Discussion Questions on Project Management and Scheduling Part 2

13. Here is a network with the activity times shown in days: a. Find the critical path. 6 14 18 B(8 ) 6 0 F (5 ) 14 6 20 14 A (6 ) 0 23 25 18 23 D ( 4) 6 14 6 11 G (3) 18 25 18 C (5) 9 28 28 25 E (7 ) 14 18 25 A-B-D-E-G b. The following table shows the normal times and the crash times, along with the associated costs for each activity. If the project is to be shortened by four days, show which activities, in order of reduction would be shortened and the resulting cost. Solution: Activity A B C D E F G Normal Time (NT) 6 8 5 4 7 5 3 Crash Time (CT) 5 7 3 3 6 4 2 Normal Cost (NC) $7,300 5,300 9,000 3,300 2,200 4,100 5,200 Crash Cost (CC) $8,300 6,700 10,600 4,100 3,600 7,300 8,300 NT-CT 1 1 2 1 1 1 1 Cost/day to expedite $1,000 1,400 800 800 1,400 3,200 3,100 First, the lowest cost activities to crash are D at $800 per day. Critical path remains the same. Second, lowest cost activity on the critical path is A. Crash activity A. Third, B and E are next lowest cost activities on the critical paths. Both have a cost of $1,400 per day. Select B then E or reverse the order (E then B). Summary of steps to reduce project by four days: Step Activity to crash 1 D Cost to crash 2 A 1,000 1 3 B (or E) 1,400 1 4 E (or B) 1,400 1 Total $4,600 $800 Days saved 1 14. The home office billing department of a chain department stores prepares monthly inventory reports for use by the stores’ purchasing agents. Given the following information, use the critical path method to determine: a. How long the total process will take. 0 10 20 b (10 ) 10 0 50 d (30 ) 20 50 80 0 80 a (0) 0 100 0 80 0 20 20 20 20 c (20 ) 0 40 40 e (20 ) 100 100 g (20 ) h ( 0) 100 100 100 80 f (40 ) 40 40 80 100 hours b. Which jobs can be delayed without delaying the early start of any subsequent activity? Activities b and d are not on the critical path. Their start can be delayed without delaying the start of any subsequent activities. Activity b can be delayed by 10 hours and d can be delayed by 30 hours without affecting the project completion date. 15. For the network shown: a. Determine the critical path and the early completion time in weeks for the project. 18 21 D (3) 6 18 18 21 B ( 12 ) 18 6 0 14 6 A( 6 ) 0 17 21 E (3) 6 18 25 G (4) 21 21 25 14 6 C (8 ) 10 18 17 14 F (3 ) 18 21 A-B-D-G, 6+12+3+4 = 25 weeks b. For the data shown, reduce the project completion time by three weeks. Assume a linear cost per week shortened, and show, step by step, how you arrived at your schedule. Solution: Activity A B C D E F G Normal Time (NT) 6 12 8 3 3 3 4 Normal Cost (NC) $7,400 12,200 5,800 4,600 3,200 7,000 8,000 Crash Time (CT) 5 7 6 1 2 2 3 Crash Cost (CC) $13,500 18,300 7,000 6,000 6,700 7,500 8,400 NT-CT 1 5 2 2 1 1 1 Cost/week to expedite $6,100 1,220 600 700 3,500 500 400 First, reduce G (lowest cost activity on the critical path) by one week. Second, crash activity D by one week. Critical paths remain the same. Third, crash activity D by one more week at a cost of $700, which is the least expensive. Summary of activities crashed: Step Activity Cost to crash Weeks reduced 1 G $400 1 2 D 700 1 3 D 700 1 Total cost $1,800 16. The following CPM network has estimates of the normal time in weeks listed for the activities: a. Identify the critical path. 5 15 8 F (6) B(3 ) 8 0 11 15 5 11 A (5 ) 0 21 21 15 21 D ( 4) 5 11 5 15 11 21 15 C (6) 5 26 G (5) 26 18 E (3) 11 18 21 A-C-D-F-G b. What is the length of time to complete the project? 