network techniques for project managements
TRANSCRIPT
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Dr. Manoj Kumar TiwariProfessor
Department of Industrial Engineering &Management
I I T Kharagpur
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This is the path that has the longest lengththrough the project.
The shortest time that a project can conceivably
be finished is the critical path. The following computational procedure are
employed for determining the critical path.
1. Earliest Occurrence time (EOT)2. Latest Occurrence time (LOT)
3. Slack
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EOT refer to the time when the event can becompleted at the earliest.
EOT (i) =Max [EOT (k)+ d(k, i) ]
EOT (1)=0
EOT (2)= EOT (1)+ d(2, 1)= 0+13=13
EOT (3)= EOT (1)+ d(3, 1)
= 0+12=12 12
15
2
3
13
2
48
15
2
Duration ofactivity (k, i)
Earliestoccurrence
time of event k( k precedes i)
0
13
12
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12
1
0
5
13
2
28
12
3
13
2
48
20
15
2
=13+2 =15
EOT (4)= Max[(EOT (2)+ d(2, 4)), (EOT(3)+ d(3,4) )]
=
=20
EOT (5)= Max[(EOT (2)+ d(2, 5)), (EOT(4)+ d(4,5) )]
=
=28
12+8=20
13+15=28 20+2=22
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Earliest Star Time (EST) of an activity isobtained as;
EST (i, j) = EOT(i)
Earliest finish Time (EFT) of an activity isobtained as:
EST (i, j) = EOT(i)+d(i,j)
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LOT refer to the latest allowable time by which anevent can occur.
LOT (i) =Min [LOT (j) - d(i, j) ]
LOT (5)=28 LOT (4)= LOT (5) - d(4, 5)
= 28-2=26 LOT (3)= LOT (4)- d(3, 4)
= 26-8=18
Latest
occurrencetime for j ( jfollows i)
Duration ofactivity (i, i)
12
1
0
5
13
2
28 28
12 18
3
13
2
48
2620
15
2
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LOT (2)=Min[(LOT (5) - d(2, 5) ), (LOT (4) - d(2, 4))]
= Min[(28-15=13), (26-2=24)]
= 13
LOT (1)= Min[(LOT (2) - d(1, 2) ), (LOT (3) - d(1, 3))]
= Min[(13-13=0), (18-12=6)]
= 0
12
100
5
13
2
13
28 28
12 18
3
13
2
48
2620
15
2
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Latest finish Time (LFT) of an activity isobtained as:
LFT (i, j) = LOT(j)
Latest Star Time (LST) of an activity is obtainedas;
LST (i, j) = LFT (i, j)-d(i,j)
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Slack is a difference of LOT and EOT for eachevent.
The critical path is marked by events whichhave 0 slack.
It stars with the beginning events andterminates with the end event.
Here, path 1-2-5 is critical path
Event LOT EOT Slack(LOT-EOT)
5 28 28 0
4 26 20 6
3 18 12 6
2 13 13 0
1 0 0 0
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Network diagram for critical path (having 0slack of events)
12
1
00
5
13
2
13
28 28
12 18
3
13
2
48
2620
15
2
0
0
0
66
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Float is the maximum amount of time that this activity can be
delay in its completion before it becomes a critical activity, i.e.,delays completion of the project
There are three measures of floats.1. Total float, 2. Free float, 3. Independent float
Total float (TF): It is the extra time available to completethe activity if it is started as early as possible, without delayingthe completion of the project.
TF (i,j)= LOT (j)- EOT (i)- d(i, j)
TF(2,4)= LOT (4)- EOT (2)- d(2, 4)=26-13-2=11
2 4
LOT(4)=26
LOT=13
d(2,4)=2
EOT=13
EOT=13
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Free float (FF): It is the float under most favorablecondition.
FF (i,j)= EOT (j)- EOT (i)- d(i, j)
FF(2,4)= EOT (4)- EOT (2)- d(2, 4)
=20-13-2=5
Independent float (TF): It is the float under mostadverse condition.
IF (i,j)= EOT (j)- LOT (i)- d(i, j)
IF(2,4)= EOT (4)- LOT (2)- d(2, 4)
=20-13-2=5
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The table summaries all types of time and floats . The critical activities (1-2) and (2-5) have zero (0) floats. This shows that
there is no flexibility associated with them.
