1 Project Management Principles ΥΠΕΥΘΥΝΟΣ: Θ. ΜΑΝΑΒΗΣ Lecture 3 – Planning and Scheduling

Project Planning, Scheduling, and Control Work Breakdown Structure 1.Project 2. Major tasks in the project 3.Subtasks in the major tasks 4. Activities to be completed 2

3 Project Scheduling and Control Techniques Gantt Chart Critical Path Method (CPM) Program Evaluation and Review Technique (PERT)

Project Network Diagram Provides a visual representation of the tasks as well as the logical sequence and dependencies among the tasks Provides information on start/finish dates and what activities may be delayed without affecting the deadline target date –Can be used to make decisions regarding scheduling and resource assignments to shorten the time required for those critical activities that will impact the project deadline 4

5  Graph or bar chart with a bar for each project activity that shows passage of time  Provides visual display of project schedule Gantt Chart

Project Management Software, Microsoft “Project” 6

7 History of CPM/PERT Critical Path Method (CPM) –E I Du Pont de Nemours & Co. (1957) for construction of new chemical plant and maintenance shut-down –Deterministic task times –Activity-on-node network construction –Repetitive nature of jobs Project Evaluation and Review Technique (PERT) –U S Navy (1958) for the POLARIS missile program –Multiple task time estimates (probabilistic nature) –Activity-on-arrow network construction (but not necessarily) –Non-repetitive jobs (R & D work)

Activity on Node Graphically represents all the project tasks as well as their logical sequence and dependencies –Activities are boxes (nodes), arrows indicate precedence and flow –Determine predecessors, successors and parallel tasks 8

Activity on Arrow Activity-on-arrow (AOA) –arrows represent activities –nodes are events for points in time Event –completion or beginning of an activity in a project 9

darla/smbs/vit10 EXAMPLE: AOA Project Network for House Lay foundation Design house and obtain financing Order and receive materials Dummy Finish work Select carpet Select paint Build house EXAMPLE: AON Project Network for House Start Design house and obtain financing Order and receive materials Select paint Select carpet Lay foundations Build house Finish work

Activity Network, AON & AOA Format Important: AOA May need dummy activities to show all relationships 11

12 Situations in network diagram A B C A must finish before either B or C can start A B C both A and B must finish before C can start D C B A both A and C must finish before either of B or D can start A C B D Dummy A must finish before B can start both A and C must finish before D can start

A Comparison of AON and AOA 13

An Example: Relocating an Office – Gantt Chart 14

9w 1w 3w 6w 11w 8w 4w Locate Facilities - 8w Interview Staff – 4w Hire and Train – 9w Order Furniture – 6w Remodel/phones - 11w Furniture set up – 3w Move in – 1w 15 An Example: Relocating an Office – AON, AOA

16 CPM – Critical Path Method

How to Find the Critical Path ACTIVITYDESCRIPTION IMMEDIATE PREDECESSORS ABuild internal components— BModify roof and floor— CConstruct collection stackA D Pour concrete and install frame B E Build high-temperature burner C FInstall control systemC GInstall air pollution deviceD, E HInspect and testF, G

How to Find the Critical Path A Build Internal Components H Inspect and Test E Build Burner C Construct Collection Stack Start F Install Control System Finish G Install Pollution Device D Pour Concrete and Install Frame B Modify Roof and Floor

How to Find the Critical Path A2A2C2C2 H2H2E4E4 B3B3D4D4G5G5 F3F3 StartFinish

How to Find the Critical Path To find the critical path, need to determine the following quantities for each activity in the network 1.Earliest start timeES 1.Earliest start time (ES): the earliest time an activity can begin without violation of immediate predecessor requirements 2.Earliest finish timeEF 2.Earliest finish time (EF): the earliest time at which an activity can end 3.Latest start timeLS 3.Latest start time (LS): the latest time an activity can begin without delaying the entire project 4.Latest finish timeLF 4.Latest finish time (LF): the latest time an activity can end without delaying the entire project

