Chapter 3 Project Management

OBJECTIVES Definition of Project Management Work Breakdown Structure Project Control Charts Structuring Projects Critical Path Scheduling 2

Project Management Defined Project is a series of related jobs usually directed toward some major output and requiring a significant period of time to perform Project Management are the management activities of planning, directing, and controlling resources (people, equipment, material) to meet the technical, cost, and time constraints of a project 3

Gantt Chart Activity 1 Activity 2 Activity 3 Activity 4 Activity 5 Vertical Axis: Always Activities or Jobs Horizontal bars used to denote length of time for each activity or job. Activity 1 Activity 2 Activity 3 Activity 4 Activity 5 Activity 6 Time Horizontal Axis: Always Time 6

Structuring Projects: Pure Project Advantages A pure project is where a self-contained team works full-time on the project Structuring Projects: Pure Project Advantages The project manager has full authority over the project Team members report to one boss Shortened communication lines Team pride, motivation, and commitment are high 8

Structuring Projects: Pure Project Disadvantages Duplication of resources Organizational goals and policies are ignored Lack of technology transfer Team members have no functional area "home" 8

Functional Project A functional project is housed within a functional division President Research and Development Engineering Manufacturing Project A B C D E F G H I Example, Project “B” is in the functional area of Research and Development. 9

Structuring Projects Functional Project: Advantages A team member can work on several projects Technical expertise is maintained within the functional area The functional area is a “home” after the project is completed Critical mass of specialized knowledge 10

Structuring Projects Functional Project: Disadvantages Aspects of the project that are not directly related to the functional area get short-changed Motivation of team members is often weak Needs of the client are secondary and are responded to slowly 11

Matrix Project Organization Structure President Research and Development Engineering Manufacturing Marketing Manager Project A Project B Project C 12

Structuring Projects Matrix: Advantages Enhanced communications between functional areas Pinpointed responsibility Duplication of resources is minimized Functional “home” for team members Policies of the parent organization are followed 13

Structuring Projects Matrix: Disadvantages Too many bosses Depends on project manager’s negotiating skills Potential for sub-optimization 14

Work Breakdown Structure A work breakdown structure defines the hierarchy of project tasks, subtasks, and work packages Program Project 1 Project 2 Task 1.1 Subtask 1.1.1 Work Package 1.1.1.1 Level 1 2 3 4 Task 1.2 Subtask 1.1.2 Work Package 1.1.1.2 4

Network-Planning Models A project is made up of a sequence of activities that form a network representing a project The path taking longest time through this network of activities is called the “critical path” The critical path provides a wide range of scheduling information useful in managing a project Critical Path Method (CPM) helps to identify the critical path(s) in the project networks 15

Prerequisites for Critical Path Methodology A project must have: well-defined jobs or tasks whose completion marks the end of the project; independent jobs or tasks; and tasks that follow a given sequence. 16

Types of Critical Path Methods CPM with a Single Time Estimate Used when activity times are known with certainty Used to determine timing estimates for the project, each activity in the project, and slack time for activities CPM with Three Activity Time Estimates Used when activity times are uncertain Used to obtain the same information as the Single Time Estimate model and probability information Time-Cost Models Used when cost trade-off information is a major consideration in planning Used to determine the least cost in reducing total project time 15

PERT and CPM Network techniques Developed in 1950’s CPM by DuPont for chemical plants (1957) PERT by Booz, Allen & Hamilton with the U.S. Navy, for Polaris missile (1958) Consider precedence relationships and interdependencies Each uses a different estimate of activity times

Six Steps PERT & CPM Define the project and prepare the work breakdown structure Develop relationships among the activities - decide which activities must precede and which must follow others Draw the network connecting all of the activities

Six Steps PERT & CPM Assign time and/or cost estimates to each activity Compute the longest time path through the network – this is called the critical path Use the network to help plan, schedule, monitor, and control the project

A Comparison of AON and AOA Network Conventions Activity on Activity Activity on Node (AON) Meaning Arrow (AOA) 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 Figure 3.5

A Comparison of AON and AOA Network Conventions Activity on Activity Activity on Node (AON) Meaning Arrow (AOA) C and D cannot begin until A and B have both been 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) C A B D Dummy activity Figure 3.5