Thus, the earliest starting time is same as the latest starting time and theearliest finishing time is the same as the latest finishing time.
Activity(i,j)
Duration Earlieststarttime(i,j)
Earliestfinish
time(i,j)
Lateststarttime(i,j)
Latestfinishtime(i,j)
Totalfloat
Freefloat
Independentfloat(weeks)
a (1,2) 13 0 13 0 13 0 0 0
b (1,3) 12 0 12 6 18 6 0 0c (2,4) 2 13 15 24 26 11 5 0
d (3,4) 8 12 20 18 26 6 0 0
e (2,5) 15 13 28 13 28 0 0 0
f (4,5) 2 20 22 26 28 6 6 0
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1. Early start schedule: The schedule in which all activitystart as early as possible. All event occurs at their earliest because all activates start at their
earliest starting time and finish at their earliest finish tine. There may be time lags b/w the completion of certain activities and
the occurrence of events which these activities lead to.
All activities emanating from an event begin at the same time. it shows a continuous attitude toward the project and a desire to
minimize the possibility of delay.
2. Late start schedule: The schedule arrived at when allactivities are started late as possible.
1. All events occurs at their latest b/c all activities start at their latest
finishing time.2. Some activities may start after a time lag subsequent to the
occurrence of the preceding events.3. All activities leading to an event are completed at the same time. It shows a desire to commit resources late as late posible.
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.
0 5 1510 20 25 30
1
2
3 4
5
Activity(i,j)
Duration Earliest start time(i,j)
Earliest finish
time (i,j)
a (1,2) 13 0 13
b (1,3) 12 0 12
c (2,4) 2 13 15
d (3,4) 8 12 20
e (2,5)15 13 28
f (4,5) 2 20 22
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.
0 5 1510 20 25 30
Activity(i,j)
Duration Latest start time(i,j) Latest finish time (i,j)
a (1,2) 13 0 13
b (1,3) 12 6 18
c (2,4) 2 24 26
d (3,4) 8 18 26
e (2,5) 15 13 28
f (4,5) 2 26 28
1
2
3 4
5
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Project Evaluation and Review Technique(PERT) U S Navy (1958) for the POLARIS missile program
Multiple task time estimates (probabilistic nature) Activity-on-arrow network construction
Non-repetitive jobs (R & D work)
PERTis based on the assumption that an activitysduration follows a probability distribution instead ofbeing a single value
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Three time estimates are required to compute the
parameters of an activitys duration distribution: pessimistic time (tp ) - the time the activity would take if
things did not go well
most likely time (tm ) - the consensus best estimate of theactivitys duration
optimistic time (to ) - the time the activity would take ifthings did go well
Mean (expected time): te =t
p
+ 4 tm
+ to
6
Variance: Vt = (stander deviation ())2 =
tp - to6
tp - to6
2
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Variability in PERT analysis is measured by variance, standerdeviation.
Variance = sum of variances of activity duration on the criticalpath
The standard deviation of the project duration probabilitydistribution is computed by adding the variances of the criticalactivities (all of the activities that make up the critical path) andtaking the square root of that sum
Stander deviation = (sum of variances of activity durationon the critical path )1/2
Thus, the stander deviation of critical path duration
= (4+5.43)1/2
=3.07
activity tp to =(tpto)/6 Variance=2
(1-2) 21 9 2 4.00
(2-5) 24 10 2.33 5.43
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Draw the network.
Analyze the paths through the network andfind the critical path.
The length of the critical path is the mean of theproject duration probability distribution whichis assumed to be normal
Probability computations can now be madeusing the normal distribution table.
