Chapter 5: Project Management Department of Business Administration Fall 2017-2018 Chapter 5: Project Management

Outline: What You Will Learn . . . Discuss the behavioral aspects of projects in terms of project personnel and the project manager. Discuss the nature and importance of a work breakdown structure in project management. Give a general description of PERT/CPM techniques. Construct simple network diagrams. List the kinds of information that a PERT or CPM analysis can provide. Analyze networks with deterministic times. Analyze networks with probabilistic times. Describe activity “crashing” and solve typical problems.

Projects Build A A Done Build B B Done Build C C Done Build D Ship JAN FEB MAR APR MAY JUN On time! Project: Unique, one-time operations designed to accomplish a specific set of objectives in a limited time frame.

Project Management How is it different? Limited time frame Narrow focus, specific objectives Less bureaucratic Why is it used? Special needs Pressures for new or improves products or services What are the Key Metrics Time Cost Performance objectives What are the Key Success Factors? Top-down commitment Having a capable project manager Having time to plan Careful tracking and control Good communication

Project Management What are the Major Administrative Issues? Executive responsibilities Project selection Project manager selection Organizational structure Organizational alternatives Manage within functional unit Assign a coordinator Use a matrix organization with a project leader What are the tools? Work breakdown structure Network diagram Gantt charts Risk management

Key Decisions Deciding which projects to implement Criteria-attractive-cost and benefit-available fund Selecting a project manager Central person Selecting a project team Person’s knowledge and skills-relationship with others Planning and designing the project Goals-timetable-budget-resources Managing and controlling project resources Personnel-equipment-budget Deciding if and when a project should be terminated Likelihood of success-costs-resources

Project Manager Responsible for: Work Quality Human Resources Time Communications Costs

Technical Information about projects Project management Institute (PMI) was established in 1969. In 1990, The institute had 17,500 members. In 2001, there were 86,000 members. Nowadays, this number reached to 100,000 members.

Technical Information about projects

Technical Information about projects https://www.wrike.com/blog/complete-collection-project-management-statistics-2015/

Technical Information about projects https://www.wrike.com/blog/complete-collection-project-management-statistics-2015/

Ethical Issues Temptation to understate costs Withhold information Misleading status reports Falsifying records Comprising workers’ safety Approving substandard work

Concept Project Life Cycle Feasibility Planning Execution Termination Management Concept: A proposal needed Feasibility: Cost, benefit and risk analyses Planning: find out the necessary human resources, time and cost Execution: control for time, available resource and cost Termination: It should be reevaluated for the sake of project’s safety

Work Breakdown Structure Project X Level 1 Level 2 Level 3 Level 4

Planning and Scheduling MAR APR MAY JUN JUL AUG SEP OCT NOV DEC Locate new facilities Interview staff Hire and train staff Select and order furniture Remodel and install phones Move in/startup Gantt Chart

The Network Diagram Network (precedence) diagram – diagram of project activities that shows sequential relationships by the use of arrows and nodes. Activity-on-arrow (AOA) – a network diagram convention in which arrows designate activities. Activity-on-node (AON) – a network diagram convention in which nodes designate activities. Activities – steps in the project that consume resources and/or time. Events – the starting and finishing of activities, designated by nodes in the AOA convention.

The Network Diagram Path Sequence of activities that leads from the starting node to the finishing node Critical path The longest path; determines expected project duration Critical activities Activities on the critical path Slack Allowable slippage for path; the difference the length of path and the length of critical path Slack is the length of the time where an activity can be delayed without interfering with the project completion. Dummy Using this variable does not cost any burden for the company. This exists solely for the purpose of establishing precedence relationships for the sake of simplicity and is not asssigned any time.