5+6+4+6+5 = 26 weeks c. Which activities have slack, and how much? B, 3 weeks; E, 3 weeks. d. Here is a table of normal and crash times and costs. Which activities would you shorten to cut two weeks from the schedule in a rational fashion? What would be the incremental cost? Is the critical path changed? Solution: Activity Normal Time (NT) Crash Time (CT) Normal Cost (NC) Crash Cost (CC) NT-CT Cost/week to expedite A 5 4 $7,600 $8,700 1 1,100 B 3 2 5,700 8,400 1 2,700 C 6 5 7,800 9,400 1 1,600 D 4 3 2,650 3,800 1 1,150 E 3 2 2,500 4,400 1 1,900 F 6 4 5,800 8,600 2 1,400 G 5 4 3,800 5,500 1 1,700 First, shorten activity A by one week at a cost of $1,100. This is the lowest cost/week activity on the critical path. Second, shorten activity D by one week at a cost of $1,150. This is the next lowest cost/week activity on the critical path. The total cost is $2,250 and the critical path remains unchanged. 17. Bragg’s Bakery is building a new automated bakery in downtown Sandusky. Here are the activities that need to be completed to get the new bakery built and the equipment installed. a. Draw the project diagram. B (3) D (10) A (4) F (2 ) C (5 ) E (4 ) b. and c. What is the normal project length? What is the project length if all activities are crashed to their minimum? Solution: Path ABDF ACDF ACEF Normal Length 19 21 15 Crashed Length 12 13 09 d. Bragg’s loses $4,000 in profit per week for every week the bakery is not completed. How many weeks will the project take if we are willing to pay crashing cost as long as it is less than $4,000? We would only crash the project until 14 weeks since the cost of crashing F is $5600 which is greater than the $4000 in additional profit. LENGTH ABDF - 19 18 17 16 15 15 15 14 13 ACDF - 21 20 19 18 17 16 15 14 13 ACEF - 15 15 15 15 15 14 13 12 11 Activity Crashed D D D D C C A F Crash cost 2000 2000 2000 2000 3800 3800 3900 5600 Cumulative Cost 2000 4000 6000 8000 11800 15600 19500 25100 18. Assume the network and data that follow: a. Construct the network diagram. Activity Normal Time (NT) Crash Time (CT) Normal Cost (NC) Crash Cost (CC) A 4 3 $140 $180 1 40 B 5 3 70 100 2 15 C 4 1 90 120 3 10 D 3 2 90 150 1 60 E 5 3 40 50 2 5 F 8 5 100 190 3 30 G 9 8 60 120 1 60 4 9 5 0 E (5) 10 10 4 4 4 11 A(4) 0 14 9 B (5) 15 8 15 15 15 C (4) 4 24 G (9) 7 7 7 7 D (3) 4 NT-CT 15 F (8 ) 15 24 Cost/week to expedite b. Indicate the critical path when normal activity times are used. The critical path is A-D-F-G, with a length of 4+3+8+9 = 24 c. Compute the minimum total direct cost for each project duration based on the cost associated with each activity. Consider durations of 19, 20, 21, 22, 23, and 24 weeks. The normal time project cost is $590 at 24 weeks. Minimum cost crashing to 19 weeks is shown below. LENGTH ABEG - 23 23 22 21 20 19 ACG - 17 17 17 17 16 15 ADFG - 24 23 22 21 20 19 Activity Crashed F Crash cost 30 E+F E+F A G 35 35 40 60 Cum. Crash Cost 30 65 100 140 200 Total Direct Cost 620 655 690 730 790 d. If the indirect costs for each project duration are $400 (24 weeks), $350 (23 weeks), $300 (22 weeks), $250 (21 weeks), $200 (20 weeks), and $150 (19 weeks), what is the total project cost for each duration? Indicate the minimum total project cost duration. Computations build off the earlier table and are shown below. The total project cost in normal time is $590 + $400 = $990. LENGTH ABEG - 23 23 22 21 20 19 ACG - 17 17 17 17 16 15 22 21 20 19 A G ADFG - 24 23 Activity Crashed F Crash cost 30 35 35 40 60 Cum. Crash Cost 30 65 100 140 200 Total Direct Cost 620 655 690 730 790 Indirect Cost 350 300 250 200 150 Total Project Cost 970 955 940 930 940 E+F E+F We can achieve minimum cost by reducing the project to 20 weeks. The duration has the total cost of $930. 19. Your project to obtain charitable donations is now 50 days into a planned 60-day project. The project is divided into 3 activities. The first activity is designed to solicit individual donations. It is scheduled to run the first 44 days of the project and to bring in $26,300. Even though we are 50 days into the project, we still see that we have only 88 percent of this activity complete. The second activity relates to company donations and is scheduled to run for 50 days starting on day 6 and extending through day 56. We estimate that even though we should have (44/50) 88 percent of this activity