What is Critical Path Analysis in the context of Edexcel International A Levels Decision Mathematics 1?

Critical Path Analysis (CPA) is a project management technique used to determine the longest sequence of dependent tasks and the minimum time needed to complete a project. In the Edexcel International A Levels Decision Mathematics 1 curriculum, CPA helps students understand how to efficiently plan and manage complex projects by identifying critical tasks that directly impact the project's completion time.

How do you identify the critical path in a project network diagram?

To identify the critical path in a project network diagram, you must first list all tasks, determine their durations, and identify dependencies. Then, create a network diagram and calculate the earliest start and finish times for each task. The critical path is the longest path through the network, determined by the tasks with zero slack time, meaning any delay in these tasks will delay the project's completion.

Why is Critical Path Analysis important in project management?

Critical Path Analysis is crucial in project management because it helps identify the most important tasks that directly affect the project's timeline. By focusing on these critical tasks, project managers can allocate resources efficiently, anticipate potential delays, and implement strategies to mitigate risks, ensuring the project is completed on time and within budget.

What are the common terms used in Critical Path Analysis?

In Critical Path Analysis, some common terms include 'activity', which refers to a task or set of tasks; 'duration', the time required to complete an activity; 'earliest start time' (EST) and 'earliest finish time' (EFT), which indicate the earliest times an activity can start and finish; 'latest start time' (LST) and 'latest finish time' (LFT), which are the latest times an activity can start and finish without delaying the project; and 'slack', the amount of time an activity can be delayed without affecting the project's completion.

How does Critical Path Analysis relate to other topics in Decision Mathematics 1?

Critical Path Analysis is closely related to other topics in Decision Mathematics 1, such as network diagrams and scheduling. Understanding CPA requires knowledge of constructing and analyzing network diagrams, which is a fundamental skill in decision mathematics. Additionally, CPA is used in conjunction with scheduling techniques to optimize project timelines and resource allocation, making it an integral part of the broader study of decision-making processes in mathematics.

Why is "Omitting a dummy activity when activities have partially overlapping predecessors" a common mistake in Critical path analysis ?

Why it happens: Trying to keep the network small. How to avoid it: Whenever two activities share SOME but not all predecessors (e.g. 'D needs A and B; C needs only A'), a dummy is required. Always check: 'does this activity depend on everything coming into this node?' If not, dummies are needed. Source: WDM01/01 Examiner Reports — flagged every session

Why is "Using MIN in the forward pass (or MAX in the backward pass)" a common mistake in Critical path analysis ?

Why it happens: Confusing the two directions. How to avoid it: Forward = MAX (the event happens once ALL incoming activities finish — the slowest one determines this). Backward = MIN (latest = earliest of (latest of successor − duration)).

Why is "Declaring an activity critical because EET = LET at both endpoints, without checking the duration condition" a common mistake in Critical path analysis ?

Why it happens: Skipping the third critical-activity check. How to avoid it: All THREE conditions are needed. The third (duration = EET-head - EET-tail) catches activities between critical NODES that are still slack themselves — e.g. activity B in the example.

Why is "Computing total float as LET(head) $-$ EET(head)" a common mistake in Critical path analysis ?

Why it happens: Confusing event float with activity float. How to avoid it: Total float of an ACTIVITY is LET(head) - EET(tail) - d. Subtract the duration. Tabulate ES, EF, LF for each activity and compute TF = LF - ES - duration.

Why is "Claiming minimum workers equals the number of activities active at the busiest instant, without giving a valid schedule" a common mistake in Critical path analysis ?

Why it happens: Skipping the schedule construction. How to avoid it: The MAXIMUM SIMULTANEITY is a lower bound. To prove it suffices, construct an actual valid schedule — Edexcel awards M1 for the schedule, not just the count.

Why is "Setting LET(sink) to something other than EET(sink)" a common mistake in Critical path analysis ?

Why it happens: Trying to allow extra time at the end. How to avoid it: The project completes at EET(sink). LET(sink) = EET(sink) BY DEFINITION. Set them equal and run the backward pass.

Decision Mathematics 1Edexcel International A Levels Mathematics (XMA01-YMA01)

Critical path analysis

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Critical Path Analysis Study Notes — Edexcel IAL Mathematics D1 (WDM01/01, 2018 spec — 2026 onwards)

Activity-on-arc networks model projects. From a precedence table, draw the network (use DUMMIES when activities partially share predecessors). FORWARD PASS (MAX over incoming arcs) gives early event times; project duration = EET at sink. BACKWARD PASS (MIN over outgoing arcs) gives late event times; LET(sink) = EET(sink). Total float of an activity = LET(head) $-$ EET(tail) $-$ duration. CRITICAL activities have float $0$ . GANTT charts (cascade) and SCHEDULING DIAGRAMS communicate the schedule and resource use.

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XMA01-YMA01 syllabus (2026 onwards).

