Dr. Ron Lembke Operations Management Project Management Dr. Ron Lembke Operations Management

What’s a Project? Changing something from the way it is to the desired state Never done one exactly like this Many related activities Focus on the outcome Regular teamwork focuses on the work process

Examples of Projects Building construction New product introduction Software implementation Training seminar Research project

Why are projects hard? Resources- Planning People, materials What needs to be done? How long will it take? What sequence? Keeping track of who is supposedly doing what, and getting them to do it

IT Projects Half finish late and over budget Nearly a third are abandoned before completion The Standish Group, in Infoworld Get & keep users involved & informed Watch for scope creep / feature creep

Project Scheduling Establishing objectives Determining available resources Sequencing activities Identifying precedence relationships Determining activity times & costs Estimating material & worker requirements Determining critical activities

Work Breakdown Structure – Scenario 1a Hierarchy of what needs to be done, in what order For me, the hardest part I’ve never done this before. How do I know what I’ll do when and how long it’ll take? I think in phases The farther ahead in time, the less detailed Figure out the tricky issues, the rest is details A lot will happen between now and then It works not badly with no deadline

Mudroom

Mudroom Remodel Big-picture sequence easy: Hard: can a sink fit? Before: Big-picture sequence easy: Demolition Framing Plumbing Electrical Drywall, tape & texture Slate flooring Cabinets, lights, paint Hard: can a sink fit? D W After: D W

Project Scheduling Techniques Gantt chart Critical Path Method (CPM) Program Evaluation & Review Technique (PERT)

Gantt Chart

PERT & CPM Network techniques Developed in 1950’s CPM by DuPont for chemical plants PERT by U.S. Navy for Polaris missile Consider precedence relationships & interdependencies Each uses a different estimate of activity times

Questions Answered by PERT & CPM Completion date? On schedule? Within budget? Probability of completing by ...? Critical activities? Enough resources available? How can the project be finished early at the least cost?

PERT & CPM Steps Identify activities Determine sequence Create network Determine activity times Find critical path Earliest & latest start times Earliest & latest finish times Slack

3 1 2 Activity on Node (AoN) Project: Obtain a college degree (B.S.) Attend class, study etc. Receive diploma Enroll 3 1 2 1 month 4? Years 1 day

1 2 3 4 Activity on Arc (AoA) Project: Obtain a college degree (B.S.) Attend class, study, etc. Receive diploma Enroll 1 2 3 4 1 month 4,5 ? Years 1 day

1 2 3 4 AoA Nodes have meaning Project: Obtain a college degree (B.S.) GraduatingSenior Applicant Student Alum

Network Example You’re a project manager for Bechtel. Construct the network. Activity Predecessors A -- B A C A D B E B F C G D H E, F

Network Example - AON D G B A E Z C H F

Network Example - AON F H C A E Z B G D

Network Example - AON E H C G A Z F B D

Network Example - AOA D G 3 6 8 B E A 1 2 5 H 7 9 C F 4

AOA Diagrams A precedes B and C, B and C precede D B A 1 2 3 D 4 C 3 B A 1 2 C 4 D 5 Add a phantom arc for clarity.

Scenario 1a Task Description Days Pred A Drawings 1 -- B Get & Dye fabric 3 -- C Build stage 4 A D Sew fabric, frame back 2 B E Rent, deliver, hang lights 1 A F Paint stage 2 C G Test assembly 1 D,F H disassemble, deliver 1 G I Assemble, touch-up 1 E,H

Scenario 1a Network Diagram F 2 Finish START C 4 B 3 H 1 G 1 D 2 3

Scenario 1a Longest Path Paths: Days AEI 3 ACFGHI 10 BDGHI 8 START A 1 B 3 E C 4 D 2 F G H I Finish 3

Critical Path Analysis Provides activity information Earliest (ES) & latest (LS) start Earliest (EF) & latest (LF) finish Slack (S): Allowable delay Identifies critical path Longest path in network Shortest time project can be completed Any delay on activities delays project Activities have 0 slack

Critical Path Analysis Example

Network Solution B D E A G 2 6 3 1 1 C F 3 4

Earliest Start & Finish Steps Begin at starting event & work forward ES = 0 for starting activities ES is earliest start EF = ES + Activity time EF is earliest finish ES = Maximum EF of all predecessors for non-starting activities

Activity A Earliest Start Solution D B C F G 1 6 2 3 4 For starting activities, ES = 0.