A Comparison of AON and AOA Network Conventions Activity on Activity Activity on Node (AON) Meaning Arrow (AOA) 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 D B Dummy activity Figure 3.5

Immediate Predecessors AON Example Milwaukee Paper Manufacturing's Activities and Predecessors Activity Description Immediate Predecessors A Build internal components — B Modify roof and floor C Construct collection stack D Pour concrete and install frame A, B E Build high-temperature burner F Install pollution control system G Install air pollution device D, E H Inspect and test F, G Table 3.1

AON Network for Milwaukee Paper Start B Activity A (Build Internal Components) Start Activity Activity B (Modify Roof and Floor) Figure 3.6

AON Network for Milwaukee Paper Activity A Precedes Activity C A Start B C D Activities A and B Precede Activity D Figure 3.7

AON Network for Milwaukee Paper G E F H C A Start D B Arrows Show Precedence Relationships Figure 3.8

AOA Network for Milwaukee Paper 1 3 2 (Modify Roof/Floor) B (Build Internal Components) A 5 D (Pour Concrete/ Install Frame) 4 C (Construct Stack) 6 (Install Controls) F (Build Burner) E (Install Pollution Device) G Dummy Activity H (Inspect/ Test) 7 Figure 3.9

Determining the Project Schedule Perform a Critical Path Analysis The critical path is the longest path through the network The critical path is the shortest time in which the project can be completed Any delay in critical path activities delays the project Critical path activities have no slack time

Determining the Project Schedule Perform a Critical Path Analysis Activity Description Time (weeks) A Build internal components 2 B Modify roof and floor 3 C Construct collection stack 2 D Pour concrete and install frame 4 E Build high-temperature burner 4 F Install pollution control system 3 G Install air pollution device 5 H Inspect and test 2 Total Time (weeks) 25 Table 3.2

Determining the Project Schedule Perform a Critical Path Analysis Activity Description Time (weeks) A Build internal components 2 B Modify roof and floor 3 C Construct collection stack 2 D Pour concrete and install frame 4 E Build high-temperature burner 4 F Install pollution control system 3 G Install air pollution device 5 H Inspect and test 2 Total Time (weeks) 25 Earliest start (ES) = earliest time at which an activity can start, assuming all predecessors have been completed Earliest finish (EF) = earliest time at which an activity can be finished Latest start (LS) = latest time at which an activity can start so as to not delay the completion time of the entire project Latest finish (LF) = latest time by which an activity has to be finished so as to not delay the completion time of the entire project Table 3.2

Determining the Project Schedule Perform a Critical Path Analysis A Activity Name or Symbol Earliest Start ES Earliest Finish EF Latest Start LS Latest Finish LF Activity Duration 2 Figure 3.10

Forward Pass Begin at starting event and work forward Earliest Start Time Rule: If an activity has only one 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)

Forward Pass 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

ES/EF Network for Milwaukee Paper ES EF = ES + Activity time Start

ES/EF Network for Milwaukee Paper 2 EF of A = ES of A + 2 ES of A A 2 Start

ES/EF Network for Milwaukee Paper Start A 2 3 EF of B = ES of B + 3 ES of B B 3

ES/EF Network for Milwaukee Paper Start A 2 C 2 4 B 3

ES/EF Network for Milwaukee Paper Start A 2 C 2 4 D 4 3 = Max (2, 3) 7 B 3

ES/EF Network for Milwaukee Paper Start A 2 C 2 4 D 4 3 7 B 3

ES/EF Network for Milwaukee Paper Start A 2 C 2 4 E 4 F 3 G 5 H 2 8 13 15 7 D 4 3 7 B 3 Figure 3.11

Backward Pass 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)

Backward Pass 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

LS/LF Times for Milwaukee Paper 4 F 3 G 5 H 2 8 13 15 7 D C B Start A 13 LS = LF – Activity time LF = EF of Project 15 Figure 3.12

LS/LF Times for Milwaukee Paper 4 F 3 G 5 H 2 8 13 15 7 D C B Start A LF = Min(LS of following activity) 10 13 Figure 3.12

LS/LF Times for Milwaukee Paper LF = Min(4, 10) 4 2 E 4 F 3 G 5 H 2 8 13 15 7 10 D C B Start A Figure 3.12