Project Network – Activity on Arrow 1 2 3 4 5 6 Locate facilities Order furniture Furniture setup Interview Hire and train Remodel Move in AOA

Project Network – Activity on Node 1 2 3 5 6 Locate facilities Order furniture Furniture setup Interview Remodel Move in 4 Hire and train 7 S AON

Network Conventions a c b Dummy activity d a and b must be completed before c can start a must be completed before b or c can start a must be completed before c can start// b and dummy must be completed before c can start a and b must be completed before b or c can start

Computing Algorithm Network activities Used to determine Forward pass Network activities ES: early start EF: early finish-EF=ES+t LS: late start-LS=LF-t LF: late finish Used to determine Expected project duration Slack time-LS-ES or LF-EF Critical path-longest period ES t EF LS LF ES t EF Backward pass

The Network Diagram- PERT and CPM PERT: Program Evaluation and Review Technique CPM: Critical Path Method Both techniques are widely used for planning and coordinating large-scale projects. Using the two techniques, manager are able to obtain: Graphically displays project activities Estimates how long the project will take Indicates most critical activities Show where delays will not affect project

Example 1-Bank Network convention- CPM The following table contains information related to the major activities of a research project. Use the information to do the following: Draw a precedence diagram using AOA and AON Find the critical path based AOA. Determine the expected length of the project. The amount of slack time for each path Activity Immediate Predecessor Expected Time (days) a - 5 c 8 d 2 b 7 e 3 f 6 i b, d 10 m f,i g 1 h k 17 end k,m

Answer-Bank Network convention Activities with no predecessors are at the beginning (life side) of the network. Activities with multiple predecessors are located at path intersections. (a) Use first AOA; the precedence diagram using AOA is constructed as follows: c a f g h d 8 2 b 7 i S 5 10 e 6 m 3 8 1 k End 17 2

Example-Bank Network convention (b)Find the critical path based AOA. a-c-d-i-m*=5+8+2+10+8=33# a-b-i-m=5+7+10+8=30 e-f-m= 3+6+8=17 g-h-k=1+2+17=20 a-c-d-i-m*-Critical path (c) Determine the expected length of the project. 33 # -Expected project duration

Answer-Bank Network convention (d) calculate the amount of slack time for each path LS LF ES t EF 13 13 c a f g h d 5 13 13 15 5 8 15 5 2 15 15 8 5 b 12 15 25 5 7 i 25 19 19 25 10 25 S 5 9 16 3 25 3 e 6 m 33 13 3 33 8 33 14 16 20 1 14 1 16 k 3 3 1 End 17 2

Example 2-Bank Network convention_ CPM The following table contains information related to the major activities of a research project. Use the information to do the following: Draw a precedence diagram using AOA Find the critical path based AOA. Determine the expected length of the project. The amount of slack time for each path Activity Immediate Predecessor Expected Time (days) a - 8 b 6 d 3 c 11 e 4 f 9 g c, d, f 1 end

Example 2-Bank Network Figure 1 2 3 4 5 6 8 weeks 6 weeks 3 weeks 4 weeks 9 weeks 11 weeks 1 week Locate facilities a Order furniture b Furniture setup d Interview e Hire and train f Remodel c Move in g (a) Bank Network question

Answer-Bank Network Figure (b) and (c) Critical Path Knowledge of slack times provides managers with information for planning allocation of scarce resources and for directing control efforts toward those activities that may be most susceptible to delaying the project.

Example-ES-EF-LS-LF-slack (d) Required: Compute slack time, ES, EF, LS and LF 1 2 3 4 5 6 Forward pass 8 6 14 14 3 17 ES t EF 8 11 19 0 8 8 19 1 20 0 4 4 4 9 13 LS LF ES t EF EF: early finish-EF=ES+t LS: late start-LS=LF-t Slack time-LS-ES or LF-EF Backward pass

Answer (d- cont...) -ES-EF-LS-LF-slack Forward pass 10 6 16 8 14 2 1 2 3 4 5 6 ES t EF 8 6 14 16 3 19 14 17 2 0 8 8 0 8 14 3 17 8 11 19 0 8 8 11 19 8 19 19 1 20 19 1 20 19 20 0 4 4 6 4 10 0 4 6 4 9 13 10 9 19 4 13 6 LS LF ES t EF EF: early finish-EF=ES+t LS: late start-LS=LF-t Slack time-LS-ES or LF-EF Backward pass

