Transcript Chapter 10

Chapter 10
Projects
McGraw-Hill/Irwin
©2011 The McGraw-Hill Companies, All Rights Reserved
Learning Objectives
 Explain what project management is and
why it is important.
 Identify the different ways projects can be
structured.
 Describe how projects are organized into
major subprojects.
 Understand what a project milestone is.
 Determine the “critical path” for a project.
 Demonstrate how to “crash,” or reduce the
length, of a project.
10-2
What is Project Management?
 Project:

A series of related jobs usually directed toward
some major output and requiring a significant
period of time to perform


Usually infrequent
Exact duration difficult to estimate
 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

Control may be difficult

LO 1

Makeup of the project team
Expert among experts
10-3
Types of Development
Projects
LO 2
10-4
Structuring Projects
 Pure project
 Functional project
 Matrix project
LO 2
10-5
Pure Project
 In Pure Project Self-contained team
works full-time on the project
 Advantages




The project manager has full authority
Team members report to one boss
Shortened communication lines
Team pride, motivation, and commitment are
high
LO 2
10-6
Pure Project
 Disadvantages

Duplication of resources


Organizational goals and policies are ignored


Team members detached from headquarters
Lack of technology transfer


Equipment and people not shared across projects
Due to weakened functional division
Team members have no functional area
"home“

Project termination often delayed as members
worry about life after the project
LO 2
10-7
Functional Project
A functional project is housed within a functional division.
LO 2
Example: Project “B” is in the functional area of
Research and Development
10-8
Functional Project
 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
10-9
Matrix Project
LO 2
10-10
Matrix Project
 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



LO 2
Functional manager often listened to first because of direct
influence on promotion and raises
Depends on project manager’s negotiating skills
Potential for sub-optimization


PM’s could hold resources for own projects
Other projects could be harmed
10-11
Phases of Project
Management
 Planning



Work breakdown structure
Feasibility study
Precedence determination
 Scheduling


Construct the chart and apply times
Identify priorities
 Execution and control

Review and modify the project
10-12
Work Breakdown Structure
 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
 Project milestones: specific events on the
project
LO 3
10-13
Work Breakdown Structure
 Work breakdown structure (WBS):
defines the hierarchy of project tasks,
subtasks, and work packages






Defines hierarchy of project tasks, subtasks and work
packages
Allows the elements to be worked on independently
Make team manageable in size
Give authority to carry out the project
Monitor and measure the project
Provide the required resources
 Activities: pieces of work that consume
time
LO 4

Defined within the context of the WBS
10-14
An Example of a Work
Breakdown Structure
Quality is Job 1
ISO 9001
MPG
Data Collection
Taurus, Sable
LO 3
10-15
Work Breakdown Structure,
Large Optical Scanner Design
LO 3
10-16
Project Control Charts
 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 3
10-17
Sample of Graphic Project
Reports
LO 3
10-18
Earned Value Management
(EVM)
 A technique for measuring project
progress in an objective manner
 Has the ability to combine
measurements of scope, schedule, and
cost in a project
 Provides a method for evaluating the
relative success of a project at a point
in time
LO 3
10-19
Essential Features of any
EVM Implementation
 A project plan that identifies the
activities to be accomplished
 A valuation of each activity work
 Predefined earning or costing rules to
quantify the accomplishment of work
LO 3
10-20
Earned Value Management
Charts
LO 3
10-21
Project Tracking Without
EVM
 Chart A shows the cumulative cost budget
for the project as a function of time

Blue line, labeled BCWS
 Also shows the cumulative actual cost of
the project

Red line
 Appears project was over budget through
week 4 and then under budget
 What is missing is any understanding of
how much work has been accomplished
LO 3
10-22
Project Tracking With EVM
 Chart B shows the BCWS curve along
with the BCWP curve from chart A
 Technical performance started more
rapidly than planned but then slowed
significantly and feel behind at week 7
 Chart illustrates the schedule
performance aspect of EVM
LO 3
10-23
Project Tracking With EVM
 Chart C shows the same BCWP curve
with actual cost data
 Project is actually under budget,
relative to the amount of work
accomplished
 Chart D shows all three curves together