Earliest Start Solution D B C F G 1 6 2 3 4

Latest Start & Finish Steps Begin at ending event & work backward LF = Maximum EF for ending activities LF is latest finish; EF is earliest finish LS = LF - Activity time LS is latest start LF = Minimum LS of all successors for non-ending activities

Earliest Start Solution D B C F G 1 6 2 3 4

Latest Finish Solution D B C F G 1 6 2 3 4 Latest Finish Solution

Compute Slack

Critical Path A E D B C F G 1 6 2 3 4

Scenario 1b ES/EF times E 1 1 2 A 1 0 1 I 1 9 10 F 2 5 7 Finish START 1 2 A 1 0 1 I 1 9 10 F 2 5 7 Finish START C 4 1 5 B 3 0 3 H 1 8 9 G 1 7 8 D 2 3 5 3

Scenario 1c LF/LS times E 1 1 2 8 9 A 1 0 1 I 1 9 10 9 10 F 2 5 7 1 2 8 9 A 1 0 1 I 1 9 10 9 10 F 2 5 7 Finish START C 4 1 5 1 5 B 3 0 3 2 5 H 1 8 9 G 1 7 8 7 8 D 2 3 5 5 7 3

Scenario 1c START A 1 0 1 B 3 0 3 2 5 E 1 2 8 9 C 4 1 5 1 5 D 2 3 5 5 7 F G 7 8 7 8 H I 9 10 9 10 Finish 3

Other notation C 7 Compute ES, EF for each activity, Left to Right LS LF Compute ES, EF for each activity, Left to Right Compute, LF, LS, Right to Left

Example #2 21 28 28 35 C 7 F 7 0 21 35 37 A 21 G 2 28 33 21 25 25 27 B 4 D 2 E 5

Example #2 21 28 28 35 C 7 F 7 0 21 35 37 A 21 G 2 28 33 21 25 25 27 B 4 D 2 E 5 F cannot start until C and D are done. G cannot start until both E and F are done.

Example #2 21 28 28 35 C 7 F 7 21 28 28 35 0 21 35 37 A 21 G 2 0 21 35 37 28 33 21 25 25 27 B 4 D 2 E 5 22 26 26 28 30 35 E just has to be done in time for G to start at 35, so it has slack. D has to be done in time for F to go at 28, so it has no slack.

Example #2 21 28 28 35 C 7 F 7 21 28 28 35 0 21 35 37 A 21 G 2 0 21 35 37 28 33 21 25 25 27 B 4 D 2 E 5 22 26 26 28 30 35 E just has to be done in time for G to start at 35, so it has slack. D has to be done in time for F to go at 28, so it has no slack.

Gantt Chart - ES A C B D E F G 0 5 10 15 20 25 30 35 40

Gantt Chart - LS A C B D E F G 0 5 10 15 20 25 30 35 40

Another Example B 4 E 6 G 7 A 1 C 3 I 4 F 2 D 7 H 9

Solved Problem 1 B 4 E 6 G 7 C 3 A 1 I 4 F 2 D 7 H 9 1 5 5 11 11 18 1 5 5 11 B 4 E 6 11 18 1 5 5 11 G 7 11 18 0 1 1 4 18 22 A 1 C 3 I 4 0 1 6 9 8 10 18 22 F 2 1 8 9 11 8 17 D 7 H 9 2 9 9 18

Can We Go Faster?

Time-Cost Models 1. Identify the critical path 2. Find cost per day to expedite each node on critical path. 3. For cheapest node to expedite, reduce it as much as possible, or until critical path changes. 4. Repeat 1-3 until no feasible savings exist.