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X +
Z
Probability
x X+2 X+3X-X-2X-3
Range Probability
0.682
0.954
0.998
x
X +
X + 2
X + 3
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The probability of completion is completed withinspecified time as follows:
1. First find,
2. Then, obtain cumulative probability up to Z bylooking probability distribution of standerdeviation.
Z =
D - T
Where, T = project mean time
= project standard mean time
D = (proposed ) specified time
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Z Cumulative Probability-3.0 0.001
-2.8 0.003
-2.6 0.005
-2.4 0.008
-2.2 0.014
-2.0 0.023
-1.8 0.036
-1.6 0.055
-1.4 0.081
-1.2 0.115
-1.0 0.159
-0.8 0.212
-0.6 0.274
-0.4 0.345
-0.2 0.421
0.0 0.500
0.2 0.579
0.4 0.655
0.6 0.726
0.8 0.788
1.0 0.841
1.2 0.885
1.4 0.919
1.6 0.945
1.8 0.964
2.0 0.977
2.2 0.986
2.4 0.992
2.6 0.995
2.8 0.997
3.0 0.999
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Given that:
project mean time (T)=28
project standard mean time ()=3.07
proposed (specified)
time (D)
Z Cumulative
probability20 (20-28)/3.07 = -2.6 0.005
25 (25-28)/3.07= -1.0 0.159
30 (30-28)/3.07= -0.6 0.726
Fined thevalue from
previoustable.
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Critical Path Method (CPM) E I Du Pont de Nemours & Co. (1957) for construction of new
chemical plant and maintenance shut-down
Deterministic task times
Activity-on-node network construction
Repetitive nature of jobs In contrast of PERT model, CPM model is developed for projects
which are relatively risk-free
PERT approach is probabilistic while CPM approach isdeterministic
In CPM network analysis we work with single time estimates. In CPM network basically we analyze variation in activity time as
a result of change in recourse assignment.
The main thrust of CPM analysis is on time cost relationship.
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1. The cost associated with a project has twocomponents :
o Direct cost: It is incurred on direct material and direct labor.
o Indirect cost: It consist of overhead items like indirect supplies, rent,insurance, managerial services etc.
2. Activity of the project can be expedited by crashingwhich involves employing more recourses:
3. Crashing reduces time but enhances direct costs b/c offactors like overtime payment, extra payment, and
wastage.The relationship b/w time and direct activity
cost cab be reasonably approximated by a downwardstraight line.
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Crashing:
reducing project time by expending additionalresources
Crash time:
an amount of time an activity is reduced
Crash cost:
cost of reducing activity time
Goal reduce project duration at minimum cost
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.Crashing activity
Crash time
Crash cost
Normal Activity
Normal
time
Normal
cost
Slope = crash cost per unit time
Activity duration
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4. Indirect costs associated with the projectincreases linearly with project duration.
project duration
Linear
relation
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CPM analysis seeks to examine the consequences of crashing ontotal cost.
Indirect cost shows linear relation with project duration thus, therelationship b/w direct cost and project duration is important andobeys following procedure.
The procedure used in this respect is generally as follows:1. Obtained the critical path in network. Determine the project
duration and direct cost.
2. Examine the cost-time slope (crash time per minute) of activitieson critical path obtained and crash the activity which has the least
slope.3. Construct new critical path after crashing as per previous step.
And further determine project duration and cost.
4. Repeat steps 2 and 3 till activities on the critical path are crashed.
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The project network depict the activities, duration and directactivity cost.
The indirect cost is Rs 2,000 per week.
1
8
42
53
7
69 3
10
5
7
9
5 6
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Normal and crash time and cost is listed in the table
Activity Time (week) Cost (Rs) Cost to expedite per week
Normal Crash Normal Crash
1-2 8 4 3,000 6,000 750
1-3 5 3 4,000 8,000 2,000
2-4 9 6 4,000 5,500 500
3-5 7 5 2,000 3,200 600
2-5 5 1 8,000 12,000 1,000
4-6 3 2 1/2 10,000 11,200 2,400
5-6 6 2 4,000 6,800 7006-7 10 7 6,000 8,700 900
5-7 9 5 4,200 9,000 1,200
45,200 70,400
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Find out critical path of the network.
Earliest Occurrence time (EOT)
1
8
42
53
7
6
9 3
10
5
7
9
5 6
8
5
17
1312
1920
30
22
Projectduration
is 30 week
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.
1
8
42
53
7
6
9 3
10
5
7
9
5 6
8
7
17
1421
20
30
9
0
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The critical path of the network is (1-2-4-6-7)
The project duration is 30 week
The total direct cost is Rs 45,200
1
8
42
53
7
6
9 3
10
5
7
9
5 6
0
2
0
1
0
00
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Examine the time-cost slope of activities on the critical path.
