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N E T W O R K C O N S T R U C T I O N –. P R E C E D E N C E D I A G R A M M E T H OD. Activity List.

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## N E T W O R K C O N S T R U C T I O N –

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**N E T W O R K C O N S T R U C T I O N –**P R E C E D E N C E D I A G R A M M E T H OD**Activity List**You’ll recall from the earlier section in the Module, the first steps for developing a schedule are to produce a list of all the activities in the project, place them in the sequence in which they will occur, and estimate the duration of each. The table below contains the activity list for this example with the estimated duration for each activity, and the relationship between the activities, e.g., finish-to-start (F-S) start-to-start (S-S), finish-to-finish (F-F), and start-to-finish (S-F). Activity Duration Relationship A 2 None B 10 A (S-S) C 13 A D 12 A E 12 B F 7 B G 10 D H 16 D I 6 F J 12 C, G (S-S) K 5 I (S-S) L 8 E, I, K M 5 H N 3 G, J, L, M**Activity “A” is the first activity in the network, but**activity “B” can also start once “A” starts (“start-to-start” relationship). 10 B 2 A**12**E 10 B 7 F Activities “C”, and “D” can both begin once “A” is completed. Activities “E”, and “F” can both begin once “B” is finished. 2 A 13 C 12 D**12**E Activity “I” can start when “F” is finished, and “K” can start once “I” starts. 10 B 7 6 F I 5 2 K A 12 13 J C Activities “G” and “H” can both start when “D” is finished, and “J” can begin once “C” finishes and “G” starts. 12 10 D G 16 H**8**12 L E 10 B Activity “L” can begin when “E”, “I” and “K” are completed. 7 6 F I 5 2 K A 12 13 J C Activity “M” can start when “H” is finished. 12 10 D G 5 16 M H**8**12 L E Finally, activity “N” can begin when “G”, “J”, “L” and “M” are finished. 10 B 7 6 F I 5 2 K A 12 3 13 J N C 12 10 D G 5 16 M H**Forward Pass**Next a “forward pass” is made through the network using the estimated durations. Begin with the earliest activity (“A” in this example), and work through to the end, to determine the earliest start and finish times for each activity. In addition to the standard “finish-to-start” relationships in the network, attention must also be paid to the other types of relationships, and they must be correctly applied to determine the actual earliest start and finish times. This is the primary difference between the forward pass in this method and when using CPM, since CPM only has “finish-to-start” relationships. In this example, activity “A” starts at time “1”, and adding the duration (2), results in it finishing at time “3”. Activity “B” can also start at time “1” (as long as activity “A” does), with a duration of “10”. It will finish at time “11”. Then, activities “C” and “D” can both begin at time “3”, and activities “E” and “F” can both begin at time “11”. Continue to apply the corresponding durations to each of the remaining activities to determine the earliest start and finish times.**Forward Pass (cont’d)**Note that activity “J” has two activities related to its start time: activity “C” (finish-to-start) and activity “G” (start-to-start), so both constraints must be considered. Activity “G” can start at time “15”, but since activity “J” must also wait for activity “C” to finish, which doesn’t happen till time “16”, we must use the later of the times (time “16”). So activity “J” cannot start till time “16”. Likewise, activity “L” has three activities related to it (“E”, “I”, and “K”). All are finish-to-start relationships, so we use the latest of the three times (activity “I” with time “24”). The next chart shows the complete forward pass and resulting early start and finish times.**24**32 11 23 8 12 1 11 L E 10 11 18 B 18 24 7 6 F I 18 23 1 3 5 2 K A 16 28 36 39 3 16 12 3 13 J N C 3 15 15 25 12 10 D G Legend: 15 31 31 36 ES EF 5 16 Duration M H Activity**Backward Pass**The next step is to complete a “backward pass,” beginning with the final activity and working backward through the network to determine the latest start and finish times. Use the earliest start and finish times for the final activity from the forward pass as the total time to complete the project. So the latest finish time is the same as the earliest finish time of the final activity. Since the duration remains the same, the earliest and latest start times will also be the same for the final activity. The latest start time of the final activity becomes the latest finish time of each immediately preceding activity. Subtract the duration of each immediately preceding activity from its latest finish time to determine its latest start time. For this example, the latest start time for activity “N” is time “36” (same as the earliest start time from the preceding chart), so the latest finish times for activities “J”, “L”, and “M” become time “36”. Subtracting the duration of activity “J” (12) from its latest finish (36) results in a latest start date of “24”. Likewise, subtracting the duration of “8” for activity “L” and “5” for activity “M” from their latest finish (36) results in latest start times of “28” and “31” respectively.**Backward Pass (cont’d)**When there is more than one successor activity, (e.g., activity “B”), the latest finish time for the preceding activity will be the earliest of the successor latest start times. The “start-to-start” relationships also require special attention. In this example, note that Activity “J” can start once “G” has started, but that “G” must be finished before “N” can start. You have to look at both impacts to determine the latest start and finish times. So “G” could finish as late as time “36” and not impact the start of “N” through that relationship, so the latest start would be time “26”. But note that a latest start time of “26” for “G” would mean that activity “J” couldn’t start until time “26” and with a duration of “12” would result in a latest finish of “38” which would delay activity “N”. In this case, the latest start time of “J” (24) takes precedence, so the latest “G” can start is time “24”, resulting in a latest finish for “G” of “34”. The next chart shows the complete forward and backward passes and resulting early and late start and finish times.**11**23 24 32 8 12 1 11 L E 10 16 28 28 36 11 18 18 24 B 7 6 5 15 F I 15 22 22 28 18 23 1 3 5 2 K A 3 16 23 28 16 28 36 39 1 3 12 3 13 J N C 24 36 11 24 36 39 3 15 15 25 12 10 Legend: D G 15 31 31 36 ES EF 24 34 3 15 5 16 Duration M H Activity 15 31 31 36 LS LF**Critical Path and Slack**Activities where the latest start and finish times are later than the earliest start and finish times have “slack” or “float.” Activities where the latest start and finish times are the same as the earliest start and finish times do not have “slack.” To determine the slack for an activity, subtract the earliest start or finish time from the latest start or finish time (e.g., LS – ES or LF – EF). You can see that activity “L” has 4 time units of slack (using the formulas above, 28 – 24 or 36 – 32). While it can start as early as time “24”, it can start as late as time “28” and not delay the overall project. The critical path is the path in the network with the longest duration. The next chart shows the project’s critical path (thicker lines) for this example (the path with activities “A-D-H-M-N”). Looking at the finish time of the project (time “39”) minus the start for the project (time “1”), we know the project duration is 38 units of time. If you add the durations for activities “A”, “D”, “H”, “M” and “N”, you’ll see they add up to “38”. Also notice there is no slack in any of the activities on the critical path. Except under unusual circumstances, this will always be the case.**11**23 24 32 8 12 1 11 L E 10 16 28 28 36 11 18 18 24 B 7 6 5 15 F I 15 22 22 28 18 23 1 3 5 2 K A 3 16 23 28 16 28 36 39 1 3 12 3 13 J N C 24 36 11 24 36 39 3 15 15 25 12 10 Legend: D G 15 31 31 36 ES EF 24 34 3 15 5 16 Duration M H Activity 15 31 31 36 LS LF

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