Multistage Knapsack

January 31, 2019 Β· Declared Dead Β· πŸ› International Symposium on Mathematical Foundations of Computer Science

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Authors Evripidis Bampis, Bruno Escoffier, Alexandre Teiller arXiv ID 1901.11260 Category cs.DS: Data Structures & Algorithms Citations 27 Venue International Symposium on Mathematical Foundations of Computer Science Last Checked 3 months ago
Abstract
Many systems have to be maintained while the underlying constraints, costs and/or profits change over time. Although the state of a system may evolve during time, a non-negligible transition cost is incured for transitioning from one state to another. In order to model such situations, Gupta et al. (ICALP 2014) and Eisenstat et al. (ICALP 2014) introduced a multistage model where the input is a sequence of instances (one for each time step), and the goal is to find a sequence of solutions (one for each time step) that are both (i) near optimal for each time step and (ii) as stable as possible. We focus on the multistage version of the Knapsack problem where we are given a time horizon t=1,2,...,T, and a sequence of knapsack instances I_1,I_2,...,I_T, one for each time step, defined on a set of n objects. In every time step t we have to choose a feasible knapsack S_t of I_t, which gives a knapsack profit. To measure the stability/similarity of two consecutive solutions S_t and S_{t+1}, we identify the objects for which the decision, to be picked or not, remains the same in S_t and S_{t+1}, giving a transition profit. We are asked to produce a sequence of solutions S_1,S_2,...,S_T so that the total knapsack profit plus the overall transition profit is maximized. We propose a PTAS for the Multistage Knapsack problem. Then, we prove that there is no FPTAS for the problem even in the case where T=2, unless P=NP. Furthermore, we give a pseudopolynomial time algorithm for the case where the number of steps is bounded by a fixed constant and we show that otherwise the problem remains NP-hard even in the case where all the weights, profits and capacities are 0 or 1.
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