A note on parallel-machine due-window assignment

Abstract

We consider a due-window assignment problem on identical parallel machines, where the jobs have equal processing times and job-dependent earliness-tardiness costs. We would like to determine a ‘due window’ during which the jobs can be completed at no cost and to obtain a job schedule in which the jobs are penalized if they finish before or after the due window. The objective is to minimize the total earliness and tardiness job penalty, plus the cost associated with the size of the due window. We present an algorithm that can solve this problem in O(n ³) time, which is an improvement of the O(n ⁴) solution procedure developed by Mosheiov and Sarig.

Appendix

Proof of Lemma 1

• Property (i) holds because P′(D ₁, D ₂) is a relaxation of P(D ₁+D ₂). For any D′₂⩾D ₂+1, problem P′(0, D′₂) differs from P′(0, D ₂) in that m(D′₂−D ₂) due-window positions are added. Thus, z′(0, D′₂)⩽z′(0, D ₂), which implies the validity of property (ii). In addition, these additional positions have the same unit cost δ/m as the unoccupied position(s) in the optimal solution of P′(0, D ₂). Because the optimal solution of P′(0, D ₂) already contains an unoccupied due-window position, there must exist an optimal solution to P′(0, D′₂) in which none of those m(D′₂−D ₂) newly added positions is occupied. Hence, property (iii) also holds.  □

Proof of Lemma 2

• Consider any integers D ₁, D ₂⩾0 such that D ₁+D ₂⩽⌊n/m⌋. Let D=D ₁+D ₂, and let

  

  be the set of all positions within the due window. P′(D ₁, D ₂) can be reformulated as follows:

  

  s.t. Constraints (2) and (3)

  
  

  In this formulation, the integer constraint ‘X _jr ∈{0, 1}’ has been relaxed to a non-negativity constraint ‘X _jr ⩾0’. This is because P′′(D ₁, D ₂) can be formulated as a minimum cost network flow problem with integer arc capacities and integer supply/demand (see Figure 1), and therefore an optimal solution with integer X _jr values can be obtained by solving P′′(D ₁, D ₂) (Ahuja et al, 1993, p 318). Constraint (A.1) ensures that at least mD ₁ positions within the due window are occupied. As all the positions within the due window have the same unit cost of δ/m, the optimal solution generated by P′′(D ₁, D ₂) can be easily transformed into an optimal solution of P′(D ₁, D ₂) by rearranging the jobs inside the due window so that the first D ₁ positions inside the due window on each machine are all occupied.

  Figure 1

  

  Minimum cost network flow representation of problem P′′(D ₁, D ₂).

  Consider any D ₂=0 ,1 ,… ,⌊n/m⌋−D ₀−1. Note that mD ₀⩽w ₀<m(D ₀+1)⩽m(D ₀+D ₂+1). The optimal solution generated by P′(0, ⌊n/m⌋) can be transformed into an optimal solution of P′(D ₀, ⌊n/m⌋−D ₀) by rearranging the jobs inside the due window so that the first D ₀ positions inside the due window on each machine are all occupied and the last ⌊n/m⌋−D ₀−1 positions inside the due window on each machine are all unoccupied. This solution becomes an optimal solution of P′(D ₀, D ₂+1) once we remove ⌊n/m⌋−D ₀−D ₂−1 unoccupied due-window positions from each machine. This implies that P′(D ₀, D ₂+1) has an optimal solution with w ₀ occupied due-window positions. In other words, P″(D ₀, D ₂+1) has an optimal solution with ∑ _r∈W ∑ _j=1 ⁿ X _jr =w ₀. Hence, if we replace the right-hand side of (A.1) by w ₀ and solve P″(D ₀, D ₂+1), then there exists an optimal solution in which Constraint (A.1) is binding. Because P″(D ₀, D ₂+1) is a linear programme, if we increase the right-hand side of (A.1) from w ₀ to a larger value, then Constraint (A.1) will remain binding at a new optimum (see Murty, 1983, Chapter 8 for a discussion of systematic changes in the right-hand-side parameters of a linear programme). Thus, if the right-hand side of (A.1) is replaced by m(D ₀+1), the constraint will remain binding at an optimum. In other words, there exists an optimal solution to P′(D ₀+1, D ₂) with exactly m(D ₀+1) occupied due-window positions. This solution becomes an optimal solution of P′(D ₀+1, 0) if we remove D ₂ unoccupied due-window positions from each machine. Hence, the optimal value of P′(D ₀+1, 0) is equal to that of P′(D ₀+1, D ₂).  □