D1 4.1 — Convert a precedence table into an activity-on-arc network using dummies where required.
D1 4.2 — Apply the forward and backward passes to find early and late event times.
D1 4.3 — Identify the critical path and compute total float, free float, and independent float.
D1 4.4 — Construct a Gantt (cascade) chart and schedule activities to a minimum number of workers.

Building the activity-on-arc network

Activities are arcs; events are nodes. Use dummies when precedences partially overlap.

Setup. A PROJECT is a set of activities; each has a duration, and some must finish before others can start (precedences).

Activity-on-arc (AOA) convention.

Each ACTIVITY is a directed arc, labelled with duration.
Each NODE is an EVENT — a moment when one or more activities complete.
TIME flows left to right (arrows point right).
A single SOURCE node (no incoming arcs) and a single SINK node (no outgoing arcs).

Building the network from a precedence table.

List all activities with no predecessors — they start at the source.
For each remaining activity, draw it after its predecessors finish.
Where multiple activities meet at a node, all of them must be predecessors of EVERY activity leaving that node.
Add DUMMY arcs (dashed, duration $0$ ) when the precedence structure doesn't naturally satisfy step 3.

When is a dummy needed?

Type 1: Partial sharing. Suppose $C$ depends on $A$ only, and $D$ depends on $A$ and $B$ .

If we draw $A$ and $B$ both ending at the same node and have $C, D$ both leave that node, $C$ would falsely depend on $B$ too.
FIX: send $A$ to its own node $2$ ; send $B$ to node $3$ ; add a dummy $2 \to 3$ to capture 'D needs A AND B'; have $C$ leave node $2$ ; have $D$ leave node $3$ .

Type 2: Uniqueness. Edexcel convention requires that no two arcs share both endpoints. If two activities have the same start and end events, split one with a dummy.

Worked example. Project:

Activity	Duration	Predecessors
A	5	—
B	3	—
C	4	A
D	6	A, B
E	2	C
F	7	D
G	3	E, F

Network: nodes $1, 2, 3, 4, 5, 6, 7$ . Arcs: $A: 1 \to 2$ ; $B: 1 \to 3$ ; dummy: $2 \to 3$ (so $D$ needs both); $C: 2 \to 4$ ; $D: 3 \to 5$ ; $E: 4 \to 6$ ; $F: 5 \to 6$ ; $G: 6 \to 7$ .

Activity-on-arc network: arcs are activities, nodes are events; the critical path (red) is the longest route from source to sink.

Checks before moving on.

Source has only outgoing arcs.
Sink has only incoming arcs.
Each activity's predecessors are exactly the activities INTO its starting node (modulo dummies).
No two arcs share both endpoints.

Arcs = activities; nodes = events.
Dummies (dashed, 0d) for partial-sharing precedences.
Dummies for uniqueness (no two arcs share both endpoints).
Single source, single sink.

Forward pass: early event times

Earliest time each event can occur. MAX over incoming.

Forward pass procedure.

EET(source) $= 0$ .
Process nodes in TOPOLOGICAL ORDER (each node only AFTER all its predecessors).
For each node $j$ : $\text{EET}(j) = \max_{i \to j} \bigl(\text{EET}(i) + d(i, j)\bigr).$
EET(sink) is the PROJECT DURATION.

Why MAX? An event 'happens' only after EVERY incoming activity has finished. The slowest incoming chain determines the event time.

Worked example.

Same network as above, with durations: $A=5, B=3,$ dummy $=0, C=4, D=6, E=2, F=7, G=3$ .

Node	Incoming arcs	Compute	EET
1	—	start	$0$
2	$A$ from 1 (5)	$0 + 5$	$5$
3	$B$ from 1 (3); dummy from 2 (0)	$\max(0+3, 5+0)$	$5$
4	$C$ from 2 (4)	$5 + 4$	$9$
5	$D$ from 3 (6)	$5 + 6$	$11$
6	$E$ from 4 (2); $F$ from 5 (7)	$\max(9+2, 11+7)$	$18$
7	$G$ from 6 (3)	$18 + 3$	$21$

Project duration $= 21$ days.

Common slip. Forgetting that the dummy contributes its $0$ duration through the MAX — students sometimes write EET(3) = $0 + 3 = 3$ , missing the $5 + 0 = 5$ alternative through the dummy.

EET(source) = 0.
Process in topological order.
EET(node) = MAX over incoming arcs.
Project duration = EET(sink).

Backward pass: late event times

Latest time each event can occur without delaying the project. MIN over outgoing.

Backward pass procedure.

LET(sink) $=$ EET(sink) $=$ project duration.
Process nodes in REVERSE topological order.
For each node $i$ : $\text{LET}(i) = \min_{i \to j} \bigl(\text{LET}(j) - d(i, j)\bigr).$
LET(source) should be $0$ — this is a sanity check.

Why MIN? An event must occur EARLY ENOUGH that ALL its outgoing activities can still finish by their late deadlines. The tightest (smallest) successor deadline determines the late time.

Worked example continued.

