The critical path

Twenty-three tasks, each 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 importing the analysers delays the opening by one day. A project manager who treats every task alike spends effort in the wrong places, and a sponsor who hears "everything is a priority" learns nothing.

The managing director promised Monday 8 March 2027. Before you ask how late the project is, you need to know where the plan itself put the finish. Was the promised date ever safe?

The concept

The critical path method

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

Float is not free time

A task with float can slip, but it uses up the float, and other tasks may share the same float. Two tasks with 8 days of float each, in the same chain, don't give you 16.

Example

Python · try it yourself

import numpy as np
import pandas as pd

base = "https://academy.cloudtechanalytics.com/datasets/lab/"
tasks = pd.read_csv(base + "tasks.csv").fillna({"predecessors": ""})
START = np.datetime64("2026-11-02")     # a Monday
PROMISE = np.datetime64("2027-03-08")   # the date promised to the commissioner

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]))

Edit the code and press Run (Ctrl + Enter). Examples on this page share their variables, like cells in a notebook.

Output

Project length: 90 working days, opening on 2027-03-05
Critical path: A1 -> A2 -> B1 -> B3 -> B4 -> B5 -> E2 -> F1 -> F2 -> F3 -> F4

The baseline finishes on Friday 5 March 2027: day 90. The promise is for Monday 8 March, day 91. The plan left one working day of cushion, and that was before anything went wrong.

Python · try it yourself

Edit the code and press Run (Ctrl + Enter). Examples on this page share their variables, like cells in a notebook.

Output

Promised date is working day 91 - cushion: 1 working day(s)
    es  ef  float                                         name
A1   0   3      0            Approve business case and charter
A2   3  11      0                  Sign lease for the premises
B1  11  19      0        Design the lab layout and clean areas
B3  19  25      0                          Order the analysers
B5  50  57      0          Install and calibrate the analysers
B4  25  50      0               Import and clear the analysers
E2  57  63      0               Stock reagents and consumables
F1  63  75      0  Validate the methods and run quality checks
F2  75  83      0   Pass the external accreditation inspection
F3  83  89      0              Trial run with sample specimens
F4  89  90      0                                  Opening day
B2  19  47      3      Fit-out: floors, power and air handling
B6  47  55      8        Install backup power and cold storage
C5  57  63     12           Test the system with the analysers

Eleven tasks have zero float. Look at what they are: the premises, the layout, ordering and importing the analysers, installing them, stocking reagents, validating the methods, the inspection, the trial run and the opening. The long-lead imported equipment runs the whole project, not the fit-out, which has 3 days of float, and not hiring, which has 30.

Walkthrough

Practice

On what date does the baseline schedule finish? Type it as YYYY-MM-DD.

tasks.csv

Practice

How many tasks are on the critical path (zero float)?

tasks.csv

Practice

How many days of float does task D1 (hire the laboratory director) have?

tasks.csv

Task

10 min

Write the baseline schedule note for the sponsor (60 to 150 words): the baseline finish date, how it compares with the promised date, which kind of tasks are critical and which have the most float, and what that means for where you'll spend your attention.

Your work is checked for

Gives the baseline finish (5 March)
Compares with the promised date (8 March)
Says the cushion is tiny (one day)
Names critical activities (analysers, installation, inspection...)
Names a high-float activity (hiring, systems)
Says where attention goes (critical path)
Between 60 and 150 words

Python · try it yourself

import numpy as np
import pandas as pd

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

Edit the code and press Run (Ctrl + Enter). Examples on this page share their variables, like cells in a notebook.

The problem

The concept