Understanding the total float of activities in a project network is crucial for effective project management. This problem requires us to calculate the total float for specific activities, 'e' and 'f', using the Critical Path Method (CPM) by first determining the earliest and latest start/finish times for all activities.

The given data for the project activities, their predecessors, and durations are presented below:

Project Network Forward Pass Calculation

The forward pass helps us determine the earliest possible start (ES) and earliest possible finish (EF) times for each project activity. We start from the beginning of the project (ES = 0 for initial activities) and move forward, calculating these times:

• Activity 'a': Since 'a' has no predecessors, its earliest start time is 0.
  • ES(a) = 0
  • EF(a) = ES(a) + Duration(a) = 0 + 2 = 2 days
• Activity 'b': Similar to 'a', 'b' also has no predecessors.
  • ES(b) = 0
  • EF(b) = ES(b) + Duration(b) = 0 + 4 = 4 days
• Activity 'c': 'c' starts after 'a' is completed.
  • ES(c) = EF(a) = 2 days
  • EF(c) = ES(c) + Duration(c) = 2 + 2 = 4 days
• Activity 'd': 'd' starts after 'b' is completed.
  • ES(d) = EF(b) = 4 days
  • EF(d) = ES(d) + Duration(d) = 4 + 3 = 7 days
• Activity 'e': 'e' starts after 'c' is completed.
  • ES(e) = EF(c) = 4 days
  • EF(e) = ES(e) + Duration(e) = 4 + 2 = 6 days
• Activity 'f': 'f' also starts after 'c' is completed.
  • ES(f) = EF(c) = 4 days
  • EF(f) = ES(f) + Duration(f) = 4 + 4 = 8 days
• Activity 'g': 'g' can only start when both 'd' and 'e' are completed. Therefore, its earliest start time is the maximum of their earliest finish times.
  • ES(g) = max(EF(d), EF(e)) = max(7, 6) = 7 days
  • EF(g) = ES(g) + Duration(g) = 7 + 5 = 12 days

The overall project completion time is 12 days, which is the earliest finish time of the final activity 'g'.

Project Network Backward Pass Calculation

The backward pass helps us determine the latest possible finish (LF) and latest possible start (LS) times for each project activity without delaying the project. We start from the project completion time and work backward.

• Activity 'g': For the last activity, its latest finish time is the project completion time.
  • LF(g) = 12 days
  • LS(g) = LF(g) - Duration(g) = 12 - 5 = 7 days
• Activity 'd': 'd' is a predecessor of 'g'. Its latest finish time is the latest start time of 'g'.
  • LF(d) = LS(g) = 7 days
  • LS(d) = LF(d) - Duration(d) = 7 - 3 = 4 days
• Activity 'e': 'e' is also a predecessor of 'g'. Its latest finish time is the latest start time of 'g'.
  • LF(e) = LS(g) = 7 days
  • LS(e) = LF(e) - Duration(e) = 7 - 2 = 5 days
• Activity 'f': Activity 'f' does not have any listed successors that are critical for the project's overall completion beyond the existing paths. In such cases, its latest finish time is assumed to be the overall project completion time.
  • LF(f) = 12 days (Project Completion Time)
  • LS(f) = LF(f) - Duration(f) = 12 - 4 = 8 days
• Activity 'c': 'c' is a predecessor to both 'e' and 'f'. Its latest finish time is the minimum of the latest start times of its successors.
  • LF(c) = min(LS(e), LS(f)) = min(5, 8) = 5 days
  • LS(c) = LF(c) - Duration(c) = 5 - 2 = 3 days
• Activity 'b': 'b' is a predecessor of 'd'.
  • LF(b) = LS(d) = 4 days
  • LS(b) = LF(b) - Duration(b) = 4 - 4 = 0 days
• Activity 'a': 'a' is a predecessor of 'c'.
  • LF(a) = LS(c) = 3 days
  • LS(a) = LF(a) - Duration(a) = 3 - 2 = 1 day

Total Float Calculation for Activities

The total float (TF) for an activity represents the maximum amount of time an activity can be delayed from its earliest start date without delaying the project completion time. It is calculated as:

\text{TF} = \text{LS} - \text{ES} \quad \text{or} \quad \text{TF} = \text{LF} - \text{EF}

Let's compile all the calculated times and total float values:

Float for Activities e and f

From the table above, we can directly find the total float for activities 'e' and 'f':

• Total float for activity e is 1 day.
• Total float for activity f is 4 days.

Therefore, the total float of activities 'e' and 'f' are 1 and 4 days, respectively.