IB Business Management HL Topic 6 — The Business Management Toolkit Papers 1, 2 & 3 HL only ~12 min read

Using Critical Path Analysis

Critical path analysis finds the shortest time a project can possibly take, and then tells you exactly which tasks would wreck that time if they slipped. Everything else has some slack. Knowing the difference is what lets a manager move staff around without missing the deadline.

📘 What you need to know

The network diagram

A network diagram with six nodes Red is the critical path. Durations are in weeks. 1 0 02 4 43 3 44 9 95 11 116 14 14A 4 B 3 C 5 D 5 E 2 F 3Critical path A, C, E, F. The project takes 14 weeks. Green is the earliest start. Red is the latest finish.
Node 3 is the only one where the two numbers differ (3 and 4). That one week of difference is the float on activity B.

The forward pass: earliest start times

Start at node 1 with zero, then add each duration as you move right. When two activities feed into the same node, take the larger number, because the node cannot be reached until the slower one has finished.

WORKED EXAMPLE

Calculate the earliest start time at every node in the diagram above. [3 marks]

Nodes 1 to 3 Node 1 = 0 Node 2 = 0 + 4 (A) = 4 Node 3 = 0 + 3 (B) = 3 Node 4: two routes arrive, take the larger via C: 4 + 5 = 9   via D: 3 + 5 = 8   so 9 Nodes 5 and 6 Node 5 = 9 + 2 (E) = 11 Node 6 = 11 + 3 (F) = 14 The project takes 14 weeks forward pass = add, and take the bigger number where paths merge

The backward pass: latest finish times

Start at the last node, copying its EST into the LFT box, then subtract each duration as you move left. When two activities lead back into the same node, take the smaller number, because that is the tighter deadline.

WORKED EXAMPLE

Calculate the latest finish time at every node, and identify the critical path. [4 marks]

Work backwards from node 6 Node 6 = 14 Node 5 = 14 − 3 (F) = 11 Node 4 = 11 − 2 (E) = 9 Node 2 = 9 − 5 (C) = 4 Node 3 = 9 − 5 (D) = 4 Node 1: two routes back, take the smaller via A: 4 − 4 = 0   via B: 4 − 3 = 1   so 0 Find where EST equals LFT Nodes 1, 2, 4, 5 and 6 all match. Node 3 does not (3 against 4). Critical path = A, C, E, F backward pass = subtract, and take the smaller number where paths merge

Float: the spare time

Total float for an activity LFT at the end node − duration − EST at the start node
WORKED EXAMPLE

Calculate the total float for activity B. [3 marks]

Step 1: gather the three numbers B ends at node 3, so LFT = 4. Duration = 3. B starts at node 1, so EST = 0. Step 2: substitute 4 − 3 − 0 = 1 Step 3: say what it means B can start or overrun by one week without delaying the project at all. Total float on B = 1 week every activity on the critical path has a float of exactly zero — a useful check
Float is not wasted time. It is where a manager finds spare staff. If activity B has a week of slack, those workers can be moved onto C for a week, and C is the task that actually controls the deadline.
Quick self-check. The LFT at node 1 must always be 0. If it is not, you have made an arithmetic error somewhere in the backward pass. Check that before you write anything else down.

Why businesses use it

BenefitLimitation
Gives the shortest possible completion timeDurations are estimates, and estimates slip
Identifies exactly which activities cannot be delayedLarge projects become huge diagrams needing software
Allows just-in-time ordering of materials, freeing up cashDoes not guarantee success; it only schedules the plan
Shows where spare resources sit, so staff can be movedStaff may need training before they can be moved onto another task
Lets managers see the knock-on effect of a delay immediatelySays nothing about cost or quality, only time

💡 Exam tips

⚠ Common mix-ups

Up next: Using Simple Linear Regression — the last tool in the box, and the one that turns two columns of data into a forecast.

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