How to Find the Critical Path In the nodes, the activity time and the early and late start and finish times are represented in the following manner ACTIVITY t ESEF LSLF  Earliest times are computed as Earliest finish time =Earliest start time + Expected activity time EF =ES + t Earliest start =Largest of the earliest finish times of immediate predecessors ES =Largest EF of immediate predecessors

How to Find the Critical Path At the start of the project we set the time to zero Thus ES = 0 for both A and B Start A t = 2 ES = 0EF= = 2 B t = 3 ES = 0EF= = 3

How to Find the Critical Path ES and EF times A202A202 C224C224 H E448E448 B303B303 D437D437 G5 813 F347F347 StartFinish

How to Find the Critical Path Latest times are computed as Latest start time =Latest finish time – Expected activity time LS = LF – t Latest finish time =Smallest of latest start times for following activities LF =Smallest LS of following activities For activity H LS = LF – t = 15 – 2 = 13 weeks

How to Find the Critical Path LS and LF times A20202A20202 C22424C22424 H E44848E44848 B30314B30314 D43748D43748 G5 813 F StartFinish

How to Find the Critical Path Once ES, LS, EF, and LF have been determined, it is a simple matter to find the amount of slack time that each activity has Slack = LS – ES, or Slack = LF – EF Activities A, C, E, G, and H have no slack time These are called critical activities and they are said to be on the critical path The total project completion time is 15 weeks

How to Find the Critical Path Schedule and slack times ACTIVITY EARLIEST START, ES EARLIEST FINISH, EF LATEST START, LS LATEST FINISH, LF SLACK, LS – ES ON CRITICAL PATH? A02020Yes B03141No C24240Yes D37481No E48480Yes F No G8138 0Yes H Yes

How to Find the Critical Path Critical path: A  C  E  G  H A20202A20202 C22424C22424 H E44848E44848 B30314B30314 D43748D43748 G5 813 F StartFinish

Another Example 29 ActivityDescriptionEstimated Duration (Days) Predecessor AEvaluate current technology platform2None BDefine user requirements5A CDesign Web page layouts4B DSet-up Server3B EEstimate Web traffic1B FTest Web pages and links4C,D GMove web pages to production environment 3D,E HWrite announcement of intranet for corp. newsletter 2F,G ITrain users5G JWrite report to management1H,I

AON Network Diagram 30

Critical Path 31 A B C D E F G H I J

Possible Activity Paths Possible PathsPathTotal Path 1A+B+C+F+H+J Path 2A+B+D+F+H+J Path 3A+B+D+G+H+J Path 4A+B+D+G+I+J19* Path 5A+B+E+G+I+J * The Critical Path 32

Critical Path Longest path – Path 4 (19 days) Shortest time project can be completed –The critical path has zero slack (or float) – any delay will impact the project completion time Slack - the amount of time an activity can be delayed before it delays the project Any change in the critical path will delay the entire project Task E can be delayed 2 days (from 8 to 10) without impacting the project completion time Must be monitored and managed! –Project manager can expedite or crash by adding resources –Fast tracking – running activities in parallel which were originally planned as sequential –The CP can change –Can have multiple CPs 33

34 PERT– Program (Project) Evaluation and Review Technique

PERT Program Evaluation and Review Technique –Developed in 1950s to help manage the Polaris Submarine Project Developed about the same time as the Critical Path Method –Often combined as PERT/CPM Employs both a project network diagram with a statistical distribution 35

Activity Times In some situations, activity times are known with certainty CPM assigns just one time estimate to each activity and this is used to find the critical path In many projects there is uncertainty about activity times PERT employs a probability distribution based on three time estimates for each activity –A weighted average of these estimates is used for the time estimate and this is used to determine the critical path

