Project Management Techniques

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Project Management Techniques

  1. 1. Project Management
  2. 2. Project Management Techniques
  3. 3. Project Management <ul><li>Questions: </li></ul><ul><li>Why do we need to study Project Management? </li></ul><ul><li>How does a project management technique work? </li></ul>( to p4) ( to p5)
  4. 4. Objective <ul><li>The main purpose is to govern the operations of a project such that all activities involved are well administrated and that we can also control its completion time </li></ul>( to p3)
  5. 5. Project management technique <ul><li>Steps to solve a project management </li></ul><ul><li>problem: </li></ul><ul><li>to represent a ‘project problem” graphically </li></ul><ul><li>to determine its completion time </li></ul><ul><li>to carry out sensitivity analysis, if any </li></ul>( to p6) ( to p12) ( to p29)
  6. 6. 1. Represent a ‘project problem” graphically <ul><li>Steps: </li></ul><ul><li>Gather all information and organize them in a table format that consists of: event, processing time, and precedent constraints as follows: </li></ul><ul><li>Draw a semantic network to represent them </li></ul><ul><li>Special case! </li></ul>( to p7) ( to p9) ( to p4) -- A B 20 30 10 A B C Precedent constraints Processing Time Event
  7. 7. Semantic network to represent them <ul><li>Here, we use three symbols: </li></ul><ul><li>node to represent stage </li></ul><ul><li>line/branch to represent event </li></ul><ul><li>arrow to represent precedent </li></ul><ul><li>constraint </li></ul><ul><li>Example </li></ul>( to p8) ( to p6)
  8. 8. Example 1 2 3 A 4 C B 20 30 10 Rule1: All nodes must starts from one Node and ends with one node ( to p7) -- A B 20 30 10 A B C 1-2 2-3 3-4 Pred Const Proc Time Event Path
  9. 9. Special case! <ul><li>When two or events taken places in the same time interval </li></ul><ul><li>(known an concurrent events) </li></ul><ul><li>Consider the following example! </li></ul><ul><li>How to draw it? </li></ul>( to p10) -- A A 3 5 7 A B C Precedent constraints Processing Time Event
  10. 10. Case 1 1 2 3 A B C 3 5 7 Wrong! Rule2: no node can have two outcomes and end with the same note Solution ( to p11)
  11. 11. Solutions for Rule 2 <ul><li>Three ways to draw it: </li></ul>1 2 3 4 5 A B C Dummy 1=0 Dummy 2 = 0 1 2 3 4 A B C Dummy = 0 1 2 3 4 A B C Dummy = 0 Solution 1: Solution 2: Solution 3: What one is better? A dummy activity shows a precedence relationship Reflects no processing time ( to p6)
  12. 12. 2. Determine its completion time <ul><li>Consider the project network as shown in next slide </li></ul><ul><li>Question: Is it an easy way to find out the </li></ul><ul><li> solution? </li></ul><ul><li>Answer: YES, it knows as </li></ul><ul><li>Critical Path Method (CPM) </li></ul>( to p15) ( to p13)
  13. 13. The Project Network All Possible Paths for Obtaining a Solution Figure 8.3 Expanded network for building a house showing concurrent activities. Table 8.1 Possible Paths to complete the House-Building Network Then the completion time for paths A, B, C and D can be computed as ( to p14)
  14. 14. The Project Network Completion time for: path A: 1  2  3  4  6  7, 3 + 2 + 0 + 3 + 1 = 9 months (Critical Path) path B: 1  2  3  4  5  6  7, 3 + 2 + 0 + 1 + 1 + 1 = 8 months path C: 1  2  4  6  7, 3 + 1 + 3 + 1 = 8 months path D: 1  2  4  5  6  7, 3 + 1 + 1 + 1 + 1 = 7 months The critical path is the longest path through the network; the minimum time the network can be completed. Figure 8.5 Alternative paths in the network This is the Solution ! ( to p12)
