3.
4.
5.
6.
1.
2.
Explain what project management is and why it is important.
Identify the different ways projects can be structured in an organization.
Describe how project objectives are set.
Describe how projects are organized into components using work breakdown structure
Determine the “critical path” for a project.
Demonstrate how to “crash,” or reduce the length, of a project.

LO 1

Building: a ship, a satellite, an oil rig, and a nuclear plant.

Developing: computer programs, an advertising campaign, a new product, a new process, and training materials.

Implementing: new technologies and work procedures.



Project: a series of related jobs usually directed toward some major output and requiring a significant period of time to perform
Project management: the management activities of planning, directing, and controlling resources (people, equipment, material) to meet the technical, cost, and time constraints of a project
LO 1

LO 1

LO 1
th

LO 1
: changing patterns of resource usage and level of activity over the course of the project

LO 1

Stages of a Conventional Project:
◦ Slow beginning
◦ Buildup of size
◦ Peak
◦ Begin a decline
◦ Termination

LO 1

LO 1
Time distribution of project effort is characterized by slow-rapid-slow


Risk during project life cycle
◦ With most projects there is some uncertainty about the ability to meet project goals
◦ Uncertainty of outcome is greatest at the start of a project
◦ Uncertainty decreases as the project moves toward completion
LO 1

LO 1
Try to avoid the “90-90 rule of project management”:

LO 1

LO 1
 During the life cycle cycle, project management is accomplished through the use of processes such as:
 Initiating, planning, executing, controlling, and closing
 Many of these processes are iterative in nature because the project is being progressively elaborated

LO 1
An Alternate View*
Stage 1: Excitement – Euphoria
Stage 2: Disenchantment
Stage 3: Search for the Guilty
Stage 4: Punishment of the Innocent
Stage 5: Distinction for the Uninvolved
*Author unknown but believed to have perished in stage 4

LO 2




Pure project
Functional project
Matrix project
LO 2



Advantages
◦ The project manager has full authority
◦ Team members report to one boss
◦ Shortened communication lines
◦ Team pride, motivation, and commitment are high
Disadvantages
◦ Duplication of resources
◦ Organizational goals and policies are ignored
◦ Lack of technology transfer
◦ Team members have no functional area "home"
LO 2

LO 2



Advantages
◦ A team member can work on several projects
◦ Technical expertise maintained in functional area
◦ Functional area is “home” after project completed
◦ Critical mass of specialized knowledge
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
LO 2

LO 2



Advantages
◦ Better communications between functional areas
◦ Project manager held responsible for success
◦ Duplication of resources is minimized
◦ Functional “home” for team members
◦ Policies of the parent organization are followed
Disadvantages
◦ Too many bosses
◦ Depends on project manager’s negotiating skills
◦ Potential for sub-optimization
LO 2

LO 3




Why Set Project Objectives
To provide direction for project activities
To enable measuring results against prior exceptions
Resource usage (manpower, materials, etc.)
◦ Schedule integrity
◦ Quality of work
To determine specific goals which will provide maximum effectiveness of project activities

LO 2







Requirements for Project Objectives
(time, resources, staff)
(vs. complex)
(vs. general, vague statements)
(“deliverables”)
(resources, schedule, quality)
(with strategy, programs, policies, procedures)
(department or individual)

LO 2

Example: D.U. Singer Project
Title: Permanent Antiseptic Production Start-
Up
Objectives:
◦ Develop a comprehensive plan for the production of a new, permanent antiseptic
◦ Complete development and testing of a manufacturing process that:
 Meets all current FDA, EPA, and OSHA regulations as well as internal specifications
 produces 95% yield of product (full packaged) at a level of 80% of full production goal of 10 million liters per year

LO 2




Statement of work (SOW): a written description of the objectives to be achieved
Task: a further subdivision of a project
◦ Usually shorter than several months
◦ Performed by one group or organization
Work package: a group of activities combined to be assignable to a single organizational unit
LO 4




Project milestones: specific events on the project
Work breakdown structure (WBS): defines the hierarchy of project tasks, subtasks, and work packages
Activities: pieces of work that consume time
◦ Defined within the context of the WBS
LO 4

LO 4

LO 4

Program: New Product Introduction
1.0 Project 1: Engineering Development
1.1 Task 1: Run pilot test
1.2 Task 2: Review process costs
1.3 Task 3: and efficiencies
Prepare Capital
Equipment List
2.0 Project 2: Market Survey
2.1 Task 1:
2.2 Task 2:
Complete Market Survey
Analyze Survey Results
2.3 Task 3: Prepare Marketing Plan
LO 4

LO 4
3.0
Project 3: Manufacturing Start-up
3.1 Task 1:
3.2 Task 2:
Install and Test New Equipment
Establish Manufacturing
Procedures
3.3 Task 3: Detailed Testing of Initial
Output
4.0 Project 4: Sales Force Training
4.1 Task 1: Select Sales People
4.2 Task 2:
4.3 Task 3:
Select Distributors
Train Sales Force and
Distributors




