Topic 2: Project Management Models
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Topic 2: Project Management Models Topic 2: Project Management Models Document Transcript

  • UCF College of Business Flexible MBA Program ISM 6407.0001 Decision Support Systems Spring 2007 Supplementary Notes Topic 2
  • Project Management
  • ISM 6407.0001 Spring 2007 Topic 2: Project Management Page 1/26 PROJECT SCHEDULING – THE EARLY DAYS Project Data: Immediate Time Activity Predecessor (weeks) A - 1 B A 2 C A 4 D B 5 E C 7 F D,E 1 Gantt Chart for tracking the progress of the project activities end of wk→ 0 1 2 3 4 5 6 7 8 9 10 11 12 13 wk→ 1 2 3 4 5 6 7 8 9 10 11 12 13 A A B B C C D D E E F F
  • ISM 6407.0001 Spring 2007 Topic 2: Project Management Page 2/26 PROJECT SCHEDULING DEMONSTRATION PROBLEM Data for initial view. (Deterministic time estimates; cost budgets) Immediate Time Budgeted Cost Budgeted Cost/Wk Activity Predecessor (weeks) (x $1000) (x $1000) A - 1 8 8 B A 2 12 6 C A 4 24 6 D B 5 25 5 E C 7 21 3 F D,E 1 5 5 A-O-A diagram 3 B,2 D,5 1 A,1 2 5 F,1 6 C,4 E,7 4 A-O-N diagram 1 3 5 7 3 8 0 1 4 0 7 12 12 13 0 1 B,2 D, 4 4 12 13 0 0 0 0 5 A, F,1 1 C,4 E,7 5 12 1 5 5 12 1 5 0 0 0 0 Results of start, finish, and slack time calculations Total Free Activity ES EF LS LF Slack Slack A 0 1 0 1 0 0 B 1 3 5 7 4 0 C 1 5 1 5 0 0 D 3 8 7 12 4 4 E 5 12 5 12 0 0 F 12 13 12 13 0 0
  • ISM 6407.0001 Spring 2007 Topic 2: Project Management Page 3/26 PROJECT MANAGEMENT: SUMMARY NOTES ON START, FINISH, AND SLACK TIME CALCULATIONS Forward Pass When making a forward pass to find earliest start times and earliest finish times, use caution when an activity has more than one immediate predecessor. The earliest start time for this activity will be the largest earliest finish time of all the activity’s immediate predecessors. Backward Pass When making a backward pass to find latest start times and latest finish times, use caution when an activity has more than one immediate successor. The latest finish time for this activity will be the smallest latest start time of all the activity’s immediate successors. Slack Our textbook uses the simple term “slack.” We will be referring to two types of slack, “total slack” and “free slack.” In our textbook, any references to the term “slack” will be identical to what we will call “total slack.” Total Slack: Total slack is a measure of how much we can delay the start of an activity and still not delay the completion of the total project. An activity’s total slack is calculated as the difference between its latest start time and its earliest start time, or TS = LS – ES Free Slack: We also described another measure of slack, one that we will call “free slack.” This concept is not mentioned in our textbook, however, since it is sometimes used in the course of project management, one should be familiar with the notion that there is more than one way to measure the amount of “cushion” that exists in a project’s schedule. Free slack is a measure of how much we can delay the completion of an activity and still not delay the start of any of that activity’s immediate successors. An activity’s free slack is calculated as the difference between its earliest finish time and the smallest of the earliest start time for its immediate successors.
