4. Planning
The overall goal of project planning is to establish a
pragmatic strategy for controlling, tracking, and
monitoring a complex technical project.
Or,
A Plan is the strategy for the successful
completion of the project. It's a description of the
project steps that produce increasing maturity of
the products or processes produced by the project.
7. Project management generally consists of three phases.
Planning:
Planning involves setting the objectives of the project. Identifying
various activities to be performed and determining the requirement of
resources such as men, materials, machines, etc.
The cost and time for all the activities are estimated, and a network diagram is
developed showing sequential interrelationships (predecessor and successor)
between various activities during the planning stage.
Scheduling:
Basd on the time estimates, the start and finish times for each activity
are worked out by applying forward and backward pass techniques, critical
path is identified, along with the slack and float for the non-critical paths.
Controlling:
Controlling refers to analyzing and evaluating the actual progress
against the plan. Reallocation of resources, crashing and review of projects
with periodical reports are carried out.
8. Management of Projects
1. Planning - goal setting, defining the
project, team organization
2. Scheduling - relates people, money,
and supplies to specific activities and
activities to each other
3. Controlling - monitors resources,
costs, quality, and budgets; revises
plans and shifts resources to meet
time and cost demands
17. GANTT Chart
o A GANTT chart is a type of bar chart that illustrates a
project schedule.
o After the PERT/CPM analysis is completed, the
following phase is to construct the GANTT chart
and then to re allocate resources and re-schedule if
necessary.
o GANTT charts have become a common technique for
representing the phases and activities of a project
work breakdown structure.
o It was introduced by Henry Gantt around 1910 -1915.
18. 18
Gantt Charts
• Gantt charts are used as a tool to monitor and
control the project progress.
• A Gantt Chart is a graphical presentation that
displays activities as follows:
– Time is measured on the horizontal axis. A horizontal
bar is drawn proportionately to an activity’ s expected
completion time.
– Each activity is listed on the vertical axis.
• In an earliest time Gantt chart each bar begins
and ends at the earliest start/finish the activity
can take place.
19. 19
• Gantt chart can be used as a visual aid for
tracking the progress of project activities.
• Appropriate percentage of a bar is shaded to
document the completed work.
• The manager can easily see if the project is
progressing on schedule (with respect to the
earliest possible completion
times).
Gantt Charts-
Monitoring Project
Progress
20. 20
The Purpose of a Gantt Charts
• To illustrate the relationship between project
activities & time.
• To show the multiple project activities on one chart
• To provide a simple & easy to understand
representation of project scheduling
-
21. 21
The Purpose of a Gantt Charts
• Determine Project start date and deadline.
• Gather all information surrounding the list of activities within
project - the Work Breakdown Structure may be useful for this
• Determine how long each activity will take
• Evaluate what activities are dependent on others
• Create Graph shell including the timeline and list of activities.
• Using either Forward Scheduling or Backward Scheduling,
• Begin to add bars ensuring to include dependencies and the ful
• duration for each activity.
22. 22
Steps for Gantt Charts
• Determine Project start date and deadline.
• Gather all information surrounding the list of activities within
project - the Work Breakdown Structure may be useful for this
• Determine how long each activity will take
• Evaluate what activities are dependent on others
• Create Graph shell including the timeline and list of activities.
• Using either Forward Scheduling or Backward Scheduling,
• Begin to add bars ensuring to include dependencies and the ful
• duration for each activity.
23. Gantt Chart
J F M A M J J
Time Period
Activity
Design
Build
Test
J F M A M J J
Time Period
Activity
Design
Build
Test
24. 24
• Advantages.
– Easy to construct
– Gives earliest completion date.
– Provides a schedule of earliest possible start and finish
times of activities.
• Disadvantages
– Gives only one possible schedule (earliest).
– Does not show whether the project is behind schedule.
– Does not demonstrate the effects of delays in any one
activity on the
start of another activity, thus on the project completion time.
Gantt Charts –
Advantages and Disadvantages
25. Gantt Chart
o Characteristics:
• The bar in each row identifies the corresponding task
• The horizontal position of the bar identifies start and end times of the
task
• Bar length represents the duration of the task
• Task durations can be compared easily
• Good for allocating resources and re-scheduling
• Precedence relationships can be represented using arrows
• Critical activities are usually highlighted
• Slack times are represented using bars with doted lines
• The bar of each activity begins at the activity earliest start time (ES)
• The bar of each activity ends at the activity latest finish time (LF).
26. Gantt Chart
o The steps to construct a GANTT chart from the
information obtained by PERT/CPM are:
1. Schedule the critical tasks in the correct position.
2. Place the time windows in which the non-critical
tasks can be scheduled.
3. Schedule the non-critical tasks according to their
earliest starting times.
