RAFSI - Ranking of Alternatives through Functional mapping of criterion sub-intervals into a Single Interval

Overview

V(A_i) ∈ (0, 1]. Higher V means better. Every value is mapped once onto the criteria functions interval [n_i, n_k] (defaults 1 and 6) between the decision-maker's anti-ideal n_j^A and ideal n_j^I. A cost criterion sends its ideal to the LOWER end n_i and is turned into higher-is-better exactly once, by the reciprocal H/(2 s_ij) in the normalisation; it is never inverted twice. Because the anchors are declared feasible bounds rather than the data's own min/max, adding or removing alternatives cannot change the other alternatives' scores (rank-reversal free). Ideal and anti-ideal must be set by the decision-maker to cover the full feasible range of each criterion.

Output

utility, higher is better

Data

Crisp, complete numeric matrix

Weights

Needs a weight source

Size

2+ alternatives, 3-12 criteria works best

Used for

Alternative selection, Supplier evaluation

How it works

1. 1

   Read the ideal n_{j}^{I} and anti-ideal n_{j}^{A} endpoints the decision maker declares for each criterion. Direction consistency (benefit: n^I > n^A, cost: n^I < n^A) and bracketing of every observed value are validated; the anchors are NOT derived from the data's own min/max.

   $$\min (n_j^I,n_j^A)\le x_{ij}\le \max (n_j^I,n_j^A)$$

   Žižović et al. 2020, Sec.3 Step 1

2. 2

   Map every value linearly onto the criteria functions interval [n_i, n_k] (defaults [1, 6]). For a benefit criterion the ideal goes to the upper end n_k; for a cost criterion the ideal goes to the LOWER end n_i, and the cost direction is corrected exactly once, by the reciprocal in Step 4.

   $$s_{ij}=n_i+\frac{x_{ij}-n_j^A}{n_j^I-n_j^A}(n_k-n_i)(benefit,Eq.9);s_{ij}=n_i+\frac{x_{ij}-n_j^I}{n_j^A-n_j^I}(n_k-n_i)(cost,Eq.10)$$

   Žižović et al. 2020, Eqs.(9)-(10)

3. 3

   Arithmetic mean A and harmonic mean H of the mapping interval itself. Both are SCALARS of [n_i, n_k] and do not depend on the alternatives (for [1, 6]: A = 3.5, H = 12/7). Computing a per-alternative A_i/H_i is the misreading removed by the 2026-08 audit.

   $$A=\frac{n_i+n_k}{2},H=\frac{2}{1/n_i+1/n_k}$$

   Žižović et al. 2020, Eqs.(11)-(12)

4. 4

   Normalise: a benefit value is divided by 2A, a cost value enters as H/(2 s_ij). Both branches land in (0, 1] and are higher-is-better; an alternative on every declared ideal scores 6/7, one on every anti-ideal scores 1/7.

   $${\hat{s}}_{ij}=s_{ij}/(2A)(benefit);{\hat{s}}_{ij}=H/(2s_{ij})(cost)$$

   Žižović et al. 2020, Eq.(13)

5. 5

   Weighted sum V(A_i) = Σ_j w_j ŝ_ij and descending ranking. On the J fixture: A1 = 0.4112, A2 = 0.4792, A3 = 0.4362; ranking A2 > A3 > A1.

   $$V(A_i)=\sum _{j=1}^n{w}_j{\hat{s}}_{ij}$$

   Žižović et al. 2020, Eq.(14)

Look elsewhere when

Assumptions to verify

• Criteria preferences are independent (no synergistic interactions)
• Compensation is acceptable: high score on one criterion can offset low on another
• Decision matrix is complete (no missing values)

Edge cases and pitfalls

• I* = I°: mapping denominator is zero - check E-2.

• Alternatives outside [I°, I*]: mapped values outside [1, 2n+1] can occur if bounds are set too tight.

Works with

How to cite