ENGI5003 · Professional Engineering Management
Scheduling: Estimation, Gantt, AON & the Critical Path
Scheduling turns a list of activities into a defensible completion date and tells you which tasks you cannot afford to slip. This chapter builds an Activity-on-Node (AON) network, runs the forward and backward pass to compute early/late dates and float (slack), identifies the critical path, and plots the result on a Gantt chart — the single most heavily weighted quantitative skill in the ENGI 5003 exam, where Part C routinely asks you to draw the network, find TE, and state the critical path.
What this chapter covers
- 01AON network conventions and the 3x3 node legend (ES|ID|EF / TS|Desc|FS / LS|Dur|LF)
- 02Forward pass: ES = max(EF of predecessors), EF = ES + Duration, T_E at the end node
- 03Backward pass: LF = min(LS of successors), LS = LF - Duration, the LS-start = 0 check
- 04Total slack vs free slack (TS = LS - ES; FS = min ES of successors - EF; TS >= FS)
- 05Identifying the critical path as the zero-slack chain
- 06Gantt charts: early-start vs late-start, dependency-driven bars, milestones
- 07How float feeds budgeting, crashing and resource levelling later in the course
Build an AON, run both passes, find the critical path
- +2Forward pass (ES = max EF of predecessors, EF = ES + Dur). A: ES 0, EF 4. B: ES 4, EF 10. C: ES 4, EF 7. D: ES 10, EF 15. E: ES 7, EF 11.
- +1Merge at F (predecessors D and E): ES = max(EF_D = 15, EF_E = 11) = 15, EF = 15 + 2 = 17. End node EF = 17, so T_E = 17 weeks.
- +2Backward pass (LF_end = T_E = 17; LF = min LS of successors, LS = LF - Dur). F: LF 17, LS 15. D: LF 15, LS 10. E: LF 15, LS 11. B: LF 10, LS 4. C: LF = LS_E = 11, LS 8. A: LF = min(LS_B = 4, LS_C = 8) = 4, LS 0 (check: LS_start = 0).
- +1Total slack TS = LS - ES: A 0, B 0, C 4, D 0, E 4, F 0. The zero-slack chain is A-B-D-F.
Key terms
- Activity-on-Node (AON) network
- A project network model in which each activity is a box (node) and each arrow is a precedence dependency; it has one start node and one end node, flows left to right, and contains no loops. Used to find the critical path and completion date.
- Forward pass
- The left-to-right sweep that computes the earliest dates: ES = max(EF of all predecessors) and EF = ES + Duration, starting from ES = 0. The EF of the end node is the project duration T_E.
- Backward pass
- The right-to-left sweep that computes the latest dates: starting from LF_end = T_E, LF = min(LS of all successors) and LS = LF - Duration. Correct work returns LS = 0 at the start node.
- Total slack (TS)
- The time an activity can be delayed without delaying project completion: TS = LS - ES = LF - EF. Activities with zero total slack lie on the critical path.
- Free slack (FS)
- The time an activity can be delayed without delaying the early start of any successor: FS = min(ES of successors) - EF. Always FS <= TS, because total slack along a chain is shared.
- Critical path
- The chain of zero-slack activities running from start to end; it determines the project duration T_E, and a delay to any activity on it delays the whole project.
Scheduling: Estimation, Gantt, AON & the Critical Path FAQ
What is the difference between total slack and free slack?
Total slack (TS = LS - ES) is how long an activity can slip without delaying the project. Free slack (FS = min ES of successors - EF) is how long it can slip without delaying any successor's earliest start. Free slack is a tighter, more local figure, so FS is always less than or equal to TS. When two non-critical activities sit in series, they share one pool of total slack: if the first uses it all, the second has none left.
How do I find the critical path on an AON network?
Run the forward pass to get every ES and EF (and the project duration T_E), run the backward pass to get every LS and LF, then compute total slack TS = LS - ES for each activity. The critical path is the continuous chain of activities whose total slack is zero, running from the start node to the end node.
What is T_E and how is it calculated?
T_E is the project's earliest completion time (its overall duration). It equals the early finish (EF) of the end node after the forward pass. When an activity has several predecessors, you take the maximum of their early finishes, so T_E is set by the longest chain through the network, which is the critical path.
Why take the maximum on the forward pass but the minimum on the backward pass?
On the forward pass an activity cannot start until all its predecessors are finished, so its ES is the maximum of the incoming EFs (wait for the slowest one). On the backward pass an activity must finish in time for its earliest-needed successor, so its LF is the minimum of the outgoing LSs. Mixing these up is the classic error and usually shows up as a critical path that does not actually have zero slack.
How does an AON network relate to a Gantt chart?
They are complements. The AON network answers how long the project takes and which activities cannot slip (precedence, critical path, float). The Gantt chart answers when each activity runs on the calendar, drawing one dependency-linked bar per activity. Build the network first to get the logic, then plot the Gantt to communicate it and to layer on budgeting and resource analysis.
Exam move
Drill the four-step machine until it is reflex, because Part C of the exam almost always hands you a precedence table and asks for the network, T_E and the critical path. The fixed workflow: (1) draw the AON straight from the precedence column, one start and one end node; (2) forward pass left to right, ES = max(EF of predecessors), EF = ES + Dur, read T_E off the end node; (3) backward pass right to left, LF = min(LS of successors), LS = LF - Dur, and finish with the LS-at-start = 0 check; (4) compute TS = LS - ES for every activity and state the zero-slack chain explicitly as the critical path. Memorise the two traps that lose the most marks: max-out / min-back (never reverse them), and that critical activities have exactly zero float (if a "critical" bar appears to have slack, your passes are wrong). The AON node legend and PERT formulae are provided in the closed-book exam, so spend your prep on the method and the max/min logic, not on rote-memorising the box layout.