4. Indicate precedence relationships between tasks.
32. Construct GANTT chart
o Resource Smoothing is a technique used to
re-allocate resources and re-schedule activities.
o In resource smoothing, non-critical tasks are re-
scheduled within their time window.
o Staff Utilization:(duration of activity x staff
requiredfor each activity,all added together)/
(maximum staff required x duration of project)
38. Construct GANTT chart
o The obvious way to reduce the overall project
duration, it is by reducing the duration of the critical
activities.
o Crashing Critical Activities refers to reducing the
duration of a critical activity by allocating more
resources to it.
o The risk is that crashing activities may actually
reduce productivity and increase costs.
40. Construct GANTT chart
Task Prec
eden
ce
Duration ES EF LS LF Slack
Time
Critical
Task
A 2 0 2 17 19 17 No
B 4 0 4 2 6 2 No
C 3 0 3 0 3 0 Yes
D C 5 3 8 3 8 0 Yes
E B 3 4 7 6 9 2 No
F D 5 8 13 8 13 0 Yes
G D,E 7 8 15 9 16 1 No
H F 3 13 16 13 16 0 Yes
I G,H 3 16 19 16 19 0 Yes
45. 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
46. • 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?
Questions Answered by
PERT & CPM
48. 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
49. 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
50. Difference between
CPM & PERT
CPM PERT
• CPM works with fixed
deterministic time
• PERT works with probabilistic
time
• CPM is useful for repetitive
and non complex projects with
a certain degree of time
estimates.
• PERT is useful for non
repetitive and complex projects
with uncertain time estimates.
• CPM includes time-cost trade
off.
• PERT is restricted to time
variable.
• CPM- for construction projects. • PERT- used for R&D
programs.
51. Activity on Node (AoN)
2
4? Years
Enroll
Receive
Certificate
Project: Obtain a college degree (B.S.)
1 month
Attend class,
study etc.
1
1 day
3
52. Activity on Arc (AoA)
4,5 ?
Years
Enroll
Receive
Certificate
Project: Obtain a college degree (B.S.)
1 month
Attend
class,
study,
etc.
1
1 day
2 3 4
54. A Comparison of AON and
AOA Network Conventions
Activity on Activity Activity on
Node (AON) Meaning Arrow (AOA)
A comes before
B, which comes
before C
(a) A B C
B
A C
A and B must both
be completed
before C can start
(b)
A
C
C
B
A
B
B and C cannot
begin until A is
completed
(c)
B
A
C
A
B
C
Figure 3.5
55. A Comparison of AON and
AOA Network Conventions
Activity on Activity Activity on
Node (AON) Meaning Arrow (AOA)
C and D cannot
begin until both
A and B are
completed
(d)
A
B
C
D B
A C
D
C cannot begin until
both A and B are
completed; D
cannot begin until B
is completed. A
dummy activity is
introduced in AOA
(e)
C
A
B D
Dummy activity
A
B
C
D
Figure 3.5
56. A Comparison of AON and
AOA Network Conventions
Activity on Activity Activity on
Node (AON) Meaning Arrow (AOA)
B and C cannot
begin until A is
completed. D
cannot begin until
both B and C are
completed. A
dummy activity is
again introduced
in AOA.
(f)
A
C
D
B A B
C
D
Dummy
activity
Figure 3.5
57. AON Example
Activity Description
Immediate
Predecessors
A Build internal components —
B Modify roof and floor —
C Construct collection stack A
D Pour concrete and install frame A, B
E Build high-temperature burner C
F Install pollution control system C
G Install air pollution device D, E
H Inspect and test F, G
Milwaukee Paper Manufacturing's
Activities and Predecessors
Table 3.1
58. AON Network for
Milwaukee Paper
A
Start
B
Start Activity
Activity A
(Build Internal Components)
Activity B
(Modify Roof and Floor)
Figure 3.6
59. AON Network for
Milwaukee Paper
Figure 3.7
C
D
A
Start
B
Activity A Precedes Activity C
Activities A and B Precede
Activity D
60. AON Network for
Milwaukee Paper
G
E
F
H
C
A
Start
D
B
Arrows Show Precedence
Relationships
Figure 3.8
62. Determining the
Project Schedule
Perform a Critical Path Analysis
The critical path is the longest path
through the network
The critical path is the shortest time in
which the project can be completed
Any delay in critical path activities delays
the project
Critical path activities have no slack time
63. 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
66. 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
67. Critical Path
Analysis Example
Event
ID
Pred. Description Time
(Wks)
A None Prepare Site 1
B A Pour fdn. & frame 6
C A Buy shrubs etc. 3
D B Roof 2
E D Do interior work 3
F C Landscape 4
G E,F Move In 1
70. 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
71. Activity ES EF LS LF Slack
A 0 1
B
C
D
E
F
Activity A
Earliest Start Solution
For starting activities, ES = 0.
