 Z = .9773 probability for “on time” or “less than” Ans = 1-.9773 for “more than” = .0228 or 2.28% Solution #3-11: (85 + 90*4 + 140) / 6 = 97.5 Page 12 A B C D E

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BA 360 Operations Management – Practice Problems Chapter 3: Project Management Solution #3-12: Only crash activities with Zero slack (which means they are on the critical path) N: 450,000 – 410,000 / 4 – 2 = $20,000 (correct answer) P: 150,000 – 100,000 / 3 – 2 = $50,000 R: 410,000 - 400,000 / 2 – 2 = cannot crash Page 13

Ch03 Practice Problems

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