project time planning and networks
DESCRIPTION
Henry Yuliando Oct 2012. Project Time Planning and Networks. Project Management. Context. A uniqueness implies that every project must be defined a new and a scheme created telling everyone involved what to do. - PowerPoint PPT PresentationTRANSCRIPT
LOGO
Project Management
Project Time Planning andNetworks
Henry YuliandoOct 2012
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Context
A uniqueness implies that every project must be defined a new and a scheme created telling everyone involved what to do.
Deciding and specifying what they have to do is the function of project definition , the output of which is a project plan.
Making sure they do it right is the function of project control.
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Learning from Past Projects
Learning from Past Projects While developing a project plan, the project
manager refer to earlier, similar projects (plans, procedures, successes, and failures).
Ideally the project manager is provided with planning assistance in the form of lessons learned, best practices, suggested methodologies and templates, and even consulting advice derived from experience in past projects.
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Planning Objectives Resources Work break-down
schedule Organization
Scheduling Project activities Start & end times Network
Controlling Monitor, compare, revise, action
Project Management Activities
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Project Planning, Scheduling, and Controlling
Figure 3.1
Before Start of project Duringproject Timeline project
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Project Planning, Scheduling, and Controlling
Figure 3.1
Before Start of project Duringproject Timeline project
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Project Planning, Scheduling, and Controlling
Figure 3.1
Before Start of project Duringproject Timeline project
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Project Planning, Scheduling, and Controlling
Figure 3.1
Before Start of project Duringproject Timeline project
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Project Planning, Scheduling, and Controlling
Figure 3.1
Before Start of project Duringproject Timeline project
BudgetsDelayed activities reportSlack activities report
Time/cost estimatesBudgetsEngineering diagramsCash flow chartsMaterial availability details
CPM/PERTGantt chartsMilestone chartsCash flow schedules
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Establishing objectives
Defining project Creating work
breakdown structure
Determining resources
Forming organization
Project Planning
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Often temporary structure Uses specialists from entire
company Headed by project manager
Coordinates activities Monitors schedule
and costs Permanent
structure called ‘matrix organization’
Project Organization
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A Sample Project Organization
TestEngineer
MechanicalEngineer
Project 1 ProjectManager
Technician
Technician
Project 2 ProjectManager
ElectricalEngineer
Computer Engineer
Marketing FinanceHumanResources Design Quality
Mgt Production
President
Figure 3.2
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Gantt chart Critical Path
Method (CPM) Program Evaluation
and Review Technique (PERT)
Project Management Techniques
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Six Steps PERT & CPM
1. Define the project and prepare the work breakdown structure
2. Develop relationships among the activities - decide which activities must precede and which must follow others
3. Draw the network connecting all of the activities
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Six Steps PERT & CPM
4. Assign time and/or cost estimates to each activity
5. Compute the longest time path through the network – this is called the critical path
6. Use the network to help plan, schedule, monitor, and control the project
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A Comparison of AON and AOA Network Conventions
Activity on Activity Activity onNode (AON) Meaning Arrow (AOA)
A comes before B, which comes before C
(a) A B CBA C
A and B must both be completed before C can start
(b)
A
CC
B
A
B
B and C cannot begin until A is completed
(c)
B
A
CA
B
CFigure 3.5
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A Comparison of AON and AOA Network Conventions
Activity on Activity Activity onNode (AON) Meaning Arrow (AOA)
C and D cannot begin until A and B have both been 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)
CA
B D
Dummy activity
A
B
C
D
Figure 3.5
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A Comparison of AON and AOA Network Conventions
Activity on Activity Activity onNode (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
DB A B
C
D
Dummy activity
Figure 3.5
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AON Example
Activity DescriptionImmediate
PredecessorsA Build internal components —B Modify roof and floor —C Construct collection stack AD Pour concrete and install frame A, BE Build high-temperature burner CF Install pollution control system CG Install air pollution device D, EH Inspect and test F, G
Milwaukee Paper Manufacturing'sActivities and Predecessors
Table 3.1
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AON Network for Milwaukee Paper
A
Start
BStart Activity
Activity A(Build Internal Components)
Activity B(Modify Roof and Floor)
Figure 3.6
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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
