Forward and backward scans | Year 12

01 / LEARN

Understand the idea.

Forward scan: all predecessors must be ready

Begin the forward scan with the start event at time zero. For each incoming arrow, add its duration to the earliest time at its tail. Record the largest arrival time at the head because every incoming activity must finish.

In the display network, event 4 receives $4 + 4 = 8$ through D and $3 + 0 = 3$ through the dummy, so it occurs at day 8. Event 5 receives 9 through C and 10 through E; F then finishes at day 13.

E_{j} = max_{i \to j} (E_{i} + d_{i j})

Backward scan: protect every successor

Begin the backward scan at the required finish time, normally the earliest project finish when finding no-delay allowances. For each outgoing arrow, subtract its duration from its head's latest time. Use the smallest candidate to protect every successor.

For event 2, C permits $10 - 5 = 5$ , but D requires $8 - 4 = 4$ . Recording 5 would delay D, so use 4. With the earliest finish as the deadline, latest event times must be at least their corresponding earliest times.

L_{i} = min_{i \to j} (L_{j} - d_{i j})

Forward and backward scans for the display network, with completion at day 13.
Event	Earliest time	Latest time
1	0	0
2	4	4
3	3	8
4	8	8
5	10	10
6	13	13

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A merge in the forward scan

Event 4 can be reached from event 2 at time 6 by an activity of duration 5, or from event 3 at time 8 by an activity of duration 2. Find its earliest time.

Worked solution

1
Calculate arrival through event 2.
$6 + 5 = 11$
The activity cannot start before event 2 occurs and then needs five time units.
2
Calculate arrival through event 3.
$8 + 2 = 10$
The second incoming activity has its own start time and duration.
3
Record the later required arrival.
$E_{4} = max (11, 10) = 11$
Event 4 cannot occur while either incoming activity is unfinished.

Answer

Event 4 has earliest time 11.

EXAMPLE 02

A split in the backward scan

An event has outgoing arrows of durations 4 and 7. Their head events have latest times 18 and 20. Find the event's latest time.

Worked solution

1
Work back through the first outgoing arrow.
$18 - 4 = 14$
Starting this arrow after time 14 would miss its head event's latest time.
2
Work back through the second outgoing arrow.
$20 - 7 = 13$
The seven-unit activity must start by time 13 to finish by time 20.
3
Satisfy both upper limits.
$L = min (14, 13) = 13$
The event must occur no later than either permitted latest start.

Answer

The latest event time is 13.

03 / YOUR TURN

Try it yourself.

Check your understanding

An event receives candidates 12, 15 and 15 in the forward scan. What earliest time is recorded?

Show answer & reasoning

15.

Record $max (12, 15, 15) = 15$ . Both arrivals of 15 control this event; choosing 12 would permit outgoing work before the other branches finish.

Keep practising

6 worksheet questions · 16 marks

Open worksheet Worked solutions

Read the questions online

Perform a complete forward scan on the displayed network.
4 marks
Perform the backward scan for the displayed network using project finish 13.
4 marks
Find the earliest time at a merge when incoming tail times and durations are $(2, 7)$ , $(6, 2)$ and $(4, 6)$ .
2 marks
Find a latest event time when outgoing head times and durations are $(15, 4)$ , $(19, 6)$ and $(14, 2)$ .
2 marks
A forward scan gives an event earliest time 9. A claimed backward scan gives latest time 7 for a finish target equal to the earliest project finish. What does this indicate?
2 marks
Why does adding all incoming durations at a merge usually give the wrong earliest time?
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Perform a complete forward scan on the displayed network.
4 marks
$E_{1} = 0$ ; $E_{2} = 4$ ; $E_{3} = 3$ .
$E_{4} = max (4 + 4, 3 + 0) = 8$ .
$E_{5} = max (4 + 5, 8 + 2) = 10$ ; $E_{6} = 10 + 3 = 13$ .
Events 1–6: 0, 4, 3, 8, 10, 13.
Perform the backward scan for the displayed network using project finish 13.
4 marks
$L_{6} = 13$ ; $L_{5} = 10$ ; $L_{4} = 8$ ; $L_{3} = 8$ .
$L_{2} = min (10 - 5, 8 - 4) = 4$ .
$L_{1} = min (4 - 4, 8 - 3) = 0$ .
Events 1–6 latest: 0, 4, 8, 8, 10, 13.
Find the earliest time at a merge when incoming tail times and durations are $(2, 7)$ , $(6, 2)$ and $(4, 6)$ .
2 marks
Candidates are 9, 8 and 10.
Their maximum is 10.
10.
Find a latest event time when outgoing head times and durations are $(15, 4)$ , $(19, 6)$ and $(14, 2)$ .
2 marks
Candidates are 11, 13 and 12.
Their minimum is 11.
11.
A forward scan gives an event earliest time 9. A claimed backward scan gives latest time 7 for a finish target equal to the earliest project finish. What does this indicate?
2 marks
The event cannot both require at least 9 and be required by 7.
Recheck durations, directions and the final-time assumption.
The scans or the model contain an error.
Why does adding all incoming durations at a merge usually give the wrong earliest time?
2 marks
Each candidate already measures elapsed time from the common start.
The event waits for the latest arrival, not the sum of arrivals.
It assumes parallel branches occurred one after another.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 General Mathematics: discrete models, project scheduling and assignment. Activity-on-arrow conventions are used for project networks; technology may verify calculations but does not replace interpreting the model.

Official curriculum source ↗

In this lesson

Understand the idea.

Forward scan: all predecessors must be ready

Backward scan: protect every successor

See the method in action.

A merge in the forward scan

Calculate arrival through event 2.

Calculate arrival through event 3.

Record the later required arrival.

A split in the backward scan

Work back through the first outgoing arrow.

Work back through the second outgoing arrow.

Satisfy both upper limits.

Try it yourself.

Check your understanding

Keep practising