project scheduling: pert-cpm
DESCRIPTION
Project Scheduling: PERT-CPM. PERT (Program evaluation and review technique) and CPM (Critical Path Method) makes a managerial technique to help planning and displaying the coordination of all the activities. Activity Description. Estimated Time. Immediate Predecessors. Activity. A B C - PowerPoint PPT PresentationTRANSCRIPT
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Project Scheduling: PERT-CPM
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PERT (Program evaluation and review
technique) and CPM (Critical Path Method)
makes a managerial technique to help
planning and displaying the coordination of
all the activities.
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ActivityActivity
DescriptionImmediate
PredecessorsEstimated
TimeABCDEFGHIJKLMN
-ABCCED
E,GCF,IJJH
K,L
2 weeks4 weeks
10 weeks6 weeks4 weeks5 weeks7 weeks9 weeks7 weeks8 weeks4 weeks5 weeks2 weeks6 weeks
ExcavateLay the foundationPut up the rough wallPut up the roofInstall the exterior plumbingInstall the interior plumbingPut up the exterior sidingDo the exterior paintingDo the electrical workPut up the wallboardInstall the flooringDo the interior paintingInstall the exterior fixturesInstall the interior fixtures
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Immediate predecessors:
For any given activity, its immediate predecessors are the activities that must be completed by no later than the starting time of the given activity.
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AOA (Activity-on-Arc):
Each activity is represented by an arc.
The arcs are used to show the precedence relationships between the activities.
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AB
C
E
M N
START
FINISH
H
G
D
J
I
F
LK
410
4 76
7
9
8
54
62
nodearc
5
0
0(Estimated)Time
2
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START A B C D G H M FINISH
2 + 4 + 10 + 6 + 7 + 9 + 2 = 40 weeks
START A B C E F J K N FINISH
2 + 4 + 10 + 4 + 5 + 8 + 4 + 6 = 43 weeks
START A B C E F J L N FINISH
2 + 4 + 10 + 4 + 5 + 8 + 5 + 6 = 44 weeks
Path and Length
Critical Path
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Critical Path:
A project time equals the length of the longest path through a project network. The longest path is called “critical path”.
Activities on a critical path are the critical bottleneck activities where any delay in their completion must be avoided to prevent delaying project completion.
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ES :
Earliest Start time for a particular activity
EF :
Earliest Finish time for a particular activity
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AB
C
E
MN
START
FINISH
H
G
D
J
I
F
LK
4
10
4 76
7
98
54
62
5
0
0
2ES=0EF=2
ES=2EF=6ES=6
EF=16
ES=16EF=20
ES=16EF=23
ES=16EF=22ES=22EF=29
ES=20EF=25
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If an activity has only a single immediate
predecessor, then ES = EF for the
immediate predecessor.
Earliest Start Time Rule:The earliest start time of an activity is equal to the largest of the earliest finish times of its immediate predecessors.
ES = largest EF of the immediate predecessors.
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AB
C
E
MN
START
FINISH
H
G
D
J
I
F
LK
4
10
4 76
7
98
54
62
5
0
0
2ES=0EF=2
ES=2EF=6
ES=6EF=1
ES=16EF=20
ES=16EF=23
ES=16EF=22ES=22EF=29
ES=20EF=25 ES=25
EF=33ES=33EF=37
ES=33EF=38
ES=38EF=44
ES=29EF=38
ES=38EF=40
ES=44EF=44
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Latest Finish Time Rule:The latest finish time of an activity is equal to the smallest of the latest finish times of its immediate successors.
LF = the smallest LS of immediate successors.