  • ISM 6407.0001 Spring 2007 Topic 2: Project Management Page 4/26 GANTT CHARTS FOR PROJECT MANAGEMENT DEMONSTRATION PROBLEM Gantt Chart for Demo Problem Using Earliest Start and Finish Times end of wk→ 0 1 2 3 4 5 6 7 8 9 10 11 12 13 wk→ 1 2 3 4 5 6 7 8 9 10 11 12 13 A 0 A 1 B 1 B 3 C 1 C 5 D 3 D 8 E 5 E 12 F 12 F 13 Gantt Chart for Demo Problem Using Latest Start and Finish Times end of wk→ 0 1 2 3 4 5 6 7 8 9 10 11 12 13 wk→ 1 2 3 4 5 6 7 8 9 10 11 12 13 A 0 A 1 B 5 B 7 C 1 C 5 D 7 D 12 E 5 E 12 F 12 F 13
  • ISM 6407.0001 Spring 2007 Topic 2: Project Management Page 5/26 Budgeted Cost Using Earliest Start Times End of week #→ 0 1 2 3 4 5 6 7 8 9 10 11 12 13 week #→ 1 2 3 4 5 6 7 8 9 10 11 12 13 Total A 8 8 B 6 6 12 C 6 6 6 6 24 D 5 5 5 5 5 25 E 3 3 3 3 3 3 3 21 F 5 5 Tot/wk 8 12 12 11 11 8 8 8 3 3 3 3 5 95 Cum. Tot. 8 20 32 43 54 62 70 78 81 84 87 90 95 Budgeted Cost Using Latest Start Times End of week #→ 0 1 2 3 4 5 6 7 8 9 10 11 12 13 week #→ 1 2 3 4 5 6 7 8 9 10 11 12 13 Total A 8 8 B 6 6 12 C 6 6 6 6 24 D 5 5 5 5 5 25 E 3 3 3 3 3 3 3 21 F 5 5 Tot/wk 8 6 6 6 6 9 9 8 8 8 8 8 5 95 Cum. Tot. 8 14 20 26 32 41 50 58 66 74 82 90 95
  • ISM 6407.0001 Spring 2007 Topic 2: Project Management Page 6/26
  • ISM 6407.0001 Spring 2007 Topic 2: Project Management Page 7/26 MONITORING AND CONTROLLING BUDGET ASPECTS OF PROJECT MANAGEMENT DEMONSTRATION PROBLEM Assume that we are at the end of week 6 and the following activity has transpired: Activity A is 100% complete, and we have spent $7,000 on it Activity B is 50% complete, and we have spent $8,000 on it Activity C is 100% complete, and we have spent $22,000 on it Activity D has not yet been started, and we have spent nothing on it Activity E has not yet been started, and we have spent nothing on it Activity F has not yet been started, and we have spent nothing on it The following table would summarize our budgetary activity: Total % Value of Actual Activity Type of Activity Budget Completed Work Done Cost Difference Difference A $8,000 100% $8,000 $7,000 -$1,000 Under budget B $12,000 50% $6,000 $8,000 $2,000 Over budget C $24,000 100% $24,000 $22,000 -$2,000 Under budget D $25,000 0% 0 0 0 On budget E $21,000 0% 0 0 0 On budget F $5,000 0% 0 0 0 On budget For the project as a whole $38,000 $37,000 -$1,000 Under budget Note that we are over budget on activity B, under budget on activities A and C, and right on budget for the remaining activities (D, E, and F). Overall for the project, thus far we are $1,000 under budget. That sounds good, but close inspection reveals that we are behind schedule and not likely to complete by the target time of 13. Even using latest start and finish times, note the following (remembering that we are at the end of week 6 in the project): A is O.K. (It is finished and should be done by now.) B is O.K. (It is half finished, and that is consistent with the LS and LF times for it.) C is O.K. (It is finished and should be done by now.) D is O.K (It has not yet been started, but its LS reveals that it could be started as late as the end of week 7.) E will cause a problem (we should have been 1 week into E, but at this point in time we haven’t even started it. This is a critical path activity, so unless we can make up some time on this, we will not complete the project by the end of week 13.) This type of monitoring would cause some fancy footwork among project managers. They might resort to shifting company resources to E in an effort to accelerate E and get the project back on track. This whole issue of shifting resources and accelerating activities will be addressed in later class meetings devoted to project management.
  • ISM 6407.0001 Spring 2007 Topic 2: Project Management Page 8/26 DETERMINISTIC (NO UNCERTAINTY) PERT/COST-- SIMPLE EXAMPLE OF ACCELERATING ACTIVITIES Normal Crash Immediate Time Cost Time Cost Acceleration Activity Predecessor (weeks) (x $1000) (weeks) (x $1000) Cost (per week) A - 1 8 1 8 - B A 2 12 1 16 4 C A 4 24 1 27 1 D B 5 25 4 27 2 E C 7 21 3 33 3 F D,E 1 5 1 5 - Path Time (using normal times) ABDF 9 ACEF 13 Time Total Cost (weeks) Activities Accelerated (x$1000) None (everything done 13 95 on a normal basis) 12 C (1st week) 95+1=96 11 C (2nd week) 96+1=97 10 C (3rd week) 97+1=98 9 E (1st week) 98+3=101 8 E (2nd week) & D 101+3+2=106 7 E (3rd week) & B 106+3+4=113 6 Can’t be done!
