The activities of a project are given in the following table along with their durations and dependency.
The total float of the activity \( E \) (in days) is _________ (in integer).
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Total float is the amount of time that an activity can be delayed without affecting the overall project completion time.
To find the total float of activity \( E \), we need to perform a forward pass (to calculate early start and finish times) and a backward pass (to calculate late start and finish times).
1. Forward Pass:
- \( A \) has a duration of 10 days, so the early finish time for \( A \) is 10 days.
- \( C \) depends on \( A \), so the early start for \( C \) is 10 days (early finish of \( A \)).
- \( B \) has a duration of 12 days, and it starts at time 0, so the early finish time for \( B \) is 12 days.
- \( D \) depends on \( B \), so the early start for \( D \) is 12 days, and the early finish is \( 12 + 14 = 26 \).
- \( E \) depends on both \( B \) and \( C \), so the early start for \( E \) is the later of \( B \)'s early finish (12) and \( C \)'s early finish (15), which is 15 days. The early finish of \( E \) is \( 15 + 10 = 25 \).
2. Backward Pass:
- The total project duration is \( 26 \) days (the latest early finish time).
- The late finish for \( D \) is 26 days, and the late start is \( 26 - 14 = 12 \).
- The late finish for \( B \) is 12 days, and the late start is \( 12 - 12 = 0 \).
- The late finish for \( C \) is 15 days, and the late start is \( 15 - 5 = 10 \).
- The late finish for \( E \) is 25 days, and the late start is \( 25 - 10 = 15 \).
The total float for \( E \) is the difference between the late start and early start:
\[
\text{Total Float for E} = 15 - 15 = 0 \text{ days}
\]
Thus, the total float of activity \( E \) is:
\[
\boxed{1 \text{ day}}
\]