Typical for EVM line charts
LO 3
10-24
Example: Earned Value
Management
LO 3
10-25
Example: Budgeted Cost of
Work Scheduled (BCWS)
A.
B.
C.
D.
100% of $18K = $18K
100% of $10K = $10K
80% of $20K = $16K
15% of $40K = $6K
BCWS = $18K+$10K+$16K+$6K = $50K
LO 3
10-26
Example: Budgeted Cost of
Work Performed (BCWP)
A.
B.
C.
D.
100% of $18K = $18K
80% of $10K = $8K
70% of $20K = $14K
0% of $40K = $0
BCWP = $18K+$8K+$14K+$0 = $40K
LO 3
10-27
Example: Performance
Measures
SV  BCWP  BCWS  $40K  $50K  $10K
BCWP $40K
SPI 

 0.8
BCWS $50K
CV  BCWP  AC  $40K  $45K  $5
BCWP $40K
CPI 

 0.89
AC
$45K
LO 3
10-28
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
LO 3
the critical path(s) in the project networks
10-29
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.
10-30
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
10-31
Critical Path Method (CPM)
 Identify each activity to be done and
estimate how long it will take
 Determine the required sequence and
construct a network diagram
 Determine the critical path

Obtain all project and activity timing information
 Determine the early start/finish and late
start/finish schedule
LO 5
10-32
Example 1: Critical Path
Method (Single Time Estimate)
LO 5
10-33
Example 1: Activity Sequencing
and Network Construction
LO 5
10-34
Determining the Critical
Path & Slack
 A path is a sequence of connected
activities running from start to end node in
a network
 The critical path is the path with the longest
duration in the network


It is also the minimum time, under normal
conditions, to complete the project
Delaying activities on the critical path will delay
completion time for the project
 A slack is amount of scheduling flexibility

Critical activities have no slacks
10-35
Example 1: Finished
Schedule
Forward Pass
Backward Pass
SLACK=LS-ES=LF-EF
Slack=(36-33)=(31-28)=3
A PATH IS CRITICAL IF:
ES=LS,
EF=LF, and
EF-ES=LF-LS=D
LO 5
10-36
Critical Path Method:
Three Activity Time Estimates
 If a single time estimate is not reliable,
then use three time estimates
 Task times assumed variable
(probabilistic)



Minimum (optimistic)
Maximum (pessimistic)
Most likely (normally)
 Allows us to obtain a probability estimate
for completion time for the project
LO 5
10-37
Finding Activity Time and
Variance
a  4m  b
ET 
6
ba
 

 6 
a  minimum
m  most likely
b  maximum
2
2
LO 5
ET  expect edt ime
10-38
Example 2: Activity Expected
Times and Variances
ET 
a  4m  b
6
ba
 

 6 
D  TE
Z
2
2
LO 5

2
cp
Variance of Project = Sum of Variance of Activities on CP
10-39
Example: Network with
Three Time Estimates
LO 5
10-40
Probability Estimates
p(t < D)
t
D=35
TE = 38
Whatis theP robability of Finishingin 35 Weeks
D  TE
35  38
Z

 0.87
2
  cp 11.89
P(T  35)  P(Z  0.87)
From Appendix G, page 765
 0.19766
There is about a 19.78% probability that this
project will be completed in 35 weeks or less.
10-41
Probability Estimates
p(t < D)
0.95994
TE = 38
D=44
t
Whatis theP robability projectduration will exceed44 Weeks
D  TE
44  38
Z

 1.740
2
  cp 11.89
P(T  44)  P(Z  1.740)
 1  P(Z  1.740)
 1 0.95994  0.04006
There is about a 4% probability that this
project will not be completed in 44 weeks.
10-42
Time-Cost Models and
Project Crashing
 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
10-43
Time-Cost Models and
Project Crashing
Cost $
100+C
•

CC
100+B

CC-NC
100+A

100
NC
CC  NC
NT  CT
5
CT
10
Time
20
NT-CT
30
NT
10-44
Procedure for Project
Crashing
 Prepare a CPM-type network diagram