LS:
Latest Start time for a particular activity
LF:
Latest Finish time for a particular activity
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AB
C
E
MN
START
FINISH
H
G
D
J
I
F
LK
4
10
4 76
7
98
54
62
5
0
0
2
LS=0LF=0
LS=0LF=2
LS=2LF=6
LS=6LF=16
LS=16LF=20
LS=20LF=25
LS=25LF=33
LS=18LF=25
LS=34LF=38
LS=33LF=38
LS=38LF=44
LS=33LF=42
LS=42LF=44
LS=26LF=33
LS=20LF=26
LS=44LF=44
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Latest
Start Time
Earliest
Start Time
S=( 2, 2 )F=( 6, 6 )
Latest
Finish Time
Earliest
Finish Time
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S=(20,20)F=(25,25)
AB
C
E
MN
START
FINISH
H
G
D
J
I
LK
4
10
4 76
7
9
8
54
62
5
0
0
2
S=(0,0)F=(0,0)S=(0,0)
F=(2,2) S=(2,2)F=(6,6)
S=(16,16)F=(20,20)
S=(25,25)F=(33,33)
S=(16,18)F=(23,25)
S=(33,34)F=(37,38)
S=(33,33)F=(38,38)
S=(38,38)F=(44,44)
S=(29,33)F=(38,42)
S=(38,42)F=(40,44)
S=(22,26)F=(29,33)
S=(16,20)F=(22,26)
S=(44,44)F=(44,44)
F
S=(6,6)F=(16,16)
Critical Path
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Slack:
A difference between the latest finish time and
the earliest finish time.
Slack = LF - EF
Each activity with zero slack is on a critical
path.
Any delay along this path delays a whole
project completion.
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Three-Estimates
Most likely Estimate (m)
= an estimate of the most likely value of time.
Optimistic Estimate (o)
= an estimate of time under the most favorable
conditions.
Pessimistic Estimate (p)
= an estimate of time under the most
unfavorable conditions.
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22
6
4
op
pmo
o pmo
Beta distribution
Mean :
Variance:
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Mean critical path:A path through the project network becomes the critical path if each activity time equals its mean.
Activity OE M PE Mean Variance
A
B
C
2
13
2
9
1
2
6
3
8
18
2
4
10
9
1
4
1
OE: Optimistic EstimateM : Most Likely EstimatePE: Pessimistic Estimate
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Activities on Mean Critical Path Mean Variance
A
B
C
E
F
J
L
N
2
4
10
4
5
8
5
6
1
4
1
1
1
91
Project Time 44p 92 p
94
94
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Approximating Probability of Meeting Deadline
T = a project time has a normal distribution
with mean and ,
d = a deadline for the project = 47 weeks.
44p 92 p
13
4447
p
pdK
Assumption:A probability distribution of project time is a normal distribution.
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Using a table for a standard normal distribution,
the probability of meeting the deadline is
P ( T d ) = P ( standard normal )
= 1 - P( standard normal )
= 1 - 0.1587
0.84.
K
K
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Time - Cost Trade - OffsCrashing an activity refers to taking special costly measures to reduce the time of an activity below its normal value.
Crash
Normal
Crashtime
Normaltime
Crash cost
Normal cost
Activitycost
Activitytime
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Activity J:
Normal point: time = 8 weeks, cost = $430,000.
Crash point: time = 6 weeks, cost = $490,000.
Maximum reduction in time = 8 - 6 = 2 weeks.
Crash cost per week saved =
= $30,000.
2
000,430$000,490$
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Cost($1,000)
Crash Costper Week
Saved
A
B
J
$100
$ 50
$ 30
Activity
Time(week)
MaximumReduction
in Time(week)N NC C
1
2
2
$180
$320
$430
$280
$420
$490
2
4
8
1
2
6
N: Normal C: Crash
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Using LP to Make Crashing Decisions
Let Z be the total cost of crashing activities.
A problem is to minimize Z, subject to the
constraint that its project duration must be less
than or equal to the time desired by a project
manager.
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= the reduction in the time of activity j
by crashing it
= the project time for the FINISH node
jx
.000,60000,50000,100 NBA xxxZ
40FINISHy
FINISHy
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= the start time of activity j
Duration of activity j = its normal time
Immediate predecessor of activity F:
Activity E, which has duration =
Relationship between these activities:
iy
ix
Ex4
.4 EEF xyy
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Immediate predecessor of activity J:
Activity F, which has time =
Activity I, which has time =
Relationship between these activities:
Fx5
Ix7
IIJ
FFJ
xyy
xyy
7
5
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.000,60000,50000,100 NBA xxxZ
Minimize
.3,,2,1 NBA xxx
.0,0,,0,0
0,,0,0
FINISHNCB
NBA
yyyy
xxx
The Complete linear programming model
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.40FINISHy
CCD
BBC
AB
xyy
xyy
xy
10
4
20
HHM xyy 9
One Immediate
Predecessor
Two ImmediatePredecessors
EEH
GGH
xyy
xyy
4
7
NNFINISH
MMFINISH
xyy
xyy
6
2
Finish Time = 40
Total Cost = $4,690,000