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You should be able to: 1. Discuss the behavioral aspects of projects in terms of project personnel and the project manager 2. Explain the nature and importance of a work breakdown structure in project management 3. Give a general description of PERT/CPM techniques 4. Construct simple network diagrams 5. List the kinds of information that a PERT or CPM analysis can provide 6. Analyze networks with deterministic times 7. Analyze networks with probabilistic times 8. Describe activity ‘crashing’ and solve typical problems 17-1

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Projects Unique, one-time operations designed to accomplish a specific set of objectives in a limited time frame Examples: The Olympic Games Producing a movie Software development Product development ERP implementation 17-2

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WBS A hierarchical listing of what must be done during a project Establishes a logical framework for identifying the required activities for the project 1. Identify the major elements of the project 2. Identify the major supporting activities for each of the major elements 3. Break down each major supporting activity into a list of the activities that will be needed to accomplish it 17-4

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PERT (program evaluation and review technique) and CPM (critical path method) are two techniques used to manage large-scale projects By using PERT or CPM Managers can obtain: 1. A graphical display of project activities 2. An estimate of how long the project will take 3. An indication of which activities are most critical to timely project completion 4. An indication of how long any activity can be delayed without delaying the project 17-7

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Network diagram Diagram of project activities that shows sequential relationships by use of arrows and nodes Activity on arrow (AOA) Network diagram convention in which arrows designate activities Activity on node (AON) Network convention in which nodes designate activities Activities Project steps that consume resources and/or time Events The starting and finishing of activities 17-8

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Activity onActivity Meaning Node (AON) A comes before B, which comes before C. (a) A B C A and B must both be completed before C can start. (b) A C B B and C cannot begin until A is completed. (c) B A C 17-9

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Activity onActivity Meaning Node (AON) C and D cannot begin until both A and B are completed. (d) A B C D C cannot begin until both A and B are completed; D cannot begin until B is completed. A dummy activity is introduced in AOA. (e) A B C D 17-10

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B and C cannot begin until A is completed. D cannot begin until both B and C are completed. A dummy activity is again introduced in AOA. (f) A C DB Activity onActivity Meaning Node (AON) 17-11

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ActivityDescription Immediate Predecessors ABuild internal components— BModify roof and floor— CConstruct collection stackA DPour concrete and install frameA, B EBuild high-temperature burnerC FInstall pollution control systemC GInstall air pollution deviceD, E HInspect and testF, G Project Activities and Predecessors 17-12

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A Start B Start Activity Activity A (Build Internal Components) Activity B (Modify Roof and Floor) 17-13

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C D A Start B Activity A Precedes Activity C Activities A and B Precede Activity D 17-14

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G E F H C A Start DB Arrows Show Precedence Relationships 17-15

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Deterministic Time estimates that are fairly certain Probabilistic Time estimates that allow for variation 17-16

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ActivityDescriptionTime (weeks) ABuild internal components2 BModify roof and floor3 CConstruct collection stack2 DPour concrete and install frame4 EBuild high-temperature burner4 FInstall pollution control system 3 GInstall air pollution device5 HInspect and test2 Total Time (weeks)

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Critical Path PathPath duration A-C-F-H = 9 A-C-E-G-H = 15 A-D-G-H = 13 B-D-G-H = 14 Critical path = Longest path A-C-E-G-H Project duration = 15 weeks 17-18

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Finding ES and EF involves a forward pass through the network diagram Early start (ES) The earliest time an activity can start Assumes all preceding activities start as early as possible For nodes with one entering arrow ES = EF of the entering arrow For activities leaving nodes with multiple entering arrows ES = the largest of the largest entering EF Early finish (EF) The earliest time an activity can finish EF = ES + t 17-19

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Finding LS and LF involves a backward pass through the network diagram Late Start (LS) The latest time the activity can start and not delay the project The latest starting time for each activity is equal to its latest finishing time minus its expected duration: LS = LF - t Late Finish (LF) The latest time the activity can finish and not delay the project For nodes with one leaving arrow, LF for nodes entering that node equals the LS of the leaving arrow For nodes with multiple leaving arrows, LF for arrows entering node equals the smallest of the leaving arrows 17-20

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Slack can be computed one of two ways: Slack = LS – ES Slack = LF – EF Critical path The critical path is indicated by the activities with zero slack 17-21

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Perform a Critical Path Analysis A Activity Name or Symbol Earliest Start ES Earliest Finish EF Latest Start LS Latest Finish LF Activity Duration

