The calculator produces a seven-day baseline, the critical path A–B–D–E, and one day of float for C. Changing C’s three estimates to 2, 3, and 4 creates two tied seven-day critical paths. Results appear in an activity table and a text dependency list; this example does not draw a node diagram.
Use the worked example
Calculate the baseline, then change C’s optimistic, most likely, and pessimistic estimates to 2, 3, and 4. Calculate again and compare its early and late times with the two critical paths. Edits clear the previous results, so recalculate before interpreting a changed row. Reset returns the five original activities.
Trace the result step by step
Use abstract elapsed days starting at zero, finish-to-start dependencies, and unlimited parallel resources. No weekends, staffing conflicts or resource limits are modeled.
Use (O + 4M + P) / 6 to calculate each activity’s duration. A starts with 1, 2, 3 and gives 2 days; B has 2, 3, 4 and gives 3; C has 1, 2, 3 and gives 2. D and E each have 1, 1, 1 and give 1. The input fields retain the three estimates; the results table shows each derived duration and its schedule times.
Early start ES is the maximum predecessor early finish; a root starts at zero. Early finish EF is ES plus duration. Rehearsal D waits for B at day 5 and C at day 4, so it starts at 5.
For the backward pass, set the final activity's late finish to the network finish, 7. Work backward using the minimum successor late start for LF, then LS = LF − duration. Total float is LS − ES.
C can start at 3 rather than its earliest 2 and still finish before D starts at 5 under these assumptions. Its one day of total float is not permission to shift every other task too, and it does not establish free float or available staffing.
A → B → D → E totals 2 + 3 + 1 + 1 = 7. A → C → D → E totals 2 + 2 + 1 + 1 = 6. The first chain controls the baseline network finish.
Now change C's three estimates to 2/3/4. Its derived duration becomes 3. C's ES/EF becomes 2/5 and LS/LF 2/5, reducing float to zero. The network still finishes at 7, but both chains now total 7. The result lists both critical paths; B’s branch no longer controls the finish alone.
Read the listed dependency chains separately. B and C are parallel branches that both feed D, followed by E. When the two paths tie, they are A–B–D–E and A–C–D–E. Combining every zero-float activity into A–B–C–D–E would invent a dependency that the activity table does not contain.
The seven-day result is a longest-path calculation using the derived activity durations. It is not an observed project duration, a confidence level or a statistically calculated expected project-completion time. Three-point inputs alone do not supply a probability forecast for the whole network.
The seven-day result depends on the stated resource and calendar assumptions. You can edit estimates and predecessor IDs for the five fixed activities, but cannot add activities or create a calendar in this example. Estimates must be finite, nonnegative, and ordered O ≤ M ≤ P. Unknown or repeated predecessor IDs and dependency cycles need correction before a schedule is shown. If a derived duration or network time exceeds the supported finite range, the calculator clears the result and asks for smaller estimates.
Copy a prompt for this example
Create a single HTML planning calculator for five fixed activity IDs with editable optimistic, most-likely, pessimistic, and predecessor fields. Use abstract elapsed days, finish-to-start dependencies, and unlimited parallel resources.
Start with A Outline, estimates 1/2/3, no predecessor; B Content, 2/3/4, after A; C Handouts, 1/2/3, after A; D Rehearsal, 1/1/1, after B and C; E Check, 1/1/1, after D. Derive each duration as (O + 4M + P) / 6.
Calculate early start as the maximum predecessor finish, early finish as start plus duration, and late times by working backward from the network finish. Show duration, ES, EF, LS, LF, and total float in a table. Include a readable text dependency list rather than promising an editable node diagram.
The baseline finishes at seven days. A-B-D-E is critical; C has one day of float. Change C to 2/3/4 and show both A-B-D-E and A-C-D-E as tied seven-day critical paths.
Reject negative or nonfinite estimates, O greater than M, M greater than P, unknown or repeated predecessor IDs, and cycles. Clear stale results on any input edit. Calculate recomputes the whole network; Reset restores the five starting rows.
Use clear field labels and a responsive layout. Explain that the longest-path result is not a calendar commitment or a project-completion probability. Return complete HTML with its calculation logic.
Take the activity table and both tied paths to your planning discussion. You can preview the prepared HTML calculator in NOS while comparing the original estimates with C’s revised values. Confirm actual staffing and calendar constraints before converting seven abstract elapsed days into a promised date.