Charts are useful because their visual presentation is easily understood
Software is available to create the charts
Gantt chart: a bar chart showing both the amount of time involved and the sequence in which activities can be performed
LO 4

LO 4

LO 5




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

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.
LO 5

LO 5



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

3.
4.
1.
2.
Identify each activity to be done and estimate how long it will take
Determine the requires sequence and construct a network diagram
Determine the critical path
Determine the early start/finish and late start/finish schedule
LO 5

Activity
Assess customer's needs
Write and submit proposal
Obtain approval
Develop service vision and goals
Train employees
Quality improvement pilot groups
Write assessment report
Designation Immed. Pred. Time (Weeks)
A None 2
B A 1
C
D
E
F
G
B
C
C
D, E
F
1
2
5
5
1
LO 5

Act.
E
F
G
C
D
A
B
Imed. Pred. Time
B
C
None
A
C
D,E
F
5
5
1
1
2
2
1
A(2) B(1) C(1)
D(2)
F(5)
G(1)
E(5)
LO 5

LO 5



Activities on the critical path cannot be delayed without delaying the completion of the project

There are two paths:
A – B – C – D – F – G: 12 weeks
A – B – C – E – F – G: 15 weeks
Activity D can be delayed by up to 3 weeks without delaying the project
The
path is critical –

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

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

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

Activity
A
B
C
D
E
F
G
H
Description
Build internal components
Modify roof and floor
Construct collection stack
Pour concrete and install frame
Build high-temperature burner
Install pollution control system
Install air pollution device
Inspect and test
LO 5
Immed. Preds.
-
-
A
A, B
C
C
D, E
F, G

2
A C
2
Start
B
3
D
4
Start is a “Dummy” activity with 0 duration
LO 5
E
4
G
5
F
3
2
H

LO 5
Earliest
Start
Latest
Start
ES
LS
EF
LF
Earliest
Finish
Latest
Finish

A
Start
B
LO 5
C
E
D G
F
H



Four paths in the network:
Path 1: Start – A – C – F – H: 9 weeks
Path 2: Start – A – C – E – G – H: 15 weeks
Path 3: Start – A – D – G – H: 13 weeks
Path 4: Start – B – D – G – H: 14 weeks
Path 2 is critical
LO 5






A, C, E, G, and H are on the critical path and so they have 0 slack
B is on path 4, so its slack is 15 – 14 = 1
D is on paths 3 and 4, so its slack is 15 – Max
(13,14) = 1
F is on path 1, so its slack is 15 – 9 = 6
An activity can be delayed by its slack and not delay the project completion
LO 5

0
0
0
Start
0
A
0
2
A
Slack=0
2
2
0
1
B
B
3
3
4
Slack=1
LO 5
2
2
C
2
H
C
4
4
Slack=0
3
4
D
H
D
7
4
8
Slack=1
4
10
4
E
8
4
4
8
Slack=0
8
F
F
H
3
7
13
Slack=6
13
8
5
13
Slack=0
13
13
H
H
2
15
15
Slack=0

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 1112 13
1415 16
LO 5

LO 5

LO 5





A-C-F-G:
A-C-E-G:
A-B-D-F-G:
A-B-D-E-G:
38
35
38
35

Paths A-C-F-G and A-B-D-F-G are

LO 5



If a single time estimate is not reliable, then use three time estimates
◦ a = Optimistic (Minimum)
◦ b = Pessimistic (Maximum)
◦ m = Most likely
Allows us to obtain a probability estimate for completion time for the project
LO 5

LO 5
Task
A
B
C
D
E
F
G
H
I
Immediate
Predecesors Optimistic Most Likely Pessimistic
None
None
3
2
6
4
15
14
A
A
C
D
6
2
5
3
12
5
11
6
30
8
17
15
B
E,F
G,H
3
1
4
9
4
19
27
7
28

ET(A)=
[3+4(6)+15]/6
Task
A
B
C
D
E
F
G
H
I
Immediate
Predecesors
None
None
A
A
C
D
B
E,F
G,H
Expected
Time
7
5.333
14
5
11
7
11
4
18
ET(A)=42/6=7
Immediate
Task Predecesors Optimistic Most Likely Pessimistic
A
B
None
None
3
2
6
4
15
14
E
F
C
D
A
A
C
D
6
2
5
3
12
5
11
6
30
8
17
15
G
H
I
B
E,F
G,H
3
1
4
9
4
19
27
7
28
Expected Time =
Opt. Time + 4(M ost Likely Time) + Pess. Time
6
LO 5