Time cost slop of activity = (crash cost-normal cost)/ (normaltime crash time)
Time cost slop of activity 1-2=(6000-3000)/(8-4)=750
Time cost slop of activity 2-4=(5500-4000)/(9-6)=500
Time cost slop of activity 4-6=(11200-1000)/(3-21/2)=2400
Time cost slop of activity 6-7=(8700-6000)/(10-7)=900
Minimum
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Activity 2-4 has lowest slop on critical path.
Now, the new critical path is 1-2-5-6-7 The project duration is 29 week
The total direct cost is Rs 46,700
1
8
42
53
7
66 3
10
5
7
9
5 6
Henceactivity 2-4is crashed to
6 (as pergiven table)
(45,200-4,000+5,500)
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Activity 5-6 has the lowest slop on the critical path.
Now the new critical path is (1-2-4-6-7)
The project duration of this path is 27 week
The total direct cost = Rs 49,500
1
8
42
53
7
66 3
10
5
7
9
5 2
Hence ,activity 5-
6 iscrashed to 2
(as pergiven datain the table)
(46,700-4,000+6,800)
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Activity1-2 has lowest slop on the new path
Now the new critical path is (1-2-4-6-7) The project duration of this path is 24 week
The total direct cost = Rs 52,500
1
4
42
53
7
66 3
10
5
7
9
5 2
Hence ,activity 1-2is crashed to
4 (as per
given data inthe table)
(49,500 -3,000+6,000)
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Activity6-7 has lowest slop on the new path
Now there are two new critical path (1-3-5-6-7) & (1-3-5-7)
The project duration of both path is 21 week
The total direct cost = Rs 55,200
1
4
42
53
7
66 3
7
5
5
9
5 2
Hence ,activity 6-7is crashed to
7 (as per
given data inthe table)
(52,500 -6,000+8,700)
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For both the critical path a common activity 3-5 has lowest slop
Now , new critical path (1-2-4-6-7) The project duration of the path is 20 week
The total direct cost = Rs 56,400
1
4
42
53
7
66 2
1/2
10
5
5
9
5 2
Hence ,activity 3-5is crashed to
5 (as per
given data inthe table)
(55,200 -2,000+3,200)
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For both the critical path a common activity 4-6 has lowest slop
Now the critical path is again (1-2-4-6-7)
The project duration of both path is 19 1/2 week
The total direct cost = Rs 57,600
1
4
42
53
7
66 2
1/2
10
5
5
9
5 2
Hence ,activity 4-6is crashed to21/2 (as per
given data inthe table)
(56,400 -10,000+11,200)
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With crashing the possible activities direct cost increases and
indirect cost decreases.
Crashed activities Project Duration Direct cost Indirect cost Total cost
None 30 45,000 60,000 105,200
(2-4) 29 46,700 58,000 104,700(2-4) and (5-6) 27 49,500 54,000 103,500
(1-2), (2-4), and (5-6) 24 52,500 48,000 100,500
(1-2), (2-4), (5-6) and (6-7) 21 55,200 42,000 97,200
(1-2), (2-4), (3-5),(5-6), and
(6-7)
20
56,400 40,000 96,400(1-2), (2-4), (3-5)(,5-6), (4-6)and (6-7)
191/2 57,600 39,000 96,600
Direct cost+ Indirect
cost
(Projectduration) *
(Indirect cost perweek=2000)
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Crashing costs increase as project duration decreases
Indirect costs increase as project duration increases
Reduce project length as long as crashing costs are less than indirect costs
Direct cost
Indirect cost
Total project cost
time
Time-Cost Tradeoff
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Useful at many stages of project management
Mathematically simple
Give critical path and slack time
Provide project documentation
Useful in monitoring costs
How long will the entire project take to be completed? What are
the risks involved?
Which are the critical activities or tasks in the project which could
delay the entire project if they were not completed on time?
Is the project on schedule, behind schedule or ahead of schedule?
If the project has to be finished earlier than planned, what is the
best way to do this at the least cost?
PERT/CPM can answer the following important questions:
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Clearly defined, independent and stable activities Specified precedence relationships Over emphasis on critical paths Deterministic CPM model Activity time estimates are subjective and depend
on judgment PERT assumes a beta distribution for these time
estimates, but the actual distribution may be
different PERT consistently underestimates the expectedproject completion time due to alternate pathsbecoming critical