of 31
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CPM Example:
CPM Network
a, 6a, 6a, 6a, 6
f, 15f, 15f, 15f, 15
b, 8b, 8b, 8b, 8
c, 5c, 5c, 5c, 5
e, 9e, 9e, 9e, 9
d, 13d, 13d, 13d, 13
g, 17g, 17g, 17g, 17 h, 9h, 9h, 9h, 9
i, 6i, 6i, 6i, 6
j, 12j, 12j, 12j, 12
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CPM Example
ES and EF Times
a, 6a, 6a, 6a, 6
f, 15f, 15f, 15f, 15
b, 8b, 8b, 8b, 8
c, 5c, 5c, 5c, 5
e, 9e, 9e, 9e, 9
d, 13d, 13d, 13d, 13
g, 17g, 17g, 17g, 17 h, 9h, 9h, 9h, 9
i, 6i, 6i, 6i, 6
j, 12j, 12j, 12j, 12
0 6
0 8
0 5
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CPM Example
ES and EF Times
a, 6a, 6a, 6a, 6
f, 15f, 15f, 15f, 15
b, 8b, 8b, 8b, 8
c, 5c, 5c, 5c, 5
e, 9e, 9e, 9e, 9
d, 13d, 13d, 13d, 13
g, 17g, 17g, 17g, 17 h, 9h, 9h, 9h, 9
i, 6i, 6i, 6i, 6
j, 12j, 12j, 12j, 12
0 6
0 8
0 5
5 14
8 21
6 23
6 21
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CPM Example
ES and EF Times
a, 6a, 6a, 6a, 6
f, 15f, 15f, 15f, 15
b, 8b, 8b, 8b, 8
c, 5c, 5c, 5c, 5
e, 9e, 9e, 9e, 9
d, 13d, 13d, 13d, 13
g, 17g, 17g, 17g, 17 h, 9h, 9h, 9h, 9
i, 6i, 6i, 6i, 6
j, 12j, 12j, 12j, 12
0 6
0 8
0 5
5 14
8 21 21 33
6 23
21 30
23 29
6 21
Projects EF = 33Projects EF = 33
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CPM Example
LS and LF Times
a, 6a, 6a, 6a, 6
f, 15f, 15f, 15f, 15
b, 8b, 8b, 8b, 8
c, 5c, 5c, 5c, 5
e, 9e, 9e, 9e, 9
d, 13d, 13d, 13d, 13
g, 17g, 17g, 17g, 17
h, 9h, 9h, 9h, 9
i, 6i, 6i, 6i, 6
j, 12j, 12j, 12j, 12
0 6
0 8
0 5
5 14
8 2121 33
6 23
21 30
23 29
6 21
21 33
27 33
24 33
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LS and LF Times
a, 6a, 6a, 6a, 6
f, 15f, 15f, 15f, 15
b, 8b, 8b, 8b, 8
c, 5c, 5c, 5c, 5
e, 9e, 9e, 9e, 9
d, 13d, 13d, 13d, 13
g, 17g, 17g, 17g, 17
h, 9h, 9h, 9h, 9
i, 6i, 6i, 6i, 6
j, 12j, 12j, 12j, 12
0 6
0 8
0 5
5 14
8 21 21 33
6 23
21 30
23 29
6 21
3 9
0 8
7 12
12 21
21 33
27 33
8 21
10 27
24 33
9 24
CPM Example
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CPM Example
Critical Path
a, 6a, 6a, 6a, 6
f, 15f, 15f, 15f, 15
b, 8b, 8b, 8b, 8
c, 5c, 5c, 5c, 5
e, 9e, 9e, 9e, 9
d, 13d, 13d, 13d, 13
g, 17g, 17g, 17g, 17 h, 9h, 9h, 9h, 9
i, 6i, 6i, 6i, 6
j, 12j, 12j, 12j, 12
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Critical Path Analysis
A critical path consists that set of dependent tasks(each dependent on the preceding one), which
together take the longest time to complete.
One way is to draw critical path tasks with a doubleline instead of a single line.
The critical path for any given method may shift as
the project progresses; this can happen when tasksare completed either behind or ahead of schedule,
causing other tasks which may still be on schedule
to fall on the new critical path
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PERT PERT is based on the assumption that an activitys duration
follows a probability distribution instead of being a single value
Three time estimates are required to compute the parameters ofan activitys duration distribution:
pessimistic time (tp ) - the time the activity would take ifthings did not go well
most likely time (tm ) - the consensus best estimate of theactivitys duration
optimistic time (to ) - the time the activity would take if thingsdid go well
Mean (expected time): te
= tp + 4 tm +to6
Variance: Vt
=W2 =tp - to
6
2
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PERT analysis
Draw the network.
Analyze the paths through the network and find thecritical path.
The length of the critical path is the mean of the projectduration probability distribution which is assumed to benormal
The standard deviation of the project duration probabilitydistribution is computed by adding the variances of the
critical activities (all of the activities that make up thecritical path) and taking the square root of that sum
Probability computations can now be made using thenormal distribution table.
