ARAS - Additive Ratio Assessment

Overview

K_i ∈ [0,1]. The optimal alternative receives K_0=1 (by definition S_0/S_0=1). All real alternatives have K_i ≤ 1 because S_0 ≥ S_i. Higher K_i means the alternative is proportionally closer to the optimal benchmark. Unlike COPRAS, ARAS uses a single additive score without splitting benefit/cost - it handles cost by inverting before normalisation.

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

   Augment matrix with optimal alternative A_0 (best per criterion).

   $$X_{0j}=\{\begin{array}{ll}{\max }_i{x}_{ij}& j\in J^{+}\\ {\min }_i{x}_{ij}& j\in J^{-}\end{array}$$

   Zavadskas-Turskis 2010, p.166 Eq.(1)

2. 2

   Linear-sum normalisation (cost criteria inverted first).

   $${\bar{x}}_{ij}=\frac{x_{ij}}{{\sum }_{k=0}^m{x}_{kj}},j\in J^{+};x_{ij}\leftarrow 1/x_{ij}\text{for}j\in J^{-}$$

   Zavadskas-Turskis 2010, p.166 Eq.(2)

3. 3

   Weighted normalised matrix d_ij = w_j · x̄_ij.

   $$d_{ij}=w_j{\bar{x}}_{ij}$$

   Zavadskas-Turskis 2010, p.167 Eq.(3)

4. 4

   Optimality function S_i = Σ d_ij (including S_0 row).

   $$S_i=\sum _{j=1}^n{d}_{ij},i=0,1,\dots ,m$$

   Zavadskas-Turskis 2010, p.167 Eq.(4)

5. 5

   Utility K_i = S_i / S_0 and descending ranking.

   $$K_i=\frac{S_i}{S_0},0\le K_i\le 1$$

   Zavadskas-Turskis 2010, p.167 Eq.(5)

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

• Optimal row not derived correctly: for cost criteria, x_0j must be min (not max) - wrong ideal inflates S_0 and deflates all K_i uniformly.

Works with

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