Question 4
A school is renovating its library. The table shows the activities involved, their duration (days) and their immediate predecessors.
| Activity | Duration (days) | Immediate predecessor/s |
|---|---|---|
| A | 4 | — |
| B | 6 | — |
| C | 3 | A |
| D | 5 | A |
| E | 2 | B, C |
| F | 7 | D |
| G | 4 | E |
| H | 3 | F, G |
Activity C is delayed and will take 5 days longer than planned.
Use a project network diagram with completed forward and backward scanning to determine the effect of this delay on the minimum time to complete the renovation.
Scan both ways, then spend the float
The network is drawn in table order and scanned forward and backward, then the float of C is compared with the delay and the network is rescanned with C taking 8 days.
Delay activity C and watch the scans rerun
Drag the slider to delay C by 0 to 6 days. Every node shows EST | LST, and thick arcs have zero float.
Each node shows EST | LST (days). Values that differ from the planned scan are bold.
Marked wrong.
Work through the solution one mark at a time
Try each step yourself before you reveal it.
Draw the network with each activity as an arc between two events: seven nodes and eight arcs, each labelled with its activity and duration. E follows both B and C, so B and C end at the same node. H follows both F and G, so F and G end at the same node.
Nodes numbered 1 to 7 here for reference.
The network shows every activity as an arc between events, so the scans can follow the dependencies.
Completing the forward scan with a duration of 8 days for activity C to find a new minimum time of 21 days is an alternative method. Method 2 shows that route.
Marker: correctly constructs the project network diagram showing all 8 activities with their durations and dependencies
Forward scan, left to right. Where arcs meet, take the largest value.
$$\begin{aligned}\text{Node 1: }&0\\\text{Node 2: }&0+4=4\\\text{Node 3: }&\max(0+6,\ 4+3)=\max(6,7)=7\\\text{Node 4: }&4+5=9\\\text{Node 5: }&7+2=9\\\text{Node 6: }&\max(9+7,\ 9+4)=\max(16,13)=16\\\text{Node 7: }&16+3=19\end{aligned}$$
The minimum time to complete the renovation is 19 days.
The forward scan gives each event its earliest start time. Every arc into a node must be finished, so the largest value is kept where arcs meet.
Marker: completes forward scanning
Backward scan, right to left. Where arcs leave, take the smallest value.
$$\begin{aligned}\text{Node 7: }&19\\\text{Node 6: }&19-3=16\\\text{Node 5: }&16-4=12\\\text{Node 4: }&16-7=9\\\text{Node 3: }&12-2=10\\\text{Node 2: }&\min(9-5,\ 10-3)=\min(4,7)=4\\\text{Node 1: }&\min(4-4,\ 10-6)=\min(0,4)=0\end{aligned}$$
Each node shows EST | LST (days).
Critical path: A–D–F–H (EST $=$ LST at every node on it), minimum time 19 days.
The backward scan gives each event its latest start time. The smallest value is kept where arcs leave, so no later activity is delayed.
An alternative representation, for example ESTs and LSTs shown in a table beside the diagram, also earns the scanning marks.
Marker: completes backward scanning
C starts at node 2 (EST 4) and ends at node 3 (LST 10), so the latest start time of C is $10-3=7$.
$$\text{Float of C}=\text{LST}-\text{EST}=7-4=3\text{ days}$$
The float of C is how long C can be delayed without delaying the project.
Follow-through applies from an error in earlier working.
Marker: determines the float time for activity C
The delay of 5 days exceeds the float of 3 days by $5-3=2$ days.
$$19+2=21\text{ days}$$
The new critical path is A–C–E–G–H ($4+8+2+4+3=21$).
Only the part of the delay beyond the float adds to the minimum time.
An equivalent statement, for example “The renovation will take 21 days.”, also earns this mark.
Marker: determines the effect of the delay on the minimum time
Give C its delayed duration, $3+5=8$ days, and draw the same network as in Method 1 with arc C labelled C, 8.
Nodes numbered 1 to 7 here for reference.
Marker: correctly constructs the project network diagram showing all 8 activities with their durations and dependencies
Forward scan with C $=8$. Where arcs meet, take the largest value.
$$\begin{aligned}\text{Node 1: }&0\\\text{Node 2: }&0+4=4\\\text{Node 3: }&\max(0+6,\ 4+8)=\max(6,12)=12\\\text{Node 4: }&4+5=9\\\text{Node 5: }&12+2=14\\\text{Node 6: }&\max(9+7,\ 14+4)=\max(16,18)=18\\\text{Node 7: }&18+3=21\end{aligned}$$
Marker: completes forward scanning
Backward scan with C $=8$. Where arcs leave, take the smallest value.
$$\begin{aligned}\text{Node 7: }&21\\\text{Node 6: }&21-3=18\\\text{Node 5: }&18-4=14\\\text{Node 4: }&18-7=11\\\text{Node 3: }&14-2=12\\\text{Node 2: }&\min(11-5,\ 12-8)=\min(6,4)=4\\\text{Node 1: }&\min(4-4,\ 12-6)=\min(0,6)=0\end{aligned}$$
Each node shows EST | LST (days).
Marker: completes backward scanning
The latest start time of C is $12-8=4$, so the float of C is $4-4=0$ days. C is now critical, on the new critical path A–C–E–G–H.
$$\text{Float of C}=4-4=0\text{ days}$$
With the delay, the float of C shows whether C now controls the project.
Marker: determines the float time for activity C
The planned minimum time is the length of the longest path A–D–F–H, $4+5+7+3=19$ days. The minimum time increases from 19 days to 21 days.
$$21-19=2\text{ days}$$
The effect is the change from the planned minimum time.
An equivalent statement, for example “The renovation will take 21 days.”, also earns this mark.
Marker: determines the effect of the delay on the minimum time
If I had to guess where you might lose marks…
- Taking the smaller value at node 3 in the forward scan: EST 6 instead of 7. The minimum time still comes out as 19, which hides the slip, but it loses Mark 2. With the delay, the same slip gives $\min(6, 12) = 6$ and “no effect, still 19 days”, which loses Mark 5 as well.
- Taking the larger value at node 2 in the backward scan: $\max(9 - 5,\ 10 - 3) = 7$ instead of 4. This loses Mark 3.
- Working out the float of C as the LST of node 3 minus the EST of node 2, $10 - 4 = 6$, which forgets the 3-day duration of C. A delay of 5 days then looks absorbed (“no effect”), which loses Marks 4 and 5.
- Adding the whole delay to the project: $19 + 5 = 24$ days. This loses Mark 5.
- Rescanning with C as 5 days instead of $3 + 5 = 8$ days. The minimum time stays at 19 days, which loses Mark 5.
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
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