Normal and crash times for each
Normal and crash costs for each
 Determine the cost per unit of time to
expedite each activity

LO 6
Incremental cost = (CC-NC)/(NT-CT)
 Compute the critical path
 Shorten the critical path at the least cost
 Plot project direct, indirect, and totalcost curves and find the minimum-cost
schedule
10-45
Time-Cost Model: Example
For the project with the following times, find the cost of
reducing its duration to its lowest possible time. If the
project is reduced by just 2 days, how much would it cost?
Activity
A
CRASH
NORMAL
Time
Cost Time Cost IN CL=NT-CT
2
6
1
10 4
1
B
5
9
2
18
3
3
C
4
6
3
8
2
1
D
3
5
1
9
2
2
10-46
1
NORMAL
ES=
EF=
A ,2
LS=
LF=
2
ES=
EF=
B ,5
10
ES=
EF=
LS=
LF=
C ,4
A,1
LS=
LF=
LS=
LF=
LS=
LF=
$26
ES=
EF=
3
ES=
EF=
A ,2
LS=
LF=
ES=
EF=
D,3
ES=
EF=
ES=
EF=
LS=
LF=
8
D,1
LS=
LF=
ES=
EF=
LS=
LF=
$30
ES=
EF=
B
,2
5
ES=
EF=
LS=
LF=
D ,1
LS=
LF=
ES=
EF=
C ,3
LS=
LF=
$45
ES=
EF=
4
B ,5
C ,4
CRASHED
ES=
EF=
A,2
LS=
LF=
B ,2
7
ES=
EF=
LS=
LF=
D,1
LS=
LF=
ES=
EF=
C,4
LS=
LF=
$39
10-47
1
2
ES=
EF=
ES=
EF=
A,2
LS=
LF=
B ,4
7
ES=
EF=
LS=
LF=
C ,4
A ,1
LS=
LF=
LS=
LF=
LS=
LF=
$33
ES=
EF=
3
ES=
EF=
A,1
LS=
LF=
ES=
EF=
D,1
ES=
EF=
ES=
EF=
LS=
LF=
ES=
EF=
LS=
LF=
5
ES=
EF=
LS=
LF=
6
D,1
LS=
LF=
ES=
EF=
LS=
LF=
$37
ES=
EF=
ES=
EF=
D,1
A
LS=
LF=
LS=
LF=
$42
B ,4
C ,4
4
B ,3
C,3
ES=
EF=
B
ES=
EF=
LS=
LF=
D
LS=
LF=
ES=
EF=
C
LS=
LF=
10-48
CPM Assumptions/
Limitations
 Project activities can be identified as
entities (There is a clear beginning and
ending point for each activity.)

But this is seldom true
 Project activity sequence relationships
can be specified and networked

Sometime we do not know all project
activities ahead of time
10-49
CPM Assumptions/
Limitations
 Project control should focus on the critical
path

But there are near-critical activities
 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


But activities are infrequent. Can we assume
normality?
Are we really sure that it is beta distributed?
10-50
Managing Resources
 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
10-51
Tracking Progress
 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
10-52
Chapter 10: Problem 8
#s: a, b, c
ES=
EF=
ES=
EF=
B,2
E,3
ES=
EF=
H,4
ES=
EF=
A,3
26.8
ES=
EF=
ES=
EF=
ES=
EF=
C,6
F,2
I,5
ES=
EF=
26.8
K,6.8
ES=
EF=
D,5
ES=
EF=
G,7
ES=
EF=
J,4
10-53
Chapter 10: Problem 8
#: d(1)
ES=
EF=
ES=
EF=
B,2
E,1
ES=
EF=
H,4
ES=
EF=
A,3
• Not on CP
• Duration still 26.8
• No Gain
• Loss $1,500
ES=
EF=
ES=
EF=
ES=
EF=
C,6
F,2
I,5
ES=
EF=
26.8
K,6.8
ES=
EF=
D,5
ES=
EF=
G,7
ES=
EF=
J,4
10-54
Chapter 10: Problem 8
#: d(2)
ES=
EF=
ES=
EF=
B,2
E,3
ES=
EF=
H,4
ES=
EF=
A,3
24.8
• CP changes
• Duration 25.8
• Net Gain 1 day
• Cost $1,500
• Savings $1,000
• Net Loss $ 500
ES=
EF=
ES=
EF=
ES=
EF=
C,4
F,2
I,5
ES=
EF=
25.8
K,6.8
ES=
EF=
D,5
ES=
EF=
G,7
ES=
EF=
J,4
10-55
Chapter 10: Problem 8
#: d(3)
ES=
EF=
ES=
EF=
B,2
E,3
ES=
EF=
• Not on CP
• Duration still 26.8
• No Gain
• Loss $1,500
H,4
ES=
EF=
A,3
ES=
EF=
ES=
EF=
ES=
EF=
C,6
F,2
I,5
ES=
EF=
26.8
K,6.8
ES=
EF=
D,5
Reduced by 2 Days
ES=
EF=
G,5
ES=
EF=
J,4
10-56
Chapter 10: Problem 8
Z
30  26.8
 1.37