Activity Times The time estimates in PERT are Optimistic time a Optimistic time ( a ) = time an activity will take if everything goes as well as possible. There should be only a small probability (say, 1 / 100 ) of this occurring. Pessimistic time b Pessimistic time ( b ) = time an activity would take assuming very unfavorable conditions. There should also be only a small probability that the activity will really take this long. Most likely time m Most likely time ( m ) = most realistic time estimate to complete the activity

Activity Times expected activity time tTo find the expected activity time ( t ), the beta distribution weights the estimates as follows variance of activity completion time To compute the dispersion or variance of activity completion time, we use the formula

Activity Analysis for PERT ActivityPredecessorOptimistic Estimates (Days) Most Likely Estima tes (Days) Pessimistic Estimates (Days) Expected Duration (a+4b+c) 6 Variance ((b-a)/6) 2 ANone BA CB DB EB FC,D GD,E HF,G IG JH,I

PERT Computations * The Critical Path 40

Possible PathsPathTotal Path 1A+B+C+F+H+J Path 2A+B+D+F+H+J Path 3A+B+D+G+H+J Path 4A+B+D+G+I+J20.5* Path 5A+B+E+G+I+J Possible PERT Activity Paths * The Critical Path 41

Probability of Project Completion critical path analysisThe critical path analysis helped determine the expected project completion time of 20.5 weeks But variation in activities on the critical path can affect overall project completion, and this is a major concern PERT uses the variance of critical path activities to help determine the variance of the overall project Project variance = ∑ variances of activities on the critical path

Probability of Project Completion We know the standard deviation is just the square root of the variance, so We assume activity times are independent and total project completion time is normally distributed

Probability of Project Completion The project’s expected completion date is 20.5 weeks. –Assume that the total project completion time follows a normal probability distribution –Chart tells us that there is a 50% chance of completing the entire project in less than 20.5 weeks and a 50% chance it will exceed 20.5 weeks Standard Deviation = 1.55 (Expected Completion Time) 20.5Weeks

Probability of Project Completion The standard normal equation can be applied as follows From the [Normal Cumulative Distribution.pdf] in the Tutorial Folder we calculate the Area Under the Standard Normal Curve table (see also automatic tools like we find the probability of that is associated with this Z value That means there is a 83.3% probability this project can be completed in 22 weeks or less The probability of completing in 23 weeks would be 94.6%

46 PERT– Another Example

Examples of the Beta Distribution Distribution 47

Project with Probabilistic Time Estimates 48 © 2014 John Wiley & Sons, Inc. - Russell and Taylor 8e

Activity Time Estimates 49 © 2014 John Wiley & Sons, Inc. - Russell and Taylor 8e TIME ESTIMATES (WKS)MEAN TIMEVARIANCE ACTIVITY ambtб 2

Activity Early, Late Times & Slack 50 © 2014 John Wiley & Sons, Inc. - Russell and Taylor 8e ACTIVITY tб  ESEFLSLFS

Earliest, Latest Times, and Slack 51 © 2014 John Wiley & Sons, Inc. - Russell and Taylor 8e

Total Project Variance 52 © 2014 John Wiley & Sons, Inc. - Russell and Taylor 8e  2 = б б б б 11 2  2 = = 6.89 weeks

Probabilistic Network Analysis 53 © 2014 John Wiley & Sons, Inc. - Russell and Taylor 8e Determine probability that project is completed within specified time where  =t p = project mean time  =project standard deviation x =proposed project time Z =number of standard deviations that x is from the mean Z = x -  

Normal Distribution of Project Time 54 © 2014 John Wiley & Sons, Inc. - Russell and Taylor 8e

Southern Textile – 30 weeks 55 © 2014 John Wiley & Sons, Inc. - Russell and Taylor 8e  2 = 6.89 weeks  = 6.89  = 2.62 weeks x -   Z= = = From the table, a Z score of 1.91 corresponds to a probability of Thus P(30) =

Southern Textile – 22 weeks 56 © 2014 John Wiley & Sons, Inc. - Russell and Taylor 8e From the Z score of corresponds to a probability of Thus P(22) =