LS/LF Times for Milwaukee Paper 4 F 3 G 5 H 2 8 13 15 7 10 D C B Start A 1 Figure 3.12

Computing Slack Time 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

Computing Slack Time A 0 2 0 2 0 Yes B 0 3 1 4 1 No C 2 4 2 4 0 Yes Earliest Earliest Latest Latest On Start Finish Start Finish Slack Critical Activity ES EF LS LF LS – ES Path A 0 2 0 2 0 Yes B 0 3 1 4 1 No C 2 4 2 4 0 Yes D 3 7 4 8 1 No E 4 8 4 8 0 Yes F 4 7 10 13 6 No G 8 13 8 13 0 Yes H 13 15 13 15 0 Yes Table 3.3

Critical Path for Milwaukee Paper 4 F 3 G 5 H 2 8 13 15 7 10 D C B Start A 1 Figure 3.13

ES – EF Gantt Chart for Milwaukee Paper A Build internal components B Modify roof and floor C Construct collection stack D Pour concrete and install frame E Build high-temperature burner F Install pollution control system G Install air pollution device H Inspect and test 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

LS – LF Gantt Chart for Milwaukee Paper A Build internal components B Modify roof and floor C Construct collection stack D Pour concrete and install frame E Build high-temperature burner F Install pollution control system G Install air pollution device H Inspect and test 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Variability in Activity Times CPM assumes we know a fixed time estimate for each activity and there is no variability in activity times PERT uses a probability distribution for activity times to allow for variability

Variability in Activity Times Three time estimates are required Optimistic time (a) – if everything goes according to plan Most–likely time (m) – most realistic estimate Pessimistic time (b) – assuming very unfavorable conditions

Variability in Activity Times Estimate follows beta distribution Expected time: Variance of times: t = (a + 4m + b)/6 v = [(b – a)/6]2

Variability in Activity Times Estimate follows beta distribution Expected time: Variance of times: t = (a + 4m + b)/6 v = [(b − a)/6]2 Probability Optimistic Time (a) Most Likely Time (m) Pessimistic Time (b) Activity Time Probability of 1 in 100 of > b occurring Probability of 1 in 100 of < a occurring

Computing Variance A 1 2 3 2 .11 B 2 3 4 3 .11 C 1 2 3 2 .11 Most Expected Optimistic Likely Pessimistic Time Variance Activity a m b t = (a + 4m + b)/6 [(b – a)/6]2 A 1 2 3 2 .11 B 2 3 4 3 .11 C 1 2 3 2 .11 D 2 4 6 4 .44 E 1 4 7 4 1.00 F 1 2 9 3 1.78 G 3 4 11 5 1.78 H 1 2 3 2 .11 Table 3.4

Probability of Project Completion Project variance is computed by summing the variances of critical activities s2 = Project variance = (variances of activities on critical path) p

Probability of Project Completion Project variance is computed by summing the variances of critical activities Project variance s2 = .11 + .11 + 1.00 + 1.78 + .11 = 3.11 Project standard deviation sp = Project variance = 3.11 = 1.76 weeks p

Probability of Project Completion PERT makes two more assumptions: Total project completion times follow a normal probability distribution Activity times are statistically independent

Probability of Project Completion Standard deviation = 1.76 weeks 15 Weeks (Expected Completion Time) Figure 3.15

Probability of Project Completion What is the probability this project can be completed on or before the 16 week deadline? Z = – /sp = (16 wks – 15 wks)/1.76 = 0.57 due expected date date of completion Where Z is the number of standard deviations the due date lies from the mean

Probability of Project Completion .00 .01 .07 .08 .1 .50000 .50399 .52790 .53188 .2 .53983 .54380 .56749 .57142 .5 .69146 .69497 .71566 .71904 .6 .72575 .72907 .74857 .75175 From Appendix I What is the probability this project can be completed on or before the 16 week deadline? Z = − /sp = (16 wks − 15 wks)/1.76 = 0.57 due expected date date of completion Where Z is the number of standard deviations the due date lies from the mean

Probability of Project Completion 0.57 Standard deviations Probability (T ≤ 16 weeks) is 71.57% 15 16 Weeks Weeks Time Figure 3.16

