Concept Design Evaluation of Sustainable Product–Service Systems: A QFD–TOPSIS Integrated Framework with Basic Uncertain Linguistic Information

Abstract

The product–service system (PSS) is a strategic design approach proposed to address sustainability in socio-economic systems amidst rapid industrialization and transition. Evaluating the concept design of a PSS is a crucial and initial step prior to implementation. This study presents an innovative framework for evaluating concept designs of sustainable PSS based on a well-defined evaluation index system via integrating quality function deployment (QFD) and the technique for order preference by similarity to ideal solution (TOPSIS) while accommodating extended basic uncertain linguistic information (EBULI). Specifically, a QFD-based framework is first developed to identify the requirements of various stakeholders and then to establish the multi-dimensional criteria for evaluating sustainable PSS. Then, a House of Quality-based relationship matrix is introduced to determine the weights of criteria more accurately. Further, an adaptive consensus-reaching process method based on an expert weighting optimization model is proposed to ensure a collective outputs recognized by multiple involved stakeholders. Finally, an improved EBULI-based TOPSIS method is presented to determine the priority ranking of alternative sustainable PSS concepts. A case study on a car-sharing PSS project demonstrates the viability and effectiveness of the proposed QFD–TOPSIS integrated approach under EBULI settings. The alternative PSS concept design, which demonstrates relatively good performance in criteria of high importance, is selected as the most suitable option. Moreover, relevant comparative and sensitivity analyses reveal that the proposed approach exhibits superiorities in appropriate criteria elicitation, accurate weights determination, and high consensus ranking outputs.

Appendix

Proofs of serveral Theorems and properties

Proof of Theorem 1

Based on the defined basic operational laws (1) and (3), we have $p\left( {{{{\widetilde{b}}}_i}} \right) = \left\langle {{g^{ - 1}}\left( {1 - {{(1 - {\alpha _i})}^p}} \right) ;{c_i}} \right\rangle$ and

$p\left( {{{{\widetilde{b}}}_i}} \right) \oplus p\left( {{{\widetilde{b}}_j}} \right) = \left\langle {{g^{ - 1}}\left( {1 - {{(1 - {\alpha _i})}^p}} \right) ;{c_i}} \right\rangle \oplus \left\langle {{g^{ - 1}}\left( {1 - {{(1 - {\alpha _j})}^p}} \right) ;{c_j}} \right\rangle = \left\langle {{g^{ - 1}}\left( {1 - {{(1 - {\alpha _i})}^p}{{(1 - {\alpha _j})}^q}} \right) ;\frac{1}{2}\left( {{c_i} + {c_j}} \right) } \right\rangle$.

Then,

${\prod\limits_{{{\begin{array}{c} \scriptstyle i,j = 1\\ \scriptstyle i \ne j \end{array}}}}^n {\left( {p({{{\widetilde{b}}}_i}) \oplus q({{\widetilde{b}}_j})} \right) } ^{\frac{1}{{n(n - 1)}}}} = \left\langle {{g^{ - 1}}\left( {\prod \limits _{\begin{array}{c} \scriptstyle i,j = 1\\ \scriptstyle i \ne j \end{array}}^n {{{\left( {1 - {{(1 - {\alpha _i})}^p}{{(1 - {\alpha _j})}^q}} \right) }^{\frac{1}{{n(n - 1)}}}}} } \right) ;\frac{1}{n}\sum \limits _{i = 1}^n {{c_i}} } \right\rangle$.

Therefore,

$\mathsf{{EBULIGBM}}{^{p,q}}\left( {{{{\tilde{b}}}_1},{{{\tilde{b}}}_2}, \cdots ,{{{\tilde{b}}}_n}} \right) = \frac{1}{{p + q}}\left( {{{\prod \limits _{\begin{array}{c} \scriptstyle i,j = 1\\ \scriptstyle i \ne j \end{array}}^n {\left( {p({{{\widetilde{b}}}_i}) \oplus q({{{\widetilde{b}}}_j})} \right) } }^{\frac{1}{{n(n - 1)}}}}} \right)$ $= \left\langle {{g^{- 1}} \left( 1 - \left( {1 - \left( {\prod \limits _{{\begin{array}{c} \scriptstyle i,j = 1\\ \scriptstyle i \ne j \end{array}}}^1 {{{(1 - {{(1 - {\alpha _i})}^p}{{(1 - {\alpha _j})}^q})}^{\frac{1}{{n(n - 1)}}}}} } \right) } \right) ^{\frac{1}{{p + q}}} \right) ;\frac{1}{n}\sum \limits _{i = 1}^n {{c_i}} } \right\rangle$

The proof is completed.

