Independent float formula with clear event times For an activity i→j of duration t, let L_i be the latest event time at its start (tail) and E_j be the earliest event time at its finish (head). If we denote t1 = L_i and t2 = E_j, what is the independent float (IF)?

Introduction / Context:
Float measures scheduling flexibility. Several flavors exist: total float, free float, and independent float. Independent float represents the amount of delay available to an activity without affecting either the earliest start of its successor or the latest finish of its predecessor—making it a conservative, “independent” buffer.

Given Data / Assumptions:

• Activity i→j with duration t.
• L_i = latest permissible time of event i (latest finish of preceding event), denoted t1.
• E_j = earliest permissible time of event j (earliest start for successors), denoted t2.

Concept / Approach:
The well-known expression for independent float is IF = E_j - L_i - t. With t1 = L_i and t2 = E_j, this becomes IF = (t2 - t1) - t. If the result is negative, independent float is taken as zero in practical scheduling, because negative float indicates infeasibility under current constraints.

Step-by-Step Solution:
Start with definition: IF = earliest at head − latest at tail − duration.Substitute symbols: IF = E_j − L_i − t.Map to problem notation: t2 = E_j, t1 = L_i → IF = (t2 − t1) − t.Hence, the correct choice is (t2 − t1) − t.

Verification / Alternative check:
Testing on a simple network confirms that delaying the activity by at most IF neither pushes the successor’s earliest start nor encroaches on the predecessor’s latest finish.

Why Other Options Are Wrong:
(t1 − t2) − t: Sign reversed; would suggest negative float for normal schedules.t − (t2 − t1): Rearrangement that does not match the definition; equals t1 − t2 + t.(t1 + t2) − t and t + (t1 − t2): Not dimensionally meaningful per the independent float definition.

Common Pitfalls:

• Confusing independent float with free float (which uses earliest times only).
• Using negative independent float directly; in practice, it is set to zero and highlights a tight network.

Final Answer:
(t2 - t1) - t