Node	Outgoing arcs	Compute	LET
7	—	sink	$21$
6	$G$ to 7 (3)	$21 - 3$	$18$
5	$F$ to 6 (7)	$18 - 7$	$11$
4	$E$ to 6 (2)	$18 - 2$	$16$
3	$D$ to 5 (6)	$11 - 6$	$5$
2	$C$ to 4 (4); dummy to 3 (0)	$\min(16-4, 5-0)$	$5$
1	$A$ to 2 (5); $B$ to 3 (3)	$\min(5-5, 5-3)$	$0$

Sanity check. LET(1) $= 0$ ✓ (matches EET(1)). If it didn't, you'd have an error.

Float meaning.

EET( $i$ ) $=$ LET( $i$ ) means node $i$ is on the critical path (no slack).
EET( $i$ ) $<$ LET( $i$ ) means the event has slack of LET( $i$ ) $-$ EET( $i$ ) days.

LET(sink) = EET(sink).
Process in REVERSE topological order.
LET(node) = MIN over outgoing arcs.
LET(source) should equal 0 — sanity check.

Critical activities, critical path, and floats

Activities with zero total float. The critical path is the longest path.

Total float of an activity. $\text{TF}(X) = \text{LET(head)} - \text{EET(tail)} - d(X).$

This is the maximum delay $X$ can absorb without delaying the project.

Critical activity. $X$ is CRITICAL iff TF( $X$ ) $= 0$ . Equivalently, all three:

EET(tail) $=$ LET(tail),
EET(head) $=$ LET(head),
$d(X) =$ EET(head) $-$ EET(tail).

The third condition matters: an activity can have critical NODES but slack ITSELF — e.g. activity $B$ in our example has EET(1) $=$ LET(1) and EET(3) $=$ LET(3), but $d(B) = 3 \neq 5 -0 = 5$ , so $B$ has float $2$ .

Critical path. A path from source to sink along which every arc is a critical activity (or a critical dummy). Project duration = length of any critical path.

Worked example. Activities in our project:

Activity	$d$	EET(tail)	LET(head)	TF	Critical?
$A$	5	0	5	$5 - 0 - 5 = 0$	YES
$B$	3	0	5	$5 - 0 - 3 = 2$	no
dummy	0	5	5	$5 - 5 - 0 = 0$	(yes — but not an activity)
$C$	4	5	16	$16 - 5 - 4 = 7$	no
$D$	6	5	11	$11 - 5 - 6 = 0$	YES
$E$	2	9	18	$18 - 9 - 2 = 7$	no
$F$	7	11	18	$18 - 11 - 7 = 0$	YES
$G$	3	18	21	$21 - 18 - 3 = 0$	YES

Critical activities: $A, D, F, G$ . Critical path: $A \to (\text{dummy}) \to D \to F \to G$ , length $5 + 0 + 6 + 7 + 3 = 21$ . ✓

Free float. $\text{FF}(X) = \text{EET(head)} - \text{EET(tail)} - d(X)$ . The delay possible without affecting any successor's earliest start. FF $\le$ TF.

Independent float. $\text{IF}(X) = \max(0, \;\text{EET(head)} - \text{LET(tail)} - d(X))$ . The delay possible WITHOUT depending on a predecessor finishing early AND WITHOUT delaying any successor. IF $\le$ FF $\le$ TF.

Why three kinds of float? TF is the most generous; FF accounts for keeping successors safe; IF accounts for both ends.

TF = LET(head) − EET(tail) − duration.
Critical activity ⇔ TF = 0 (all three conditions).
Critical path = a longest source-to-sink path.
Free float ≤ total float; independent float ≤ free float.

Gantt charts and scheduling

Visualise schedule; allocate workers to the minimum count.

Gantt (cascade) chart.

A bar chart with TIME on the horizontal axis and ACTIVITIES stacked vertically. Each activity is shown as a solid bar from its earliest start (EET of tail) to its earliest finish (EET of tail + duration). For non-critical activities, a DASHED extension shows the float — i.e. the activity could be slid right by up to TF days.

Worked example (text form).

Time:    0    5    9   11      18  21
  A:  ████████              (critical, no float)
  B:  ██████░░               (TF = 2)
  C:        █████████░░░░░░░ (TF = 7)
  D:        ████████████     (critical)
  E:                █████░░░░░░░░ (TF = 7)
  F:                ██████████████ (critical)
  G:                              ██████ (critical)

Critical bars run unbroken from time $0$ to $21$ .

Reading the chart.

Bar start = earliest start time.
Bar end (solid) = earliest finish.
Dashed extension to LET(head) = total float.

Scheduling diagram (worker allocation).

Given the Gantt chart at earliest starts, identify the MAXIMUM number of activities simultaneously in progress — this is a LOWER BOUND on the number of workers required.

To prove the bound suffices, construct a worker-by-worker schedule. Each worker handles one activity at a time; precedences must be respected.

Worked example. From the Gantt chart above, look at concurrent activities at each interval:

Time	Active activities	Count
0-3	$A, B$	2
3-5	$A$	1
5-9	$D, C$	2
9-11	$D, E$	2
11-18	$F$	1
18-21	$G$	1

Maximum concurrent = $2$ . So $2$ workers SUFFICE if we schedule sensibly.