  15. 15. Critical Path Method (CPM) <ul><li>General concepts: </li></ul><ul><ul><li>For each branch of the project network, we firstly determine four values of ES, EF, LS and LF </li></ul></ul><ul><ul><li>For each branch, we compute their slack time, </li></ul></ul><ul><ul><ul><li>Slack time = (LS-ES) or (LF-EF) </li></ul></ul></ul><ul><ul><li>The critical path is located at branch that has </li></ul></ul><ul><ul><li>slack time = 0 </li></ul></ul><ul><ul><li>(Do you know the reason why?) </li></ul></ul><ul><ul><li>How it works? </li></ul></ul>( to p16)
  16. 16. How CPM works? <ul><li>Steps: </li></ul><ul><li>Prepare the project network </li></ul><ul><li>Construct a table as follows: </li></ul><ul><li>Compute ES and EF </li></ul><ul><li>Compute LS and LF </li></ul><ul><li>Compute LS-ES or LF-EF </li></ul>ES ij = max (EF i ) EF ij = ES i + t ij with EF 1 =0 Critical path when LS-ES=0 ( to p4) ( to p17) ( to p26) ( to p22) LF LS EF ES Branch
  17. 17. Compute ES and EF <ul><li>Note: </li></ul><ul><li>When computing these values, the pattern is like moving zic-zac format by firstly computer ES 12 and then adding it to EF 12 and move to next branch by copying the max values of the branch 1-2 to say, 2-3 </li></ul><ul><li>We compute them from top to bottom! </li></ul><ul><li>Their relationship : </li></ul><ul><li>Example 1: </li></ul>( to p18) ( to p22) ( to p19)
  18. 18. The starting point of ES and EF <ul><li>Consider: </li></ul><ul><li>Then </li></ul><ul><li>EF 1 = 0 </li></ul><ul><li>ES 12 = max (EF 1 ) EF 12 = ES 12 + t 12 </li></ul><ul><li>= 0 = 0 + t 12 </li></ul>1 2 t 12 ( to p17)
  19. 19. The overall computation is shown in next slide ( to p20) EF 12 =ES 12 +t 12 = EF 23 =ES 23 +t 23 = EF 24 = EF 34 = EF 45 = EF 46 = EF 56 = EF 67 = ES 12 = max(EF 1 )= ES 23 =max(EF 2 )= ES 24 =max(EF 2 )= ES 34 =max(EF 3 )= ES 45 =max(EF 4 )= ES 46 =max(EF 4 )= ES 56 =max(EF 5 )= ES 67 =max(EF 6 )= 1-2 2-3 2-4 3-4 4-5 4-6 5-6 6-7 EF ij =ES ij +t ij ES ij = max(EF i ) Branches
  20. 20. - ES is the earliest time an activity can start. ES ij = Maximum (EF i ) - EF is the earliest start time plus the activity time. EF ij = ES ij + t ij (note: you can compute these values and show in the network diagram as well) Add all t to note 4 and take the longest time Max (node 3+t34, node2+t24) max (5+0, 3+1) =max(5,4)=5 add all t i for note 2 Max(node4+t46,node5+t56 =max(5+3,5+1)=8 Complete solution ( to p4) ( to p21)
  21. 21. The Project Network Activity Scheduling- Earliest Times - ES is the earliest time an activity can start. ES ij = Maximum (EF i ) - EF is the earliest start time plus the activity time. EF ij = ES ij + t ij Figure 8.6 Earliest activity start and finish times ( to p20)
  22. 22. Compute LS and LF <ul><li>Note: We compute these values from the bottom to top, with assigning: </li></ul><ul><li>LS ij = LF i -t ij LF ij = min LS j </li></ul><ul><li>with </li></ul><ul><li>the end of LF ij = EF ij </li></ul><ul><li>Example: computing Figure 8.3 </li></ul>( to p23)
  23. 23. The overall computational is shown in next slide ( to p24) LF 12 =min(LS 2 )= LF 23 =min(LS 3 )= LF 24 =min(LS 4 )= LF 34 =min(LS 4 )= LF 45 =min(LS 5 )= LF 46 =min(LS 6 )= LF 56 =min(LS 6 )= LF 67 =min(LS 7 )= LS 12 = L i12 -t 12 = LS 23 = LF 23 -t 23 = LS 24 = LF 24 -t 24 = LS 34 = LF 34 -t 34 = LS 45 = LF 45 -t 45 = LS 46 = LF 46 - i46 = LS 56 = LF 56 -t 56 = LS 67 = LF 67 -t 67 = 1-2 2-3 2-4 3-4 4-5 4-6 5-6 6-7 LF ij =min(LS j ) LS ij = LF ij -t ij Branches
