Topics to cover in 2 nd part 1

Topics to cover in 2 nd part 1 (to p 2)

Chapter 8 - Project Management Chapter Topics 2 (to p 3)

Project Management Questions: 1. Why do we need to study Project (to p 4) Management? 2. How does a project management (to p 5) technique work? 3

Objective • 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 4 (to p 3)

Project management technique Steps to solve a project management problem: 1. to represent a ‘project problem” (to p 6) graphically (to p 12) 2. to determine its completion time (to p 29) 3. to carry out sensitivity analysis, if any 5

1. Represent a ‘project problem” graphically Steps: 1. Gather all information and organize them in a table format that consists of: event, processing time, and precedent constraints as follows: 2. Draw a semantic network to represent them (to p 7) Special case! Event Processing Time Precedent constraints A B C 20 30 10 -A B (to p 9) (to p 4) 6

Semantic network to represent them Here, we use three symbols: node to represent stage line/branch to represent event arrow to represent precedent constraint Example (to p 8) (to p 6) 7

Example Path Event Proc Time Pred Const 1 -2 2 -3 3 -4 A B C 20 30 10 -A B A 1 2 20 C B 3 30 4 10 Rule 1: All nodes must starts from one Node and ends with one node (to p 7) 8

Special case! Event Processing Precedent • When two or events Time constraints taken places in the same time interval • (known an concurrent A 3 -events) B 5 A • Consider the following C 7 A example! (to p 10) • How to draw it? 9

Case 1 A 1 B 2 3 5 3 C 7 Wrong! Rule 2: no node can have two outcomes and end with the same note Solution (to p 11) 10

Solutions for Rule 2 A dummy activity shows a precedence relationship Reflects no processing time Three ways to draw it: B Solution 1: A 1 3 C 2 5 4 Solution 2: A 1 2 What one is better? 4 3 B A 1 Dummy 2 = 0 B C Solution 3: Dummy 1=0 2 Dummy = 0 3 C Dummy = 0 (to p 6) 4 11

2. Determine its completion time Consider the project network as shown in next slide (to p 13) Question: Is it an easy way to find out the solution? Answer: YES, it knows as (to p 15) Critical Path Method (CPM) 12

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 (to p 14) Then the completion time for paths A, B, C and D can be computed as 13

The Project Network Completion time for: path A: 1 2 3 4 6 7, 3 + 2 + 0 + 3 + 1 = 9 months (Critical Path) This is the Solution! path B: 1 2 3 4 5 6 7, 3 + 2 + 0 + 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 = 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 (to p 12) 14

Critical Path Method (CPM) • General concepts: – For each branch of the project network, we firstly determine four values of ES, EF, LS and LF – For each branch, we compute their slack time, • Slack time = (LS-ES) or (LF-EF) – The critical path is located at branch that has slack time = 0 (Do you know the reason why? ) How it works? (to p 16) 15

How CPM works? Steps: Branch ES EF LS LF 1. Prepare the project network 2. Construct a table as follows: (to p 17) 3. Compute ES and EF (to p 22) 4. Compute LS and LF 5. Compute LS-ES or LFESij = max (EFi) EF (to p 26) Critical path when LS-ES=0 with EFij = ESi + tij EF 1=0 16 (to p 4)

Compute ES and EF Note: 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 We compute them from top to bottom! (to p 18) Their relationship : (to p 19) (to p 22) Example 1: 17

The starting point of ES and EF Consider: 1 Then t 12 2 EF 1 = 0 ES 12 = max (EF 1) =0 EF 12 = ES 12 + t 12 = 0 + t 12 (to p 17) 18

Branches ESij = max(EFi) EFij=ESij+tij 1 -2 2 -3 2 -4 3 -4 4 -5 4 -6 5 -6 6 -7 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)= EF 12=ES 12+t 12= EF 23=ES 23+t 23= EF 24= EF 34= EF 45= EF 46= EF 56= EF 67= The overall computation is shown in next slide 19 (to p 20)

Complete solution - ES is the earliest time an activity can start. ESij = Maximum (EFi) - EF is the earliest start time plus the activity time. EFij = ESij + tij add all ti for note 2 Add all t to note 4 and take the longest time Max (node 3+t 34, node 2+t 24) max (5+0, 3+1) =max(5, 4)=5 Max(node 4+t 46, node 5+t 56 (note: you can compute these values and show in the network diagram as well) (to p 21) =max(5+3, 5+1)=8 20 (to p 4)

The Project Network Activity Scheduling- Earliest Times - ES is the earliest time an activity can start. ESij = Maximum (EFi) - EF is the earliest start time plus the activity time. EFij = ESij + tij Figure 8. 6 Earliest activity start and finish times (to p 20) 21

Compute LS and LF Note: We compute these values from the bottom to top, with assigning: LSij = LFi -tij LFij = min LSj with the end of LFij = EFij Example: computing Figure 8. 3 (to p 23) 22