Valid $2$ -worker schedule.

Worker 1: $A (0-5) \to D (5-11) \to F (11-18) \to G (18-21)$ .
Worker 2: $B (0-3) \to C (5-9) \to E (9-11)$ , then idle.

Total time = $21$ days. ✓

Minimum-worker question pattern. Exam questions often ask:

'How many workers are needed to complete the project in the minimum time?' — Identify maximum simultaneity; construct a valid schedule.
'With only $N$ workers, what is the minimum completion time?' — May need to shift non-critical activities to later starts.

Gantt: solid bars at earliest start; dashed = float.
Critical bars span the entire project duration.
Maximum simultaneity = lower bound on workers.
Validate via an explicit worker-by-worker schedule.

How it’s examined

Critical Path Analysis is a $14$ - $18$ -mark question on every WDM01/01 paper. Standard format: (a) construct the AOA network from a precedence table with dummies ( $4$ - $5$ marks); (b) forward and backward pass to give EET/LET at every node ( $5$ - $6$ marks); (c) state critical activities and critical path with project duration ( $2$ - $3$ marks); (d) compute total/free float for specified activities ( $2$ marks); (e) construct a Gantt chart OR find minimum workers for the minimum time ( $3$ - $5$ marks). Marks lost: missing dummy (B2), MIN/MAX swap (multiple A marks), failing the LET(1) = 0 sanity check, listing critical nodes instead of activities. Edexcel rewards careful tabulation — present all working in tables.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WDM01/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Step-by-step worked examples — Critical path analysis

Step-by-step solutions to past-paper-style questions on critical path analysis , written exactly the way a tutor would explain them at the board.

1Construct an activity-on-arc network from a precedence table (6 marks, D1)
Extended• Adapted from WDM01/01 January 2024 Q7• AOA network, precedence, dummy
Question
Construct an activity-on-arc network for the following project. Use dummy activities where necessary.

Activity Duration (days) Immediate predecessors
A 5 —
B 3 —
C 4 A
D 6 A, B
E 2 C
F 7 D
G 3 E, F
Step-by-step solution
1. Step 1
  Identify start and finish. $A, B$ have no predecessors → they start from node $1$ . $G$ has no successors → it ends at the final node.
2. Step 2
  Map predecessors.
  
  $A$ precedes $C$ AND $D$ .
  
  $B$ precedes $D$ ONLY (not $C$ ).
  
  If we draw $A$ and $B$ both ending at the SAME node before $C$ and $D$ , that would falsely make $B$ a predecessor of $C$ . To fix this we use a DUMMY activity.
3. Step 3
  Network structure.
  
  Node $1$ (start).
  
  Arc $A: 1 \to 2$ .
  
  Arc $B: 1 \to 3$ .
  
  Dummy arc: $2 \to 3$ (duration $0$ , dashed). This forces $D$ to wait for both $A$ AND $B$ , but $C$ depends only on $A$ .
  
  Arc $C: 2 \to 4$ (after $A$ ).
  
  Arc $D: 3 \to 5$ (after $A$ AND $B$ ).
  
  Arc $E: 4 \to 6$ (after $C$ ).
  
  Arc $F: 5 \to 6$ (after $D$ ).
  
  Arc $G: 6 \to 7$ (after $E$ AND $F$ ).
4. Step 4
  Why the dummy is needed. Without it, drawing $A, B$ both into the same node would force any successor of that node to require BOTH $A$ and $B$ . But $C$ only needs $A$ . The dummy $2 \to 3$ lets the constraint 'D needs A AND B' flow without polluting $C$ 's dependencies.
5. Step 5
  Final network nodes: $1, 2, 3, 4, 5, 6, 7$ . Arcs: $A(1{-}2), B(1{-}3), \text{dummy}(2{-}3), C(2{-}4), D(3{-}5), E(4{-}6), F(5{-}6), G(6{-}7)$ .
Answer
Network: start node $1$ ; $A: 1 \to 2$ (5d); $B: 1 \to 3$ (3d); dummy $2 \to 3$ (0d); $C: 2 \to 4$ (4d); $D: 3 \to 5$ (6d); $E: 4 \to 6$ (2d); $F: 5 \to 6$ (7d); $G: 6 \to 7$ (3d).
Examiner tip
B1 all activities included. B1 dummy correctly placed. B1 dummy direction correct. M1 valid AoA structure. A1 each correct precedence is preserved (D depends on A AND B; C depends on A only). Edexcel mark schemes ALWAYS reward correct use of dummies — drawing a network without a dummy when one is needed is a $2$ -mark loss.
2Forward and backward pass — event times (8 marks, D1)
Extended• Adapted from WDM01/01 June 2024 Q8• forward pass, backward pass, early/late times
Question
For the network from the previous example, compute the early event time (EET) and late event time (LET) at every node, and identify the project duration.
Step-by-step solution
1. Step 1
  Forward pass (EET). Start at node $1$ with EET $= 0$ . For each later node, EET = MAX over incoming arcs of (EET of tail $+$ duration).
2. Step 2
  EET computations.
  