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AON Network for Milwaukee Paper
G
E
F
H
CA
Start
DB
Arrows Show Precedence Relationships
Figure 3.8
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H(Inspect/
Test)
7Dummy Activity
AOA Network for Milwaukee Paper
6
F(Install
Controls)E(B
u ild Bur ne r)
G
(Insta
ll
Pollutio
n
Device)
5D
(Pour Concrete/
Install Frame)
4C
(Construct Stack)
1
3
2
B(Modify
Roof/Floor)
A(B
uild I
nterna
l
Compo
nents
)
Figure 3.9
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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
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Determining the Project Schedule
Perform a Critical Path AnalysisActivity Description Time (weeks)
A Build internal components 2B Modify roof and floor 3C Construct collection stack 2D Pour concrete and install frame 4E Build high-temperature burner 4F Install pollution control system 3G Install air pollution device 5H Inspect and test 2
Total Time (weeks) 25
Table 3.2
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Determining the Project Schedule
Perform a Critical Path Analysis
Activity Description Time (weeks)A Build internal components 2B Modify roof and floor 3C Construct collection stack 2D Pour concrete and install frame 4E Build high-temperature burner 4F Install pollution control system 3G Install air pollution device 5H Inspect and test 2
Total Time (weeks) 25
Earliest start (ES) = earliest time at which an activity can start, assuming all predecessors have been completedEarliest finish (EF) = earliest time at which an activity can be finishedLatest start (LS) = latest time at which an activity can start so as to not delay the completion time of the entire projectLatest finish (LF) = latest time by which an activity has to be finished so as to not delay the completion time of the entire project
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Determining the Project Schedule
Perform a Critical Path Analysis
A
Activity Name or Symbol
Earliest Start ES
Earliest FinishEF
Latest Start
LS Latest Finish
LF
Activity Duration
2
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Activity DescriptionImmediate
PredecessorsA Build internal components —B Modify roof and floor —C Construct collection stack AD Pour concrete and install frame A, BE Build high-temperature burner CF Install pollution control system CG Install air pollution device D, EH Inspect and test F, G
Milwaukee Paper Manufacturing'sActivities and Predecessors
AON Example
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Forward Pass
Begin at starting event and work forwardEarliest Start Time Rule:
If an activity has only one immediate predecessor, its ES equals the EF of the predecessor
If an activity has multiple immediate predecessors, its ES is the maximum of all the EF values of its predecessors
ES = Max (EF of all immediate predecessors)
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Forward Pass
Begin at starting event and work forwardEarliest Finish Time Rule:
The earliest finish time (EF) of an activity is the sum of its earliest start time (ES) and its activity time
EF = ES + Activity time
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ES/EF Network for Milwaukee Paper
Start
0
0
ES
0
EF = ES + Activity time
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Start0
0
0
A
2
2
EF of A = ES of A + 2
0
ESof A
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B
3
Start0
0
0
A
2
20
3
EF of B = ES of B + 3
0
ESof B
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C
2
2 4
B
3
0 3
Start0
0
0
A
2
20
...
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C
2
2 4
B
3
0 3
Start0
0
0
A
2
20
D
4
73= Max (2, 3)
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4
3 7
C
2
2 4
B
3
0 3
Start0
0
0
A
2
20
...
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E
4
F
3
G
5
H
2
4 8 13 15
4
8 13
7
D
4
3 7
C
2
2 4
B
3
0 3
Start0
0
0
A
2
20
Figure 3.11
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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)
Backward Pass
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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
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E
4
F
3
G
5
H
2
4 8 13 15
4
8 13
7
D
4
3 7
C
2
2 4
B
3
0 3
Start0
0
0
A
2
20
Figure 3.12
LF = EF of Project
1513
LS = LF – Activity time
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E
4
F
3
G
5
H
2
4 8 13 15
4
8 13
7
13 15
D
4
3 7
C
2
2 4
B
3
0 3
Start0
0
0
A
2
20
LF = Min(LS of following activity)
10 13
Figure 3.12
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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
Start0
0
0
A
2
20
LF = Min(4, 10)
42
Figure 3.12
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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
Start0
0
0
A
2
20
42
84
20
41
00
Figure 3.12
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After computing the ES, EF, LS, and LF times for all activities, compute the slack or free time for each activity
Slack is the length of time an activity can be delayed without delaying the entire project
Slack = LS – ES or Slack = LF – EF
Computing Slack time
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Earliest Earliest Latest Latest OnStart Finish Start Finish Slack Critical
Activity ES EF LS LF LS – ES Path
A 0 2 0 2 0 YesB 0 3 1 4 1 NoC 2 4 2 4 0 YesD 3 7 4 8 1 NoE 4 8 4 8 0 YesF 4 7 10 13 6 NoG 8 13 8 13 0 YesH 13 15 13 15 0 Yes
Table 3.3
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Figure 3.13
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
Start0
0
0
A
2
20
42
84
20
41
00
...