  • ISM 6407.0001 Spring 2007 Topic 2: Project Management Page 9/26 DETERMINISTIC (NO UNCERTAINTY) PERT/COST-- SOME ADDITIONAL PROBLEM SCENARIOS Embellishments to problem on page 6, Module 2. Normal Crash Immediate Time Cost Time Cost Acceleration Activity Predecessor (weeks) (x $1000) (weeks) (x $1000) Cost (per week) A - 1 8 1 8 - B A 2 12 1 16 4 C A 4 24 1 27 1 D B 5 25 4 27 2 E C 7 21 3 33 3 F D,E 1 5 1 5 - Path Time (using normal times) ABDF 9 ACEF 13 Embellishment #1. Suppose the client offers to give us a $2,000 bonus for each week we come in under 13 weeks. We can see with incremental analysis on the total cost column that it is worth spending an extra $1000 to cut the project to 12 weeks (we get a bonus of $2000). It is worth spending another $1000 to cut the project to 11 weeks (we get another bonus of $2000). The same holds true for cutting the project to 10 weeks. However, if we want to cut the project to 9 weeks we have to spend another $3000. This is not worth it, for it will only return a bonus of another $2000. We would come to the same conclusion if we placed another column on the table and tallied total cost minus bonus ($ outflow - $ inflow) and then picked the time with the lowest total. Time Total Cost Total Cost – Bonus (weeks) Activities Accelerated (x$1000) (x$1000) None (everything done 13 95 95 on a normal basis) 12 C (1st week) 95+1=96 96-2=94 11 C (2nd week) 96+1=97 97-4=93 10 C (3rd week) 97+1=98 98-6=92 (the least costly option) 9 E (1st week) 98+3=101 101-8=93 8 E (2nd week) & D 101+3+2=106 106-10=96 7 E (3rd week) & B 106+3+4=113 113-12=101 6 Can’t be done!
  • ISM 6407.0001 Spring 2007 Topic 2: Project Management Page 10/26 Embellishment #2. Ignore the previous scenario. Suppose the deal we strike with the client calls for a penalty of $2000 for each week the project is delayed beyond week 8 (apparently the client would really like delivery of the finished product by that time). We could do an incremental analysis and decide that it is wise to shorten from 13 to 12 (spend $1000, avoid $2000 penalty), wise to shorten from 12 to 11 (spend another $1000, avoid another $2000 penalty), wise to shorten from 11 to 10 (spend another $1000, avoid another $2000 penalty), but unwise to shorten from 10 to 9 (spend another $3000, avoid another $2000 penalty). We would come to the same conclusion if we placed another column on the table and tallied the total cost plus the penalty cost, giving us a slightly different look to our table from page 16 of the course pack. Time Total Cost Total Cost + Penalty Cost (weeks) Activities Accelerated (x$1000) (x$1000) None (everything done 13 95 95+10=105 on a normal basis) 12 C (1st week) 95+1=96 96+8=104 11 C (2nd week) 96+1=97 97+6=103 10 C (3rd week) 97+1=98 98+4+102 (the least costly option) 9 E (1st week) 98+3=101 101+2=103 8 E (2nd week) & D 101+3+2=106 106 7 E (3rd week) & B 106+3+4=113 113 6 Can’t be done!
  • ISM 6407.0001 Spring 2007 Topic 2: Project Management Page 11/26 Nonlinear tradeoff embellishment to the problem on page 6, Module 2. Normal Crash Immediate Time Cost Time Max. Acceleration Cost per Activity Predecessor (weeks) (x $1000) (weeks) Accel. week (x$1000) A - 1 8 1 0 - B A 2 12 1 1 4 C A 4 24 1 3 1, 2, 7 D B 5 25 4 1 2 E C 7 21 3 4 3, 3, 4, 8 F D,E 1 5 1 0 - Path Time (using normal times) ABDF 9 ACEF 13 The following table shows the full range of possibilities for project time and project cost. We could embellish this nonlinear situation with bonuses or penalties much like we did the linear case above. Time Total Cost (weeks) Activities Accelerated (x$1000) None (everything done 13 95 on a normal basis) 12 C (1st week) 95+1=96 11 C (2nd week) 96+2=98 10 E (1st week) 98+3=101 9 E (2nd week) 101+3=104 8 E (3rd week) & D 104+4+2=110 7 C (3rd week) & B 110+7+4=121 6 Can’t be done!