complete, it is actually only 55 percent complete. This part of the project was scheduled to bring in $151,300 in donations. The final activity is for matching funds. This activity is scheduled to run the last 10 days of the project and has not started. It is scheduled to bring in an additional $52,600. So far $175,600 has actually been brought in on the project. Calculate the schedule variance, schedule performance index, and cost (actually value in this case) performance index. How is the project going? Hint: Note that this problem is different since revenue rather than cost is the relevant measure. Use care in how the measures are interpreted. Solution: Activity 1. Solicit 2. Donations 3. Matching Funds Total Expected Revenue $26,300 $151,300 Planned Duration 44 50 Planned Start Date 0 6 Planned Comp. Date 44 56 Expected % Complete 100% 88% Actual % Complete 88% 55% $52,600 10 50 60 0% 0% $230,200 $175,600 BCWS Activity 1 Activity 2 Activity 3 100% of $26,300 = 88% of $151,300 = 0 % of 52,600 = 26,300 133,144 0 $159,444 BCWP Activity 1 Activity 2 Activity 3 88% of 26,300 55% of 151,300 0% of $52,600 Actual Rev. to Date = 23,144 = 83,215 =0 106,359 Scheduled Variance = 106,359 - 159,444 = -53,085 Scheduled Performance Index = 106,359/159,444 = .667 Cost Variance = 106,359 - 175,600 = -69,241 Cost Performance = 106,359/175,000 = .606 Because we are working with revenues instead of costs, we have to invert the evaluation rules listed in the text. Our performance measures here are actually good. Although we are behind schedule on completing tasks 1 and 2, we have brought in more money than expected for the amount of work we have completed. 20. A project to build a new bridge seems to be going very well since the project is well ahead of schedule and costs seem to be running very low. A major milestone has been reached where the first two activities have been totally completed and the third activity is 70 percent complete. The planners were only expecting to be 57 percent through the third activity at this time. The first activity involves prepping the site for the bridge. It was expected that this would cost $1,426,000 and it was done for only $1,306,000. The second activity was the pouring of concrete for the bridge. This was expected to cost $10,506,000 but was actually done for $9,006,000. The third and final activity is the actual construction of the bridge superstructure. This was expected to cost a total of $8,506,000. To date they have spent $5,006,000 on the superstructure. Calculate the schedule variance, schedule performance index, and cost index for the project to date. How is the project going? Solution: Activity Expected Cost Expected % Complete Actual % Complete Actual Cost to Date 1. Site preparation 2. Pour concrete $1,426,000 $10,506,000 100% 100% 100% 100% $1,306,000 $9,006,000 3. Construction Total $8,506,000 $20,438,000 57% 70% $5,006,000 $15,318,000 BCWS Activity 1 Activity 2 Activity 3 100% of $1,426,000 100% of $10,506,000 57 % of $8,506,000 = = = $1,426,000 $10,506,000 $4,848,420 $16,780,420 BCWP Activity 1 Activity 2 Activity 3 100% of $1,426,000 100% of $10,506,000 70 % of $8,506,000 = = = $1,426,000 $10,506,000 $5,954,200 $17,886,200 Scheduled Variance = 17,886,200-16, 780,420 = 1,105,780 Scheduled Performance Index = 17,886,200/16,780,420 = 1.066 Cost Performance = 17,886,200/15,318,000 = 1.168 Ahead of schedule and under budget. 21. What feature in project management information systems can be used to resolve overallocation of project resources? Leveling 22. What was the first major project management information system that is now commonly used for managing very large projects? Primavera Project Planner 23. What type of chart compares the current project schedule with the original baseline schedule so that deviations from the original plan can be easily noticed? Tracking Gantt chart Analytics Exercise: Product Design Project 1, 2. 