PERT Model-Time Estimates The main determinant of the way PERT and CPM networks are analysed and interpreted is whether activity time estimates are probabilistic or deterministic. Deterministic Time estimates that are fairly certain Probabilistic Estimates of times that allow for variation

Probabilistic Time Estimates Optimistic time Time required under optimal conditions Pessimistic time Time required under worst conditions Most likely time Most probable length of time that will be required

Probabilistic Estimates Beta Distribution is generally used to describe the inherent variability in timeEstimates. Although there is no real theoretical justification for using the Beta Distribution, it has certain features that make it attractive in practice. Activity start Optimistic time Most likely time (mode) Pessimistic time to tp tm te

te = to + 4tm +tp 6 te = expected time to = optimistic time tm = most likely time tp = pessimistic time The knowledge of the expected path times and their std. Deviation enables a manager to compute probabilistic estimates of the project completion time as such specific time and scheduled time

(tp – to)2 2 = 36 Variance 2 = variance to = optimistic time 2 = (tp – to)2 36 2 = variance to = optimistic time tp = pessimistic time The size of Variance reflects the degree of uncertainty associated with an activity’s time: The large the variance, the greater the uncertainty.

Example 3-Probabilistic Time Estimates The following table contains information related to the major activities of a research project. Use the information to do the following: Draw its precedence diagram. Find the critical Path. Determine the expected time of the project. Calculate variance or standart deviation of the project Compute the probability that the project can be completed within 15 and 17 weeks of its start. Activity Immediate Predecessor O-M-P a - 1-3-4 b 2-4-6 c 2-3-5 d 3-4-5 e 3-5-7 f 5-7-9 g 2-3-6 h 4-6-8 i 3-4-6 end c, f, i

Answer-Probabilistic Time Estimates (a) Draw its precedence diagram 1-3-4 a 3-4-5 d 3-5-7 e 5-7-9 f 2-4-6 b 4-6-8 h 2-3-6 g 3-4-6 i 2-3-5 c Optimistic time Most likely Pessimistic

Answer b,c and d-Probabilistic Time Estimates

Answer-Probabilistic Time Estimates σabc = 0.97(critical path) σdef = 1.0 σghi = 1.07 Tabc = 10.0 Tdef = 16.0 (critical path) Tghi = 13.50 2.83 a 4.00 d 5.0 e 7.0 f b 6.0 h 3.33 g 4.17 i 3.17 c Based on your estimations, what would your comment be?

Specified time – Path mean Path standard deviation Path Probabilities Z = Specified time – Path mean Path standard deviation Z indicates how many standard deviations of the path distribution the specified time is beyond the expected path duration. The more positive the value, the better. A negative value of z indicates that the specified time is earlier than the expected path duration. Z=+3.00-probability 100%- From the relevant table +3.00 is almost equal to 0.9987.

Example-The Path probability Given the information on the example of probabilistic time estimates (part (e) in the previous example): e) Determine The probability that the project can be completed within 17 weeks of its start. The probability that the project will be completed within 15 weeks of its start. The probability that the project will not be completed within 15 weeks of its start.

Answer-The Path probability Determine (e) The probability that the project can be completed within 17 weeks of its start. Path: a-b-c 17 – 10 0.97 Z = =7.22 Prob.comp in 17 week=1.00 Appendix B, Table B, p.p 884/5 Determine The probability that the project will be completed within 17 weeks of its start. Path: d-e-f 17 – 16 1 =1 Prob.comp in 17 week=0.8413 Appendix B, Table B, p.p 885 Z =

Answer-The Path probability Determine The probability that the project will be completed within 17 weeks of its start. Path: g-h-i 17 – 13.5 1.07 Prob.comp in 17 week=1.00 Appendix B, Table B, p.p 884/5 Z = =3.27 Prob finish in 17 week=1.00 X 0.8413 X 1.00= 0.8413

Answer-The Path probability Determine The probability that the project can be completed within 15 weeks of its start. Path: a-b-c 15 – 10 0.97 Z = =5.15 Prob.comp in 15 week=1.00 Appendix B, Table B, p.p 884/5 Determine The probability that the project will be completed within 15 weeks of its start. Path: d-e-f 15 – 16 1 =-1.00 Prob.comp in 15 week=0.1587 Appendix B, Table B, p.p 885 Z =