Determining Project Completion Time Probability of 0.99 Z Probability of 0.01 2.33 Standard deviations 2.33 From Appendix I Figure 3.17

Variability of Completion Time for Noncritical Paths 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

What Project Management Has Provided So Far The project’s expected completion time is 15 weeks There is a 71.57% chance the equipment will be in place by the 16 week deadline Five activities (A, C, E, G, and H) are on the critical path Three activities (B, D, F) have slack time and are not on the critical path A detailed schedule is available

Trade-Offs And Project Crashing It is not uncommon to face the following situations: The project is behind schedule The completion time has been moved forward Shortening the duration of the project is called project crashing

Factors to Consider When Crashing A Project The amount by which an activity is crashed is, in fact, permissible Taken together, the shortened activity durations will enable us to finish the project by the due date The total cost of crashing is as small as possible

Steps in Project Crashing Compute the crash cost per time period. If crash costs are linear over time: Crash cost per period = (Crash cost – Normal cost) (Normal time – Crash time) Using current activity times, find the critical path and identify the critical activities

Steps in Project Crashing If there is only one critical path, then select the activity on this critical path that (a) can still be crashed, and (b) has the smallest crash cost per period. If there is more than one critical path, then select one activity from each critical path such that (a) each selected activity can still be crashed, and (b) the total crash cost of all selected activities is the smallest. Note that a single activity may be common to more than one critical path.

Steps in Project Crashing Update all activity times. If the desired due date has been reached, stop. If not, return to Step 2.

Crashing The Project A 2 1 22,000 22,750 750 Yes Time (Wks) Cost ($) Crash Cost Critical Activity Normal Crash Normal Crash Per Wk ($) Path? A 2 1 22,000 22,750 750 Yes B 3 1 30,000 34,000 2,000 No C 2 1 26,000 27,000 1,000 Yes D 4 2 48,000 49,000 1,000 No E 4 2 56,000 58,000 1,000 Yes F 3 2 30,000 30,500 500 No G 5 2 80,000 84,500 1,500 Yes H 2 1 16,000 19,000 3,000 Yes Table 3.5

Crash and Normal Times and Costs for Activity B | | | 1 2 3 Time (Weeks) $34,000 — $33,000 — $32,000 — $31,000 — $30,000 — — Activity Cost Crash Normal Crash Cost/Wk = Crash Cost – Normal Cost Normal Time – Crash Time = $34,000 – $30,000 3 – 1 = = $2,000/Wk $4,000 2 Wks Crash Time Normal Time Crash Cost Normal Cost Figure 3.18

Critical Path And Slack Times For Milwaukee Paper 4 F 3 G 5 H 2 8 13 15 7 10 D C B Start A 1 Slack = 1 Slack = 0 Slack = 6 Figure 3.19

Steps in the CPM with Single Time Estimate 1. Activity Identification 2. Activity Sequencing and Network Construction 3. Determine the critical path From the critical path all of the project and activity timing information can be obtained 15

CPM with Single Time Estimate Consider the following consulting project: Activity Designation Immed. Pred. Time (Weeks) Assess customer's needs A None 2 Write and submit proposal B 1 Obtain approval C Develop service vision and goals D Train employees E 5 Quality improvement pilot groups F D, E Write assessment report G Develop a critical path diagram and determine the duration of the critical path and slack times for all activities.

First draw the network D(2) E(5) F(5) A(2) B(1) C(1) G(1) Act. Imed. Pred. Time A None 2 B A 1 C B 1 D C 2 D(2) E(5) E C 5 F D,E 5 G F 1 F(5) A(2) B(1) C(1) G(1) 18

Determine early starts and early finish times EF=6 D(2) ES=0 EF=2 ES=2 EF=3 ES=3 EF=4 ES=9 EF=14 ES=14 EF=15 A(2) B(1) C(1) F(5) G(1) ES=4 EF=9 Hint: Start with ES=0 and go forward in the network from A to G. E(5) 21

Determine late starts and late finish times Hint: Start with LF=15 or the total time of the project and go backward in the network from G to A. ES=4 EF=6 D(2) E(5) ES=0 EF=2 ES=2 EF=3 ES=3 EF=4 ES=9 EF=14 ES=14 EF=15 LS=7 LF=9 F(5) A(2) B(1) C(1) G(1) ES=4 EF=9 LS=3 LF=4 LS=2 LF=3 LS=0 LF=2 LS=9 LF=14 LS=14 LF=15 LS=4 LF=9 22