  Node 1: $0$ .
  
  Node 2 (from $A$ ): $0 + 5 = 5$ .
  
  Node 3 (from $B$ or dummy): $\max(0 + 3, \;5 + 0) = 5$ .
  
  Node 4 (from $C$ ): $5 + 4 = 9$ .
  
  Node 5 (from $D$ ): $5 + 6 = 11$ .
  
  Node 6 (from $E$ or $F$ ): $\max(9 + 2, \;11 + 7) = \max(11, 18) = 18$ .
  
  Node 7 (from $G$ ): $18 + 3 = 21$ .
  
  Project duration = EET at sink = $21$ days.
3. Step 3
  Backward pass (LET). Start at the sink with LET $=$ EET $= 21$ . For each earlier node, LET = MIN over outgoing arcs of (LET of head $-$ duration).
4. Step 4
  LET computations.
  
  Node 7: $21$ .
  
  Node 6 (from $G$ ): $21 - 3 = 18$ .
  
  Node 5 (from $F$ ): $18 - 7 = 11$ .
  
  Node 4 (from $E$ ): $18 - 2 = 16$ .
  
  Node 3 (from $D$ ): $11 - 6 = 5$ .
  
  Node 2 (from $C$ or dummy): $\min(16 - 4, \;5 - 0) = \min(12, 5) = 5$ .
  
  Node 1 (from $A$ or $B$ ): $\min(5 - 5, \;5 - 3) = \min(0, 2) = 0$ . ✓
5. Step 5
  Summary table.
  
  Node EET LET
  1 0 0
  2 5 5
  3 5 5
  4 9 16
  5 11 11
  6 18 18
  7 21 21
  
  Project duration $= 21$ days.
Answer
EET / LET: 1: 0/0, 2: 5/5, 3: 5/5, 4: 9/16, 5: 11/11, 6: 18/18, 7: 21/21. Project duration = $21$ days.
Examiner tip
M1 forward pass attempt. A1 each EET ( $\times 6$ ). M1 backward pass attempt. A1 each LET ( $\times 6$ ). B1 project duration. Edexcel deducts for taking MIN instead of MAX in the forward pass (and vice versa). Tabulate cleanly. Always do node $1$ check (final LET should equal $0$ ); if not, error.
3Identify the critical path and total floats (5 marks, D1)
Extended• Adapted from WDM01/01 June 2024 Q8(c)• critical path, total float
Question
Using the EET/LET values above, identify all critical activities, state the critical path, and compute the total float for activities $C$ and $E$ .
Step-by-step solution
1. Step 1
  Critical activity condition. Activity $X$ from node $i$ to node $j$ is CRITICAL iff:
  
  EET $(i)$ = LET $(i)$ ,
  
  EET $(j)$ = LET $(j)$ , AND
  
  EET $(j)$ $-$ EET $(i)$ = duration( $X$ ).
  
  All three conditions must hold.
2. Step 2
  Check each activity.
  
  $A$ ( $1 \to 2$ , $5$ d): $0=0$ , $5=5$ , $5-0=5$ ✓ CRITICAL.
  
  $B$ ( $1 \to 3$ , $3$ d): $0=0$ , $5=5$ , $5-0 = 5 \neq 3$ ✗ NOT critical.
  
  dummy ( $2 \to 3$ , $0$ d): $5=5$ , $5=5$ , $5-5=0$ ✓ CRITICAL.
  
  $C$ ( $2 \to 4$ , $4$ d): $5=5$ , $9 \neq 16$ ✗ NOT critical.
  
  $D$ ( $3 \to 5$ , $6$ d): $5=5$ , $11=11$ , $11-5=6$ ✓ CRITICAL.
  
  $E$ ( $4 \to 6$ , $2$ d): $9 \neq 16$ ✗ NOT critical.
  
  $F$ ( $5 \to 6$ , $7$ d): $11=11$ , $18=18$ , $18-11=7$ ✓ CRITICAL.
  
  $G$ ( $6 \to 7$ , $3$ d): $18=18$ , $21=21$ , $21-18=3$ ✓ CRITICAL.
3. Step 3
  Critical path: $A \to \text{dummy} \to D \to F \to G$ , total duration $5 + 0 + 6 + 7 + 3 = 21$ ✓.
  
  In written form: $A, D, F, G$ (the dummy is not an actual activity).
4. Step 4
  Total float. $\text{TF}(X) = \text{LET(head)} - \text{EET(tail)} - \text{duration}(X)$ .
  
  $C$ ( $2 \to 4$ , $4$ d): $\text{TF} = 16 - 5 - 4 = 7$ days.
  
  $E$ ( $4 \to 6$ , $2$ d): $\text{TF} = 18 - 9 - 2 = 7$ days.
  