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Begin at starting event and work forward Earliest Start Time Rule: If an activity has only a single immediate predecessor, its ES equals the EF of the predecessor If an activity has multiple immediate predecessors, its ES is the maximum of all the EF values of its predecessors ES = Max {EF of all immediate predecessors} 17-23

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Begin at starting event and work forward Earliest Finish Time Rule: The earliest finish time (EF) of an activity is the sum of its earliest start time (ES) and its activity time EF = ES + Activity time 17-24

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Start 0 0 ES 0 EF = ES + Activity time 17-25

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Start A2A2 2 EF of A = ES of A ES of A 17-26

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B3B3 Start A2A EF of B = ES of B ES of B 17-27

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C2C2 24 B3B3 03 Start A2A

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C2C2 24 B3B3 03 Start A2A2 20 D4D4 7 3 = Max (2, 3) 17-29

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D4D4 37 C2C2 24 B3B3 03 Start A2A

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E4E4 F3F3 G5G5 H2H D4D4 37 C2C2 24 B3B3 03 Start A2A

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Begin with the last event and work backwards Latest Finish Time Rule: If an activity is an immediate predecessor for just a single activity, its LF equals the LS of the activity that immediately follows it If an activity is an immediate predecessor to more than one activity, its LF is the minimum of all LS values of all activities that immediately follow it LF = Min {LS of all immediate following activities} 17-32

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Begin with the last event and work backwards Latest Start Time Rule: The latest start time (LS) of an activity is the difference of its latest finish time (LF) and its activity time LS = LF – Activity time 17-33

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E4E4 F3F3 G5G5 H2H D4D4 37 C2C2 24 B3B3 03 Start A2A2 20 LF = EF of Project 1513 LS = LF – Activity time 17-34

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E4E4 F3F3 G5G5 H2H D4D4 37 C2C2 24 B3B3 03 Start A2A2 20 LF = Min(LS of following activity)

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E4E4 F3F3 G5G5 H2H D4D4 37 C2C2 24 B3B3 03 Start A2A2 20 LF = Min(4, 10)

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E4E4 F3F3 G5G5 H2H D4D4 37 C2C2 24 B3B3 03 Start A2A

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After computing the ES, EF, LS, and LF times for all activities, compute the slack or free time for each activity Slack is the length of time an activity can be delayed without delaying the entire project Slack = LS – ES or Slack = LF – EF 17-38

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EarliestEarliestLatestLatestOn StartFinishStartFinishSlackCritical ActivityESEFLSLFLS – ESPath A02020Yes B03141No C24240Yes D37481No E48480Yes F No G Yes H Yes 17-39

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Knowledge of slack times provides managers with information for planning allocation of scarce resources Control efforts will be directed toward those activities that might be most susceptible to delaying the project Activity slack times are based on the assumption that all of the activities on the same path will be started as early as possible and not exceed their expected time If two activities are on the same path and have the same slack, this will be the total slack available to both 17-40

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The beta distribution is generally used to describe the inherent variability in time estimates The probabilistic approach involves three time estimates: Optimistic time, (t o ) The length of time required under optimal conditions Pessimistic time, (t p ) The length of time required under the worst conditions Most likely time, (t m ) The most probable length of time required 17-41

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The expected time, t e,for an activity is a weighted average of the three time estimates: The expected duration of a path is equal to the sum of the expected times of the activities on that path: 17-42

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A project is not complete until all project activities are complete It is risky to only consider the critical path when assessing the probability of completing a project within a specified time. To determine the probability of completing the project within a particular time frame Calculate the probability that each path in the project will be completed within the specified time Multiply these probabilities The result is the probability that the project will be completed within the specified time 17-44

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The standard deviation of each activity’s time is estimated as one-sixth of the difference between the pessimistic and optimistic time estimates. The variance is the square of the standard deviation: Standard deviation of the expected time for the path 17-45

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MostExpected OptimisticLikelyPessimisticTimeVariance Activityt o t m t p t = (t o + 4 t m + t p )/6[(t p – t o )/6] 2 A12324/36 B23434/36 C12324/36 D246416/36 E147436/36 F129364/36 G /36 H12324/

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Knowledge of expected path times and their standard deviations enables managers to compute probabilistic estimates about project completion such as: The probability that the project will be completed by a certain time The probability that the project will take longer than its expected completion time 17-47

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Path variance is computed by summing the variances of activities on the path 2 = Project variance = (variances of activities on the path) p 17-48

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Variance of critical path 2 = 4/36 + 4/ / /36 + 4/36 = 112/36 = Standard deviation of critical path p = Variance of critical path = = weeks p 17-49

