4. C RITICAL P ATH B ASED T IME A NALYSIS

1. 4. CRITICAL PATH BASED TIME ANALYSIS Objective: 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.

2. Summary: 4.1 Computing the Project Duration 4.2 Determining the Critical Path(s) 4.3 Determining the Activity Floats 4.4 Lead and Lag Times and Ladder Constructs 4.5 Representing Time Graphically 4.6 Determining Activity Durations

3. 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.

4. The floats assist in scheduling activities in a way that satisfies all project objectives, taking into account all resource constraints.

5. constr. temp. haul road pour conc. excav. pad found. clear site clean up position form & fix steel constr. form Fig. 4-1: Addition of Durations to Foundation Network • The duration of a project is given by the longest time path through the network: 6 5 7 0 0 5 10 7 10 add activity durations

6. 6 5 7 0 0 5 10 7 10 Fig. 4-2: Computation of Early Event Times 5 3 event number 25 12 1 2 7 0 32 5 early event time project duration = 32 0 + 5 = 5 4 6 25 15 add event numbers merge events use largest computed value calculate early event times

7. 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.

8. 5 3 25 12 6 5 7 1 2 7 0 0 0 32 5 5 10 7 4 6 25 15 10 Fig. 4-3: Computation of Late Event Times 32 - 5 = 27 15 27 0 32 5 late event time 25 15 burst events use smallest computed value calculate late event times

9. 5 3 evlf - eves - d = 0 ? if yes then critical activity 25 15 27 12 6 5 7 1 2 7 0 0 0 0 32 32 5 5 5 10 7 4 6 25 25 15 15 10 = critical path Fig. 4-4: Identification of Critical Path eve = evl? if yes then critical event

10. 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).

11. Activity A Finish Event Start Event d • Activity times and event times should not be confused. early event time (eventef) late event time (eventlf) early event time (eventes) late event time (eventls) early activity start = eventes early activity finish = eventes +d late activity start = eventlf -d late activity finish = eventlf

12. Table 4-I: Activity Start and Finish Times for the Foundation Operation 0055 551515 12151215 12211827 15152525 25273032 25272527 25253232 Act EarlyLateEarly Late ID StartStartFinish Finish 1-2 2-4 3-4 3-5 4-6 5-7 6-5 6-7

13. 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.

14. Total Float. The maximum amount of time by which an activity’s completion can be delayed without extending the completion date of the project.

15. START EVENT FINISH EVENT early late early late TIME TOTAL FLOAT Activity Duration = d (a)interpretationof total float 5 3 25 15 27 12 6 5 7 1 2 7 0 0 0 0 32 32 5 5 5 10 7 dummy activities can have TF > 0 4 6 25 25 15 15 10 (b)total floats for foundation operation Fig. 4-5: Computation of Total Float TF = 9 TF = 2 TF = 3 TF = 0 TF = 2 TF = 3 TF = 0 TF = 0 TF = 0 critical activities have zero or -ve TF

16. Free Float. The maximum amount of time by which the activity’s completion can be delayed without delaying succeeding activities.

17. START EVENT FINISH EVENT early late early late TIME FREE FLOAT Activity Duration = d (a)interpretationof free float 5 3 25 15 27 12 6 5 7 1 2 7 0 0 0 0 32 32 5 5 5 10 7 4 6 25 25 15 15 10 (b)free floats for foundation operation Fig. 4-6: Computation of Free Float FF = 7 FF = 2 FF = 0 FF = 0 FF = 0 FF = 3 FF = 0 FF = 0 FF = 0

18. 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.

19. START EVENT FINISH EVENT early late early late TIME Activity Duration = d INDEPENDENT FLOAT (a)interpretationof independent float 5 3 25 15 27 12 6 5 7 1 2 7 0 0 0 0 32 32 5 5 5 10 7 4 6 25 25 15 15 10 (b)independent floats for foundation operation Fig. 4-7: Computation of Independent Float IF = 4 Independent float can be -ve even if there are no delays IF = 0 IF = 0 IF = 0 IF = 0 IF = 0 IF = 0 IF = 0 IF = 0

20. 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.

21. 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.

22. Lead and lag times can be used in a ladder to simplify representation of phased sequential activities.

23. excav. trn. 2 excav. trn. 1 excav. trn. 3 shore 1 shore 2 shore 3 lay pipe 1 lay pipe 2 lay pipe 3 (a) phased lengthy sequential activities Fig. 4-8 The Use of Lead and Lag Dummies to Simplify Network Construction continued... 0 1 3 0 1 3 1 day 2 days 2 days 2 days total duration = 10 3 3 6 6 3 3 6 6 3 days 2 days 2 days 6 8 10 6 8 10 2 days 2 days

24. excav. trn. lag 1 lead 1 shore lag 2 lead 2 lay pipe (b) ladder construction Fig. 4-8 The Use of Lead and Lag Dummies to Simplify Network Construction Some loss of logic: In (a), excav. and lay pipe are partially critical 0 0 6 5 5 days 2 days Again total duration = 10 1 day 1 1 8 8 7 days 2 days 2 days 10 10 4 3 6 days

25. 4.5 Representing Time Graphically • Activity-on-the-arrow networks can be conveniently scaled to represent time graphically:

26. 5 3 25 15 27 12 6 5 7 1 2 7 0 0 0 0 32 32 5 5 5 10 7 4 6 25 25 15 15 10 (a) original activity-on-the-arrow-network TIME 0 5 12 15 18 25 30 32 1 5 2 3 7 6 5 7 5 4 6 10 10 7 (b) time scaled activity-on-the-arrow-network Fig. 4-9: Time-Scaled Representation of Activity Network ... Free Float

27. TIME 0 5 12 15 18 25 30 32 progress can be conveniently indicated 1-2 (5) 2-3 (7) 3-5 (6) 5-7 (5) 2-4 (10) 4-6 (10) 6-7 (7) (c) linked bar chart Fig. 4-9: Time-Scaled Representation of Activity Network • Alternatively, activity networks (including precedence networks) can be converted into linked bar charts to show time graphically.

28. 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)

29. 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. ft Production 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).

30. 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.

31. 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;

32. 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.

33. An activity’s duration will vary from repetition to repetition. • The reasons for this can be divided into two categories: • stochastic causes of variance: • these are random and thus impossible to predict; • (2)deterministiccauses 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).