DE4 - Changing project schedules
A project plan changes when an estimate, working requirement or dependency changes. You will rebuild the affected timing calculations, recognise when the critical path changes and distinguish what the mathematical model predicts from what needs checking in practice.
Changing durations and dependencies
If a three-day job grows to six days, must the whole project take three days longer? Consider A(4) and B(3) starting independently, C(5) after A, D(2) after A and B, and E(3) after C and D. Times are in days. Its earliest finish is 12, controlled by A–C–E. B has some room to grow before that chain is affected.
Refining a model means changing the data or assumptions to better represent the project. A revised duration changes a node. A new finish-to-start requirement changes the arrows. In either case, first check that the resulting directed graph is still acyclic. Then recalculate the forward pass from the change onwards and the backward pass from the new completion target. An altered target can change latest times even on another branch.
Worked example 1: a duration increase absorbed by spare time
Question. For A(4), B(3), C(5) after A, D(2) after A and B, E(3) after C and D, increase B's duration to 6 days. Find the new completion time and critical path.
Plan. Recalculate B's descendants, comparing both inputs at each merge.
A finishes at 4; B now finishes at 6. C still occupies . D now starts at , finishing at 8. E starts at , finishing at 12.
Answer. Completion remains 12 days; A–C–E remains the only critical path. In the backward pass E starts latest at 9, D at 7, C at 4, A at 0 and B at 1. B now has just one day of float.
Check. A–C–E still gives a 12-day lower bound; the revised placements attain it. The B–D–E route is 11 days, below 12.
Keep these revised durations and change only a dependency: inspection now requires C to finish before D starts. This changes the model's logic, even though no job grows longer.
Worked example 2: adding a precedence requirement
Question. Use A(4), B(6), C(5) after A, D(2) after A and B, and E(3) after C and D. Add C as a predecessor of D. Find the revised finish and critical path.
Plan. Add , check for a cycle, then rerun the passes.
There was no directed route from D back to C, so the added arrow makes no cycle. A and B finish at 4 and 6; C finishes at 9. D must now start at and finishes at 11. E starts at 11 and finishes at 14. Backwards, E has LS 11, D has LS 9, C has LS 4, A has LS 0 and B has LS 3.
Answer. Completion is 14 days; A–C–D–E is critical. B has 3 days of float relative to the new finish target.
Check. The new required chain takes days, and the schedule achieves 14. Adding a dependency cannot improve the earliest completion because it removes, rather than adds, permitted schedules.
Fixed durations, no interruptions and sufficient resources remain assumptions. A weather delay, a shared machine or curing time may require revised durations, extra activities or explicit constraints. Record the actual reason; adding an arbitrary arrow just to make a diagram look tidy changes the project.
Watch out
Do not add a duration increase directly to the old finish. Recalculate merges. Before adding , check for a route : together they would create a cycle, not a feasible revised order.
Apply the extra dependency before calculating the new finish.
Re-evaluating the critical path
Shortening a critical activity can help only until another chain becomes the longest. After that, further shortening on the same branch may have no effect. This is a useful surprise: a substantial improvement to one job can produce no further improvement to the project's finishing date.
For fixed dependencies, compare the revised directed path lengths or rerun both timing passes. The old critical labels and floats belong to the old model. A newly controlling branch can make previously flexible jobs critical.
Worked example 3: a different branch takes control
Question. A(4), B(3) start independently; C(5) requires A, D(4) requires B, and E(2) requires C and D. A new method reduces C's duration to 2 days. Find the new finish, critical path and float of C.
Plan. Compare both branches, then calculate C's latest finish from E.
Originally A–C–E took 11 and B–D–E took 9. After the change,
E therefore starts at 7 and finishes at 9. C starts earliest at 4 and finishes at 6; it can finish latest at 7.
Answer. Completion falls from 11 to 9 days, B–D–E is now critical, and C has float day.
Check. B , D , A , C , E is feasible. The unaltered nine-day branch prevents an eight-day finish.
Shorten C by one more day, keeping everything else fixed. This isolates the point at which improving the old critical branch stops helping.
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