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Note: Total project completion times follow a normal probability distribution Activity times are statistically independent 17-50

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Standard deviation = weeks 15 Weeks (Expected Completion Time) 17-51

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What is the probability this project can be completed on or before the 16 week deadline? Z = T M = Path mean completion time T = Due date Z= (16 – 15)/ = 0.57 Where Z is the number of standard deviations the due date or target date lies from the mean or expected date 17-52

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Time Probability (T ≤ 16 weeks) is Standard deviations 1516 WeeksWeeks Probability (T > 16 weeks) is 1 – =

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Determine a due date for the project completion time that will have a 99% probability of meeting. Due date = T E + Z p 17-54

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Probability of 0.01 Z From Appendix I Probability of Standard deviations

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Probability of 0.01 Z From Appendix I Probability of Standard deviations Due date (T) = T M + z p 17-56

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Probability of 0.01 Z From Appendix I Probability of Standard deviations Due date (T) = (1.7638) 17-57

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Probability of 0.01 Z From Appendix I Probability of Standard deviations Due date (T) = 19.1 weeks 17-58

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Variability of times for activities on noncritical paths must be considered when finding the probability of finishing in a specified time Variation in noncritical activity may cause change in critical path 17-59

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1. The project’s expected completion time is 15 weeks 2. There is a 71.57% chance the equipment will be in place by the 16 week deadline 3. Five activities (A, C, E, G, and H) are on the critical path 4. Three activities (B, D, F) are not on the critical path and have slack time 5. A detailed schedule is available 17-60

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Budget control is an important aspect of project management Costs can exceed budget Overly optimistic time estimates Unforeseen events Unless corrective action is taken, serious cost overruns can occur 17-61

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Activity time estimates are made for some given level of resources It may be possible to reduce the duration of a project by injecting additional resources Motivations: To avoid late penalties Monetary incentives Free resources for use on other projects 17-62

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Crashing Shortening activity durations Typically, involves the use of additional funds to support additional personnel or more efficient equipment, and the relaxing of some work specifications The project duration may be shortened by increasing direct expenses, thereby realizing savings in indirect project costs 17-63

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To make decisions concerning crashing requires information about: 1. Regular time and crash time estimates for each activity 2. Regular cost and crash cost estimates for each activity 3. A list of activities that are on the critical path Critical path activities are potential candidates for crashing Crashing non-critical path activities would not have an impact on overall project duration 17-64

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General procedure: 1. Crash the project one period at a time 2. Crash the least expensive activity that is on the critical path 3. When there are multiple critical paths, find the sum of crashing the least expensive activity on each critical path If two or more critical paths share common activities, compare the least expensive cost of crashing a common activity shared by critical paths with the sum for the separate critical paths 17-65

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Among the most useful features of PERT: 1. It forces the manager to organize and quantify available information and to identify where additional information is needed 2. It provides the a graphic display of the project and its major activities 3. It identifies a. Activities that should be closely watched b. Activities that have slack time 17-67

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Potential sources of error: 1. The project network may be incomplete 2. Precedence relationships may not be correctly expressed 3. Time estimates may be inaccurate 4. There may be a tendency to focus on critical path activities to the exclusion of other important project activities 5. Major risk events may not be on the critical path 17-68

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Technology has benefited project management CAD To produce updated prototypes on construction and product- development projects Communication software Helps to keep project members in close contact Facilitates remote viewing of projects Project management software Specialized software used to help manage projects Assign resources Compare project plan versions Evaluate changes Track performance 17-69

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Advantages include: Imposes a methodology and common project management terminology Provides a logical planning structure May enhance communication among team members Can flag the occurrence of constraint violations Automatically formats reports Can generate multiple levels of summary and detail reports Enables “what if” scenarios Can generate a variety of chart types 17-70

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Risks are an inherent part of project management Risks relate to occurrence of events that have undesirable consequences such as Delays Increased costs Inability to meet technical specifications Good risk management involves Identifying as many risks as possible Analyzing and assessing those risks Working to minimize the probability of their occurrence Establishing contingency plans and budgets for dealing with any that do occur 17-71

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Projects present both strategic opportunities and risks It is critical to devote sufficient resources and attention to projects Projects are often employed in situations that are characterized by significant uncertainties that demand Careful planning Wise selection of project manager and team Monitoring of the project Project software can facilitate successful project completion Be careful to not focus on critical path activities to the exclusion of other activities that may become critical It is not uncommon for projects to fail When that happens, it can be beneficial to examine the probable reasons for failure 17-72

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