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Activity on Arrow
1
2
3
4
5
A
B
C
D
dummy EF
6G
7H
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A Build internal components
B Modify roof and floorC Construct collection
stackD Pour concrete and
install frameE Build high-temperature
burnerF Install pollution control
systemG Install air pollution
deviceH Inspect and test
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
ES – EF Gantt Chartfor Milwaukee Paper
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A Build internal components
B Modify roof and floorC Construct collection
stackD Pour concrete and
install frameE Build high-temperature
burnerF Install pollution control
systemG Install air pollution
deviceH Inspect and test
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
LS – LF Gantt Chartfor Milwaukee Paper
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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
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Three time estimates are requiredOptimistic time (a) – if everything goes
according to planMost–likely time (m) – most realistic
estimatePessimistic time (b) – assuming very
unfavorable conditions
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Estimate follows beta distributionExpected time:
Variance of times: t = (a + 4m + b)/6
v = [(b – a)/6]2
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Estimate follows beta distribution
t = (a + 4m + b)/6
v = [(b − a)/6]2Probability of 1 in 100 of > b occurring
Probability of 1 in 100 of < a occurring
Pro
babi
lity
Optimistic Time (a)
Most Likely Time (m)
Pessimistic Time (b)
Activity Time
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Most ExpectedOptimistic Likely Pessimistic Time Variance
Activity a m b t = (a + 4m + b)/6 [(b – a)/6]2
A 1 2 3 2 .11B 2 3 4 3 .11C 1 2 3 2 .11D 2 4 6 4 .44E 1 4 7 4 1.00F 1 2 9 3 1.78G 3 4 11 5 1.78H 1 2 3 2 .11
Table 3.4
Computing Variance
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Project variance is computed by summing the variances of critical activities
s2 = Project variance
= (variances of activities on critical path)
p
Probability of Project Completion
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Project variance is computed by summing the variances of critical activitiesProject variance
s2 = .11 + .11 + 1.00 + 1.78 + .11 = 3.11
Project standard deviationsp = Project variance
= 3.11 = 1.76 weeks
p
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PERT makes two more assumptions:
Total project completion times follow a normal probability distribution
Activity times are statistically independent
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Standard deviation = 1.76 weeks
15 Weeks
(Expected Completion Time)Figure 3.15
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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 datedate of completion
Where Z is the number of standard deviations the due date lies from the
mean
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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 datedate of completion
Where Z is the number of standard deviations the due date lies from the
mean
.00 .01 .07 .08.1 .50000 .50399 .52790 .53188.2 .53983 .54380 .56749 .57142
.5 .69146 .69497 .71566 .71904
.6 .72575 .72907 .74857 .75175
From Appendix I
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Time
Probability(T ≤ 16 weeks)is 71.57%
Figure 3.16
0.57 Standard deviations
15 16Weeks Weeks
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Probability of 0.01
Z
Figure 3.17
From Appendix I
Probability of 0.99
2.33 Standard deviations
0 2.33
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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
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The project’s expected completion time is 15 weeks
There is a 71.57% chance the equipment will be in place by the 16 week deadline
Five activities (A, C, E, G, and H) are on the critical path
Three activities (B, D, F) have slack time and are not on the critical path
A detailed schedule is available
What Project Management Has Provided So Far
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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
Trade-Offs And Project Crashing
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The amount by which an activity is crashed is, in fact, permissible
Taken together, the shortened activity durations will enable us to finish the project by the due date
The total cost of crashing is as small as possible
Factors to Consider When Crashing A Project
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1. Compute the crash cost per time period. If crash costs are linear over time:
Crash costper period =
(Crash cost – Normal cost)(Normal time – Crash time)
2. Using current activity times, find the critical path and identify the critical activities
Steps in Project Crashing
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3. If there is only one critical path, then select the activity on this critical path that (a) can still be crashed, and (b) has the smallest crash cost per period. If there is more than one critical path, then select one activity from each critical path such that (a) each selected activity can still be crashed, and (b) the total crash cost of all selected activities is the smallest. Note that a single activity may be common to more than one critical path.
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4. Update all activity times. If the desired due date has been reached, stop. If not, return to Step 2.
