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NTU Management Review Vol. 33 No. 3 Dec. 2023




                   Kang et al. (2023) proposes a way to transform the benefit evaluation problem
               to a maximum flow problem. We now show how to extend this method to incorporate
               time dependency, which is not considered in his work. Given a construction plan y, we
               construct a directed acyclic graph whose structure is similar to that in Figure 1. For each

               customer location i, we add a customer node C into the graph. A source node S is created
                                                         i
               and connected to customer node C  with capacity d  for all i ∈ I. For facility j, if under
                                              i
                                                             i
               the given construction plan it is built at scale level k (i.e., y =1), we add a node F  for
                                                                      jk
                                                                                          jt
               each of activity session t. A destination node D is created and linked from node F  with
                                                                                         jt
               capacity q . Finally, a link from node C  to node F  is added with infinite capacity if p > 0.
                                                           jt
                                                                                          ijt
                        jk
                                                 i
               Figure 1 is an example for three customers, two activity sessions, and a construction plan
               that builds three facilities. According to the way we connect C  and F , we find customer
                                                                             jt
                                                                       i
               1 is unwilling to visit facility 2 in activity session 1. Once the graph is constructed, its
               maximum flow may be solved. For the maximum flow instance made from a construction
               plan y, we denote the maximized flow value as w(y).
                   Note that the information about preference levels p  is largely omitted in the
                                                                     ijt
               constructed maximum flow problem: It only matters whether p > 0 or not. It is thus not
                                                                       ijt
               surprising that the solutions to the two problems are not the same. Nevertheless, as for the
               benefit evaluation problem all we need is the objective value, the constructed maximum
               flow problem is enough. We now show in Theorem 1 that the maximized flow value of the
               maximum flow problem constructed given a construction plan y is always identical to z(y),

               the objective value of the benefit evaluation problem.
                   Theorem 1: For any construction plan y, we have z(y)=w(y).



                   Proof. Given any construction plan y, each flow on the graph we constructed is
               bounded by the demand of customers and capacity of facilities. Therefore, the capacity
               constraints in benefit evaluation problem are obeyed. For every node C , if the maximum
                                                                               i
               flow flows out according to the preference levels in equilibrium, the customers’ decisions
               in graph are exactly the same as that in the benefit evaluation problem. It turns out

               z(y)=w(y). Based on this maximum flow instance, assume that there is at least one flow
               flowing out some node C  changes to the less preferred edge. If this change does not
                                      i
               decrease w(y), i.e., does not make any edge out of capacity, z(y)=w(y). If it does, this new

               instance is impossible to be an outcome of the maximum flow problem. In conclusion,


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