q-Rung orthopair PROMETHEE (Akram and Shumaiza, 2021)

A construction employer evaluates eight subcontractor candidates (a1 to a8) on ten criteria: technical competence, training skills, reputation, financial standing, organisational skills, experience, reliability, safety, past performance record and quality assurance. All criteria are "more is better." Every cell is a (μ, ν) pair drawn from pooled expert assessments; because the most extreme pairs exceed q = 2, rung q = 3 was chosen.

Three of the ten criteria are shown below as an example; the engine's input is all ten criteria.

The method first reduces every cell to a Yager score, then compares each candidate pair with the plainest preference function, builds the weighted overall preference index, and sums the incoming and outgoing flows.

The result reads as follows. a8 ranks first because it carries high support degrees on heavily-weighted criteria such as safety and past performance. a3 ranks second, and the gap to a1 (0.268 against 0.055) is marked. a7, with the lowest net flow, carries a lower support or higher rejection degree than the others on most criteria.

The employer's hesitation: two cells in the paper's own table, between a1 and a7 and between a1 and a5, contradict another of the paper's own tables. Once these two cells are corrected according to the paper's own internal consistency and the calculation independently redone, the net flow of a1, a5 and a7 changes by only 0.014, and the ranking does not change. This shows that small data errors do not immediately upset the result, but they must still be checked.

In the report: "Eight subcontractor candidates were evaluated on ten criteria at q = 3; a8 comes out clearly first (Φ = 0.388); a3 is second and the gap is not small."

Source: Akram, M., & Shumaiza (2021), Section 4 subcontractor-selection example, Tables 5-8, pp. 117-121. The weighted matrix is taken from the paper's own Table 5; net flows were recomputed independently by running the engine, and matched exactly the ranking published by the paper (a8, a3, a1, a5, a6, a2, a4, a7).

A clothing manufacturer will choose its main supplier for the year among three fabric suppliers. Criteria: production capacity, quality consistency and delivery performance; all three are "more is better." The expert team has given every supplier both a support and a rejection degree on every criterion; one supplier's capacity criterion was reported with both a very high support and a marked rejection degree, and since this pair exceeds the Pythagorean constraint, q = 3 was chosen.

The method reduces every cell to a score, builds the weighted preference index and computes the net flows. Suppose the supplier with the highest capacity but also the highest quality reservation stays second, because the weight on quality consistency is kept high; the supplier with middling capacity but balanced support on every criterion comes out first.

The manufacturer's hesitation: would the order change if the capacity weight were raised slightly? The strength of the supplier with the highest capacity is concentrated in a single criterion; the balanced supplier's advantage is spread across more than one criterion and appears more resilient to weight changes.

In the report: "The supplier with a balanced support profile comes out clearly first; the advantage of the supplier with the highest capacity rests on a single criterion and is therefore more fragile to weight changes."

In the subcontractor table, entering a1's technical-competence pair as (0.734; 0.579) and choosing q = 2 based on this single cell, while ignoring that the constraint at q = 2 is exceeded elsewhere in the matrix (for instance in a8's safety pair), is wrong; q is chosen once for the whole matrix. The second error is calculating the scores and then rounding the difference between them as "small, negligible" and treating them as equal; the plainest preference function counts every positive difference as total preference, and rounding can change the ranking. The third error is deriving the rejection degree from the support degree (ν = 1 − μ). In that case the sum is always exactly 1 and q = 1 already suffices; raising q then serves no purpose.