Time-Cost Example ABC is critical path=30 Crash cost Crash D 8 A 10 B 10 C 10 ABC is critical path=30 Crash cost Crash per week wks avail A 500 2 B 800 3 C 5,000 2 D 1,100 2 Cheapest way to gain 1 Week is to cut A

Time-Cost Example ABC is critical path=29 Crash cost Crash D 8 A 9 B 10 C 10 ABC is critical path=29 Crash cost Crash per week wks avail A 500 1 B 800 3 C 5,000 2 D 1,100 2 Wks Incremental Total Gained Crash $ Crash $ 1 500 500 Cheapest way to gain 1 wk Still is to cut A

Time-Cost Example ABC is critical path=28 Crash cost Crash D 8 A 8 B 10 C 10 ABC is critical path=28 Crash cost Crash per week wks avail A 500 0 B 800 3 C 5,000 2 D 1,100 2 Wks Incremental Total Gained Crash $ Crash $ 1 500 500 2 500 1,000 Cheapest way to gain 1 wk is to cut B

Time-Cost Example ABC is critical path=27 Crash cost Crash D 8 A 8 B 9 C 10 ABC is critical path=27 Crash cost Crash per week wks avail A 500 0 B 800 2 C 5,000 2 D 1,100 2 Wks Incremental Total Gained Crash $ Crash $ 1 500 500 2 500 1,000 3 800 1,800 Cheapest way to gain 1 wk Still is to cut B

Time-Cost Example Critical paths=26 ADC & ABC Crash cost Crash per week wks avail A 500 0 B 800 1 C 5,000 2 D 1,100 2 Wks Incremental Total Gained Crash $ Crash $ 1 500 500 2 500 1,000 3 800 1,800 4 800 2,600 To gain 1 wk, cut B and D, Or cut C Cut B&D = $1,900 Cut C = $5,000 So cut B&D

Time-Cost Example Critical paths=25 ADC & ABC Crash cost Crash per week wks avail A 500 0 B 800 0 C 5,000 2 D 1,100 1 Wks Incremental Total Gained Crash $ Crash $ 1 500 500 2 500 1,000 3 800 1,800 4 800 2,600 5 1,900 4,500 Can’t cut B any more. Only way is to cut C

Time-Cost Example Critical paths=24 ADC & ABC Crash cost Crash per week wks avail A 500 0 B 800 0 C 5,000 1 D 1,100 1 Wks Incremental Total Gained Crash $ Crash $ 1 500 500 2 500 1,000 3 800 1,800 4 800 2,600 5 1,900 4,500 6 5,000 9,500 Only way is to cut C

Time-Cost Example Critical paths=23 ADC & ABC Crash cost Crash per week wks avail A 500 0 B 800 0 C 5,000 0 D 1,100 1 Wks Incremental Total Gained Crash $ Crash $ 1 500 500 2 500 1,000 3 800 1,800 4 800 2,600 5 1,900 4,500 6 5,000 9,500 7 5,000 14,500 No remaining possibilities to reduce project length

Time-Cost Example D 7 A 8 B 7 C 8 Now we know how much it costs us to save any number of weeks Customer says he will pay $2,000 per week saved. Only reduce 5 weeks. We get $10,000 from customer, but pay $4,500 in expediting costs Increased profits = $5,500 Wks Incremental Total Gained Crash $ Crash $ 1 500 500 2 500 1,000 3 800 1,800 4 800 2,600 5 1,900 4,500 6 5,000 9,500 7 5,000 14,500 No remaining possibilities to reduce project length

Ex. AOA

AON Paths: ABF: 18 CDEF: 20 C5 B 10 A 6 D 4 E 9 F2 Activity Avail Cost

Gantt charts to keep track B10 A 6 D 4 E 9 F2 Original Activity Avail Cost A 0 ---- B 2 500 C 1 300 D 3 700 E 2 600 F 1 800 Consider CDEF 2 C is cheapest, crash 1 day A 6 B 10 F 2 C 5 D 4 E 9 Revised Pictures: Activity Avail Cost A 0 ---- B 2 500 C 0 D 3 700 E 2 600 F 1 800 Consider DEF E cheapest Crash E 1 day A 6 B 10 C 4 D 4 E 9 1 F 2 C4 B10 A 6 D 4 E 9 F2

Keep Track with Gantt Charts New Pictures: C4 B10 A 6 D 4 E8 F2 All Activities Critical ABF: 18 CDEF: 18 A 6 B 10 C 4 D 4 E 9 F 2 A 6 B 10 C 4 D 4 E 9 F 1 Options: Crash: F $800 – last one B, AND D or E E is cheaper than D $500+$600 = $1,100 > 1,000 Cost > Benefit Activity Avail Cost A 0 ---- B 2 500 C 0 D 3 700 E 1 600 F 1 800 C4 B10 A 6 D 4 E8 F1