$\square$

Proof of Property 1

As ${{\widetilde{b}}_i} = {\widetilde{b}}$ for all $i = 1,2, \ldots ,n$, then we have $\mathsf{{EBULIGBM}}{^{p,q}}\left( {\widetilde{{b_1}},{{{\widetilde{b}}}_2}, \cdots ,{{{\widetilde{b}}}_n}} \right) = {\mathsf{{EBULIGBM}}^{p,q}}\left( {{\widetilde{b}},{\widetilde{b}}, \cdots ,{\widetilde{b}}} \right)$.

Furtherly, based on the Definition 7, we have $\mathsf{{EBULIGBM}}{^{p,q}}\left( {\widetilde{{b_1}},{{\widetilde{b}}_2}, \cdots ,{{{\widetilde{b}}}_n}} \right) = \frac{1}{{p + q}} {\prod \limits _{{\begin{array}{c} \scriptstyle i,j = 1\\ \scriptstyle i \ne j \end{array}}}^n {\left( {p({{{\widetilde{b}}}_i}) \oplus q({{{\widetilde{b}}}_j})} \right) } ^{\frac{1}{{n(n - 1)}}}} = \frac{1}{{p +q}}{\left( {(p + q)({\widetilde{b}})} \right) ^{\frac{{n(n - 1)}}{{n(n - 1)}}}} = {\widetilde{b}}$

Thus, ${\mathsf{{EBULIGBM}}^{p,q}}\left( {{\widetilde{b}},{\widetilde{b}}, \cdots ,{\widetilde{b}}} \right) = \mathsf{{EBULIGBM}}{^{p,q}}\left( {\widetilde{{b_1}},{{{\widetilde{b}}}_2}, \cdots ,{{{\widetilde{b}}}_n}} \right) = {\widetilde{b}}$

The proof is completed.

Proof of Property 2

Based on the Theorem 1, we have

$\begin{array}{l} \mathsf{{EBULIGBM}}{^{p,q}}\left( {{\tilde{b}}_1^1,\tilde{b}_2^1, \cdots ,{\tilde{b}}_n^1} \right) = \frac{1}{{p + q}}{\prod \limits _{\begin{array}{c} \scriptstyle i,j = 1\\ \scriptstyle i \ne j \end{array}}^n {\left( {p({\widetilde{b}}_i^1) \oplus q({\widetilde{b}}_j^1)} \right) } ^{\frac{1}{{n(n - 1)}}}}\\ \mathrm{{ }} = \left\langle {{g^{ - 1}}\left( {1 - {{\left( {1 - \left( {\prod \limits _{{\begin{array}{c} \scriptstyle i,j = 1\\ \scriptstyle i \ne j \end{array}}}^n {{{(1 - {{(1 - \alpha _i^1)}^p}{{(1 - \alpha _j^1)}^q})}^{\frac{1}{{n(n - 1)}}}}} } \right) } \right) }^{\frac{1}{{p + q}}}}} \right) ;\frac{1}{n}\sum \limits _{i = 1}^n {c_i^1} } \right\rangle \end{array}$ and

$\begin{array}{l} {\mathsf{{EBULIGBM}}^{p,q}}\left( {{\tilde{b}}_1^2,\tilde{b}_2^2, \cdots ,{\tilde{b}}_n^2} \right) = \frac{1}{{p + q}}{\prod \limits _{\begin{array}{c} \scriptstyle i,j = 1\\ \scriptstyle i \ne j \end{array}}}^n {\left( {p({\tilde{b}}_i^2) \oplus q({\tilde{b}}_j^2)} \right) } ^{\frac{1}{{n(n - 1)}}}= \left\langle {{g^{ - 1}}\left( {1 - {{\left( {1 - \left( {\prod \limits _{\begin{array}{c} \scriptstyle i,j = 1\\ \scriptstyle i \ne j \end{array}}}^n {{{(1 - {{(1 - \alpha _i^2)}^p}{{(1 - \alpha _j^2)}^q})}^{\frac{n}{{n(n - 1)}}}}} \right) } \right) }^{\frac{1}{{p + q}}}}} \right) ;\frac{1}{n}\sum \limits _{i = 1}^n {c_i^2} } \right\rangle \end{array}$.

Since ${\widetilde{b}}_i^1 \le {\widetilde{b}}_i^2$ for all $i = 1,2, \ldots ,n$, and ${\tilde{b}}_i^1 \le {\tilde{b}}_i^2 \Leftrightarrow \left( {\alpha _i^1c_i^1 \le \alpha _i^2c_i^2} \right) \vee \left( {(\alpha _i^1c_i^1 = \alpha _i^2c_i^2) \wedge (\alpha _i^1 \le \alpha _i^2)} \right)$.

Therefore, ${\mathsf{{EBULIGBM}}^{p,q}}\left( {{\tilde{b}}_1^1,\tilde{b}_2^1, \cdots ,{\tilde{b}}_n^1} \right) \le {\mathsf{{EBULIGBM}}^{p,q}}\left( {{\tilde{b}}_1^2,{\tilde{b}}_2^2, \cdots ,{\tilde{b}}_n^2} \right)$

The proof is completed. $\square$