  24. 24. - LS is the latest time an activity can start without delaying critical path time. LS ij = LF ij - t ij - LF is the latest finish time LF ij = Minimum (LS j ) Start with the end node first Same as EF67 from the previous slide Again, you can place these values onto the branches Min(node 6-t46,node5-t45) =Min(8-3,7-1) =Min(5,6)=5 Min(node3-t23,node4-t24) =Min(5-2,5-1)=Min(3,4)=3 Min(node 7-t67) =Min(9-1)=8 ( to p25) ( to p22)
  25. 25. The Project Network Activity Scheduling - Latest Times - LS is the latest time an activity can start without delaying critical path time. LS ij = LF ij - t ij - LF is the latest finish time LF ij = Minimum (LS j ) Figure 8.7 Latest activity start and finish times ( to p24)
  26. 26. Compute LS-ES or LF-EF <ul><li>Two ways you can achieve it: </li></ul><ul><li>by compiling slack, S ij </li></ul><ul><li>by showing branches </li></ul>( to p27) ( to p28) ( to p16)
  27. 27. The Project Network Calculating Activity Slack Time - Slack, S ij , computed as follows: S ij = LS ij - ES ij or S ij = LF ij - EF ij Table 8.2 Activity Slack   Figure 8.9 Activity Slack * What does it mean? ( to p26)
  28. 28. The Project Network Activity Slack <ul><li>Slack is the amount of time an activity can be delayed without delaying the project. </li></ul><ul><li>Slack time exists for those activities not on the critical path for which the earliest and latest start times are not equal. </li></ul><ul><li>Shared slack is slack available for a sequence of activities. </li></ul>Figure 8.8 Earliest activity start and finish times ( to p26)
  29. 29. Sensitivity Analysis <ul><li>Today, we only consider one case – </li></ul><ul><li> “ Probabilistic Activity Times” </li></ul><ul><li>Refer to activity time estimates usually can not be made with certainty </li></ul><ul><li>PERT is known as the solution method </li></ul>( to p30)
  30. 30. PERT <ul><li>In PERT, three different time estimations are applied: </li></ul><ul><li>most likely time (m), </li></ul><ul><li>the optimistic time (a) , and </li></ul><ul><li>the pessimistic time (b). </li></ul><ul><li>How do we make use of these three values? </li></ul>( to p31)
  31. 31. Probabilistic Activity Times <ul><li>We used these values to estimate the mean and variance of a beta distribution: </li></ul><ul><li>mean (expected time): </li></ul><ul><li>variance: </li></ul><ul><li>How to use these values to solve a project network problem? </li></ul>( to p32)
  32. 32. PERT <ul><li>We simply apply t values in CPM and determine the values of: </li></ul><ul><ul><ul><li>ES </li></ul></ul></ul><ul><ul><ul><li>EF </li></ul></ul></ul><ul><ul><ul><li>LS </li></ul></ul></ul><ul><ul><ul><li>LF </li></ul></ul></ul><ul><ul><ul><li>S </li></ul></ul></ul><ul><ul><li>and branches with slack = 0 still consider as critical paths </li></ul></ul><ul><li>Example. </li></ul>( to p33)
  33. 33. Procedures for PERT <ul><li>Step 1: based on the values of a, b and m, determine the t and v values for each path </li></ul><ul><li>Step 2: determine the critical path by using t values in the CPM </li></ul><ul><li>Step 3: compute its corresponding means and standard deviations according. </li></ul><ul><li>Example </li></ul><ul><li>Result implication </li></ul><ul><li>Applications </li></ul>( to p34) ( to p38) ( to p39)