  Both have $7$ days of float — $C$ and $E$ lie on the same NON-CRITICAL chain.
Answer
Critical activities: $A, D, F, G$ . Critical path: $A \to D \to F \to G$ (project duration $21$ days). Total floats: $\text{TF}(C) = 7$ , $\text{TF}(E) = 7$ .
Examiner tip
B1 critical-activity criterion stated. A1 critical activities correctly identified. A1 critical path stated. M1 total-float formula. A1 each correct float. Critical activities have TOTAL FLOAT zero — verify this on any critical activity. The dummy is critical here but is not listed as a 'critical activity' in the final answer (it has no real duration).
4Construct a Gantt (cascade) chart (5 marks, D1)
Extended• Adapted from WDM01/01 January 2024 Q7(d)• Gantt chart, cascade chart
Question
Using the network from the previous worked examples, draw a Gantt chart showing each activity at its earliest possible start. Indicate the total float for non-critical activities.
Step-by-step solution
1. Step 1
  Tabulate. For each activity: earliest start $=$ EET(tail); earliest finish $=$ EET(tail) $+$ duration; latest finish $=$ LET(head); total float $=$ latest finish $-$ earliest start $-$ duration.
  
  Activity Tail Head Duration ES EF LF Float
  $A$ 1 2 5 0 5 5 0
  $B$ 1 3 3 0 3 5 2
  $C$ 2 4 4 5 9 16 7
  $D$ 3 5 6 5 11 11 0
  $E$ 4 6 2 9 11 18 7
  $F$ 5 6 7 11 18 18 0
  $G$ 6 7 3 18 21 21 0
2. Step 2
  Cascade chart. Time axis from $0$ to $21$ days. For each activity, draw a solid bar from ES to EF; for non-critical activities, draw a dashed extension from EF to LF showing the available float.
  
  Visually (text representation):
  
  Time: 0 5 9 11 18 21 A: ████████ B: ██████░░ C: █████████░░░░░░░ D: ████████████ E: █████░░░░░░░░ F: ██████████████ G: ██████
  
  Critical: $A, D, F, G$ (no dashed float).
3. Step 3
  Reading off the chart. Critical activities form an unbroken backbone from time $0$ to time $21$ . Non-critical activities ( $B, C, E$ ) have visible float shown by the dashed extensions.
Answer
See cascade chart above. Critical activities $A, D, F, G$ shown as solid bars; non-critical $B$ (float $2$ ), $C$ (float $7$ ), $E$ (float $7$ ) shown with dashed extensions.
Examiner tip
B1 correct time axis. M1 attempt to plot activities at earliest start. A1 each activity correctly positioned ( $\times 3$ — critical bars). A1 float shown on non-critical activities. Edexcel mark schemes accept either solid+dashed OR labelled bars showing the float amount. The cascade chart is essentially a visual restatement of the EET/LET table.
5Scheduling diagram: minimum number of workers (6 marks, D1)
Challenge• Adapted from WDM01/01 June 2023 Q8• scheduling, workers, resource
Question
Using the activity data from the previous examples, determine the minimum number of workers needed to complete the project in the minimum time of $21$ days, assuming each worker performs one activity at a time and any worker can perform any activity.
Step-by-step solution
1. Step 1
  Critical path forces sequencing. $A \to D \to F \to G$ are sequential and span all $21$ days. AT LEAST one worker must continuously process them.
2. Step 2
  Identify overlaps. Time intervals:
  
  Day $0$ - $3$ : $A$ (worker 1), $B$ (worker 2) — $2$ workers.
  
  Day $3$ - $5$ : $A$ (worker 1) — $B$ done by day $3$ ; $1$ worker active.
  
  Day $5$ - $9$ : $D$ (worker 1), $C$ (worker 2) — $2$ workers.
  
  Day $9$ - $11$ : $D$ (worker 1), $E$ (worker 2 — starts at day $9$ , finishes day $11$ ) — $2$ workers.
  
  Day $11$ - $18$ : $F$ (worker 1) only ( $E$ done by day $11$ ) — $1$ worker.
  
  Day $18$ - $21$ : $G$ (worker 1).
3. Step 3
  Maximum simultaneity. Two activities are simultaneously active in intervals (0,3), (5,9), (9,11). MAXIMUM concurrent activities $= 2$ .
  
  So $2$ workers SUFFICE if we schedule each non-critical activity at its earliest start.
4. Step 4
  Construct a $2$ -worker schedule.
  
  Worker 1: $A$ (0-5), $D$ (5-11), $F$ (11-18), $G$ (18-21).
  
  Worker 2: $B$ (0-3), idle (3-5), $C$ (5-9), $E$ (9-11), idle (11-21).
  
