An accelerated approach for solving fuzzy relation equations with a linear objective function

In this paper, we have studied a linear optimization problem subject to a system of fuzzy relation equations and presented a procedure to nd an optimal solution. Because of the non-convexity nature of its feasible domain, we tend to believe that there is no polynomial-time algorithm for this problem. The best we can do here is that, after analyzing the properties of its feasible domain, we convert the original problem into a 0-1 integer programming problem, then apply the well known branch-and-bound method to nd one solution. The question of how to generate the whole optimal solution set is yet to be investigated. 10 the objective value Z. For example, at node 3, we compute Z c 0 5 b 1 + c 0 4 b 2 = 0:6. Hence any choice associated with node 3 must have Z 0:6. Similar reasoning shows that Z 3:0 for node 4. Again, we do not have any reason to exclude any of nodes 3 and 4 from consideration, so we need to branch on one node. The jump-tracking technique directs us to branch node 3. Any choice associated with node 3 must satisfy either x 13 = 1 or x 23 = 1, since I 3 = f1; 2g. This yields nodes 5 and 6 in Figure 1. Same reasoning as before, any choice associated with node 5 must have Z 2:1 and any choice associated with node 6 must have Z 2:6. Of course, we are interested in node 5. To branch furthermore on node 5, any choice associated with node 5 must have x 14 Note that node 7 corresponds to the sequence 5-4-1-1. This sequence leads a value of Z = 2:1. Therefore, node 7 is a feasible sequence which may be viewed as a candidate solution with Z = 2:1. Because the Z value for nodes from 8 to 11 can not be lower than 2.1, these nodes can be eliminated from further consideration. Similarly, node 4 (Z 3:0), node 6 (Z 2:6) can be eliminated. However, node 1 can not be eliminated yet, because it is still possible for node 1 to yield a sequence having Z < 2:1. Hence we now branch on node 1. Since I 2 = f4; 6g, any sequence associated with node 1 must have either x 42 = 1 or x 62 = 1. Correspondingly , we create nodes 12 and …