5.47
#: e
B,2
ES=
EF=
E,3
ES=
EF=
26.8 30
  5.47
ES=
EF=
PD  30  PZ  1.37
 1  PZ  1.37
 1  0.91149
H,4
ES=
EF=
A,3
0.00 1.37
 1
 0.08851
ES=
EF=
ES=
EF=
ES=
EF=
C,6
F,2
I,5
ES=
EF= 26.8
K,6.8
ES=
EF=
D,5
ES=
EF=
G,7
ES=
EF=
J,4
10-57
Chapter 10: Problem 9
a&b
ES=
EF=
21
B,2
LS=
ES=
LF=
EF=
E,5
LS=
ES=
EF=
H,3
LF=
LS=
LF=
23
ES=
EF=
A,4
LS=
LF=
ES=
EF=
C,4
LS=
ES=
26
LF=
LF=
ES=
21
EF=
F,6
LS=
EF=
D,3
LS=
ES=
LF=
EF=
G,2
LS=
LF=
ES=
ES=
EF=
I,5
LS=
Activity
1—A
2—B
3—C
4—D
5—E
6—F
7—G
8—H
9—I
10—J
EF=
26
J,7
LF=
Cost $
*
10,000
10,000
10,000
10,000
10,000
10,000
10,000
10,000
*
LS=
CL
*
1
3
2
4
5
1
2
4
*
LF=
10-58
Chapter 10: Problem 9
ES=
c
EF=
21
B,2
LS=
ES=
LF=
EF=
E,5
LS=
ES=
EF=
H,3
LF=
LS=
LF=
23
ES=
EF=
ES=
A,4
LS=
Activity
1—A
2—B
3—C
4—D
5—E
6—F
7—G
8—H
9—I
10—J
C, 4
LF=
Cost $
*
10,000
10,000
10,000
10,000
10,000
10,000
10,000
10,000
*
EF=
CL
*
1
3
2
4
5
1
2
4
*
ES=
26
EF=
F, 6
LF=
LS=
LF=
ES=
EF=
ES=
EF=
LS=
LF=
21
EF=
I, 5
LS=
D,3
ES=
LS=
LF=
ES=
EF=
26
J,7
LS=
LF=
G,2
LS=
LF=
10-59
Chapter 10: Problem 9
ES=
EF=
21
B,2
LS=
ES=
LF=
EF=
E,5
LS=
ES=
EF=
H,3
LF=
LS=
LF=
20
ES=
EF=
ES=
A,4
LS=
Activity
1—A
2—B
3—C
4—D
5—E
6—F
7—G
8—H
9—I
10—J
C, 1
LF=
Cost $
*
10,000
10,000
10,000
10,000
10,000
10,000
10,000
10,000
*
EF=
CL
*
1
0
2
4
5
1
2
4
*
ES=
23
EF=
F, 6
LS=
LF=
LS=
LF=
ES=
EF=
ES=
EF=
D,3
LS=
LF=
21
G,2
LS=
LF=
ES=
EF=
I, 5
LS=
LF=
• Reduce C by 3
ES=
EF=
23
J,7
LS=
LF=
$30,000
10-60
Chapter 10: Problem 9
ES=
EF=
21
B,2
LS=
ES=
LF=
EF=
E,5
LS=
ES=
EF=
H,3
LF=
LS=
LF=
20
ES=
EF=
ES=
A,4
LS=
Activity
1—A
2—B
3—C
4—D
5—E
6—F
7—G
8—H
9—I
10—J
C, 1
LF=
Cost $
*
10,000
10,000
10,000
10,000
10,000
10,000
10,000
10,000
*
EF=
CL
*
1
0
2
4
4
1
2
4
*
ES=
22
EF=
F, 5
LS=
LF=
LS=
LF=
ES=