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Time (Wks) Cost ($) Crash Cost CriticalActivity Normal Crash Normal Crash Per Wk ($) Path?
A 2 1 22,000 22,750 750 YesB 3 1 30,000 34,000 2,000 NoC 2 1 26,000 27,000 1,000 YesD 4 2 48,000 49,000 1,000 NoE 4 2 56,000 58,000 1,000 YesF 3 2 30,000 30,500 500 NoG 5 2 80,000 84,500 1,500 YesH 2 1 16,000 19,000 3,000 Yes
Table 3.5
Crashing The Project
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| | |1 2 3 Time (Weeks)
$34,000 —
$33,000 —
$32,000 —
$31,000 —
$30,000 —
—
Activity Cost Crash
Normal
Crash Time Normal Time
Crash Cost
Normal Cost
Crash Cost/Wk = Crash Cost – Normal CostNormal Time – Crash Time
= $34,000 – $30,0003 – 1
= = $2,000/Wk$4,0002 Wks
Figure 3.18
Crash and Normal Times and Costs for Activity B
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Example : Desired completion time from 32 days to 29 day
Original Project Network – Crashing ExampleCritical path 1367 (B F I) shown in boldface, 32 days length
Picture 11.
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Table 4. Normal and crash data
Activity Time (days) Cost ($) Crash Cost
per DayNormal Crash Normal CrashABCDEFGHI
789
118
10121314
667857
101110
600750900
1100850
1000130014001500
750900
1100140012001300150015002000
15075
100100
116.6610010050
125Total $9400
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Based on the critical path, the activity that can be shortened in the cheapest manner is activity B, at an incremental cost $75 per day.
(see table 4). Crashing activity to B to the maximum extent
possible (i.e. normal time – crash time = 8 – 6 = 2 days) would reduce the completion time for activities B F I to 32 – 2 = 30 days, at a total project cost of $9400 + $150 = $9550.
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Project Network – Activity B crashed 2 daysThe new critical path 1257 (A D H) shown in boldface, 31 days length
Picture 12.
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Project Network – Activity B crashed 2 days; activity H crashed 2 daysThe new critical path 1367 (A D H) shown in boldface, 30 days length
Picture 13.
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Picture 14.
Project Network – Activity B crashed 2 days; activity H crashed 2 days; activity F crashed 1 day. The new critical path 1257 and 1367 shown in boldface, 29 days length (desired completion time)
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Table 5. Summary of time-cost tradeoffs
Step Action Critical PathTotal Project Completion
TimeTotal Cost
($)
01
2
3
No crashing in networkActivity B crashed by 2 daysActivity H crashed by 2 daysActivity F crashed by 1 day
1367 1257
1367
12571367
32
31
30
29
9400
9550
9650
9750
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Making crashing decision using LP
Based on the previous case:
Minimize (crash cost) Z = $150yA + $75yB + $100yC + $100yD + $116.66yE +
$100yF + $100yG + $50yH + $125yI Project completion-date constraint
x7 29 Activity crash-time constraint
yA 1 ; yB 2 ; yC 2 ; yD 3 ; yE 3 ; yF 3 ; yG 2 ;
yH 2 ; yI 4
IB,...,A, ;activity for crashed timeofamount y1,2,...,7 ;event of occurence of time
i
jjiixi
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… Constraint describing the network :
1. The occurrence time for an event must be equal to, or greater than, the activity completion time for all activities leading into the node that represents that event.
2. The start time for an activity is equal to the occurrence time of its preceding event.3. The time required to complete an activity is equal to its normal time minus the length of time it is
crashed. Begin with the event occurrence time for event 1 = 0 or x1 = 0
For event 2Occurrence time time required to + start time for for event 2 complete activity A activity A
(x1 = 0) Normal time – Crash time for activity A for activity A
x2 7 – yA + 0
or x2 + yA 7
And for following events are
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… Event 3 : x3 8 - yB + 0 = x3 + yB 8
Event 4 : x4 9 - yC + x2 = x4 - x2 + yC 9 (note that activity C begins with event 2, x2)
Event 5 : x5 11 - yD + x2 = x5 - x2 + yD 11 (for the path from activity D) : x5 8 - yE + x3 = x5 - x3 + yE 8 (for the path from activity E)
Event 6 : x6 10 - yF + x3 = x6 - x3 + yF 10
Event 7 : x7 12 - yG + x4 = x7 - x4 + yG 12 (for the path from activity G) : x7 13 - yH + x5 = x7 - x5 + yH 13 (for the path from activity H)
: x7 14 - yI + x6 = x7 - x6 + yI 14 (for the path from activity I)
With xi 0 for i = 1,2,..,7 yj 0 for j = A,B,..,I
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… The optimal solution
The solution values of yB = 2, yF = 1, and yH = 2, indicate that activity B, F and H must be crashed by 2, 1, 2 days, respectively.
