Question 12
The activity times in the project network shown are in days.
The greatest float time for a non-critical activity in this network is
That’s it. Step through the hints below to see why.
Not quite. One way to get 4 is to add the floats of C and E, $2+2$. But C and E sit on the same path, so they share the same 2 spare days: if C uses them up, E has none left. 4 is also the duration of C, which isn’t a float at all.
Not quite. 5 is how long E takes, read from the label “E, 5”. Float is spare time: how long an activity can be delayed without delaying the whole project. E has $17-10-5=2$ days of it.
Not quite. 12 is the latest start time for E, read straight from the node before E. Float compares the latest start with the earliest start, so E’s float is $12-10=2$ days.
Pick an activity to find its float. Then delay it with the slider and watch when the finish time starts to move.
Latest time at the end node (blue), minus earliest time at the start node (orange), minus the duration.
Float in days, filled in as you check each activity with no delay.
Read the key first. The left half of each node is the earliest start time (EST) and the right half is the latest start time (LST). Where the two halves match, that node sits on the critical path.
The critical path is A, B, D, F, G. Its durations add to the finish time:
$$6+5+3+3+5=22\text{ days}$$
That leaves C and E as the non-critical activities. The node between them, 10 and 12, is the only one whose halves differ.
Float is how long an activity can be delayed without delaying the project. For an activity on an arc:
$$\text{float}=\text{LST}_{\text{end}}-\text{EST}_{\text{start}}-\text{duration}$$
$$\begin{aligned}\text{C: }&12-6-4=2\text{ days}\\\text{E: }&17-10-5=2\text{ days}\end{aligned}$$
For E you can also read it off the key: latest start 12 minus earliest start 10 gives 2 days.
Don’t add the two floats. C and E are on the same path, A, C, E, G, which takes
$$6+4+5+5=20\text{ days}$$
That is 2 days less than the critical path, so the whole path has 2 spare days, shared between C and E. Delay C by 2 days on the left and watch E’s float drop to 0.
So the greatest float is 2 days. The 5 and the 12 in the other options are a duration and a start time, read off the diagram rather than calculated.
The critical path A, B, D, F, G takes 22 days, so those five activities have zero float. The non-critical activities are C and E:
$$\begin{aligned}\text{float of C}&=12-6-4=2\text{ days}\\\text{float of E}&=17-10-5=2\text{ days}\end{aligned}$$
Both are 2 days, and they share the same 2 days because they are on the same path. The greatest float for a non-critical activity is 2 days.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2020 QCAA General Mathematics external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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