A resource conflict occurs when overlapping assignments exceed the capacity of the same named resource. Audit a four-task proposal, list every conflicting pair and intersection, then save one checked alternative using only the owner’s supplied order. Passing dependency checks does not establish that a schedule can use its equipment as proposed.
Updated: 18 October 2026. This is a fictional planning exercise, not a client case, measured production study or scheduling-software result. It establishes feasibility under stated assumptions, not an optimal finish or a delivery promise.
What does a resource conflict check decide?
Picture a small workshop reviewing proposed equipment reservations. Two independent tasks may both be ready, but that does not mean they can run together on one machine. Dependencies control what must finish first. Resource capacity controls which assignments may coexist. Inspect those rules separately before calling the proposal workable.
A resource is the means assigned to perform a task. M1 and M2 are two distinct equipment identifiers, each with capacity one: one task at a time. This is not a model with only one machine in total. Each task uses exactly its listed machine, and machines have no other reservations in this exercise.
The dependency audit covers circular waits and incomplete task references. Here the predecessor list is supplied complete and acyclic. We already have proposed starts and ends; the new decision is whether those reservations collide on the same machine despite respecting the predecessor relationships.
Recording individual working hours is a separate task covered by the employee time-tracking guide. Do not derive payroll hours from equipment conflicts.
Keep this separate from critical-path analysis, which assumes independently available resources when assessing duration changes and float. No float calculation or fastest-finish search follows here. Your deliverable is a failed-proposal audit and one alternative checked against an explicit owner instruction.
Freeze the four-task proposal before inspecting it
Use continuous abstract time units from zero. They are not calendar hours, business days or dates. All tasks are released at zero, although D still has to wait for its predecessors. There are no setup periods, breaks, failures, other reservations or interruptions. Once a task starts, it runs to completion without splitting.
The proposed start values below are positions in a candidate schedule, not immutable release times. Preserve that distinction when changing them later. Unknown duration, resource identity or capacity makes a real decision unresolved; it is not an invitation to insert a convenient estimate.
| Task | Duration | Resource | Must finish first | Start | End |
|---|---|---|---|---|---|
| A | 3 | M1 | - | 0 | 3 |
| B | 2 | M1 | - | 1 | 3 |
| C | 2 | M1 | - | 2 | 4 |
| D | 1 | M2 | A, B, C | 4 | 5 |
Check end minus start against duration in every row: three minus zero is three; three minus one is two; four minus two is two; five minus four is one. D starts at four. A and B end at three and C ends at four, so all three predecessor conditions pass. Resource feasibility is still untested.
The owner has separately supplied the order C, A, B, D for constructing an alternative. We are not choosing it from alphabetical order, revenue or a scheduling heuristic. For authority and responsibilities, see the task-delegation guide. An absent priority instruction must remain a question, not be disguised as an approved decision.
Check every task pair on its named resource
Treat intervals as start-included and end-excluded. For example, [0,3) includes zero but stops at three. To find an intersection, take the larger start and the smaller end. There is a positive overlap only if that end is strictly greater than that start. Equal boundaries are adjacent, not conflicting.
A machine conflict requires both a matching resource ID and a positive time intersection. Simultaneous work on different machines is allowed by this model. Four tasks produce six unordered pairs. Check all six; reversing A/B into B/A does not create a new pair.
| Pair | Resources | Intersection | Length | Conflict |
|---|---|---|---|---|
| A/B | M1 / M1 | [1,3) | 2 | Yes |
| A/C | M1 / M1 | [2,3) | 1 | Yes |
| A/D | M1 / M2 | None | 0 | No |
| B/C | M1 / M1 | [2,3) | 1 | Yes |
| B/D | M1 / M2 | None | 0 | No |
| C/D | M1 / M2 | None | 0 | No |
For A/B, the later start is one and the earlier end is three, giving [1,3), length two. A/C and B/C each give [2,3), length one. A/D and B/D do not overlap in time. C ends at four exactly when D starts, so even their time intersection has zero length; they also use different resources.
Record the conclusion as three separate findings: duration checks pass, dependency checks pass and M1 resource checks fail. Include all three positive conflict pairs, their machine and their intervals. Stopping at the first conflict would leave the owner with an incomplete description of the proposed reservations.
Separate pairwise overlap from elapsed overloaded time
The pairwise intersection lengths are two, one and one. Their sum is four, but the machine is not overloaded for four distinct elapsed units. The union of the conflict intervals is [1,3), length two. Repeated portions of time must be counted once when reporting elapsed overloaded time.