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Probability computation
Determine probability that project is completed within specified time
Z =
x - Q
W
where Q = tp = project mean time
W = project standard mean time
x = (proposed ) specified time
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PERT Example
Immed. Optimistic Most Likely Pessimistic
Activity Predec. Time (Hr.) Time (Hr.) Time (Hr.)
A -- 4 6 8
B -- 1 4.5 5
C A 3 3 3D A 4 5 6
E A 0.5 1 1.5
F B,C 3 4 5
G B,C 1 1.5 5H E,F 5 6 7
I E,F 2 5 8
J D,H 2.5 2.75 4.5
K G,I 3 5 7
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PERT Example
AA
DD
CC
BB
FF
EE
GG
II
HH
KK
JJ
PER
T Network
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PERT Example
Activity Expected Time Variance
A 6 4/9B 4 4/9
C 3 0
D 5 1/9
E 1 1/36F 4 1/9
G 2 4/9
H 6 1/9
I 5 1J 3 1/9
K 5 4/9
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PERT ExampleActivity ES EF LS LF Slack
A 0 6 0 6 0 *critical
B 0 4 5 9 5
C 6 9 6 9 0 *
D 6 11 15 20 9
E 6 7 12 13 6
F 9 13 9 13 0 *
G 9 11 16 18 7
H 13 19 14 20 1I 13 18 13 18 0 *
J 19 22 20 23 1
K 18 23 18 23 0 *
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PERT Example
Vpath = VA + VC + VF + VI + VK
= 4/9 + 0 + 1/9 + 1 + 4/9
= 2
Wpath = 1.414
z = (22 - 23)/W!(22-23)/1.414 = -0.71
From the Standard Normal Distribution table:
P(z < 0.71) = .5 + .2612 = .7612
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We will use PERT/CPM
Analysis to determine TaskSecondary properties:
Tail Event and Head Event
Earliest Start, Earliest Complete
Latest Start, Latest Complete
Critical / Non-Critical Status
Total Float, Free Float
Scheduled Start, Scheduled Complete
Actual Staffing, Duration, and Variable Costs
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We will then use Task Secondary
Properties to generate Project
Management Tools:
Gantt Chart (Project Schedule)
Manpower Chart
Expenditure Curves Project Completion (PC)
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Generate Initial CPM Diagram Must strictly enforce all prerequisite relationships.
Number of events is initially unknown
Critical path is initially unknown
Iterative Process
Try to minimize number of Dummy Tasks
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CPM Hint #1
Add or remove events at your pleasure.
Do not number events until last.
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CPM Hint #2
The initial event is the Tail Event for all
tasks which have empty prerequisite sets(Initial Tasks).
The Final Event is the Head Event for all
tasks which are not members of any
prerequisite set (Final Tasks).
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CPM Hint #3
Tasks which have identical prerequisite sets
have the same Tail Event
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CPM Hint #4 Starting with the Final Tasks, work backwards,
enforcing the smallest prerequisite sets first.
Use Dummy Tasks to enforce any prerequisitesin large sets which have already been enforced
in a smaller set.
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Finish CPM Diagram
Remove all redundant Dummy Tasks
Remove all redundant Events Number all remaining events
Not really finished .. havent identified critical
tasks yet.
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Generate PERT Chart:
Enter Data for Each Task
Task Symbol
Tail Event
Head Event
Task Duration (TD)
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Forward Pass:
Determine Earliest Start (ES) andEarliest Complete (EC)
for each Task
For all Initial Tasks, ES = 0
Once ES is Determined, EC equals ES plus TD.
The ES for all tasks with tail [i] is equal to the
largest value of EC for all tasks with head [i].
PC is the largest value of EC for all Final Tasks.
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Backward Pass:
Determine Latest Start (LS) andLatest Complete (LC)
for each Task
For all Final Tasks, LC = PC
Once LC is Determined, LS equals LC minus TD.
The LC for all tasks with head [j], is equal to the
smallest value ofLS for all tasks with tail [j]. At least one Initial Task must have LS = 0; none
may be negative.
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Determine Total Float (TF):
Allowable delay in start of task whichwill not delay Project Completion
For task with tail [i] and head [j],
TF[i,j] = (LC[j] ES[i]) TD[i,j]
ES[i] is earliest start for all tasks with tail [i].
LC[j] is latest complete for all tasks with head [j].
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Determine Free Float (FF):
Allowable delay in start of task whichwill not delay start of any other task.
For task with tail [i] and head [j],
FF[i,j] = ES[j] - ES[i] - TD[i, j]
= ES[j] - EC[i,j]
If [j] is the final event, use PC for ES[j]
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Determine Critical Path
All Tasks with zero Total Float are Critical.
Any delay in these Tasks will delay ProjectCompletion.
Darken these Tasks to finish CPM Diagram.