  Total duration = $21$ days. ✓
5. Step 5
  Could we do it with $1$ worker? Total work $= 5+3+4+6+2+7+3 = 30$ activity-days; a single worker would take $30$ days. So $1$ worker is INSUFFICIENT. Minimum $= 2$ workers.
Answer
Minimum $= 2$ workers. Worker 1: $A (0-5) \to D (5-11) \to F (11-18) \to G (18-21)$ . Worker 2: $B (0-3) \to C (5-9) \to E (9-11)$ .
Examiner tip
M1 attempt schedule at earliest starts. A1 identify maximum simultaneity ( $2$ ). M1 valid worker assignment. A1 each worker schedule correct. B1 confirm $1$ worker insufficient via total-work argument. Edexcel scheduling questions reward clarity — present worker schedules as time-labelled lines.

Node	EET	LET
1	0	0
2	5	5
3	5	5
4	9	16
5	11	11
6	18	18
7	21	21

Activity	Tail	Head	Duration	ES	EF	LF	Float
$A$	1	2	5	0	5	5	0
$B$	1	3	3	0	3	5	2
$C$	2	4	4	5	9	16	7
$D$	3	5	6	5	11	11	0
$E$	4	6	2	9	11	18	7
$F$	5	6	7	11	18	18	0
$G$	6	7	3	18	21	21	0

Key Formulae — Critical path analysis

The formulae you need to memorise for critical path analysis on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Forward pass (early event time)
$\text{EET}(j) \;=\; \max_{i \to j} \bigl(\text{EET}(i) + d(i, j)\bigr)$
$i \to j$
arcs into node $j$
$d(i, j)$
duration of activity on arc $i \to j$
When to use
Compute the earliest event time at each node. Start at source with $0$ ; proceed in topological order. NOT in the IAL formula booklet.
Example
Node $6$ has incoming arcs from $4$ (EET 9, duration 2) and $5$ (EET 11, duration 7). EET $(6) = \max(11, 18) = 18$ .
Backward pass (late event time)
$\text{LET}(i) \;=\; \min_{i \to j} \bigl(\text{LET}(j) - d(i, j)\bigr)$
$i \to j$
arcs out of node $i$
$d(i, j)$
duration of activity on arc $i \to j$
When to use
Compute the latest event time. Start at sink with LET $=$ EET; proceed in reverse topological order. NOT in the IAL formula booklet.
Example
Node $2$ has outgoing arcs to $3$ (LET 5, duration 0) and $4$ (LET 16, duration 4). LET $(2) = \min(5, 12) = 5$ .
Total float
$\text{TF}(X) \;=\; \text{LET(head)} - \text{EET(tail)} - \text{duration}(X)$
$\text{LET}, \text{EET}$
late and early event times
When to use
Slack in an activity's schedule. Critical activities have TF $= 0$ . NOT in the IAL formula booklet.
Example
Activity $C$ from $2$ to $4$ , duration $4$ , EET $(2) = 5$ , LET $(4) = 16$ : $\text{TF} = 16 - 5 - 4 = 7$ .
Critical activity test
$X \text{ critical} \;\iff\; \text{EET(tail)} = \text{LET(tail)},\; \text{EET(head)} = \text{LET(head)},\; \text{EET(head)} - \text{EET(tail)} = d(X)$
$d(X)$
duration of activity $X$
When to use
Determine whether an activity is critical. All THREE conditions must hold. Equivalent to total float $= 0$ .
Example
$D$ ( $3 \to 5$ , $6$ d): EET $(3) = 5 =$ LET $(3)$ ; EET $(5) = 11 =$ LET $(5)$ ; $11 - 5 = 6 = d(D)$ . Critical. ✓

Key Definitions and Keywords — Critical path analysis

Definitions to memorise and the exact keywords mark schemes credit for critical path analysis answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Activity-on-arc (AOA) network
Examiner keyword
A directed graph in which each ARC represents an activity (labelled with duration) and each NODE represents an event (a moment when one or more activities complete). Used to model project precedence.
Precedence table
Examiner keyword
A table listing each activity with its duration and its IMMEDIATE PREDECESSORS (the activities that must finish before it can start). Equivalent input to a network.
Dummy activity
Examiner keyword
An arc of duration $0$ (drawn dashed) used to (a) preserve correct precedence when two activities partially share predecessors, or (b) ensure two arcs do not share both endpoints (uniqueness convention).
Early event time (EET)
Examiner keyword
The EARLIEST time at which an event (node) can occur. EET(source) $= 0$ . Found by FORWARD PASS using MAX.
Late event time (LET)
Examiner keyword
The LATEST time an event can occur without delaying the project. LET(sink) $=$ EET(sink) $=$ project duration. Found by BACKWARD PASS using MIN.
Critical activity
Examiner keyword
An activity with zero total float; any delay extends the project. Identified by the three-condition test (EET = LET at both endpoints, and duration equals EET-difference).
Critical path
Examiner keyword
A longest path from source to sink. All critical activities lie on at least one critical path. Project duration = length of any critical path.
Total float
Examiner keyword
Maximum amount by which an activity can be delayed (or extended) WITHOUT delaying the project. $\text{TF}(X) = \text{LET(head)} - \text{EET(tail)} - d(X)$ .
Free float
Maximum delay of $X$ without affecting the EARLIEST start of any SUCCESSOR. $\text{FF}(X) = \text{EET(head)} - \text{EET(tail)} - d(X)$ . Free $\le$ total float.
Independent float
Delay of $X$ available without delaying any successor and without depending on any predecessor finishing early. $\text{IF}(X) = \max(0, \text{EET(head)} - \text{LET(tail)} - d(X))$ .
Gantt (cascade) chart
Examiner keyword
A horizontal bar chart in which each activity is a bar from its earliest start to its earliest finish, with dashed extensions showing total float for non-critical activities.
Scheduling diagram
A worker-by-time diagram showing the assignment of activities to workers, used to determine the minimum number of workers required.

