**Appendix A**

The following is a description of the procedure followed to solve the optimisation problem posed in Step 6 and given by expression (9):

$$\min\_{a\_{rs}} \max\_{k} \pi\_{k} \left( GCI^{(k)} + \frac{2}{n(n-1)} \left[ \left( a\_{rs} - a\_{rs}^{(k)} \right)^{2} + \frac{2n}{n-2} \left( a\_{rs} - a\_{rs}^{(k)} \right) \varepsilon\_{rs}^{(k)} \right] \right)$$

with *αrs* ∈ )log *<sup>a</sup>trs*, log *atrs* \*, where *αrs* = log *ars*, *α*(*k*) *rs* = log *a* (*k*) *rs* and *ε* (*k*) *rs* = log *e* (*k*) *rs*

This problem can be written as:

$$\min\_{\alpha\_{rs}\in[l,u]} \max\_{k} p\_k(\alpha\_{rs})$$

where *pk*(*<sup>α</sup>rs*) = <sup>π</sup>*k GCI*(*k*) + 2 *<sup>n</sup>*(*<sup>n</sup>*−<sup>1</sup>) ' *αrs* − *α*(*k*) *rs* 2 + 2*n n*−2 *αrs* − *α*(*k*) *rs ε* (*k*) *rs* ( are second degree polynomials with *αrs*the variable and the coefficient for the quadratic term positive.

Therefore, the problem can be rewritten as:

$$\min\_{\mathbf{x}\in[l\_{\prime},u]} \max\_{k} p\_{k}(\mathbf{x})\tag{A1}$$

where *pk*(*x*) = *akx*<sup>2</sup> + *bkx* + *ck* with *ak* > 0. Therearethree cases:

a. The polynomial that is dominant in *x* = *l* is an increasing function at this point. In this case, the solution to the optimisation problem (9) is *x*<sup>∗</sup> = *l* (Figure A1a).


**Figure A1.** Graphical representation of the three cases.

**Case a.** *When we say that a polynomial pr is dominant in x* = *l*, *we refer to the situation in which the polynomial pr provides the maximum value in a neighbourhood* [*l*, *l* + *ε*):

$$p\_r(\mathbf{x}) = \max\_k p\_k(\mathbf{x}) \\ \text{with} \\ \mathbf{x} \in [l, l + \varepsilon)$$

*In order to obtain this dominant polynomial, we start by analysing the values* [*l*, *l* + *<sup>ε</sup>*). *The polynomial pr*(*x*) *that provides the maximum value at this point is the polynomial that we were looking for. Nevertheless, it is possible that some ties exist. In this case, we should look for the polynomial that is dominant in the neighbourhood* [*l*, *l* + *<sup>ε</sup>*).

*If pi*(*l*) = *pj*(*l*), *in order to determine which of the two polynomials is dominant in* [*l*, *l* + *<sup>ε</sup>*), *we examine the first derivative: if <sup>p</sup> i*(*l*) > *<sup>p</sup> <sup>j</sup>*(*l*), *the polynomial pi is dominant*.

*If pi*(*l*) = *pj*(*l*) *and <sup>p</sup> i*(*l*) = *<sup>p</sup> <sup>j</sup>*(*l*), *the dominant polynomial will be that one with the maximum value in the second derivative*.

**Figure A2.** Several domination relationships.

*In the Figure A2 it can be appreciated that p*1(*l*) = 1 *and p*2(*l*) = *p*3(*l*) = *p*4(*l*) = 4. *It can also be observed that <sup>p</sup>* 2(*l*) < *<sup>p</sup>* 3(*l*) *and <sup>p</sup>* 2(*l*) < *<sup>p</sup>* <sup>4</sup>(*l*), *thus the polynomial p*2(*x*) *is not dominant. Polynomials p*3(*x*) *and p*4(*x*) *have the same value as their derivative in x* = *l*, *but the polynomial p*4(*x*) *has a greater value in the second derivative (greater coefficient of x*2), *it is therefore the dominant polynomial*.

*In short, if pr is the polynomial that is dominant in x* = *l*, *the following condition should be fulfilled*:

$$(p\_r(l), p\prime\_r(l), p\prime\_r^{\prime\prime}(l)) = \text{lex}\_k \max\_k (p\_k(l), p\prime\_k(l), p\prime\_k^{\prime\prime}(l))$$

**Case b.** *The polynomial that is dominant in x* = *u* can be determined in an analogous way:

$$\mathbb{P}\left(p\_{\mathbb{S}}(\mathsf{u}), -p\mathsf{t}\_{\mathbb{S}}(\mathsf{u}), \ p\mathsf{\prime\prime}\_{\mathbb{S}}(\mathsf{u})\right) = \operatorname\*{lex}\_{\mathbb{S}}\max\_{k} \left(p\_{k}(\mathsf{u}), -p\mathsf{\prime\prime}\_{k}(\mathsf{u}), \ p\mathsf{\prime\prime}\_{k}(\mathsf{u})\right)$$

*By determining the polynomials which are dominant in l and u*, *and calculating their respective derivatives we will be able to identify cases a and b*.