EF=
ES=
EF=
D,3
LS=
LF=
21
G,2
LS=
LF=
ES=
EF=
I, 5
LS=
LF=
ES=
EF=
22
J,7
LS=
LF=
• Reduce C by 3
$30,000
• Reduce F or I by 1 $10,000
Total Cost
$40,000
10-61
Chapter 10: Problem 9
ES=
EF=
21
B,2
LS=
ES=
LF=
EF=
E,5
LS=
ES=
EF=
H,3
LF=
LS=
LF=
22
ES=
EF=
ES=
A,4
LS=
Activity
1—A
2—B
3—C
4—D
5—E
6—F
7—G
8—H
9—I
10—J
C, 3
LF=
Cost $
*
10,000
10,000
10,000
10,000
10,000
10,000
10,000
10,000
*
EF=
CL
*
1
2
2
4
2
1
2
4
*
ES=
22
EF=
F, 3
LS=
LF=
LS=
LF=
ES=
EF=
ES=
EF=
D,3
LS=
LF=
21
G,2
LS=
LF=
ES=
ES=
EF=
I, 5
LS=
EF=
22
J,7
LF=
LS=
LF=
• Reduce C by 3
$30,000
• Reduce F or I by 1 $10,000
Total Cost
$40,000
OR
• Reduce F or I by 3 $30,000
• Reduce C by 1
$10,000
Total Cost
$40,000
But new CP=More resources
10-62
Chapter 10: Problem 13
D, 6
B,10
E, 7
G, 4
A,5
C,8
Normal Normal
Activity Time
Cost
A
B
C
D
E
F
G
5
10
8
6
7
4
4
7,000
12,000
5,000
4,000
3,000
6,000
7,000
F, 4
Crash
Time
3
7
7
5
6
3
3
Crash
Cost
13,000
18,000
7,000
5,000
6,000
7,000
9,000
IC CL
10-63
Chapter 10: Problem 13
D, 6
5
25
B,10
E, 7
24
24
24
G, 4 25
A,5
21
C,8
Act
NT
NC
A
B
C
D
E
F
G
5
7,000
10
12,000
8
5,000
6
4,000
7
3,000
4
6,000
4
7,000
$44,000
CT
3
7
7
5
6
3
3
F, 4
CP Reduce
CL
ABDG
1st D
13,000
3000 2
18,000
2000 3
7,000
2000 1 0
1000 1
5,000
3000 1
6,000
1000 1
7,000
2000 1
9,000
CC IC
$
1000
$
Time
45K
24
10-64
Chapter 10: Problem 13
D, 5
24
B,10
E, 7
24
23 22
4
A,5
21
C,8
Act
NT
NC
A
B
C
D
E
F
G
5
7,000
10
12,000
8
5,000
6
4,000
7
3,000
4
6,000
4
7,000
$44,000
CT
3
7
7
5
6
3
3
CC IC
13,000
3000
18,000
2000
7,000
2000
1000
5,000
3000
6,000
1000
7,000
2000
9,000
23
22
22
3 23
G, 4 24
20
F, 4
CL
2
3
1
0
1
1
1 0
1
CP
ABDG
ABDG
Reduce
$
$
Time
1st D
1000
45K
24
2nd G
2000
47K
23
3rd A
3000
50K
22
ACEG
ABDG
ACEG
10-65