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
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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)
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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. 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)
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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. - 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. 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)
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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. - 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. 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)
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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)
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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. 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)
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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)
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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)
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43. Probability Analysis of the Project Network CPM/PERT Analysis with QM for Windows Exhibit 8.1 (to p16)
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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 *