  • ISM 6407.0001 Spring 2007 Topic 2: Project Management Page 12/26 ANOTHER EXAMPLE OF PERT/COST AND ACTIVITY ACCELERATION Normal Crash Immediate Time Time Max. Acceleration Cost Activity Predecessor (weeks) Cost (weeks) Accel. per week A - 4 $4,000 2 2 $100, $350 B A 3 $3,000 2 1 $200 C A 4 $3,000 2 2 $200, $200 D B, C 6 $6,000 4 2 $1200, $1600 B,3 A, D, 4 6 C,4 Path Time ABD 13 ACD 14 Time Total Cost (weeks) Activities Accelerated (x$1000) None (everything done 14 $16,000 on a normal basis) 13 A (1st) $16,000 + $100 = $16,100 12 C (1st) $16,100 + $200 = $16,300 11 A (2nd) $16,300 + $350 = $16,650 B (1st) 10 $16,650 + $200 + $200 = $17,050 C (2nd) 9 D (1st) $17,050 + $1,200 = $18,250 8 D (2nd) $18,250 + $1,600 = $19,850 7 Can’t be done! -----------
  • ISM 6407.0001 Spring 2007 Topic 2: Project Management Page 13/26 PROJECT SCHEDULING WITH UNCERTAINTY IN TIME ESTIMATES Data for initial view. (Probabilistic time estimates) Optimistic Most likely Pessimistic Expected Immediate time time time time Variance Activity Predecessor a m b te σ2 A - .5 1 1.5 1 1/36 B A 1 2 3 2 4/36 C A 3 4 5 4 4/36 D B 3 5 7 5 16/36 E C 6 7 14 8 64/36 F D,E .5 1 1.5 1 1/36 (a + 4m + b) b-a 2 te = 6 σ2 = ( 6 ) Results of start, finish, and slack time calculations using mean time Total Free Activity ES EF LS LF Slack Slack A 0 1 0 1 0 0 B 1 3 6 8 5 0 C 1 5 1 5 0 0 D 3 8 8 13 5 5 E 5 13 5 13 0 0 F 13 14 13 14 0 0 Path Expected time, Te Variance, σ2 Standard Deviation, σ ABDF 9 ACEF 14 70/36 1.3944 or 1.9444 Critical path (ACEF) has an expected time of 14 and standard deviation of 1.3944
  • ISM 6407.0001 Spring 2007 Topic 2: Project Management Page 14/26 SOME ILLUSTRATIONS OF PROBABILITY CALCULATIONS WITH OUR DEMONSTRATION PROJECT MANAGEMENT PROBLEM Recall that the demonstration problem had a critical path with a mean time, t, of 14 weeks and a standard deviation, σ, of 1.394 1. What is the probability that the project can be completed within 16 weeks (i.e., 16 weeks or less)? z = (16-14)/1.394 = 1.43 Probability to the left of this z from the normal table = .92364 2. What is the probability that the project can be completed within 18 weeks (i.e., 18 weeks or less)? z = (18-14)/1.394 = 2.87 Probability to the left of this z from the normal table = .99795 3. What is the probability that the project can be completed within 13 weeks (i.e., 13 weeks or less)? z = (13-14)/1.394 = -.72 There are no negative z values in the table, but we can make use of the symmetry in the normal curve. Probability to the left of z = .72 from the normal table = .76424 Probability in the tail to the right of z = .72 is 1-.76424 = .23576 This tail to the right of z = .72 is exactly the same size as the tail to the left of z = -.72 Therefore, the probability of completing the project with in 13 weeks is .23576 4. What is the probability that the project will be completed in somewhere between 16 weeks and 18 weeks? This probability will be the area between a z value of 1.43 (the z for 16 weeks from above) and a z value of 2.87 (the z for 18 weeks from above). This probability is .99795-.92364 = .07431 5. What is the probability of being able to start activity F by week 14 of the project? Before you can start F you must finish all the activities on the network that lead up to F. You can treat those activities as a mini-project with two paths (ABD with a mean time of 8 and ACE with a mean time of 13). The critical path here is ACE, and it has a variance of 1.91667 (i.e., 69/36). The standard deviation would be the square root of this, or standard deviation = 1.384 z = (14-13)/1.384 = .72 Probability to the left of this z value from the normal table = .76424