Critical Path & Slack ES=4 EF=6 D(2) ES=0 EF=2 ES=2 EF=3 ES=3 EF=4 Slack=(7-4)=(9-6)= 3 Wks D(2) ES=0 EF=2 ES=2 EF=3 ES=3 EF=4 ES=9 EF=14 ES=14 EF=15 LS=7 LF=9 F(5) A(2) B(1) C(1) G(1) ES=4 EF=9 LS=3 LF=4 LS=2 LF=3 LS=0 LF=2 LS=9 LF=14 LS=14 LF=15 E(5) LS=4 LF=9 Duration=15 weeks 23

Example 2. CPM with Three Activity Time Estimates

Example 2. Expected Time Calculations ET(A)= 3+4(6)+15 6 ET(A)=42/6=7

Ex. 2. Expected Time Calculations ET(B)= 2+4(4)+14 6 ET(B)=32/6=5.333

Ex 2. Expected Time Calculations ET(C)= 6+4(12)+30 6 ET(C)=84/6=14

Example 2. Network A(7) B C(14) D(5) E(11) F(7) H(4) G(11) I(18) (5.333) C(14) D(5) E(11) F(7) H(4) G(11) I(18) Duration = 54 Days

Example 2. Probability Exercise What is the probability of finishing this project in less than 53 days? D=53 p(t < D) t TE = 54

(Sum the variance along the critical path.)

p(t < D) t D=53 TE = 54 p(Z < -.156) = .438, or 43.8 % (NORMSDIST(-.156)) There is a 43.8% probability that this project will be completed in less than 53 weeks.

Ex 2. Additional Probability Exercise What is the probability that the project duration will exceed 56 weeks? 30

Example 2. Additional Exercise Solution TE = 54 p(t < D) D=56 p(Z > .312) = .378, or 37.8 % (1-NORMSDIST(.312)) 31

Time-Cost Models Basic Assumption: Relationship between activity completion time and project cost Time Cost Models: Determine the optimum point in time-cost tradeoffs Activity direct costs Project indirect costs Activity completion times 32

CPM Assumptions/Limitations Project activities can be identified as entities (There is a clear beginning and ending point for each activity.) Project activity sequence relationships can be specified and networked Project control should focus on the critical path The activity times follow the beta distribution, with the variance of the project assumed to equal the sum of the variances along the critical path 15

Answer: d. All of the above Question Bowl Which of the following are examples of Graphic Project Charts? Gantt Bar Milestone All of the above None of the above Answer: d. All of the above 7

Answer: d. All of the above Question Bowl Which of the following are one of the three organizational structures of projects? Pure Functional Matrix All of the above None of the above Answer: d. All of the above 7

Question Bowl Answer: a. SOW (or Statement of Work) A project starts with a written description of the objectives to be achieved, with a brief statement of the work to be done and a proposed schedule all contained in which of the following? SOW (statement of work) WBS (work breakdown structure) Early Start Schedule Late Start Schedule None of the above Answer: a. SOW (or Statement of Work) 7

Question Bowl Answer: c. Slack time For some activities in a project there may be some leeway from when an activity can start and when it must finish. What is this period of time called when using the Critical Path Method? Early start time Late start time Slack time All of the above None of the above Answer: c. Slack time 7

Question Bowl How much “slack time” is permitted in the “critical path” activity times? Only one unit of time per activity No slack time is permitted As much as the maximum activity time in the network As much as is necessary to add up to the total time of the project None of the above Answer: b. No slack time is permitted (All critical path activities must have zero slack time, otherwise they would not be critical to the project completion time.) 7

Question Bowl When looking at the Time-Cost Trade Offs in the Minimum-Cost Scheduling time-cost model, we seek to reduce the total time of a project by doing what to the least-cost activity choices? Crashing them Adding slack time Subtracting slack time Adding project time None of the above Answer: a. Crashing them (We “crash” the least-cost activity times to seek a reduced total time for the entire project and we do it step-wise as inexpensively as possible.) 7

End of Chapter 3