Branches LSij = LFij-tij LFij=min(LSj) 1 -2 2 -3 2 -4 3 -4 4 -5 4 -6 5 -6 6 -7 LS 12 = Li 12 -t 12 = LS 23 = LF 23 -t 23 = LS 24 = LF 24 -t 24 = LS 34 = LF 34 -t 34 = 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 45 = LF 45 -t 45 = LS 46 = LF 46 -i 46 = LS 56 = LF 56 -t 56 = LS 67 = LF 67 -t 67 = The overall computational is shown in next slide 23 (to p 24)

- LS is the latest time an activity can start without delaying critical path time. LSij = LFij - tij - LF is the latest finish time LFij = Minimum (LSj) Min(node 3 -t 23, node 4 -t 24) =Min(5 -2, 5 -1)=Min(3, 4)=3 Min(node 6 -t 46, node 5 -t 45) =Min(8 -3, 7 -1) =Min(5, 6)=5 Min(node 7 -t 67) =Min(9 -1)=8 Start with the end node first Again, you can place these values onto the branches Same as EF 67 from the previous slide 24 (to p 25) (to p 22)

The Project Network Activity Scheduling - Latest Times - LS is the latest time an activity can start without delaying critical path time. LSij = LFij - tij - LF is the latest finish time LFij = Minimum (LSj) Figure 8. 7 Latest activity start and finish times (to p 24) 25

Compute LS-ES or LF-EF Two ways you can achieve it: 1. 2. by compiling slack, Sij by showing branches (to p 27) (to p 28) 26 (to p 16)

The Project Network Calculating Activity Slack Time - Slack, Sij, computed as follows: Sij = LSij - ESij or Sij = LFij - EFij Table 8. 2 Activity Slack * Figure 8. 9 Activity Slack What does it mean? 27 (to p 26)

The Project Network Activity Slack • Slack is the amount of time an activity can be delayed without delaying the project. • Slack time exists for those activities not on the critical path for which the earliest and latest start times are not equal. • Shared slack is slack available for a sequence of activities. Figure 8. 8 Earliest activity start and finish times 28 (to p 26)

Sensitivity Analysis • Today, we only consider one case – “Probabilistic Activity Times” • Refer to activity time estimates usually can not be made with certainty • PERT is known as the solution method (to p 30) 29

PERT • In PERT, three different time estimations are applied: most likely time (m), the optimistic time (a) , and the pessimistic time (b). • How do we make use of these three values? (to p 31) 30

Probabilistic Activity Times • We used these values to estimate the mean and variance of a beta distribution: mean (expected time): variance: How to use these values to solve a project network problem? (to p 32) 31

PERT • We simply apply t values in CPM and determine the values of: • • • ES EF LS LF S and branches with slack = 0 still consider as critical paths • Example. (to p 33) 32

Procedures for PERT Step 1: based on the values of a, b and m, determine the t and v values for each path Step 2: determine the critical path by using t values in the CPM Step 3: compute its corresponding means and standard deviations according. (to p 34) Example Result implication (to p 39) Applications (to p 38) 33

PERT Example • Step 1: computer t and v values • Step 2: determine the CPM • Step 3: determine v value (to p 35) (to p 36) (to p 37) (to p 33) 34

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 35 (to p 34)

Step 2: determine the CPM Figure 8. 12 Earliest and latest activity times Table 8. 4 Activity Earliest and Latest Times and Slack 36 (to p 34)

Step 3: determine v value • The expected project time is the sum of the expected times of the critical path activities. • The project variance is the sum of the variances of the critical path activities. • The expected project time is assumed to be normally distributed (based on central limit theorum). In example, expected project time (tp) and variance (vp) interpreted as the mean ( ) and variance ( 2) of a normal distribution: = 25 weeks 2 = 6. 9 weeks 37 (to p 34)

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 38 (to p 33)

Consider when x = 30 (to p 40) x = 22 (to p 41) Tutorial Assignment (to p 42) 39

Probability Analysis of the Project Network Example 1 2 = 6. 9 = 2. 63 Z = (x- )/ = (30 -25)/2. 63 = 1. 90 -Z value of 1. 90 corresponds to probability of. 4713 in Appendix A of p 715. Probability of completing project in 30 weeks or less : (. 5000 +. 4713) =. 9713, or 97. 13% (Why so high a probability rate? ) Figure 8. 14 Probability the network will be completed in 30 weeks or less 40 (to p 39)

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 41 (to p 39)

Tutorial Assignment • Try to use QM to solve CPM/PERT (to p 43) problems (see slide 19) • Exercises (Chapter 8) – Old: 8, 10, 17 – New: 4, 6, 11 42

Probability Analysis of the Project Network CPM/PERT Analysis with QM for Windows Exhibit 8. 1 43 (to p 16)

The Project Network Activity Slack • Slack is the amount of time an activity can be delayed without delaying the project. • Slack time exists for those activities not on the critical path for which the earliest and latest start times are not equal. • Shared slack is slack available for a sequence of activities. Figure 8. 8 Earliest activity start and finish times 44

The Project Network Calculating Activity Slack Time - Slack, Sij, computed as follows: Sij = LSij - ESij or Sij = LFij - EFij Table 8. 2 Activity Slack * Figure 8. 9 Activity Slack 45