**Case c.** *Starting from point x*0 = *l and from the corresponding polynomial that is dominant at this point, we determine point x*1 *in which this polynomial becomes dominated. If the following polynomial that is dominant is a decreasing function at this point, we continue the process, updating x*0 = *x*1. *At the moment in which we go out from the interval under study, or when the new polynomial that is dominant is an increasing function, we have finished the iterative stage and we only have to calculate the minimum of the present dominant polynomial in the interval under study*.

In Figure A1c we can see that p1 is the polynomial that is dominant in x = l. It continues to be the dominant polynomial until x1 where the new polynomial that is dominant is p2. Again, this polynomial is dominant until point x2. But the polynomial that is dominant from this point is an increasing function at x2, and therefore it is not necessary to continue. The solution to the optimisation problem is given by the minimum of the polynomial p2 in the interval [x1, x2].

Finding the points where a polynomial ceases to be dominant is determined by exploring the possible cut points with the rest of the polynomials in the problem (those that are within the interval under study) as following Algorithm A1.

**Algorithm A1**

$$\min\_{\mathbf{x}\in[l\_r u]} \max\_k p\_k(\mathbf{x}) \text{ where } p\_k(\mathbf{x}) = a\_k \mathbf{x}^2 + b\_k \mathbf{x} + c\_k \text{ with } a\_k > 0$$

Step 1: Find *r* and with

$$(p\_{\boldsymbol{\nu}}(\boldsymbol{l}), \ p\boldsymbol{\nu}\_{\boldsymbol{\nu}}(\boldsymbol{l}), \ p\boldsymbol{\nu}\_{\boldsymbol{\nu}}(\boldsymbol{l})) = \operatorname\*{lex}\max\_{k}(p\_{k}(\boldsymbol{l}), \ p\boldsymbol{\nu}\_{k}(\boldsymbol{l}), \ p\boldsymbol{\nu}\_{k}(\boldsymbol{l})) $$
 
$$(p\_{\boldsymbol{\nu}}(\boldsymbol{u}), -p\boldsymbol{\nu}\_{\boldsymbol{\nu}}(\boldsymbol{u}), \ p\boldsymbol{\nu}\_{\boldsymbol{\nu}}(\boldsymbol{u})) = \operatorname\*{lex}\max\_{k}(p\_{k}(\boldsymbol{u}), -p\boldsymbol{\nu}\_{k}(\boldsymbol{u}), \ p\_{k}^{\boldsymbol{\nu}}(\boldsymbol{u})) $$

If *p r*(*l*) ≥ 0 then *x*<sup>∗</sup> = *l*. STOP

If *p s*(*u*) ≤ 0 then *x*<sup>∗</sup> = *u*. STOP

Step 2: Start from point *x*0 = *l*, where the polynomial that is dominant, *pr*(*x*), is a decreasing function at this point.

Step 3: Calculate *I* = {*i* such that ∃ *ti* ∈ (*<sup>x</sup>*0, *u*] with*pi*(*ti*) = *pr*(*ti*)and*pi* (*ti*) > *pr* (*ti*)} Step 4: If *I* = ∅ the optimal solution is given by:

$$\min\_{\alpha \in [\chi\_0, w]} p\_{\mathcal{V}}(\mathbf{x})$$

Step 5: Let *j* be the value such that

$$(-t\_{\boldsymbol{f}}, \ p\iota\_{\boldsymbol{f}}(t\_{\boldsymbol{f}}), \ p\_{\boldsymbol{f}}^{\prime\prime}(t\_{\boldsymbol{f}})) = \operatorname\*{le\kappa}\max\_{i \in \boldsymbol{I}}(-t\_{\boldsymbol{i}}, \ p\iota\_{i}(t\_{\boldsymbol{i}}), \ p\_{\boldsymbol{i}}^{\prime\prime}(t\_{\boldsymbol{i}})) $$

If *<sup>p</sup> j*(*tj*) > 0 the optimal solution is given by:

$$\min\_{\mathbf{x} \in [\mathbf{x}\_{0\prime}, t\_{\dagger}]} p\_{\mathbf{r}}(\mathbf{x})$$

Otherwise, update and *r* = *j* and go to Step 3.