A
E
D
B
C F
G
1
6 2 3
1
4
3
72. Activity ES EF LS LF Slack
A 0 1
B 1 7
C 1 4
D 7 9
E 9 12
F 4 8
G 12 13
Earliest Start Solution
A
E
D
B
C F
G
1
6 2 3
1
4
3
73. 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
74. Activity ES EF LS LF Slack
A 0 1
B 1 7
C 1 4
D 7 9
E 9 12
F 4 8
G 12 13 13
Earliest Start Solution
A
E
D
B
C F
G
1
6 2 3
1
4
3
75. Activity ES EF LS LF Slack
A 0 1 0 1
B 1 7 1 7
C 1 4 4 7
D 7 9 7 9
E 9 12 9 12
F 4 8 7 12
G 12 13 12 13
Latest Finish Solution
A
E
D
B
C F
G
1
6 2 3
1
4
3
76. Activity ES EF LS LF Slack
A 0 1 0 1 0
B 1 7 1 7 0
C 1 4 5 8 4
D 7 9 7 9 0
E 9 12 9 12 0
F 4 8 8 12 4
G 12 13 12 13 0
Compute Slack
77. Determining the
Project Schedule
Perform a Critical Path Analysis
Activity Description Time (weeks)
A Build internal components 2
B Modify roof and floor 3
C Construct collection stack A 2
D Pour concrete and install frame A,B 4
E Build high-temperature burner C 4
F Install pollution control system C 3
G Install air pollution device D,E 5
H Inspect and test F,G 2
Total Time (weeks) 25
Table 3.2
78. Determining the
Project Schedule
Perform a Critical Path Analysis
Figure 3.10
A
Activity Name or
Symbol
Earliest
Start ES
Earliest
Finish
EF
Latest
Start
LS Latest
Finish
LF
Activity Duration
2
81. E
4
F
3
G
5
H
2
4 8 13 15
4
8 13
7
D
4
3 7
C
2
2 4
ES/EF Network for
Milwaukee Paper
B
3
0 3
Start
0
0
0
A
2
2
0
Figure 3.11
82. Backward Pass
Begin with the last event and work backwards
Latest Finish Time Rule:
If an activity is an immediate predecessor for just a single
activity, its LF equals the LS of the activity that immediately
follows it
If an activity is an immediate predecessor to more than one
activity, its LF is the minimum of all LS values of all activities
that immediately follow it
LF = Min {LS of all immediate following activities}
83. Backward Pass
Begin with the last event and work backwards
Latest Start Time Rule:
The latest start time (LS) of an activity is the difference of its
latest finish time (LF) and its activity time
LS = LF – Activity time
84. LS/LF Times for
Milwaukee Paper
E
4
F
3
G
5
H
2
4 8 13 15
4
8 13
7
13 15
10 13
8 13
4 8
D
4
3 7
C
2
2 4
B
3
0 3
Start
0
0
0
A
2
2
0
4
2
8
4
2
0
4
1
0
0
85. Computing Slack Time
Earliest Earliest Latest Latest On
Start Finish Start Finish Slack Critical
Activity ES EF LS LF LS – ES Path
A 0 2 0 2 0 Yes
B 0 3 1 4 1 No
C 2 4 2 4 0 Yes
D 3 7 4 8 1 No
E 4 8 4 8 0 Yes
F 4 7 10 13 6 No
G 8 13 8 13 0 Yes
H 13 15 13 15 0 Yes
Table 3.3
86. Critical Path for
Milwaukee Paper
E
4
F
3
G
5
H
2
4 8 13 15
4
8 13
7
13 15
10 13
8 13
4 8
D
4
3 7
C
2
2 4
B
3
0 3
Start
0
0
0
A
2
2
0
4
2
8
4
2
0
4
1
0
0
87. LS – LF Gantt Chart
for Milwaukee Paper
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 11 12 13 14 15 16
88. CPM assumes we know a fixed time
estimate for each activity and there is
no variability in activity times
PERT uses a probability distribution for
activity times to allow for variability
Variability in Activity
Times
89. Three time estimates are required
Optimistic time (a) – if everything goes
according to plan
Pessimistic time (b) – assuming very
unfavorable conditions
Most likely time (m) – most realistic
estimate
Variability in Activity
Times
90. Estimate follows beta distribution
Variability in Activity
Times
Expected time:
Variance of times:
t = (a + 4m + b)/6
v = [(b – a)/6]2
91. Estimate follows beta distribution
Variability in Activity
Times
Expected time:
Variance of times:
t = (a + 4m + b)/6
v = [(b − a)/6]2
Probability of
1 in 100 of > b
occurring
Probability of
1 in 100 of
< a occurring
Probability
Optimistic
Time (a)
Most Likely Time
(m)
Pessimistic Time
(b)
Activity
Time
Figure 3.12
92. Computing Variance
Most Expected
Optimistic Likely Pessimistic Time Variance
Activity a m b t = (a + 4m + b)/6 [(b – a)/6]2
A 1 2 3 2 .11
B 2 3 4 3 .11
C 1 2 3 2 .11
D 2 4 6 4 .44
E 1 4 7 4 1.00
F 1 2 9 3 1.78
G 3 4 11 5 1.78
H 1 2 3 2 .11
Table 3.4
93. Probability of Project
Completion
Project variance is computed by
summing the variances of critical
activities
s2 = Project variance
= (variances of activities
on critical path)
p
94. Probability of Project
Completion
Project variance is computed by
summing the variances of critical
activities
Project variance
s2 = .11 + .11 + 1.00 + 1.78 + .11 = 3.11
Project standard deviation
sp = Project variance
= 3.11 = 1.76 weeks
p
95. Probability of Project
Completion
PERT makes two more assumptions:
Total project completion times follow a
normal probability distribution
Activity times are statistically independent
97. Probability of Project
Completion
What is the probability this project can be
completed on or before the 16 week
deadline?