Inspect the occupancy directly. From zero to one, M1 has A. From one to two, it has A and B. From two to three, it has A, B and C. From three to four, it has only C. The triple-assignment segment [2,3) appears in three different pairwise findings. That explains why summing pair lengths answers a different question.
Do not infer missing staff, payroll amounts, penalties or lost sales from these intervals. They establish only overlapping assignments beyond the supplied machine capacity. The product-mix comparison makes another decision: quantities under a period-level capacity budget and contribution objective. This audit does not select product quantities or change an objective to maximize machine use.
Build one alternative using the supplied owner order
Follow the supplied order C, A, B, D. A task may start only when its predecessors have finished and its own machine is free. Set its start to the greatest of those times and zero. Keep every duration and machine assignment unchanged. Do not split tasks or invent extra equipment to make the alternative fit.
C starts at zero and ends at two. A waits for M1, starts at two and ends at five. B then occupies M1 from five to seven. M2 has been free throughout, but D must still wait for A, B and C. The latest predecessor finish is seven, so D runs from seven to eight.
| Task | Duration | Resource | Revised start | Revised end |
|---|---|---|---|---|
| C | 2 | M1 | 0 | 2 |
| A | 3 | M1 | 2 | 5 |
| B | 2 | M1 | 5 | 7 |
| D | 1 | M2 | 7 | 8 |
Recheck all durations: two, three, two and one in this displayed order. All three predecessors of D finish by seven. The three same-M1 pairs C/A, C/B and A/B have no positive intersections. The other pairs use different machines. The alternative therefore meets all written duration, dependency and resource constraints.
Eight is the finish of this checked alternative, not a proven shortest finish. The original finish of five belonged to an infeasible proposal, so “the real project was delayed by three” would also be misleading. If an owner order requires D before C while D must wait for C, stop and return that contradiction. Do not quietly reorder the instruction and label the result compliant.
A useful owner question is whether the listed capacities and priority order are confirmed for the intended planning boundary. Real calendars and reservations would require additional evidence. A mathematically feasible teaching card does not establish permission to reserve equipment, promise a date or change someone’s assignment.
Save the audit and keep source claims narrow
On Windows, use Start to open Notepad and create a blank document. Copy the four input rows, all six pair checks and the alternative table. Add the compact conclusion below. Press Ctrl+S, choose Documents and save as resource-feasibility-V1.txt. Close the document, then reopen it from Documents in File Explorer.
Fictional V1; abstract continuous units; all released0 M1=1; M2=1; no task splitting; start included/end excluded Input A3/M1/0–3; B2/M1/1–3; C2/M1/2–4 D1/M2/4–5; D waits for A,B,C Proposal: durationPASS; precedencePASS; resourceFAIL All conflicts: A/B[1,3); A/C[2,3); B/C[2,3) Pair-length sum4; elapsed union[1,3), length2 Supplied owner order C,A,B,D Alternative C0–2; A2–5; B5–7; D7–8 Alternative duration/precedence/resourcePASS; no optimality claim Unknown data or incompatible order: stop and ask the owner
Check after reopening that both machine identifiers, the supplied order and all three conflicts remain visible. The suggested fifteen-minute core is the small four-task audit and save route, not a measured participant completion time. Do not expand the exercise until another person could inspect the original inputs and your reasoning.
In his published Azerbaijani reflection, Anar describes projects delayed by pursuing perfection. Applying that documented principle here means completing one inspectable record first. It is not a claim that he tested this machine schedule, ran a factory experiment or achieved a client result.
Google’s Job Shop documentation distinguishes predecessor constraints from machine exclusivity. Microsoft explains that resolving resource overallocation can change task timing. Those are the narrow conceptual attributions used here. No Google solver or Microsoft Project interface was run. Our fixed owner-order calculation is not a claim about either product’s automatic ordering, default settings or optimized output.
Frequently asked questions
Does passing dependencies make a schedule feasible?
Not by itself. In the proposed example, D waits for every predecessor, but M1 still has three conflicting task pairs.
Do tasks conflict when one starts exactly as another ends?
Not under the stated start-included, end-excluded convention. C ending at two and A starting at two have no positive intersection.
Why do pair lengths total four while elapsed overlap is two?
The triple-assignment segment [2,3) appears in multiple pairwise intersections. Their union is [1,3), with elapsed length two.
Is eight the shortest possible finish?
No such claim is made. Eight is the finish of one checked alternative using the supplied C,A,B,D order; no optimal-order search was performed.