  • ISM 6407.0001 Spring 2007 Topic 2: Project Management Page 15/26 A TWIST ON PROBABILISTIC PERT Return to our demonstration problem, which had a mean time of 14 and a standard deviation of 1.394 for the completion of the project. Suppose that, as a project manager, you are wondering what promise completion date would have about an 80% chance of being achieved. To solve this you would have to find the z value that corresponds to an 80% probability (which means you use the z table in a reverse mode). Search around in the z table to find a probability of .8000 (or one pretty close to that value). If you go into the table, you can find a probability of .79955 (that’s pretty close to .8000, so we’ll use that rather than interpolating between values in the table). The z value that corresponds to this probability is .84 The only unknown in the z calculation is the promise date (or due date, as the book calls it): Z = (due date – mean)/standard deviation or .84 = (due date – 14)/1.394 solve this and you find that due date = 14 + (.84)(1.394) = 15.17 weeks (which is a little over 15 weeks and 1 day)
  • ISM 6407.0001 Spring 2007 Topic 2: Project Management Page 16/26 AND YET ANOTHER TWIST ON PROBABILISTIC PERT Return to our demonstration problem, which had a mean time of 14 and a standard deviation of 1.394 for the completion of the project. Suppose that the potential customer has a strong desire to have this project completed by week 13. We have already seen (page 12 of this module) that there is only a .23576 probability (i.e., a 23.576% chance) that we will be finished by week 13. We have decided that we would like to promise that the project will be done by week 13, but we would like to improve the odds that we can accomplish this to about 98.5%. In order to increase the odds, we are going to have to shorten the project mean time by accelerating some activities. This should be apparent by looking at the formula for the Z calculation: Z = (due date – project mean time)/σ In this formula we know the Z value (Z = 2.17, which is the Z value that would correspond to a probability of .9850), we know the due date (this is the 13 weeks that the customer wants), we know the standard deviation (σ = 1.394). What we don’t know is what the project mean time should be reduced to in order to satisfy the equation. This is a simple calculation, as follows: 2.17 = (13 – project mean time)/1.394 (2.17)(1.394) = 13 – project mean time project mean time = 13 – (2.17)(1.394) = 13 – 3.02 = 9.98 weeks (let’s call it 10 weeks) So, the project mean time (which was 14 weeks if we did no acceleration) will have to be reduced to 10 weeks. This means we will have to accelerate the project a total of 4 weeks. To do that, we do as we did with our earlier acceleration exercises, i.e., shorten the critical path(s) down to 10 weeks. One question that often comes up in this context is, “Won’t the process of accelerating activities change the standard deviation in addition to the mean?” The answer is NO. Accelerating activities will lower the mean time for the critical path, but it will have no impact on the standard deviation. Here is the reason: The three time estimates we made for each activity were based on the premise that we would work on these activities at a normal pace (9-5 Monday through Friday, no overtime, no extra shifts, no extra equipment, etc.). If we choose to work at a faster pace (using some of these options), we still are not sure how long each activity will take, so we will still have three time estimates; they will just be lower across the board. So, the impact of accelerating an activity will be to lower its three time estimates across the board. This will certainly lower the mean times for accelerated activities, but their variances will remain the same due to the computational formula (the difference between the pessimistic and optimistic times is the same before and after the acceleration, so the variance calculation will yield the same result).