0 2 P1 (2) 0 2 2 6 P2 (4) 2 16 22 22 S2 (6) 19 32 35 V1 (10) 25 25 35 35 6 6 11 21 P3 (5) 6 16 23 16 21 16 16 24 21 S1 (5) P4 (5) 11 22 D1 (1) 11 11 22 23 D3 (1) 21 23 27 27 D4 (4) 24 21 24 30 31 I1 (3) 28 23 31 23 27 27 31 27 27 I2 (4) 23 23 35 I4 (4) 28 D2 (2) 21 35 31 I3 (4) 31 The project will take 37 weeks to complete. The critical path is P1-P2-P3-P4-S1-D2-I2-I3-I4-V2. 3. Slack for each activity is listed in the following table. Major Subprojects/Activities Project Specifications (P) Market research Overall product specification Hardware Software Supplier specifications (S) Hardware Software Product design (D) Battery Display Camera Outer cover Product integration (I) Hardware User interface Software coding Prototype testing Subcontracting (V) Supplier selection Contract negotiation 37 V2 (2) Activity ID Dependencies P1 P2 P3 P4 -P1 P2 P3 S1 S2 P4 P4 D1 D2 D3 D4 S1 S1 S1 D1, D2, D3 I1 I2 I3 I4 D4 D2 I2 I1, I3 V1 V2 S1, S2 I4, V1 Duration (Weeks) 16 2 4 5 5 6 5 6 6 1 2 1 4 12 3 4 4 4 12 10 2 ES, LS Slack 0, 0 2, 2 6, 6 11, 11 ----- 16, 16 16, 19 -3 21, 23 21, 21 21, 23 23, 24 2 -2 1 27, 28 23, 23 27, 27 31, 31 1 ---- 22, 25 35, 35 3 -- 37 4. Assume that the activity lengths remain the same, but the precedence relationships within subprojects no longer apply as all activities in a subproject will be worked on in parallel. Therefore, the length of each subproject will now be equal to the length of the longest activity in the subproject. The subprojects are outlined in dashed lines in the network drawing above. Subproject P will take 5 weeks to complete; subproject S, 6 weeks; D, 4 weeks; I, 4 weeks; and V, 10 weeks. Since all of these subprojects will completed in series now, the length of the project is the sum of the subproject times: 5+6+4+4+10 = 29 weeks, a decrease of 8 weeks. 5. The revised network diagram is shown below. Adding P5 extends the time of subproject P to 12 weeks while subprojects D and I remain at 4 weeks each. By eliminating subprojects S and V, the project length is now down to 20 weeks – 17 weeks less than the original schedule. 0 2 P1 (2) 10 12 0 12 4 P2 (4) 13 D1 (1) 16 8 12 0 0 5 0 15 16 12 13 12 16 12 0 15 5 12 12 14 16 16 20 I4 (4) 17 20 16 20 16 20 16 16 16 16 I2 (4) 14 D2 (2) 12 weeks 6. D4 (4) 12 12 P4 (5) 7 19 I1 (3) 16 P5 (12) D3 (1) P3 (5) 7 12 20 20 I3 (4) 20 4 weeks 4 weeks Having the team focus on a single subproject at a time will allow more collaboration on each subproject as opposed to the team being split across several subprojects. This might result in higher quality of work as the entire project team is focused on a single subproject at a time and you will have more input and a wider variety of experiences working on each subproject than you would otherwise. By combining subprojects S and V with P, Nokia can perhaps take advantage of supplier expertise in designing the new phone. Nokia would better understand the technological capabilities of their suppliers and include them in the phone design from the beginning. One concern might be the feasibility of eliminating the original precedence relationships when changing to the new project structures in parts 4 and 5. Assuming there were good reasons for those relationships originally, eliminating them might cause problems in the project if Nokia does not fully address those reasons in the new project structure. For example, if they select suppliers and negotiate contracts before the product specifications are complete, and they do not include their suppliers in the product specification process, they might end up with a supplier that cannot supply the needed materials nor do so at the proper level of quality. Assuming the technical/managerial precedence issues are properly addressed, the new project structures make sense. In addition to reducing project time, there are other possible benefits to be gained from the increased collaboration the new structures would bring.