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Transcript of 1 4. C RITICAL P ATH B ASED T IME A NALYSIS Objective: To learn the principles of activity network...
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4. CRITICAL PATH BASED TIME
ANALYSISObjective:
To learn the principles of activity network based preliminary time analysis, calculating: – project duration, – critical path, – activity floats, and – event times.
In addition, introduce:– lead and lag times;– conversion into time-scaled charts.
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Summary:
4.1 Computing the Project Duration4.2 Determining the Critical Path(s)4.3 Determining the Activity Floats4.4 Lead and Lag Times and Ladder Constructs4.5 Representing Time Graphically4.6 Determining Activity Durations
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4.1 COMPUTING THE PROJECT
DURATION
Once the logical dependencies between the activities have been established, a time analysis can be performed.
The preliminary time analysis will consider only logical constraints on the timing of activities, and determines:– preliminary project duration, and
– activity floats.
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The floats assist in scheduling activities in a way that satisfies all project objectives, taking into account all resource constraints.
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• The duration of a project is given by the longest time path through the network:
Fig. 4-1: Addition of Durations to Foundation Network
clear site
excav. pad found.
constr. temp. haul road
constr. form
position form & fix steel
clean up
pour conc.
add activity durations
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Fig. 4-2: Computation of Early Event Times
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0
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add event numbers
eventnumber
eventnumber
1 2
3
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calculate early event times
earlyevent time
earlyevent time
0
0 + 5 = 50 + 5 = 5
5
12
merge events use largestcomputed value
merge events use largestcomputed value
15 25
25
32
projectduration
= 32
projectduration
= 32
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4.2 Determining the Critical Path(s)
The next step is to determine which activities are critical.
The critical activities will always form at least one path connecting the initial and final events.
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Fig. 4-3: Computation of Late Event Times
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0
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0
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1 2
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70 5
12
15 25
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calculate late event times
lateeventtime
lateeventtime
32
32 - 5 = 27 32 - 5 = 27
27
burst events use smallestcomputed value
burst events use smallestcomputed value
2515
15
50
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Fig. 4-4: Identification of Critical Path
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eve = evl?
if yes then critical event
eve = evl?
if yes then critical event
evlf - eves - d = 0 ?
if yes thencritical activity
evlf - eves - d = 0 ?
if yes thencritical activity
= critical path = critical path
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• Knowledge of the critical path is useful for:– reducing the project duration;
– scheduling activities to meet resource constraints; and
– focusing management efforts to minimize the possibility of delay to the project.
Note, a non-critical activity could be very susceptible to delays and thus easily become critical (eg: activities susceptible to inclement weather).
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• Activity times and event times should not be confused.
StartEvent
FinishEvent
Activity A
d
early event time
(eventes)
late event time
(eventls)
early event time
(eventef)
late event time
(eventlf)
early activity start = eventes
early activity finish = eventes + d
late activity start = eventlf - d
late activity finish = eventlf
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Table 4-I: Activity Start and Finish Times for the Foundation Operation
ActAct EarlyEarly LateLate EarlyEarly LateLateIDID StartStart StartStart FinishFinish FinishFinish1-21-22-42-43-43-43-53-54-64-65-75-76-56-56-76-7
00 00 55 55 55 55 1515 1515 1212 1515 1212 1515 1212 2121 1818 2727 1515 1515 2525 2525 2525 2727 3030 3232 2525 2727 2525 2727 2525 2525 3232 3232
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4.3 Determining the Activity Floats
Non critical activities can experience some delay before they will cause other activities to be delayed and/or the project completion time to be delayed.
– This leeway is termed float or slack.
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• Total Float. The maximum amount of time by which an activity’s completion can be delayed without extending the completion date of the project.
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15Fig. 4-5: Computation of Total Float
(a) interpretation of total float
(b) total floats for foundation operation
START EVENTSTART EVENT FINISH EVENTFINISH EVENTearly late early late
TIME
Activity Duration = d TOTALFLOAT
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TF = 0
TF = 3
TF = 0
TF = 9
TF = 3
TF = 0
TF = 2
TF = 2
TF = 0
criticalactivitieshave zeroor -ve TF
criticalactivitieshave zeroor -ve TF
dummy activitiescan have TF > 0
dummy activitiescan have TF > 0
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• Free Float. The maximum amount of time by which the activity’s completion can be delayed without delaying succeeding activities.
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17Fig. 4-6: Computation of Free Float
(a) interpretation of free float
(b) free floats for foundation operation
START EVENTSTART EVENT FINISH EVENTFINISH EVENTearly late early late
TIME
Activity Duration = d FREEFLOAT
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0
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70 5
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32 32
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FF = 0
FF = 0
FF = 0
FF = 7
FF = 3
FF = 0
FF = 2
FF = 0
FF = 0
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• Independent Float. The maximum amount of time by which the activity’s duration can be extended without delaying other activities, even if all float in the preceding activities has been consumed.
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19Fig. 4-7: Computation of Independent Float
(a) interpretation of independent float
(b) independent floats for foundation operation
START EVENTSTART EVENT FINISH EVENTFINISH EVENTearly late early late
TIME
Activity Duration = d
INDEPENDENTFLOAT
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IF = 0
IF = 0
IF = 0
IF = 4
IF = 0
IF = 0
IF = 0
IF = 0
IF = 0
Independent floatcan be -ve even ifthere are no delays
Independent floatcan be -ve even ifthere are no delays
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• Shared Float. Shared float is that which is common to connected activities.