  34. 34. PERT Example <ul><li>Step 1: computer t and v values </li></ul><ul><li>Step 2: determine the CPM </li></ul><ul><li>Step 3: determine v value </li></ul>( to p35) ( to p36) ( to p37) ( to p33)
  35. 35. Step 1: computer t and v values Figure 8.11 Network with mean activity times and variances Table 8.3 Activity Time Estimates for Figure 8.10 ( to p34)
  36. 36. Step 2: determine the CPM Figure 8.12 Earliest and latest activity times Table 8.4 Activity Earliest and Latest Times and Slack ( to p34)
  37. 37. Step 3: determine v value <ul><li>The expected project time is the sum of the expected times of the critical path activities. </li></ul><ul><li>The project variance is the sum of the variances of the critical path activities. </li></ul><ul><li>The expected project time is assumed to be normally distributed (based on central limit theorum). </li></ul><ul><li>In example, expected project time (t p ) and variance (v p ) interpreted as the mean (  ) and variance (  2 ) of a normal distribution: </li></ul><ul><li> = 25 weeks </li></ul><ul><li> 2 = 6.9 weeks </li></ul>( to p34)
  38. 38. Probability Analysis of the Project Network - Using normal distribution, probabilities are determined by computing number of standard deviations (Z) a value is from the mean. - Value is used to find corresponding probability in Table A.1, App. A. Figure 8.13 Normal distribution of network duration Critical value ( to p33)
  39. 39. <ul><li>Consider when </li></ul><ul><ul><ul><li>x = 30 </li></ul></ul></ul><ul><ul><ul><li>x = 22 </li></ul></ul></ul><ul><li>Tutorial Assignment </li></ul>( to p40) ( to p41) ( to p42)
  40. 40. Probability Analysis of the Project Network Example 1 <ul><li> 2 = 6.9  = 2.63 </li></ul><ul><li>Z = (x-  )/  = (30 -25)/2.63 = 1.90 </li></ul><ul><li>Z value of 1.90 corresponds to probability of .4713 in Appendix A of p715. Probability of completing project in 30 weeks or less : (.5000 + .4713) = .9713, </li></ul><ul><li>or 97.13% (Why so high a probability rate?) </li></ul>Figure 8.14 Probability the network will be completed in 30 weeks or less ( to p39)
  41. 41. Probability Analysis of the Project Network Example 2 Z = (22 - 25)/2.63 = -1.14 Z value of 1.14 (ignore negative) corresponds to probability of .3729 in Table A.1, appendix A. Probability that customer will be retained is .1271 (= 0.5- 0.3729) , or 12.71% (Again, why so low probability rate?) Figure 8.15 Probability the network will be completed in 22 weeks or less ( to p39)
  42. 42. Tutorial Assignment <ul><li>Try to use QM to solve CPM/PERT problems (see slide 19) </li></ul><ul><li>Exercises (Chapter 8) </li></ul><ul><ul><li>Old: 8, 10, 17 </li></ul></ul><ul><ul><li>New: 4, 6, 11 </li></ul></ul>(to p43)
  43. 43. Probability Analysis of the Project Network CPM/PERT Analysis with QM for Windows Exhibit 8.1 (to p16)
  44. 44. The Project Network Activity Slack <ul><li>Slack is the amount of time an activity can be delayed without delaying the project. </li></ul><ul><li>Slack time exists for those activities not on the critical path for which the earliest and latest start times are not equal. </li></ul><ul><li>Shared slack is slack available for a sequence of activities. </li></ul>Figure 8.8 Earliest activity start and finish times
  45. 45. The Project Network Calculating Activity Slack Time - Slack, S ij , computed as follows: S ij = LS ij - ES ij or S ij = LF ij - EF ij Table 8.2 Activity Slack   Figure 8.9 Activity Slack *

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