ET(B)=
[2+4(4)+14]/6
Task
A
B
C
D
E
F
G
H
I
Immediate
Predecesors
None
None
A
A
C
D
B
E,F
G,H
Expected
Time
7
5.333
14
5
11
7
11
4
18
ET(B)=32/6=5.333
Immediate
Task Predecesors Optimistic Most Likely Pessimistic
A
B
None
None
3
2
6
4
15
14
E
F
C
D
A
A
C
D
6
2
5
3
12
5
11
6
30
8
17
15
G
H
I
B
E,F
G,H
3
1
4
9
4
19
27
7
28
Expected Time =
Opt. Time + 4(M ost Likely Time) + Pess. Time
6
LO 5

Task
A
B
C
D
E
F
G
H
I
Immediate
Predecesors
None
None
A
A
C
D
B
E,F
G,H
Expected
Time
7
5.333
14
5
11
7
11
4
18
ET(C)=
[6+4(12)+30]/6
ET(C)=84/6=14
Immediate
Task Predecesors Optimistic Most Likely Pessimistic
A
B
None
None
3
2
6
4
15
14
E
F
C
D
A
A
C
D
6
2
5
3
12
5
11
6
30
8
17
15
G
H
I
B
E,F
G,H
3
1
4
9
4
19
27
7
28
Expected Time =
Opt. Time + 4(M ost Likely Time) + Pess. Time
6
LO 5

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

What is the probability of finishing this project in less than 53 days?
LO 5 p(t < D)
D=53 T
E
= 54
Z =
D - T
E
  cp
2 t

LO 5

2
= (
P essim . - O p tim .
)
2
6
Task
A
B
C a m b
Optimistic Most Likely Pessimistic Variance
4 3
2
6
6
4
12
15
14
30 16
D
E
F
G
H
3
3
2
5
1
5
11
6
9
4
I 4 19
(Sum the variance along the critical path .)
8
17
15
27
7
28
 
2
4
1
16
= 41

p(t < D)
Z =
D=53
D - T
E
  cp
2
T
E
= 54
=
53 - 54
= -.156
41 p(Z < -.156) = .438, or 43.8 % (NORMSDIST(-.156)) t
There is a 43.8% probability that this project will be completed in less than 53 weeks.
LO 5


What is the probability that the project duration will exceed 56 weeks?
LO 5

p(t < D) t
T
E
= 54
D=56
Z =
D - T
E
  cp
2
=
56 - 54
= .312
41 p(Z > .312) = .378
, or 37.8 % (1-NORMSDIST(.312))
LO 5

LO 5

LO 5
Two critical paths:
A-C-F-G: var. = 11 8/9 = 11.89
A-B-D-F-G var. = 10 2/9 = 10.33
Take larger var.

LO 5
Probabilit y of Finishing
Z

D

 
T
E
2 cp

So prob

0.1922
35

38
11 .
89 in 35 Weeks
 
0 .
87


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
LO 6

3.
4.
5.
1.
2.
Prepare a CPM-type network diagram
Determine the cost per unit of time to expedite each activity
Compute the critical path
Shorten the critical path at the least cost
Plot project direct, indirect, and total-cost curves and find the minimum-cost schedule
LO 6

LO 6

LO 6

LO 6

Assumes indirect cost is a constant
$10 up to 8 days and increases
$5/day thereafter
8 days is lowest total cost
LO 6

LO 6




Four paths in the network:
Path 1: Start – A – C – F – H: 9 weeks
Path 2: Start – A – C – E – G – H: 15 weeks
Path 3: Start – A – D – G – H: 13 weeks
Path 4: Start – B – D – G – H: 14 weeks
Path 2 is critical
Desire to crash this project by two weeks at the smallest direct cost

Act. NT CT NC CC CC/WK CP?
A 2 1 22,000 22,750 750 Y
B 3 1 30,000 34,000 2000 N
C 2 1 26,000 27,000 1,000 Y
D 4 3 48,000 49,000 1,000 N
E 4 2 56,000 58,000 1,000 Y
F 3 2 30,000 30,500 500 N
G 5 2 80,000 84,500 1,500 Y
H 2 1 16,000 19,000 3,000 Y
Total: 308,000 to complete the project in 15 weeks – called the normal cost
LO 6

LO 6




Select the activity with smallest crash cost per week that is on the critical path – activity A at a cost of $750
Start – B – D – G – H is also critical (14 wks)
Crash G by 1 week at a cost of $1,500 to reduce the project by an additional week
(vs. crashing C and D at a combined cost of $2,000)
Cost to complete project in 13 weeks =
$308,000 + 750 + 1,500 = $310,250




In addition to scheduling each task, must assign resources
Software can spot over-allocation
◦ Allocations exceed resources
Must either add resources or reschedule
◦ Moving a task within slack can free up resources
LO 1




Actual progress on a project will be different from the planned progress
◦ Planned progress is called the baseline
A tracking Gantt chart superimposes the current schedule onto a baseline so deviations are visible
Project manager can then manage the deviations
LO 1