• Shared float is computed as the difference between the late and early event times at an event.
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4.4 Lead and Lag Times and Ladder Constructs
Sometimes, it is necessary to impose a delay between events using dummy activities:– Lead time when the delay follows the start of
an activity, and
– Lag time where the delay follows the finish of an activity.
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• Lead and lag times can be used in a ladder to simplify representation of phased sequential activities.
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Fig. 4-8 The Use of Lead and Lag Dummies to Simplify Network Construction continued...
(a) phased lengthy sequential activities
excav. trn. 1
excav. trn. 2 excav. trn. 3
shore 1 shore 2
shore 3 lay pipe 1 lay
pipe 3
laypipe 2
1 day 2 days
2 days2 days
3 days2 days
2 days
2 days 2 days
0 1 3
3 3 6
6
6
8 10 108
6
6
633
310
totalduration
= 10
totalduration
= 10
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Fig. 4-8 The Use of Lead and Lag Dummies to Simplify Network Construction
(b) ladder construction
excav. trn.
shore
lay pipe
lead 1lag 1
lead 2
lag 2
5 days
7 days
6 days
1 day
2 days
2 days
2 days
0 5
1 8
3 10 10
8
4
1
60
Againtotal
duration= 10
Againtotal
duration= 10
Some loss of logic:In (a), excav. and lay pipe
are partially critical
Some loss of logic:In (a), excav. and lay pipe
are partially critical
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4.5 Representing Time Graphically
• Activity-on-the-arrow networks can be conveniently scaled to represent time graphically:
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26Fig. 4-9: Time-Scaled Representation of Activity Network ...
(b) time scaled activity-on-the-arrow-network
(a) original activity-on-the-arrow-network
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0 5 12 15 18 25 30 32TIME
5 71 2 3
4
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10 7
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Free Float
Free Float
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• Alternatively, activity networks (including precedence networks) can be converted into linked bar charts to show time graphically.
Fig. 4-9: Time-Scaled Representation of Activity Network(c) linked bar chart
0 5 12 15 18 25 30 32TIME
1-2 (5)1-2 (5)
2-3 (7)2-3 (7)
3-5 (6)3-5 (6)
5-7 (5)5-7 (5)
2-4 (10)2-4 (10)
4-6 (10)4-6 (10)
6-7 (7)6-7 (7)
progress can beconveniently indicated
progress can beconveniently indicated
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4.6 Determining Activity Durations
• An accurate estimate of project duration requires accurate estimates of the activity durations.
• The duration for an activity is dependent on many things.• Often, a good approximation for an activity duration can be estimated
from just 3 factors:– the quantity of work to be performed;– the production rates of the productive resources (crews and equipment); and– the numbers of productive resources employed on the task.
• The data for this can be based on:– personal experience;– company historic data;– published data (for example, R.S. Means)
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• Example 1: Determine the time required to drive 25 no. 12” diameter 50 ft steel piles (step tapered, round, and concrete filled).
• An approximation:Duration = Quantity of work per crew / (Production rate per crew × No. of crews)Quantity of work = 50 (v.l. ft / pile) × 25 piles = 1250 v.l. ftProduction rate per crew = 630 (v.l. ft / (crew ∙ day)) (RS Means)No. of crews = 2 (available)
• note:Quantity of work per crew = 1250 / 2 = 625 v.l ft per crew.Each pile = 50 v.l. ft.So, one crew would sink 650 v.l. ft, and the other 600 v.l. ft.
• therefore:Duration (to complete all piles) = 650 / 630 = 1.032 days (approximately 1 day).
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• Note, this does not include:– mobilization;– demobilization;– moving equipment; and setting-out.
• Such factors would be significant and need to be taken into account
• Also, the more crews you have operating in an area, the greater the interference leading to extensions in duration.
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• Example 2: Determine the time required to excavate 2,000 cubic yards of earth using a scraper-based system:
– An approximate estimate requires a lot more information than in the previous example, most notably:
• number of scrapers and their capacities;• policy on % of bowl to be filled at each load operation;• load growth curves;• power of the tractor and whether or not bulldozers are used to assist scraper
loading;• distance the scrapers have to travel to dump their loads;• slopes on the haul roads;• type of soil to be excavated and its moisture content;
other factors that are important but are more difficult to quantify include:• condition of the haul road;• experience of the operator;• balance in the numbers of scrapers and bulldozers;
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• Problems of this type can be solved by:- Tables published by equipment manufacturers,
such as Caterpillar Handbook.- Simulation software:
generic construction simulation software, such as CYCLONE; or
manufacturer specific (again, such as that provided by Caterpillar).
- Beware, the data published by some companies represents idealized rates exclusive of unavoidable inefficiencies:
fueling; start-up conditions; and interference between items of equipment.
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• An activity’s duration will vary from repetition to repetition.
• The reasons for this can be divided into two categories:(1) stochastic causes of variance:
these are random and thus impossible to predict;
(2) deterministic causes of variance:these can be predicted, at least in principle. For example:
• patterns have been observed between the day of the week when a task is performed and the rate at which that task progresses; and
• learning effects whereby, the time required to repeat a task decreases that task (discussed in a later lecture).