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Project Management Techniques
 

Project Management Techniques

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    Project Management Techniques Project Management Techniques Presentation Transcript

    • Project Management
    • Project Management Techniques
    • Project Management
      • Questions:
      • Why do we need to study Project Management?
      • How does a project management technique work?
      ( to p4) ( to p5)
    • 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
      ( to p3)
    • Project management technique
      • Steps to solve a project management
      • problem:
      • to represent a ‘project problem” graphically
      • to determine its completion time
      • to carry out sensitivity analysis, if any
      ( to p6) ( to p12) ( to p29)
    • 1. Represent a ‘project problem” graphically
      • Steps:
      • Gather all information and organize them in a table format that consists of: event, processing time, and precedent constraints as follows:
      • Draw a semantic network to represent them
      • Special case!
      ( to p7) ( to p9) ( to p4) -- A B 20 30 10 A B C Precedent constraints Processing Time Event
    • 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 p8) ( to p6)
    • 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
    • Special case!
      • When two or events taken places in the same time interval
      • (known an concurrent events)
      • Consider the following example!
      • How to draw it?
      ( to p10) -- A A 3 5 7 A B C Precedent constraints Processing Time Event
    • 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)
    • Solutions for Rule 2
      • Three ways to draw it:
      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)
    • 2. Determine its completion time
      • Consider the project network as shown in next slide
      • Question: Is it an easy way to find out the
      • solution?
      • Answer: YES, it knows as
      • Critical Path Method (CPM)
      ( to p15) ( to p13)
    • 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)
    • 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)
    • 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 p16)
    • How CPM works?
      • Steps:
      • Prepare the project network
      • Construct a table as follows:
      • Compute ES and EF
      • Compute LS and LF
      • Compute LS-ES or LF-EF
      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
    • 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!
      • Their relationship :
      • Example 1:
      ( to p18) ( to p22) ( to p19)
    • The starting point of ES and EF
      • Consider:
      • Then
      • EF 1 = 0
      • ES 12 = max (EF 1 ) EF 12 = ES 12 + t 12
      • = 0 = 0 + t 12
      1 2 t 12 ( to p17)
    • 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
    • - 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)
    • 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)
    • Compute LS and LF
      • Note: We compute these values from the bottom to top, with assigning:
      • LS ij = LF i -t ij LF ij = min LS j
      • with
      • the end of LF ij = EF ij
      • Example: computing Figure 8.3
      ( to p23)
    • 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
    • - 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)
    • 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)
    • Compute LS-ES or LF-EF
      • Two ways you can achieve it:
      • by compiling slack, S ij
      • by showing branches
      ( to p27) ( to p28) ( to p16)
    • 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)
    • 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 ( to p26)
    • 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 p30)
    • 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 p31)
    • 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 p32)
    • 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 p33)
    • 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.
      • Example
      • Result implication
      • Applications
      ( to p34) ( to p38) ( to p39)
    • PERT Example
      • Step 1: computer t and v values
      • Step 2: determine the CPM
      • Step 3: determine v value
      ( to p35) ( to p36) ( to p37) ( to p33)
    • 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)
    • 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)
    • 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 (t p ) and variance (v p ) interpreted as the mean (  ) and variance (  2 ) of a normal distribution:
      •  = 25 weeks
      •  2 = 6.9 weeks
      ( to p34)
    • 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)
      • Consider when
          • x = 30
          • x = 22
      • Tutorial Assignment
      ( to p40) ( to p41) ( to p42)
    • 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 p715. 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 ( to p39)
    • 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)
    • Tutorial Assignment
      • Try to use QM to solve CPM/PERT problems (see slide 19)
      • Exercises (Chapter 8)
        • Old: 8, 10, 17
        • New: 4, 6, 11
      (to p43)
    • Probability Analysis of the Project Network CPM/PERT Analysis with QM for Windows Exhibit 8.1 (to p16)
    • 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
    • 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 *