The backward pass

Covers AS91576

What the backward pass finds

The backward pass works right to left through the network to find, for every activity:
- its latest finish time (LFT) — the latest it can finish without delaying the project,
- its latest start time (LST) — the latest it can start and still finish by its LFT.
These go in the bottom row of each node box (LST bottom-left, LFT bottom-right).

Latest finish and latest start

For each activity:

$\text{LST} = \text{LFT} - \text{duration}$

LFT — latest finish time; the last moment it can finish without pushing the project out.
duration — the fixed length of the activity.
LST — latest start time; subtract the duration from the latest finish.

The backward pass rule

Set the final activity's LFT equal to the project completion time found by the forward pass (so the project is not delayed).
Work backwards: an activity's LFT is the latest start time of the activity that follows it:

$\text{LFT of an activity} = \text{LST of its successor}$

Then apply $\text{LST} = \text{LFT} - \text{duration}$ and carry the LST back along each arrow.

Merging paths — take the smallest

If an activity has several successors (its arrow splits), it must finish in time for all of them.
Its LFT is the smallest (earliest) of the successors' latest start times:

$\text{LFT at a split} = \min(\text{LSTs of all successors})$

Taking the largest by mistake would let the activity finish too late for one of its successors — the classic backward-pass error.

The completed backward pass

Community hall network with earliest and latest times filled in

Reading the bottom rows of the hall network: G has $\text{LFT}=19$ , $\text{LST}=15$ . E has $\text{LST}=12$ , F has $\text{LST}=13$ . D splits to E and F, so $\text{LFT}_D = \min(12, 13) = 12$ and $\text{LST}_D = 7$ . B has $\text{LST}=3$ ; C has $\text{LFT}=7$ , $\text{LST}=5$ ; A has $\text{LST}=0$ .

Worked ExampleBackward pass on the hall build

Using the project duration of 19 days, carry out the backward pass for the hall network to find the LST and LFT of activities D, C and A.

Step 1 — Set the final activity

G is last, so its latest finish equals the project duration:

$\text{LFT}_G = 19, \qquad \text{LST}_G = 19 - 4 = 15$

Step 2 — Step back to E and F

Both feed G, so each takes $\text{LFT} = \text{LST}_G = 15$ :

$\text{E: LST}=15-3=12 \qquad \text{F: LST}=15-2=13$

Step 3 — Split at D (take the smallest)

D feeds E and F, so its latest finish is the smaller successor latest start:

$\text{LFT}_D = \min(12, 13) = 12, \qquad \text{LST}_D = 12 - 5 = 7$

Step 4 — Step back to C and A

C feeds D, so $\text{LFT}_C = \text{LST}_D = 7$ and $\text{LST}_C = 7 - 2 = 5$ .

B feeds D, so $\text{LFT}_B = 7$ and $\text{LST}_B = 3$ . A feeds B, so $\text{LFT}_A = \text{LST}_B = 3$ and $\text{LST}_A = 3 - 3 = 0$ .

$\boxed{\text{D: } 7/12, \quad \text{C: } 5/7, \quad \text{A: } 0/3 \ \ (\text{LST}/\text{LFT})}$

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

An activity has latest finish time 15 and duration 3. Find its latest start time and state the formula.

Merit

Activity V feeds two successors whose latest start times are 12 and 13. V has duration 5. Find V's latest finish and latest start, and explain the choice at the split.

Excellence

Explain why, for the final activity of any project, the latest finish time set at the start of the backward pass must equal the project completion time found by the forward pass — no larger and no smaller — and what each error would do.

The backward pass

Covers AS91576

What the backward pass finds

The backward pass works right to left through the network to find, for every activity:
- its latest finish time (LFT) — the latest it can finish without delaying the project,
- its latest start time (LST) — the latest it can start and still finish by its LFT.
These go in the bottom row of each node box (LST bottom-left, LFT bottom-right).

Latest finish and latest start

For each activity:

$\text{LST} = \text{LFT} - \text{duration}$

LFT — latest finish time; the last moment it can finish without pushing the project out.
duration — the fixed length of the activity.
LST — latest start time; subtract the duration from the latest finish.

The backward pass rule

Set the final activity's LFT equal to the project completion time found by the forward pass (so the project is not delayed).
Work backwards: an activity's LFT is the latest start time of the activity that follows it:

$\text{LFT of an activity} = \text{LST of its successor}$

Then apply $\text{LST} = \text{LFT} - \text{duration}$ and carry the LST back along each arrow.

Merging paths — take the smallest

If an activity has several successors (its arrow splits), it must finish in time for all of them.
Its LFT is the smallest (earliest) of the successors' latest start times:

$\text{LFT at a split} = \min(\text{LSTs of all successors})$

Taking the largest by mistake would let the activity finish too late for one of its successors — the classic backward-pass error.

The completed backward pass

Community hall network with earliest and latest times filled in

Reading the bottom rows of the hall network: G has $\text{LFT}=19$ , $\text{LST}=15$ . E has $\text{LST}=12$ , F has $\text{LST}=13$ . D splits to E and F, so $\text{LFT}_D = \min(12, 13) = 12$ and $\text{LST}_D = 7$ . B has $\text{LST}=3$ ; C has $\text{LFT}=7$ , $\text{LST}=5$ ; A has $\text{LST}=0$ .

Worked ExampleBackward pass on the hall build

Using the project duration of 19 days, carry out the backward pass for the hall network to find the LST and LFT of activities D, C and A.

Step 1 — Set the final activity

G is last, so its latest finish equals the project duration:

$\text{LFT}_G = 19, \qquad \text{LST}_G = 19 - 4 = 15$

Step 2 — Step back to E and F

Both feed G, so each takes $\text{LFT} = \text{LST}_G = 15$ :

$\text{E: LST}=15-3=12 \qquad \text{F: LST}=15-2=13$

Step 3 — Split at D (take the smallest)

D feeds E and F, so its latest finish is the smaller successor latest start:

$\text{LFT}_D = \min(12, 13) = 12, \qquad \text{LST}_D = 12 - 5 = 7$

Step 4 — Step back to C and A

C feeds D, so $\text{LFT}_C = \text{LST}_D = 7$ and $\text{LST}_C = 7 - 2 = 5$ .

B feeds D, so $\text{LFT}_B = 7$ and $\text{LST}_B = 3$ . A feeds B, so $\text{LFT}_A = \text{LST}_B = 3$ and $\text{LST}_A = 3 - 3 = 0$ .

$\boxed{\text{D: } 7/12, \quad \text{C: } 5/7, \quad \text{A: } 0/3 \ \ (\text{LST}/\text{LFT})}$

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

An activity has latest finish time 15 and duration 3. Find its latest start time and state the formula.

Merit

Activity V feeds two successors whose latest start times are 12 and 13. V has duration 5. Find V's latest finish and latest start, and explain the choice at the split.

Excellence