Z= – /sp
= (16 wks – 15 wks)/1.76
= 0.57
due expected date
date of completion
Where Z is the number of standard
deviations the due date or target date lies
from the mean or expected date
98. Variability of Completion
Time for Noncritical Paths
Variability of times for activities on
noncritical paths must be considered
when finding the probability of
finishing in a specified time
Variation in noncritical activity may
cause change in critical path
99. Trade-Offs And Project
Crashing
The project is behind schedule
The completion time has been
moved forward
It is not uncommon to face the
following situations:
Shortening the duration of the
project is called project crashing
100.
101.
102. An R & D project has a list of tasks to be performed whose time estimates are
given in the Table 8.11, as follows.
Example
a. Draw the project network.
b. Find the critical path.
c. Find the probability that the project is completed in 19 days. If the
probability is less that 20%, find the probability of completing it in 24 days.
103. calculate the time earliest, TE and
time Latest TL for all the activities.
Construct a network diagram:
From the network diagram Figure 8.24, the critical path is identified as
1-4, 4-6, 6-7, with a project duration of 22 days.
104.
105. The probability of completing the project within 19 days is given by, P (Z< Z0)
To find Z0 ,
we know, P (Z <Z Network Model 0) = 0.5 – z (1.3416) (from normal tables, z (1.3416) = 0.4099)
= 0.5 – 0.4099
= 0.0901
= 9.01% Thus, the probability of completing the R & D project in 19 days is 9.01%.
Since the probability of completing the project in 19 days is less than 20% As in
question, we find the probability of completing it in 24 days.
106. Project Crashing
• Projects will sometimes have deadlines that are
impossible to meet using normal procedures
• By using exceptional methods it may be possible
to finish the project in less time than normally
required
• However, this usually increases the cost of the
project
• Reducing a project’s completion time is called
crashing
Crash cost/Time period =
Crash cost – Normal cost
Normal time – Crash time
107. 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.
108. Time-Cost Example
• 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
C 10
B 10
A 10
D 8
Cheapest way to gain 1
Week is to cut A
109. Time-Cost Example
• 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
C 10
B 10
A 9
D 8
Cheapest way to gain 1 wk
Still is to cut A
Wks Incremental Total
Gained Crash $ Crash $
1 500 500
110. Time-Cost Example
• 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
C 10
B 10
A 8
D 8
Cheapest way to gain 1 wk
is to cut B
Wks Incremental Total
Gained Crash $ Crash $
1 500 500
2 500 1,000
111. Time-Cost Example
• 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
C 10
B 9
A 8
D 8
Cheapest way to gain 1 wk
Still is to cut B
Wks Incremental Total
Gained Crash $ Crash $
1 500 500
2 500 1,000
3 800 1,800
112. 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
C 10
B 8
A 8
D 8
To gain 1 wk, cut B and D,
Or cut C
Cut B&D = $1,900
Cut C = $5,000
So cut B&D
Wks Incremental Total
Gained Crash $ Crash $
1 500 500
2 500 1,000
3 800 1,800
4 800 2,600
113. 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
C 10
B 7
A 8
D 7
Can’t cut B any more.
Only way is to cut C
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
114. 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
C 9
B 7
A 8
D 7
Only way is to cut C
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
115. 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
C 8
B 7
A 8
D 7
No remaining possibilities to
reduce project length
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
116. Time-Cost Example
C 8
B 7
A 8
D 7
No remaining possibilities to
reduce project length
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
• Now we know how much it
costs us to save any
number of days
• Customer says he will pay
$2,000 per wks saved.
• Only reduce 5 wkss.
• We get $10,000 from
customer, but pay $4,500 in
expediting costs
• Increased profits = $5,500