New method for solving reviewer assignment problem using type-2 fuzzy sets and fuzzy functions

Abstract

Reviewer Assignment Problem (RAP) is one of the cardinal problems in Government Funding agencies where the expertise level of the referee reviewing a proposal needs to be optimised to guarantee the selection of good R&D projects. Although many solutions have been proposed for RAP in the past, none of them deals with the inherent imprecision associated with the problem. For instance, it is not possible to determine the “exact expertise level” of a particular reviewer in a particular domain. In this paper, we propose a novel approach for assigning reviewers to proposals. To calculate the expertise of a reviewer in a particular domain, we create a type-2 fuzzy set by assigning relevant weights to the various factors that affect the expertise of the reviewer in that domain. We also create a fuzzy set of the proposal by selecting three keywords that best represent the proposal. We then use a fuzzy functions based equality operator to compute the equality of the type-2 fuzzy set of experts and the fuzzy set of proposal keywords, which is then subjected to a set of relevant constraints to optimize the solution. We consider the four important aspects: workload balancing of reviewers, avoiding Conflicts of Interest, considering individual preferences by incorporating bidding and mapping multiple keywords of a proposal. As an extension to this approach, we further consider the relative importance of each keyword with respect to the submitted proposal by using representative percentage weights to create the FUZZY sets which represent the keywords. Hence, we propose an integrated solution based on the strong mathematical foundation of fuzzy logic, comprised of all the different aspects of expertise modeling and reviewer assignment. An Expert System has also been developed for the same.

Appendix: Example for calculation of fuzzy equality of 2 type-2 fuzzy sets

Let c ₁,c ₂∈I ^domC be fuzzy sets.

Let domC={d ₁,d ₂,d ₃,d ₄,d ₅}, let the membership function of c ₁ and c ₂ be given as follows

$$\mu_{{c}_{1}} ( x ) = \left \{ \begin{array}{l@{\quad}l} 1 & \mbox{if}\ x = d _{1},\ d _{1} \in{\operatorname{dom}C} \\0.8 & \mbox{if}\ x = d _{2},\ d _{2} \in{\operatorname{dom}C} \\0 & \mbox{if}\ x = d _{3},\ d _{3} \in{\operatorname{dom}C} \\0.2 & \mbox{if}\ x = d _{4},\ d _{4} \in{\operatorname{dom}C} \\0.7 & \mbox{if}\ x = d _{5},\ d _{5} \in{\operatorname{dom}C} \end{array} \right . $$

$$\mu_{{c}_{2}} ( x ) = \left \{ \begin{array}{l@{\quad}l} 0.5 & \mbox{if}\ x = d _{1},\ d _{1} \in{\operatorname{dom}C} \\1 & \mbox{if}\ x = d _{2},\ d _{2} \in{\operatorname{dom}C} \\0 & \mbox{if}\ x = d _{3},\ d _{3} \in{\operatorname{dom}C} \\0.4 & \mbox{if}\ x = d _{4},\ d _{4} \in{\operatorname{dom}C} \\0.6 & \mbox{if}\ x = d _{4},\ d _{4} \in{\operatorname{dom}C} \end{array} \right . $$

Now we use the equality operator discussed in Sect. 2.2 to calculate the equality for c ₁ and c ₂ i.e. E _domC (c ₁,c ₂) as:

$$I(c _{1}, c _{2} )= \bigl\{x \in {\operatorname{dom}}C: \mu_{{c}_{1}}(x ) \leq\mu _{{c}_{2}}(x) \bigr\} $$

which is calculated as for

$$x= d _{1}:\mu_{{c}_{1}} (d _{1} ) = 1 \quad \mbox{and} \quad\mu_{{c}_{2}} (d _{1} ) =0.5 $$

So

$$\mu_{{c}_{1}} (x) \nleq\mu_{c_{2}} (x) \quad\mbox{for}\ x = d _{1} $$

For

$$x= d _{2}:\mu_{{c}_{1}} (d _{2} ) = 0.8 \quad \mbox{and} \quad \mu_{{c}_{2}} (d _{2} ) = 1 $$

So

$$\mu_{{c}_{1}} (x ) \leq\mu_{{c}_{2}} (x) \quad\mbox{for}\ x = d _{2} $$

Similarly,

Thus

Now

For

We can calculate similar values for x=d ₂, x=d ₃, x=d ₄, x=d ₅.

Therefore

$$\bigl[{\widetilde{\subseteq}}(c _{1}, c _{2} ) \bigr] = 0.5 \wedge1 \wedge1 \wedge0.8 \wedge0.6 = 0.5 $$

Similarly,

$$\bigl[{\widetilde{\subseteq}}(c _{2}, c _{1} ) \bigr] = 0.6 $$

Now by the definition of

$$\bigl\{ \chi_{I ( c_{1}, c_{2} )} (x) \bigr\} = \left \{ \begin{array}{l@{\quad}l} 1 & \mbox{if}\ c _{1} \subseteq c _{2} \\0 & \mbox{if}\ c _{1} \not\subset c _{2} \end{array} \right . $$

Therefore

$$\bigwedge_{\mathrm{dom}C} \bigl\{ \chi_{I ( c_{1}, c_{2} )} ( x ) \bigr\} = \bigwedge_{\mathrm{dom}C} \bigl\{\chi_{ \{ x \in {\operatorname{dom}C}: \mu_{c _{1}} ( x ) \leq\mu_{c _{2}} ( x ) \} } (x) \bigr\} $$

which is given by

Similarly

$$\bigwedge_{\mathrm{dom}C} \bigl\{ \chi_{I ( c_{1}, c_{2} )} \ ( x ) \bigr\} = 0 $$

Hence

Similarly

So

$$E _{\mathrm{dom}C}(c _{1}, c _{2} ) = 0.5 \wedge0.6 = 0.5 $$

Hence the fuzzy equality between sets c ₁ and c ₂=0.5.