Scheduling and the critical path

01The problem

Twenty-three tasks, each with an owner pushing to finish theirs. But they don't all matter equally for the opening date. Hiring staff can slip by weeks without delaying anything; one day lost on the permits delays the whole launch by a day. Knowing which is which tells the project manager where to spend attention.

02The concept

The critical path method (CPM)

1. Forward pass: each task's earliest start is the latest finish of everything it depends on; earliest finish = earliest start + duration. The project's length is the last earliest finish.
2. Backward pass: working back from the end, each task's latest finish is the earliest latest start of the tasks that depend on it.
3. Float (slack) = latest start − earliest start: how long a task can slip without delaying the end.
4. The critical path is the chain of tasks with zero float. Any delay on it delays the project.

Dates

Durations are working days. Convert with a working calendar (Monday to Friday here; real plans also remove public holidays).

03Example

A scheduling function you'll reuse in later lessons:

import numpy as np
import pandas as pd

base = "https://academy.cloudtechanalytics.com/datasets/project/"
tasks = pd.read_csv(base + "tasks.csv").fillna({"predecessors": ""})
START = np.datetime64("2026-06-01")      # a Monday

def schedule(tasks, durations):
    preds = {t: [p for p in ps.split(";") if p] for t, ps in zip(tasks["task_id"], tasks["predecessors"])}
    es, ef = {}, {}
    for t in tasks["task_id"]:                       # tasks are listed after their predecessors
        es[t] = max((ef[p] for p in preds[t]), default=0)
        ef[t] = es[t] + durations[t]
    end = max(ef.values())
    ls, lf = {}, {}
    for t in reversed(list(tasks["task_id"])):
        successors = [s for s in tasks["task_id"] if t in preds[s]]
        lf[t] = min((ls[s] for s in successors), default=end)
        ls[t] = lf[t] - durations[t]
    out = pd.DataFrame({"es": es, "ef": ef, "ls": ls, "lf": lf})
    out["float"] = out["ls"] - out["es"]
    return out, end

def finish_date(days):
    return np.busday_offset(START, int(np.ceil(days)) - 1, roll="forward")

likely = dict(zip(tasks["task_id"], tasks["likely_days"]))
plan, end = schedule(tasks, likely)
print("Project length:", end, "working days, opening on", finish_date(end))
print("Critical path:", " -> ".join(plan.index[plan["float"] == 0]))

Project length: 75 working days, opening on 2026-09-11
Critical path: A1 -> A2 -> A3 -> B2 -> C3 -> C5 -> D3 -> F2 -> F3

Nine of the 23 tasks are critical. The path runs from the charter through the lease, permits and fit-out, then the network, system testing, training, the trial run and the opening. Now the tasks with the most float:

plan.sort_values("float", ascending=False).head(5)

es  ef  ls  lf  float
C4  10  20  48  58     38
C1   3  10  41  48     38
E3  21  31  59  69     38
E1  13  21  51  59     38
C2  10  20  48  58     38

Choosing and configuring the warehouse system, buying equipment, agreeing supplier deliveries and setting up routes can each slip by 38 working days without moving the opening. Meanwhile the permits, the fit-out and the network and system testing have none. The plan says the depot opens on 11 September, comfortably before the 30 September deadline. But this plan uses most likely durations everywhere. Redo it with the PERT expected durations:

expected = dict(zip(tasks["task_id"], (tasks["optimistic_days"] + 4 * tasks["likely_days"] + tasks["pessimistic_days"]) / 6))
plan_expected, end_expected = schedule(tasks, expected)
print("With expected durations:", round(end_expected, 1), "working days, opening on", finish_date(end_expected))

With expected durations: 80.3 working days, opening on 2026-09-21

The opening moves later, and the deadline gets closer. Lesson 5 asks the better question: what's the chance of opening by 30 September?

04Walkthrough

1. Run the cells. Change the permits (A3) to 20 days. What happens to the opening date?
2. Change hiring the manager (D1) to 30 days. Does the opening move? Why not?
3. Draw the critical path as boxes and arrows.
4. Explain float to the HR lead (the task below).

05Practice

06Check your understanding