1. Unique, one-time operations designed to
accomplish a specific set of objectives in a
limited time frame.
2. 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
3. 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 communications
4. 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
5. What are the tools?
Work breakdown structure
Network diagram
Gantt charts
Risk management
6. Project X
Level 1
Level 2
Level 3
Level 4
7. 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
8. PERT: Program Evaluation and
Review Technique
CPM: Critical Path Method
Graphically displays project activities
Estimates how long the project will take
Indicates most critical activities
Show where delays will not affect project
9. Network (precedence) diagram
Diagram of project activities that shows a sequential
relationships by use of arrows and nodes
Activity-on-arrow (AOA)
Network diagram convention in which arrows designate
activities.
Activity-on-node (AON)
Network diagram convention in which nodes designate
activities.
10. 1
2
3
4
5 6
Locate
facilities
Order
furniture
Furniture
setup
Interview
Hire and
train
Remodel
Move
in
AOA
11. 1
2
3
5
6
Locate
facilities
Order
furniture
Furniture
setup
Interview
Remodel
Move
in
4
Hire and
train
7S
AON
12. Activities
Project steps that consume resources and/or time.
Events
The starting and finishing of activities, designated by nodes AOA
convention.
Path
A 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
13. Deterministic
Time estimates that are fairly certain
Probabilistic
Estimates of times that allow for variation
14. 1 65
4
3
2
1 WEEKS
MOVE IN
Determine the ff:
a. The length of each path c. The expected length of time
b. The critical path d. The amount of slack time for each path
16. Network activities
ES: early start
EF: early finish
LS: late start
LF: late finish
Used to determine
Expected project duration
Slack time
Critical path
17. FOUR PIECES OF INFORMATION ABOUT
THE NETWORK ACTIVITIES:
ES-the earliest time activity can start
EF-the earliest time the activity can finish
LS-the latest time the activity can start and not
delay the project
LF-the latest time the activity can finish and not
delay the project
18. Computation of earliest starting and finishing time is aided
by two simple rules:
1. The earliest finish time for any activity is equal to its
earliest start time plus its expected duration, t:
EF = ES + t
2. ES for activities at nodes with one entering arrow is
equal to EF of the entering arrow.
ES for activities leaving nodes with multiple entering
arrows is equal to the largest EF of the entering arrow.
19. Computation of latest starting and finishing time is aided
by the use two rules:
1. The latest starting time for each activity is equal to its
latest finishing time minus its expected duration, t:
LS = LF - t
2. For nodes with one leaving arrow, LF for arrows
entering that node equals the LS of the leaving arrow.
For nodes with multiple leaving arrows, LF for arrows
entering that node equals the smallest LS of leaving
arrows.
20. 1 65
4
3
2
1 WEEKS
MOVE IN
Determine the ff:
a. ES c. EF
b. LS d. LF
24. Consider the following consulting project:
Develop a critical path diagram and determine
the duration of the critical path and slack times
for all activities.
Activity Designation Immed. Pred. Time (Weeks)
Assess customer's needs A None 2
Write and submit proposal B A 1
Obtain approval C B 1
Develop service vision and goals D C 2
Train employees E C 5
Quality improvement pilot groups F D, E 5
Write assessment report G F 1
25. A(2) B(1) C(1)
D(2)
E(5)
F(5) G(1)
A None 2
B A 1
C B 1
D C 2
E C 5
F D,E 5
G F 1
Act. Imed. Pred. Time
26. ES=9
EF=14
ES=14
EF=15
ES=0
EF=2
ES=2
EF=3
ES=3
EF=4
ES=4
EF=9
ES=4
EF=6
A(2) B(1) C(1)
D(2)
E(5)
F(5) G(1)
Hint: Start with ES=0
and go forward in the
network from A to G.
27. ES=9
EF=14
ES=14
EF=15
ES=0
EF=2
ES=2
EF=3
ES=3
EF=4
ES=4
EF=9
ES=4
EF=6
A(2) B(1) C(1)
D(2)
E(5)
F(5) G(1)
LS=14
LF=15
LS=9
LF=14
LS=4
LF=9
LS=7
LF=9
LS=3
LF=4
LS=2
LF=3
LS=0
LF=2
Hint: Start with LF=15 or
the total time of the project
and go backward in the
network from G to A.
