cases, the run-time of the saturation algorithm can be improved. Instead of finding all saturated objectives, we can break out of the inner loop after finding one saturated objective; the algorithm still stops after at most n iterations, and may reduce the number of linear programs (P2) we need to solve. Furthermore, instead of looping over all objectives to find a saturated one, the algorithm can find a saturated objective using the dual problem of (P1). In some cases, the dual variables are given as a byproduct of solving (P1), for example, when the objectives and constraints are linear and the solver is the simplex algorithm. In this case, (P2) is not needed at all, and the run-time of the algorithm is at most times the run-time of the solver of (P1). All these variants work only for convex problems. For non-convex problems, there might be no saturated objective, so the algorithm might not stop. Ordered Outcomes Algorithm for general problems The Ordered Outcomes Algorithm works in arbitrary domains (not necessarily convex). It was developed by Ogryczak and Śliwiński and also presented in the context of telecommunication networks by Ogryczak, Pioro and Tomaszewski, and in the context of location problems by Ogryczak. The algorithm reduces lexmaxmin optimization to the easier problem of lexicographic optimization. Lexicographic optimization can be done with a simple sequential algorithm, which solves at most n linear programs. The reduction starts with the following presentation of lexmaxmin: This problem cannot be solved as-is, because (the t-th smallest value in ) is not a simple function of x. The problem (L1) is equivalent to the following problem, where the sum of the t smallest values in : For every t, the sum can be computed as the optimal value to the following problem, with n+1 auxiliary variables (an unbounded variable , and non-negative variables for all j in 1,...,n), and n additional constraints:Proof. Let us compute the values of the auxiliary variables in the optimal solution. For all j, should be at least as large as both 0 and , and subject to this, it should be minimized, since it appears in the objective with a minus sign. Therefore, . So the objective can be written as: . For any k between 0 and n, if is larger than the smallest k objective values (that is, ), then the sum at the right-hand side contains exactly k positive