x1 = 10x2 = 7x3 = 6x4 = 17 (the slack for activity C = 1 day)x5= 18x6 = 15x7 = 15
yA = 0yB = 2yC = 0yD = 0yE = 0yF = 1yG = 0yH = 2yI = 0
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PERT/COST The first step in PERT/COST procedure is to
subdivide the project into components that can be used to plan and schedule the cost associated the project (budgeting process).
Major steps :1. For each activity in the project, determine the aggregate cost
associated with the activities. This will be the budget for that activity.
2. Given the expected activity time for each activity, convert the budgeted cost for each activity into a cost per unit time period. (assumed at uniform rate over time)
3. Using the expected activity times, perform the critical path calculations to determine the critical path for the project.
4. Using the earliest and latest start times from the critical path calculations, determine the amount of money that should be spent during each time period in order to complete the project by a desired date.
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Picture 15. Project network – PERT/COST example
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Table 6. Expected activity times and cost estimates
Activity Expected activity time (months)
Estimated cost(Budget $)
Budgeted cost per month
ABCDEFGH
42383267
20,00020,00012,00024,00021,00018,00036,00014,000
$ 5,00010,0004,0003,0007,0009,0006,0002,000
Total budgeted cost = 165,000
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…
Table 7. Activity schedule and slack time
ActivityEarliest Start Time (ES)
Latest Start Time(LS)
Earliest FinishTime (EF)
Latest FinishTime(LF)
Slack(S)
On Critical Path?
ABCDEFGH
004477
129
076499
1211
427
1212111816
499
1212111818
07202202
YesNoNoYesNoNoYesNo
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…Table 8. Budgeted Costs ($000), using earliest start time
Activity Month Totals 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18A 5 5 5 5 20B 10 10 20C 4 4 4 12D 3 3 3 3 3 3 3 3 24E 7 7 7 21F 9 9 18G 6 6 6 6 6 6 36H 2 2 2 2 2 2 2 14
Total Cost/ Month
15 15 5 5 7 7 7 19 19 12 5 5 8 8 8 8 6 6 165
Total Cost to Date
15 30 35 40 47 54 61 80 99 111
116
121
129
137
145
153
159
165
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them
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lery
.com
…Table 9. Budgeted Costs ($000), using latest start time
Activitiy Month Totals 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18A 5 5 5 5 20B 10 10 20C 4 4 4 12D 3 3 3 3 3 3 3 3 24E 7 7 7 21F 9 9 18G 6 6 6 6 6 6 36H 2 2 2 2 2 2 2 14
Total Cost/ Month
5 5 5 5 3 3 7 17 17 19 19 12 8 8 8 8 8 8 165
Total Cost to Date
5 10 15 20 23 26 33 50 67 86 105
117
125
133
141
149
157
165
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them
egal
lery
.com
…
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 180
20
40
60
80
100
120
140
160
180
Budgeted cost earliest start times
Budgeted cost lat-est start times
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Monitoring and Controlling Project Costs
Value of work completed = (percent of completion ) x (total budgeted cost)
Activity cost difference = total actual cost – value of work completed
Ex :
Table 10. Activity cost and completion : end of month 9
ActivityTotal
Budgeted Cost ($)
Percent of Completion
Value of Work
Completed ($)
Total Actual Cost ($)
Activity Cost Difference ($)
ABCDEFGH
Total
20,00020,00012,00024,00021,00018,00036,00014,000
165,000
1001001005025000
20,00020,00012,00012,0005250
000
18,00022,00015,00013,0005000
000
-2000200030001000-250
000
3750
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them
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Problem in the application of PERT/CPM
It may be difficult to divide a project into a set of independent activities.
It may be difficult to firmly establish the precedence relationship among various activities, and not all precedence relationship can be anticipated before a project begins. (for R&D projects)
The PERT procedure is highly dependent on being able to accurately make activity time estimates.
The theoretical foundation of the PERT statistical procedure is subject to question regarding to assumptions in expected activity time and its variance. (not nearly as important as the practical problems associated with making accurate time estimates)