  • ISM 6407.0001 Spring 2007 Topic 2: Project Management Page 17/26 PROJECT MANAGEMENT – REVIEW PROBLEMS USING SAME NETWORK STRUCTURE Problem 1: Deterministic Mode (No Uncertainty in the Time Estimates) Data gathered in preliminary steps are in bold face. Nonbold figures were calculated using these initial estimates. Immediate Time Budgeted Cost Budgeted Cost/Wk Activity Predecessor (weeks) (x $1000) (x $1000) A - 2 4 2 B A 3 12 4 C A 2 6 3 D A 2 8 4 E B 1 5 5 F C 3 9 3 G D 2 10 5 H E, F, G 2 4 2 Resulting project network is drawn on next page Results of start, finish, and slack time calculations Total Free Activity ES EF LS LF Slack Slack A 0 2 0 2 0 0 B 2 5 3 6 1 0 C 2 4 2 4 0 0 D 2 4 3 5 1 0 E 5 6 6 7 1 1 F 4 7 4 7 0 0 G 4 6 5 7 1 1 H 7 9 7 9 0 - Budgeted Cost Using Earliest Start and Finish Times End of week #→ 0 1 2 3 4 5 6 7 8 9 10 11 12 13 week #→ 1 2 3 4 5 6 7 8 9 10 11 12 13 Total A 2 2 4 B 4 4 4 12 C 3 3 6 D 4 4 8 E 5 5 F 3 3 3 9 G 5 5 10 H 2 2 4 Tot/wk 2 2 11 11 12 13 3 2 2 58 Cum. Tot. 2 4 15 26 38 51 54 56 58
  • ISM 6407.0001 Spring 2007 Topic 2: Project Management Page 18/26 Budgeted Cost Using Latest Start and Finish Times End of week #→ 0 1 2 3 4 5 6 7 8 9 10 11 12 13 week #→ 1 2 3 4 5 6 7 8 9 10 11 12 13 Total A 2 2 4 B 4 4 4 12 C 3 3 6 D 4 4 8 E 5 5 F 3 3 3 9 G 5 5 10 H 2 2 4 Tot/wk 2 2 3 11 11 12 13 2 2 58 Cum. Tot. 2 4 7 18 29 41 54 56 58 A-O-A diagram B E A C F H D G A-O-N diagram B E A C F H D G Path Time ABEH 8 ACFH 9 * Critical Path ADGH 8
  • ISM 6407.0001 Spring 2007 Topic 2: Project Management Page 19/26 Problem 2: Introduce Data for Time/Cost Tradeoffs (Still Deterministic Mode) Data gathered in preliminary steps are in bold face. Nonbold figures were calculated using these initial estimates. Normal Crash Immediate Time Cost Time Cost Acceleration Activity Predecessor (weeks) (x $1000) (weeks) (x $1000) Cost (per week) A - 2 4 2 4 - B A 3 12 2 18 6 C A 2 6 1 8 2 D A 2 8 2 8 - E B 1 5 1 5 - F C 3 9 2 13 4 G D 2 10 1 15 5 H E, F, G 2 4 1 16 12 Path Time ABEH 8 ACFH 9 * Critical Path ADGH 8 Find the activities to accelerate and the project cost if we were to try to get the project done in 9 weeks, 8 weeks, and 7 weeks. Time Total Cost (weeks) Activities Accelerated (x$1000) None (everything done on 9 58 a normal basis) Must reduce path ACFH 8 58 + 2 = 60 Choose C (1st and only week of acceleration) Must reduce paths ABEH, ACFH, and ADGH 7 60 + 12 = 72 st Choose H (1 and only week of acceleration)
  • ISM 6407.0001 Spring 2007 Topic 2: Project Management Page 20/26 Problem 3: No Time/Cost Tradeoffs, but We Have Probabilistic (Stochastic) Mode (There is Uncertainty in the Time Estimates) Data gathered in preliminary steps are in bold face. Nonbold figures were calculated using these initial estimates. Optimistic Most likely Pessimistic Expected Immediate time time time time Variance Activity Predecessor a m b te σ2 A - 1 2 3 2 4/36 B A 2 3 4 3 4/36 C A 1 2 6 2.5 25/36 D A 1 2 3 2 4/36 E B .5 1 1.5 1 1/36 F C 1 3 8 3.5 49/36 G D 1 2 3 2 4/36 H E, F, G 1 2 3 2 4/36 Time Path (using expected times) ABEH 8 ACFH 10* Critical Path σ2 = 2.28 σ = 1.51 ADGH 8 Find: a. Probability of completing the project within 12 weeks. Z = (12-10)/1.51 = 1.32 Probability = .9066 b. Probability that it will take more than 11 weeks to complete the project. Z = (11-10)/1.51 = .66 Probability = .2546 (note that we want the area in the right tail) c. Probability that the project will be completed in 7 weeks or less. Z = (7-10)/1.51 = -1.99 Probability = .0233 (we must use symmetry of table to get this)