Scenario 2 Task Normal Crash $/day B 3 2 $1,000 C 4 1 2,000 D 2 1 500 Now, need to do it in 7 days! Task Normal Crash $/day B 3 2 $1,000 C 4 1 2,000 D 2 1 500 F 2 1 800 Not critical $800 gain 1 day Not critical Cheapest START A 1 0 1 B 3 0 3 2 5 E 1 2 8 9 C 4 1 5 1 5 D 2 3 5 5 7 F G 7 8 7 8 H I 9 10 9 10 3

Scenario 2 Task Normal Crash $/day B 3 2 $1,000 C 4 1 2,000 D 2 1 500 Now, need to do it in 7 days! Task Normal Crash $/day B 3 2 $1,000 C 4 1 2,000 D 2 1 500 F 2 1 800 Not critical Day 1 $800 Day 2 $2,000 Only option Not critical Used up E 1 1 2 8 8 A 1 0 1 I 1 8 9 8 9 F 1 5 6 START C 4 1 5 1 5 B 3 0 3 1 4 H 1 7 8 G 1 6 7 D 2 3 5 4 6 3

Scenario 2 Task Normal Crash $/day B 3 2 $1,000 C 3 1 2,000 D 2 1 500 Now, need to do it in 7 days! Task Normal Crash $/day B 3 2 $1,000 C 3 1 2,000 D 2 1 500 F 2 1 800 Day 1 $800 Day 2 $2,000 Day 3 $2,500 Total $5,300 Used up Something on both paths C, and B or D, D cheaper E 1 1 2 8 8 A 1 0 1 I 1 7 8 7 8 F 1 4 5 START C 3 1 4 1 4 B 3 0 3 H 1 6 7 G 1 5 6 D 2 3 5 3

Scenario 2 Task Normal Crash $/day B 3 2 $1,000 C 2 1 2,000 D 1 1 500 Now, Done in 7 days! Task Normal Crash $/day B 3 2 $1,000 C 2 1 2,000 D 1 1 500 F 2 1 800 Day 1 $800 Day 2 $2,000 Used up Day 3 $2,500 Total $5,300 Used up E 1 1 2 5 6 A 1 0 1 I 1 6 7 6 7 F 1 3 4 START C 2 1 3 1 3 B 3 0 3 H 1 5 6 G 1 4 5 D 1 3 4 3

What about Uncertainty?

 PERT Activity Times 3 time estimates Follow beta distribution Optimistic times (a) Most-likely time (m) Pessimistic time (b) Follow beta distribution Expected time: t = (a + 4m + b)/6 Variance of times: v = (b - a)2/36 

Example Activity a = 2, m = 4, b = 6 E[T] = (2 + 4*4 + 6)/6 = 24/6 = 4.0 σ2 = (6 – 2)2 / 36 = 16/36 = 0.444

Example C B A Activity a m b E[T] variance A 2 4 8 4.33 1 6.48 7.67 Activity a m b E[T] variance A 2 4 8 4.33 1 B 3 6.1 11.5 6.48 2 C 4 8 10 7.67 1 Project 18.5 4 Complete in 18.5 days, with a variance of 4.

Sum of 3 Normal Random Numbers Average value of the sum is equal to the sum of the averages Variance of the sum is equal to the sum of the variances Notice curve of sum is more spread out because it has large variance 10 20 30 40 50 60

Back to the Example: Probability of <= 21 wks Average time = 18.5, st. dev = 2 21 is how many standard deviations above the mean? 21-18.5 = 2.5. St. Dev = 2, so 21 is 2.5/2 = 1.25 standard deviations above the mean Book Table says area between Z=1.25 and –infinity is 0.8944 Probability <= 21 wks = 0.8944 = 89.44% 18.5 21

Benefits of PERT/CPM Useful at many stages of project management Mathematically simple Use graphical displays Give critical path & slack time Provide project documentation Useful in monitoring costs

Limitations of PERT/CPM Clearly defined, independent, & stable activities Specified precedence relationships Activity times (PERT) follow beta distribution Subjective time estimates Over emphasis on critical path

Conclusion Explained what a project is Summarized the 3 main project management activities Drew project networks Compared PERT & CPM Determined slack & critical path Found profit-maximizing crash decision Computed project probabilities