Common Mistakes and Misconceptions — Critical path analysis

The traps other students keep falling into on critical path analysis questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Omitting a dummy activity when activities have partially overlapping predecessors
WDM01/01 Examiner Reports — flagged every session
Why it happens
Trying to keep the network small.
How to avoid it
Whenever two activities share SOME but not all predecessors (e.g. 'D needs A and B; C needs only A'), a dummy is required. Always check: 'does this activity depend on everything coming into this node?' If not, dummies are needed.
✕Using MIN in the forward pass (or MAX in the backward pass)
Why it happens
Confusing the two directions.
How to avoid it
Forward = MAX (the event happens once ALL incoming activities finish — the slowest one determines this). Backward = MIN (latest = earliest of (latest of successor − duration)).
✕Declaring an activity critical because EET = LET at both endpoints, without checking the duration condition
Why it happens
Skipping the third critical-activity check.
How to avoid it
All THREE conditions are needed. The third (duration = EET-head $-$ EET-tail) catches activities between critical NODES that are still slack themselves — e.g. activity $B$ in the example.
✕Computing total float as LET(head) $-$ EET(head)
Why it happens
Confusing event float with activity float.
How to avoid it
Total float of an ACTIVITY is $\text{LET(head)} - \text{EET(tail)} - d$ . Subtract the duration. Tabulate ES, EF, LF for each activity and compute TF $=$ LF $-$ ES $-$ duration.
✕Claiming minimum workers equals the number of activities active at the busiest instant, without giving a valid schedule
Why it happens
Skipping the schedule construction.
How to avoid it
The MAXIMUM SIMULTANEITY is a lower bound. To prove it suffices, construct an actual valid schedule — Edexcel awards M1 for the schedule, not just the count.
✕Setting LET(sink) to something other than EET(sink)
Why it happens
Trying to allow extra time at the end.
How to avoid it
The project completes at EET(sink). LET(sink) = EET(sink) BY DEFINITION. Set them equal and run the backward pass.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Critical path analysis — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Activity	Tail	Head	Duration	ES	EF	LF	Float
$A$	1	2	5	0	5	5	0
$B$	1	3	3	0	3	5	2
$C$	2	4	4	5	9	16	7
$D$	3	5	6	5	11	11	0
$E$	4	6	2	9	11	18	7
$F$	5	6	7	11	18	18	0
$G$	6	7	3	18	21	21	0

Activity	Tail	Head	Duration	ES	EF	LF	Float
$A$	1	2	5	0	5	5	0
$B$	1	3	3	0	3	5	2
$C$	2	4	4	5	9	16	7
$D$	3	5	6	5	11	11	0
$E$	4	6	2	9	11	18	7
$F$	5	6	7	11	18	18	0
$G$	6	7	3	18	21	21	0

Critical path analysis

Detailed Notes

What you’ll learn

How it’s examined

1Construct an activity-on-arc network from a precedence table (6 marks, D1)

2Forward and backward pass — event times (8 marks, D1)

3Identify the critical path and total floats (5 marks, D1)

4Construct a Gantt (cascade) chart (5 marks, D1)

5Scheduling diagram: minimum number of workers (6 marks, D1)

Forward pass (early event time)

Backward pass (late event time)

Total float

Critical activity test

Activity-on-arc (AOA) network

Precedence table

Dummy activity

Early event time (EET)

Late event time (LET)

Critical activity

Critical path

Total float

Free float

Independent float

Gantt (cascade) chart

Scheduling diagram

✕Omitting a dummy activity when activities have partially overlapping predecessors

✕Using MIN in the forward pass (or MAX in the backward pass)

✕Declaring an activity critical because EET = LET at both endpoints, without checking the duration condition

✕Computing total float as LET(head) −-− EET(head)

✕Claiming minimum workers equals the number of activities active at the busiest instant, without giving a valid schedule

✕Setting LET(sink) to something other than EET(sink)

Past paper style quiz

Critical path analysis — frequently asked questions

✕Computing total float as LET(head) $-$ EET(head)

Activity	Tail	Head	Duration	ES	EF	LF	Float
$A$	1	2	5	0	5	5	0
$B$	1	3	3	0	3	5	2
$C$	2	4	4	5	9	16	7
$D$	3	5	6	5	11	11	0
$E$	4	6	2	9	11	18	7
$F$	5	6	7	11	18	18	0
$G$	6	7	3	18	21	21	0