Critical path analysis: does one delay change the finish?

Critical path analysis shows whether a supplied duration change alters the calculated finish. Build a six-task baseline and one comparison record, keeping calendar and resource assumptions separate from any real delivery promise.

Content last updated: 17 October 2026.

Which delay changes the project finish?

Critical path analysis answers a narrower question than deciding which task deserves attention. With supplied durations and dependencies, it identifies the chain controlling the earliest calculated finish. You will produce a six-task baseline and one duration-change record, rather than a complete operating schedule for your business.

Imagine an independent exhibition team preparing an internal sample pack. Its coordinator wants to know whether extra layout work necessarily postpones completion. The example below is fictional teaching material. The durations are stipulated, not measured, and no customer delivery commitment follows from them.

Start by separating structural validity from timing. The dependency audit checks circular waits, unresolved references and an incomplete predecessor list. This exercise starts with those relationships already supplied correctly. Its output is the effect of a duration change on the calculated finish, not another list of blocked tasks.

Use continuous abstract hours measured from time zero. Every activity has its own available people and equipment. There are no weekends, breaks, resource conflicts, waiting lags or uncertain durations. A successor may begin immediately when all its predecessors finish. Those are explicit exercise assumptions, not observations about a real team. If one designer owns two simultaneous activities, the arithmetic alone cannot establish your delivery date.

Keep six supplied activities separate from their interpretation

Use A for the brief, B for copy, C for layout, D for copy review, E for layout review and F for assembly. Names are memory aids; they do not create additional dependencies. F requires both reviews. Replacing that requirement with whichever review finishes first changes the problem.

Copy the input before calculating. Keep the duration in its own column and preserve the original version. A missing value is a question for the plan owner, not permission to estimate what a familiar task usually takes. For actual responsibility rather than timing arithmetic, use the task delegation guide.

This is a complete directed graph without a cycle.

Work forward using the last predecessor to finish

ES means earliest start and EF means earliest finish. Begin the first activity at zero. Add its duration to obtain EF. For a later activity, take the largest EF among its immediate predecessors as its ES, then add its own duration. Every required predecessor must have finished.

A occupies zero to two. Copy B occupies two to six, while layout C occupies two to four. Copy review D runs six to eight; layout review E runs four to six. Assembly F waits for both eight and six, so its earliest start is eight and its finish is nine.

Adding every duration gives thirteen, but thirteen is not the finish of this model. The two branches can overlap under the unlimited-resource assumption. The complete paths are A-B-D-F with length nine and A-C-E-F with length seven. The longer branch controls the shared assembly step.

Write the reason beside the number: “Finish nine because F waits for D at eight, then takes one.” No software is required. See the artificial intelligence guide for separate tool choices.

Work backward to find total float

LS means latest start without extending the calculated finish; LF means latest finish on that basis. Give terminal activity F an LF of nine, then subtract its duration. Its LS is eight. Moving backward, an activity must finish by the smallest LS among its immediate successors. Subtract duration again to find its LS.

D must finish at eight and can start at six. E must also finish at eight but can start at six because it takes two. B can finish at six and start at two. C can finish at six and start at four. A must support both B starting at two and C starting at four, so its latest finish is two.

Total float, TF, is LS minus ES, equivalently LF minus EF. In this unconstrained exercise, zero-float activities form the controlling path A-B-D-F. C and E each have two hours of total float. The table preserves early and late values rather than replacing one with the other.

Total float concerns the project finish. It does not mean an activity can move without affecting its immediate successor. C could push E later while assembly still begins at eight. Keep that distinction visible instead of calling every kind of spare time the same thing.

The PMI conceptual reference explains these scheduling calculations. Our zero-float result belongs to this written mathematical contract. It is not a claim about a scheduling tool's default screen colour or a report of testing software.

Recompute one layout duration change from the baseline

For the core practice, change only C from two to three. This models additional activity duration. It is not an imposed actual-start restriction, an absent employee or a calendar exception. Preserve every dependency and every other duration, then run the complete calculation again.

A still ends at two, B at six and D at eight. C now ends at five, so E ends at seven. F compares eight with seven and still finishes at nine. The layout branch now totals eight, leaving one hour before it matches the copy branch. C and E each have one hour of remaining total float.

The late values below were recalculated too. Do not alter one early-finish cell and leave the rest of the previous table labelled current. A changed merge can propagate through the plan, and a changed project finish changes the reference used for the backward pass.

Your saved conclusion is modest: “Only C duration increased by one; finish remains nine; controlling path A-B-D-F; exercise assumptions unchanged.” This is a useful decision record even though no real activity has occurred.

Optional stress checks: ties, switching paths and shared float

Each row below starts independently from the original six inputs. They are not cumulative edits. All activity times were recomputed for each case. You can stop after the preceding one-change exercise; the extra cases test whether you understand the limits of the result.

At C equal to four, both complete paths have length nine. Listing only the original copy path would miss a tied critical branch. At C equal to five, the layout branch reaches ten and becomes controlling. Increasing the originally critical B by one instead makes the copy branch ten.

The final row rejects a common interpretation of float. C and E each initially show two, but they share the same branch. Spending two on each adds four to a seven-hour branch, producing eleven. Their displayed float is not four independently available hours. Recalculate combined changes rather than adding spare-time labels.

The fixed-kit capacity exercise concerns a different limiting factor: component quantities, not duration along task paths. This timing note neither reserves stock nor selects a production quantity. It also does not optimize staff allocation or choose which activity should be shortened for a commercial project.

Save an inspectable result without promising a delivery date

On Windows, search Start for Notepad, open it and create a new blank document headed “Timing practice V1”. Copy the six source rows, model assumptions, baseline table and comparison where only C changes from two to three. Press Ctrl+S, select Documents, enter critical-path-V1.txt and choose Save. Close the editor; open Documents in File Explorer and double-click that named file. Check that original C remains two and the changed case remains three.

Include the path, finish difference and unresolved real-world questions. If working hours, shared staff or duration confidence are unknown, say so. The SOP guide covers repeatable operating instructions; this note is only a calculation record.

Source: fictional V1; six supplied activities
Model: time0; continuous hours; separate resources; zero lags
Baseline: finish9, path A-B-D-F
Only change: C duration2→3
Recomputed: finish9, difference0
Real calendar/resource questions and owner: [complete]
Not a customer deadline or report of completed work.

A short practice session can finish the baseline and one comparison. Completing every optional case within fifteen minutes is not a measured promise. For a wider rollout, the digital transformation roadmap addresses pilot acceptance and adoption separately.

In his published Azerbaijani reflection on perfection and effectiveness, Anar describes delaying his own projects while pursuing perfection. Applying that documented principle here means finishing one inspectable note before expanding the model. It does not claim he ran this timing exercise with a client, tested scheduling software or achieved a measured project improvement.