Answer-The Path probability Determine The probability that the project will be completed within 15 weeks of its start. Path: g-h-i 15 – 13.5 1.07 Prob.comp in 15 week=0.9192 Appendix B, Table B, p.p 884/5 Z = =1.40 Prob finish in 15 week=1.00 X 0.1587 X 0.9192= 0.1459 The probability that the project will not be completed within 15 weeks of its start: 1- 0.1459=0.8541

Answer-The Path probability-Graphically 17 Weeks 1.00 a-b-c Weeks 10.0 0.8413 d-e-f Weeks 16.0 1.00 g-h-i Weeks 13.5

Answer-The Path probability-Graphically 15 Weeks 1.00 a-b-c Weeks 10.0 0.1587 d-e-f Weeks 16.0 0.9192 g-h-i Weeks 13.5

Table 12.3, page 569

Time-cost Trade-offs: Crashing Excluded from the exam topics In many projects, it is possible to reduce the length of a project by injecting additional resources. The impetus to shorten projects may reflect efforts to avoid late penalties, or/ to take advantage of monetary incentives for timely completion of a project, or/ to free resources for use on other projects. This is called crashing. Crash – briefly, shortening activity duration Procedure for crashing Crash the project one period at a time Only an activity on the critical path Crash the least expensive activity Multiple critical paths: find the sum of crashing the least expensive activity on each critical path

Time-Cost Trade-Offs: Crashing Excluded from the exam topics Total cost Shorten Cumulative (direct) cost of crashing Expected indirect costs Optimum CRASH

Example-Crashing Using the following information, develop the optimal time cost solution. Indirect costs are $ 1000 per day. Determine which activities are on the critical path, its length, and the length of the other path Rank the critical activities in order of lowest crashing cost, and determine the number of days each can be crashed. Determine the critical path after each reduction by shortening the project. Excluded from the exam topics Activity Normal time Crash time Cost per day to crash a 6 c 10 8 $500 d 5 4 300 b 1 700 e 9 7 600 f 2 800

10 6 b a 2 f 5 9 c e 4 d Determine the expected length of the project. Excluded from the exam topics 6 a 4 d 5 c 10 b 9 e 2 f

Answer-Crashing Excluded from the exam topics Determine which activities are on the critical path, its length, and the length of the other path Path length a-b-f 18 c-d-e-f 20 (critical path) (b) Rank the critical activities in order of lowest crashing cost, and datermine the number of days each can be crashed. Activity Cost per day to crash Available days c $ 300 1 e 600 2 d 700 3 f 800 1

Answer-Crashing (c) Determine the critical path after each reduction by shortening the project. (1) Shorten activity c one day at a cost of $ 300. The length of the critical path becomes 19 days. (2) Activity c cannot be shorten any more. Shorten activity e one day at cost of $ 600. The length of the critical path c-d-e-f becomes 18 days which is the same as length of path a-b-f. (3) The path are now both critical, further improvement will necesitate shortening both paths. Path Activity Cost per day to crash a-b-f a no reduction possible b $ 500 f 800 c-d-e-f c no further reduction possible d $ 700 e 600 f 800 Excluded from the exam topics

Answer-Crashing At the first glance, it would seem that crashing f would not be advantageous, because it has the highest crashing cost. However, f is on both paths, so shortening f by one day would shorten both paths by one day for a cost of $ 800. The option of shortening the least expensive activity on each path would cost $ 500 for b and $ 600 for e or $ 1100. Thus shorten f by one day. The project duration is now 17 days. (4) At this point, no additional improvement is feasible. The cost to crash b is $ 500 and the cost to crash e is $ 600, for a total of $ 1100 and that would exceed the indirect costs of $ 100 per day. (5) The crashing sequence is summarized below: Length after crashing n days Path n=0 1 2 3 a-b-f 18 18 18 17 c-d-e-f 20 19 18 17 activity crashed c e f cost $300 600 800 Excluded from the exam topics

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