C.28.CM:5.1 - Separate an omission, its record and the incident sample

Four incident tickets mention missing order identifiers after a reporting-template change. A planner asks whether supplying the identifier before import would remove manual matching. All four tickets came from problematic reports; the frequency among all reports is unknown.

Choose M for an actually missing identifier and Y for manual matching after import. X denotes template use and L high process load. One account proposes X → M → Y, with load affecting template deployment and identifier omission. A second account proposes that load produces both missing identifiers and an additional wrong join key K; K causes matching work even after the identifier is supplied. The second account has L → M and L → K → Y without an M → Y mechanism.

These accounts make different conditional predictions. If missing identifiers are the operative joining failure and supplying one changes no other input, supplying it removes that modeled obstacle. If K remains wrong, the same correction leaves the K-related work. A report with its identifier restored but the same failed join can challenge the first account’s claim of sufficiency. It does not establish the second account merely by eliminating one rival.

Now recover the observing process. Let R be a log saying the identifier is missing. A logging defect can make R=1 while M=0. Inspecting the submitted report and the importer’s actual inputs can therefore resolve a recording question before further causal research is useful.

Let S=1 mean a report entered the incident sample, either because R=1 or because matching work was severe. The structure R → S ← Y means that analyzing only S=1 conditions on a collider. The four tickets cannot by themselves establish the population association or the effect of correcting identifiers. Obtain the needed comparison cases if they could change the decision; retaining a qualified unresolved answer is also possible.

The first return is specific: determine which joining inputs were actually missing or wrong, then distinguish the two modeled failure mechanisms. If all retained accounts support a cheap temporary manual check under the decision’s own conditions, using it does not establish which account caused the incident.

Onset, persistence and amplification. Suppose a queue began during a specialist’s absence. After their return, ten new items and capacity for ten items arrive each day. Under the simplified balance B(t+1)=max(0, B(t)+A(t)+R(t)−C(t)), with backlog B(0)=12, A=10, R=0 and C=10, the backlog remains 12. Two daily rework items, R=2, increase it by two per day. The absence explains the initial loss of service; the present flow balance explains persistence; rework explains growth under these premises. Restoring attendance alone need not clear the backlog. A different arrival pattern, capacity or feedback from delay to rework reopens the corresponding part of the model.