29. 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
30. Activity
start
Optimistic
time
Most likely
time (mode)
Pessimistic
time
to tptm te
31. te = to + 4tm +tp
6
te = expected time
to = optimistic time
tm = most likely time
tp = pessimistic time
32. 2 = (tp – to)2
36
2 = variance
to = optimistic time
tp = pessimistic time
33. The network diagram for a project is shown in the
accompanying figure, with three time estimates for each
activity. Activity times are in weeks. Do the following:
a. Compute the expected time for each activity and the
expected duration for each path
b. Identify the critical path
c. Compute the variance of each activity and the
variance and standard deviation of each path
34. 3-4-5
d
3-5-7
e
5-7-9
f
2-4-6
b
4-6-8
h
Optimistic
time
Most likely
time
Pessimistic
time
35. a.
36. b. The path that has the longest expected duration is the
critical path. Because path d-e-f has the largest path total, it
is the critical path.
c.
37. Z =
Specified time – Path mean
Path standard deviation
Z indicates how many standard deviations
of the path distribution the specified tine
is beyond the expected path duration.
38. Using information given from the previous example answer the ff:
a. What is the probability that the project can be completed within 17
weeks of its start?
b. What is the probability that the project will be completed within 15
weeks of its start?
c. What is the probability that the project will not be completed within
15 weeks of its start?
40. a.
Path Probability of Completion in 17 weeks
a-b-c + 7.22 1.00
d-e-f +1.00 .8413
g-h-I +3.27 1.00
P(Finish by week 17) = 1.00 x .8413 x 1.00 = .8413
Z = 17– Path Duration
Path standard deviation
41. b.
Path Probability of Completion in 17 weeks
a-b-c + 5.15 1.00
d-e-f - 1.00 .1587
g-h-I + 1.40 .9192
P(Finish by week 15) = 1.00 x .1587 x .9192 = .1459
c. 1 - .1459 = .8541
Z = 15– Path Duration
Path standard deviation
42. Task
Immediate
Predecesors Optimistic Most Likely Pessimistic
A None 3 6 15
B None 2 4 14
C A 6 12 30
D A 2 5 8
E C 5 11 17
F D 3 6 15
G B 3 9 27
H E,F 1 4 7
I G,H 4 19 28
43. ET(A)= 3+4(6)+15
6
ET(A)=42/6=7
Task
Immediate
Predecesors
Expected
Time
A None 7
B None 5.333
C A 14
D A 5
E C 11
F D 7
G B 11
H E,F 4
I G,H 18
Task
Immediate
Predecesors Optimistic Most Likely Pessimistic
A None 3 6 15
B None 2 4 14
C A 6 12 30
D A 2 5 8
E C 5 11 17
F D 3 6 15
G B 3 9 27
H E,F 1 4 7
I G,H 4 19 28
Expected Time =
Opt. Time + 4(Most Likely Time) + Pess. Time
6
44. Example 2. Network
A(7)
B
(5.333)
C(14)
D(5)
E(11)
F(7)
H(4)
G(11)
I(18)
Duration = 54 Days
45. Example 2. Probability Exercise
What is the probability of finishing this project in
less than 53 days?
p(t < D)
TE = 54
Z =
D - TE
cp
2
t
D=53
46. Activity variance, = (
Pessim. - Optim.
6
)2 2
Task Optimistic Most Likely Pessimistic Variance
A 3 6 15 4
B 2 4 14
C 6 12 30 16
D 2 5 8
E 5 11 17 4
F 3 6 15
G 3 9 27
H 1 4 7 1
I 4 19 28 16
(Sum the variance along the critical path.) 2
= 41
47. There is a 43.8% probability that this project will be
completed in less than 53 weeks.
p(Z < -.156) = .438, or 43.8 % (NORMSDIST(-.156)
Z =
D - T
=
53- 54
41
= -.156E
cp
2
TE = 54
p(t < D)
t
D=53
48. What is the probability that the project
duration will exceed 56 weeks?
49. t
TE = 54
p(t < D)
D=56
Z =
D - T
=
56 - 54
41
= .312E
cp
2
p(Z > .312) = .378, or 37.8 % (1-NORMSDIST(.312))
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