Abstract

The scheduling problem of minimizing total tardiness on a single machine is known to be N P -hard in the ordinary sense. In this paper, we consider the special case of the problem when the processing times p j and the due dates d j of the jobs j , j ∈ N = { 1 , 2 , … , n } , are oppositely ordered: p 1 ≥ p 2 ≥ ⋯ ≥ p n and d 1 ≤ d 2 ≤ ⋯ ≤ d n . It is shown that already this special case is N P -hard in the ordinary sense, too. The set of jobs N is partitioned into k , 1 ≤ k ≤ n , subsets M 1 , M 2 , … , M k , M ν ⋂ M μ = 0̸ for ν ≠ μ , N = M 1 ⋃ M 2 ⋃ ⋯ ⋃ M k , such that max i , j ∈ M ν | d i − d j | ≤ min j ∈ M ν p j for each ν = 1 , 2 , … , k . We propose algorithms which solve the problem: in O ( k n ∑ p j ) time if 1 ≤ k < n ; in O ( n 2 ) time if k = n ; and in O ( n 2 ) time if max i , j ∈ N | d i − d j | ≤ 1 . The polynomial algorithms do neither require the conditions p 1 ≥ p 2 ≥ ⋯ ≥ p n mentioned above nor integer processing times to construct an optimal schedule. Finally, we apply the idea of the presented algorithm for the case k = 1 to the even–odd partition problem.

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