  • ISM 6407.0001 Spring 2007 Topic 2: Project Management Page 21/26 SOLUTIONS TO SOME TEXTBOOK PROJECT MANAGEMENT PROBLEMS 13-12 & 13-13 Critical path is BDEG; Time is 26. Network diagram using activity on node convention A E B D G F C Network diagram using activity on arrow convention 2 B D E G 1 A 3 5 6 F C 4 13-14 & 13-15 Critical path is BDEG; Time is 17 Network diagram using activity on node convention A F C G E B D
  • ISM 6407.0001 Spring 2007 Topic 2: Project Management Page 22/26 Network diagram using activity on arrow convention 2 F A C 1 4 5 6 E G B D 3 13-16 & 13-17 Critical paths are ACG and BEG; Time is 19 Network diagram using activity on node convention C A G D B E H F Network diagram using activity on arrow convention C 2 4 A G D 1 6 E B H 3 5 F
  • ISM 6407.0001 Spring 2007 Topic 2: Project Management Page 23/26 13.19 Critical path is ADFHJK, Mean time is 68.7, standard deviation. is 3.5; P(70)=.6443; P(80)=.9994; P(90)=.9999 Network diagram using activity on node convention I A D F H B K J G C E L Network diagram using activity on arrow convention 2 8 D A L G J K 1 4 6 7 B H C 3 E F 5 I 13-21 Activity Total Budgeted Percent of Value of Work Actual Cost Activity Cost Completion Completed Difference A 22,000 100 22,000 20,000 -2,000 B 30,000 100 30,000 36,000 +6,000 C 26,000 100 26,000 26,000 0 D 48,000 100 48,000 44,000 -4,000 E 56,000 50 28,000 25,000 -3,000 F 30,000 60 18,000 15,000 -3,000 G 80,000 10 8,000 5,000 -3,000 H 16,000 10 1,600 1,000 -600 After 8 weeks: Value of work completed = $181,600 Actual cost = $172,000 Cost under run = $9,600
  • ISM 6407.0001 Spring 2007 Topic 2: Project Management Page 24/26 13-23 Path Normal Time Time Activities Accelerated Total Cost ACFH 9 15 None $308,000 ACEGH 15 13 1A+1C+1D or 1A+1E+1D or 1C+1E+1D or $311,000 BDGH 14 2E+1D or 1A+1G or 1C+1G or 1E+1G Network diagram using activity on node convention F A C H E B D G Network diagram using activity on arrow convention C 2 4 F A E H 1 6 7 G B 3 5 D 13-24 Path Normal Time Time Activities Accelerated Cost ADG 14 14 None $10,950 BEG 12 13 1st D $10,950+75 = $11,025 CF 3 12 2nd D $11,025+75 = $11,110 11 3rd D & 1st E $11,110+75+50 = $11,225 10 4th D & 2nd E $11,225+75+50 = $11,350
  • ISM 6407.0001 Spring 2007 Topic 2: Project Management Page 25/26 Network diagram using activity on node convention A D G B E C F Network diagram using activity on arrow diagram 2 A D B E G 1 4 5 6 C 3 F
  • ISM 6407.0001 Spring 2007 Topic 2: Project Management Page 26/26 HERE IS SOME ADDITIONAL DETAIL ON THE SOLUTIONS TO PROBLEMS 13-13 AND 13-15 13-13 Critical Path Time Path? AEG 13 BDEG 26 * C.P. CFG 15 Critical Activity Time ES EF LS LF TS FS Activity? A 2 0 2 13 15 13 13 no B 5 0 5 0 5 0 0 yes C 1 0 1 11 12 11 0 no D 10 5 15 5 15 0 0 yes E 3 15 18 15 18 0 0 yes F 6 1 7 12 18 11 11 no G 8 18 26 18 26 0 - yes 13-15 Critical Path Time Path? AFG 12 ACEG 15 BDEG 17 * C.P Critical Activity Time ES EF LS LF TS FS Activity? A 3 0 3 2 5 2 0 No B 7 0 7 0 7 0 0 Yes C 4 3 7 5 9 2 2 No D 2 7 9 7 9 0 0 Yes E 5 9 14 9 14 0 0 Yes F 6 3 9 8 14 5 5 No G 3 14 17 14 17 0 - Yes