A self-adaptation algorithm for quay crane scheduling at a container terminal

Abstract

Quay cranes scheduling at container terminals is a fertile area of study that is attracting researchers as well as practitioners in different parts of the world, especially in OR and artificial intelligence (AI). This process efficiency may affect the accomplishment and the competitive merits. As such, four local search algorithms (LSs) are utilized in the current work. These are hill climbing (HC), simulated annealing (SA), tabu search (TS), and iterated local search (ILS). The results obtained demonstrated that none of these LSs succeeded to achieve good results on all instances. This is because different QCSP instances have different characteristics with NP-hardness nature. Therefore, it is difficult to define which LS can yield the best outcomes for all instances. Consequently, appropriate LS selection should be governed by the type of problem and search status. The current work proposes to achieve this, the self-adaptation heuristic (self-H). The self-H is composed of two separate stages: The upper (LS-controller) and the lower (QCSP-solver). The LS-controller embeds an adaptive selection mechanism to adaptively select which LS is to be adopted by the QCSP-solver to solve the given problem. The results revealed that the self-H outperformed others as it attained better results over most instances and competitive results.

Solving the trapezoidal fuzzy assignment problem using the novel Dhouib-Matrix-AP1 heuristic

“…Figure 10 depicts the generated solution using Python programming language. The total trapezoidal fuzzy assignment cost is 𝐹 ̃= (17,22,37,49). This solution is computed by: 𝐹 ̃= (3,4,6,9) + (2,3,5,7) + (6,7,9,10) + (4,5,7,9) + (2,3,10,14) = (17,22,37,49) Thus, for the decision maker, the total allocation cost will lie at [56] with a 100% level of satisfaction (lying at [61].…”

“…The total trapezoidal fuzzy assignment cost is 𝐹 ̃= (17,22,37,49). This solution is computed by: 𝐹 ̃= (3,4,6,9) + (2,3,5,7) + (6,7,9,10) + (4,5,7,9) + (2,3,10,14) = (17,22,37,49) Thus, for the decision maker, the total allocation cost will lie at [56] with a 100% level of satisfaction (lying at [61]. Also, increasing and decreasing levels for the remaining values of the minimal cost as shown in (3).…”