DE2, DE3 - Critical paths and float

Forward pass

For duration $d$, earliest start ES and earliest finish EF,

For an activity with no predecessor, use $\mathrm{ES}=0$. The earliest project finish is the maximum EF of all terminal activities, or the time of a zero-duration common Finish. These scheduling rules are not supplied in the AQA formula booklet.

Worked example 1: calculate the earliest schedule

Question. Find all ES and EF values for the five-activity table above.

Plan. Start A and B at zero, then work through C and D to E.

Answer. Earliest completion is 12 days; the ES values in A–E order are 0, 0, 4, 4, 9.

Check. The chain A–C–E takes $4+5+3=12$ days and must be sequential. It gives a lower bound of 12, and the displayed schedule reaches it.

Worked example 2: a different predecessor controls a merge

Question. In the same project, B now takes 6 days; all other durations and dependencies are unchanged. Find ES and EF.

Plan. Recalculate downstream of B, taking a maximum at each merge.

A remains $[0,4)$, B is $[0,6)$, C remains $[4,9)$. D must start at $\max(4,6)=6$, so finishes at 8. E starts at $\max(9,8)=9$ and finishes at 12.

Answer. ES values are 0, 0, 4, 6, 9; EF values are 4, 6, 9, 8, 12. Completion is still 12 days.

Check. A–C–E still needs 12 days. The other controlling branch B–D–E takes $6+2+3=11$, so it can fit before that finish.

Watch out

At a merge, use the largest predecessor finish, not their sum or their smallest value. Independent predecessor activities can overlap, but every one must finish.

Backward pass

For latest finish LF and latest start LS,

Set LF equal to the project target $T$ at every terminal activity, or start from a zero-duration common Finish at $T$. Here $T$ is the earliest completion from the forward pass. These rules are not supplied in the AQA formula booklet.

Worked example 3: reverse the five-activity project

Question. A(4) and B(3) start independently; C(5) requires A; D(2) requires A and B; E(3) requires C and D. Find LS and LF for a target of 12 days.

Plan. Give E the target finish, then ensure every predecessor finishes by its earliest successor's latest start.

E has LF 12, LS $12-3=9$. Therefore C and D have LF 9, giving LS $9-5=4$ and $9-2=7$. A must finish by both C's LS 4 and D's LS 7:

B has only D as successor, so LF 7 and LS $7-3=4$.

Answer. In A–E order, LF is 4, 7, 9, 9, 12; LS is 0, 4, 4, 7, 9.

Check. Placing every activity at LS gives A $[0,4)$, B $[4,7)$, C $[4,9)$, D $[7,9)$, E $[9,12)$. Every predecessor finishes by its successor's start, and all durations are correct.

Worked example 4: a longer predecessor in the backward pass

Question. B takes 6 days; the other data in Worked example 3 and the 12-day target are unchanged. Find the revised LS and LF values.

Plan. Work backwards again; a duration change may affect both its own LS and preceding activities.

E, C and D still give LS 9, 4 and 7 respectively. A still has LF $\min(4,7)=4$, LS 0. B must still finish by 7, so its LS becomes $7-6=1$.

Answer. LF remains 4, 7, 9, 9, 12; LS becomes 0, 1, 4, 7, 9.

Check. B placed at $[1,7)$ meets D at time 7. Its earlier placement $[0,6)$ is also feasible, consistent with its forward-pass ES of zero.

Watch out

Use a minimum when working backwards through a split. A must meet the earliest of its successors' latest starts; meeting only the later one may delay another branch.