Comparable functional results, fewer postoperative complications, and improved quality of life can be achieved by reducing the amount of PP in meshes used for laparoscopic hernia repair by TAPP procedure.

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Abstract. Online makespan minimization is a classical problem in which a sequence of jobs σ = J1, . . . , Jn has to be scheduled on m identical parallel machines so as to minimize the maximum completion time of any job. In this paper we investigate the problem with an essentially new model of resource augmentation. More specifically, an online algorithm is allowed to build several schedules in parallel while processing σ. At the end of the scheduling process the best schedule is selected. This model can be viewed as providing an online algorithm with extra space, which is invested to maintain multiple solutions. The setting is of particular interest in parallel processing environments where each processor can maintain a single or a small set of solutions. As a main result we develop a (4 3 + ε)-competitive algorithm, for any 0 < ε ≤ 1, that uses a constant number of schedules. The constant is 1 ε O(log(1 ε)) . We also give a (1 + ε)-competitive algorithm, for any 0 < ε ≤ 1, that builds a polynomial number of (m ε) O(log(1 ε) ε) schedules. This value depends on m but is independent of the input σ. The performance guarantees are nearly best possible. We show that any algorithm that achieves a competitiveness smaller than 4 3 must construct Ω(m) schedules. Our algorithms make use of novel guessing schemes that (1) predict the optimum makespan of a job sequence σ to within a factor of 1+ε and (2) guess the job processing times and their frequencies in σ. In (2) we have to sparsify the universe of all guesses so as to reduce the number of schedules to a constant. The competitive ratios achieved using parallel schedules are considerably smaller than those in the standard problem without resource augmentation. Furthermore they are at least as good and in most cases better than the ratios obtained with other means of resource augmentation for makespan minimization.

Makespan minimization on identical parallel machines is a classical scheduling problem. We consider the online scenario where a sequence of n jobs has to be scheduled non-preemptively on m machines so as to minimize the maximum completion time of any job. The best competitive ratio that can be achieved by deterministic online algorithms is in the range [1. 88, 1.9201]. Currently no randomized online algorithm with a smaller competitiveness is known, for general m.In this paper we explore the power of job migration, i.e. an online scheduler is allowed to perform a limited number of job reassignments. Migration is a common technique used in theory and practice to balance load in parallel processing environments. As our main result we settle the performance that can be achieved by deterministic online algorithms. We develop an algorithm that is α m -competitive, for any m ≥ 2, where α m is the solution of a certain equation. For m = 2, α 2 = 4/3 and lim m→∞ α m = W −1 (−1/e 2 )/(1 + W −1 (−1/e 2 )) ≈ 1.4659. Here W −1 is the lower branch of the Lambert W function. For m ≥ 11, the algorithm uses at most 7m migration operations. For smaller m, 8m to 10m operations may be performed. We complement this result by a matching lower bound: No online algorithm that uses o(n) job migrations can achieve a competitive ratio smaller than α m . We finally trade performance for migrations. We give a family of algorithms that is c-competitive, for any 5/3 ≤ c ≤ 2. For c = 5/3, the strategy uses at most 4m job migrations. For c = 1.75, at most 2.5m migrations are used. Algorithm ALG(α m ):Job arrival phase. Each J t , 1 ≤ t ≤ n, is scheduled as follows.• J t is small: Assign J t to an M j with ℓ s (j, t) ≤ β(j)L * t . • J t is large: Assign J t to a least loaded machine. Lemma 11In the job removal step ALG(1.75) removes at most four jobs from each machine M j ∈ A.Proof. We show that, for any M j ∈ A, it suffices to remove at most four jobs from M j such that the resulting load is upper bounded by 0.75L.First assume that ℓ s (j, n+1) ≤ 0.75L. Then it suffices to remove all jobs that are large at time n+1. Each such job has a processing time greater than 0.5L and was large at the time it was assigned to M j . Consider the last time when such a job was assigned to M j . At that time M j had a load of at most 1.25L and hence could contain no more than two jobs of processing time greater than 0.5L. Thus at time n + 1 machine M j contains at most three of these large jobs.Next assume ℓ s (j, n + 1) > 0.75L. If ℓ s (j, n) ≤ 0.75L n , then J n is assigned to M j because L = L n . Hence it suffices to remove J n and, as shown in the last paragraph, three additional jobs of processing time greater than 0.5L n = 0.5L.We concentrate on the case that ℓ s (j, n + 1) > 0.75L and ℓ s (j, n) > 0.75L n . Let t * be the earliest time such that ℓ s (j, t) > 0.75L t holds for all times t ≥ t * . We partition the jobs that reside on M j at time n + 1 into three sets. Set T 1 (set T 2 ) contains those jobs that were assigned to M j at or before time t * −...

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