**Appendix C**

**Proof.** For *λ* ∈ (0, <sup>∞</sup>), since *α*<sup>−</sup> ≤ *αj* ≤ *<sup>α</sup>*+, then

$$\begin{split} a\_{j}^{\lambda} \geq \left(\boldsymbol{\alpha}^{-}\right)^{\lambda}, 1 - a\_{j}^{\cdot} \overset{\scriptstyle \lambda}{\leq} & 1 - \left(\boldsymbol{\alpha}^{-}\right)^{\lambda}, \left(1 - a\_{j}^{\cdot\lambda}\right)^{w\_{j}} \leq \left(1 - \left(\boldsymbol{\alpha}^{-}\right)^{\lambda}\right)^{w\_{j}}, \\ & \prod\_{j=1}^{n} \left(1 - a\_{j}^{\cdot\lambda}\right)^{w\_{j}} \leq \prod\_{j=1}^{n} \left(1 - \left(\boldsymbol{\alpha}^{-}\right)^{\lambda}\right)^{w\_{j}}, \\ & 1 - \prod\_{j=1}^{n} \left(1 - a\_{j}^{\cdot\lambda}\right)^{w\_{j}} \geq 1 - \prod\_{j=1}^{n} \left(1 - \left(\boldsymbol{\alpha}^{-}\right)^{\lambda}\right)^{w\_{j}}, \\ & \left(1 - \prod\_{j=1}^{n} \left(1 - a\_{j}^{\cdot\lambda}\right)^{w\_{j}}\right)^{1/\lambda} \geq \left(1 - \prod\_{j=1}^{n} \left(1 - \left(\boldsymbol{\alpha}^{-}\right)^{\lambda}\right)^{w\_{j}}\right)^{1/\lambda} = \boldsymbol{\alpha}^{-}. \end{split}$$

similarly, we have

$$\left(1 - \prod\_{j=1}^{n} \left(1 - a\_j^{\cdot \lambda} \right)^{w\_j} \right)^{1/\lambda} \le \left(1 - \prod\_{j=1}^{n} \left(1 - \left(a^{+} \right)^{\lambda} \right)^{w\_j} \right)^{1/\lambda} = a^{+} \cdot \lambda$$

As *β*− ≤ *βj* ≤ *β*<sup>+</sup>, then

1 − *βj* ≤ 1 − *β*<sup>−</sup>, -1 − *<sup>β</sup>j*.*<sup>λ</sup>* ≤ (1 − *β*−)*<sup>λ</sup>*, 1 − -1 − *<sup>β</sup>j*.*<sup>λ</sup>* ≥ 1 − (1 − *β*−)*<sup>λ</sup>*, 1 − -1 − *<sup>β</sup>j*.*<sup>λ</sup>wj* ≥ 1 − (1 − *<sup>β</sup>*−)*<sup>λ</sup>wj* , *n* ∏ *<sup>j</sup>*=<sup>1</sup><sup>1</sup> − -1 − *<sup>β</sup>j*.*<sup>λ</sup>wj* ≥ *n*∏*<sup>j</sup>*=<sup>1</sup><sup>1</sup> − (1 − *<sup>β</sup>*−)*<sup>λ</sup>wj* , 1 − *n* ∏ *<sup>j</sup>*=<sup>1</sup><sup>1</sup> − -1 − *<sup>β</sup>j*.*<sup>λ</sup>wj* ≤ 1 − *n*∏*<sup>j</sup>*=<sup>1</sup><sup>1</sup> − (1 − *<sup>β</sup>*−)*<sup>λ</sup>wj* , 1 − *n*∏*<sup>j</sup>*=<sup>1</sup><sup>1</sup> − -1 − *<sup>β</sup>j*.*<sup>λ</sup>wj*1/*<sup>λ</sup>* ≤ 1 − *n*∏*<sup>j</sup>*=<sup>1</sup><sup>1</sup> − (1 − *<sup>β</sup>*−)*<sup>λ</sup>wj*1/*λ*, 1 − 1 − *n*∏*<sup>j</sup>*=<sup>1</sup><sup>1</sup> − -1 − *<sup>β</sup>j*.*<sup>λ</sup>wj*1/*<sup>λ</sup>* ≥ 1 − 1 − *n*∏*<sup>j</sup>*=<sup>1</sup><sup>1</sup> − (1 − *<sup>β</sup>*−)*<sup>λ</sup>wj*1/*<sup>λ</sup>* = *β*<sup>−</sup>.

similarly, we have

$$1 - \left(1 - \prod\_{j=1}^{n} \left(1 - \left(1 - \beta\_j\right)^{\lambda}\right)^{w\_j}\right)^{1/\lambda} \le 1 - \left(1 - \prod\_{j=1}^{n} \left(1 - \left(1 - \beta^{+}\right)^{\lambda}\right)^{w\_j}\right)^{1/\lambda} = \beta^{+} \cdot \lambda$$

and, as *γ*<sup>−</sup> ≤ *γj* ≤ *<sup>γ</sup>*+, we have

$$\gamma^- \le 1 - \left(1 - \prod\_{j=1}^n \left(1 - \left(1 - \gamma\_j\right)^{\lambda}\right)^{w\_j}\right)^{1/\lambda} \le \gamma^+.$$

let *GPHFWAλ*(*n*1, *<sup>n</sup>*2,..., *<sup>n</sup>n*) = *n* = {{*α*}, {*β*}, {*γ*}}, then

$$s(\hat{n}) = \frac{1 + \frac{1}{l} \sum\_{i=1}^{l} a\_i - \frac{1}{p} \sum\_{i=1}^{p} \beta\_i - \frac{1}{q} \sum\_{i=1}^{q} \gamma\_i}{2} \ge \frac{1 + \frac{1}{l} \sum\_{i=1}^{l} a^- - \frac{1}{p^-} \sum\_{i=1}^{p^-} \beta^+ - \frac{1}{q} \sum\_{i=1}^{q^-} \gamma^+}{2} = s(\hat{n}^-), \quad s(\hat{n}^+) = s(\hat{n}^-)$$

and

$$s(\vec{n}) = \frac{1 + \frac{1}{T}\sum\_{i=1}^{l} a\_i - \frac{1}{p}\sum\_{i=1}^{p} \beta\_i - \frac{1}{q}\sum\_{i=1}^{q} \gamma\_i}{2} \le \frac{1 + \frac{1}{T^+}\sum\_{i=1}^{l^+} a^+ - \frac{1}{p^+}\sum\_{i=1}^{p^+} \beta^- - \frac{1}{q^+}\sum\_{i=1}^{q^+} \gamma^-}{2} = s(\vec{n}^+).$$

we obtain

$$
\widetilde{n}^- \le \text{GPHFVA}\_{\lambda}(\widetilde{n}\_1, \widetilde{n}\_2, \dots, \widetilde{n}\_{\text{tr}}) \le \widetilde{n}^+, \lambda \in (0, \infty).
$$

Similarly, we have

$$
\hat{n}^- \le \text{GPHFVA}\_{\lambda}(\hat{n}\_1, \hat{n}\_2, \dots, \hat{n}\_n) \le \hat{n}^+, \lambda \in (-\infty, 0).
$$

