How do we schedule a project and find the shortest time in which it can finish?
Use activity networks, forward and backward scanning, and float to identify the critical path of a project.
Activity networks, earliest and latest start times, float, and the critical path that fixes the minimum project completion time in TCE Mathematics Applications.
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What this dot point is asking
Large projects break into activities, some of which must wait for others to finish first. Critical path analysis finds the longest unavoidable chain of activities, which fixes how quickly the whole project can possibly be done.
Forward scan: earliest start times
Working from the start, the earliest start time (EST) of an activity is the latest of the earliest finish times of all its immediate predecessors. The earliest finish time of an activity is its EST plus its duration. The largest earliest finish time at the end is the minimum completion time.
Float and the critical path
The float (slack) of an activity is how long it can be delayed without delaying the whole project. It equals the latest start time minus the earliest start time. Activities with zero float cannot be delayed at all; together they form the critical path.
Why it matters
Knowing the critical path tells a manager exactly which activities to watch. Speeding up a non-critical activity such as C saves nothing, but any delay on a critical activity such as B pushes the whole finish date out day for day.
The backward scan and float in detail
After the forward scan fixes the project completion time, the backward scan finds the latest each activity can start without pushing the finish out. Set the final activity's latest finish time equal to the project completion time, then work right to left: the latest finish time of an activity is the smallest of the latest start times of the activities that follow it, and its latest start time is that latest finish minus its duration. Float is then latest start minus earliest start. An activity on a non-critical branch has positive float; you can delay or extend it by up to its float before it begins to delay the project.
A delay beyond an activity's float adds the excess to the completion time and can change which chain is critical, which is exactly what the multi-part exam question above tests when it adds four weeks to tasks G and E.
A complete answer shows the forward scan to find the minimum completion time, the backward scan for latest start times, calculates float as latest minus earliest start, and names the critical path as the zero-float chain whose total duration equals that completion time.
Exam-style practice questions
Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
TCE 20248 marksQuestions about Critical Path Analysis. a) Complete a forward scan on a precedence graph. b) Complete a backwards scan on another precedence graph. Figure 11 is a precedence graph for a shed, times in weeks. c) Write the Critical Path as a sequence of letters. d) Calculate the float for task G and use it to determine how the completion time is affected if G takes four weeks more than expected. e) Calculate the float for task E and determine how the critical path and completion time change if E takes four weeks more than expected.Show worked answer →
a) (2 marks) Forward scan: the earliest start time (EST) of each activity is the largest of the earliest finish times of its immediate predecessors. Fill the top box of each node working left to right.
b) (2 marks) Backwards scan: the latest finish time (LFT) of each activity is the smallest of the latest start times of the activities that follow it. Fill the lower boxes working right to left.
c) (1 mark) The critical path is the chain of activities with zero float, written as a sequence of letters (the longest path through the network).
d) (1.5 marks) Float for G = LFT - EST - duration. If G is on a non-critical path, the project completion time is unchanged provided the four-week increase does not exceed G's float; if it does, the overrun beyond the float adds to the completion time.
e) (1.5 marks) Compute E's float the same way. If adding four weeks exceeds E's float, E becomes critical, the critical path is redrawn through E, and the new completion time increases by the amount the delay exceeds the old float.
TCE 20194 marksA precedence graph is given for building a bridge. b) Complete the critical path analysis by detailing the EST and LFT of each activity. c) State the critical path. d) What is the shortest possible completion time for building the bridge? e) What is the float associated with making the cables, and what does this mean?Show worked answer →
b) (part of the question) Forward scan to fill each EST (largest earliest finish of predecessors), then backwards scan to fill each LFT (smallest latest start of successors).
c) (1 mark) The critical path is the sequence of activities whose float is zero, that is, where EST equals LFT minus duration along the longest chain.
d) (1 mark) The shortest possible completion time equals the earliest finish time of the final activity, which is the total duration along the critical path.
e) (2 marks) Float for making the cables LFT EST duration for that activity. A float means making the cables can be delayed (or take longer) by up to that many time units without delaying the whole project; it is slack that is not on the critical path.