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U = π(r ) and v = π(r + 1). Let permutation π be obtained from π by swapping the jobs u and v. For a single machine problem, let Cπ(h) and Cπ (h) denote the completion time of the job sequenced in the hth position in permutation π and π , respectively, 1 ≤ h ≤ n. 13) hold. Proof It is convenient to represent permutation π as π = (π1 , u, v, π2 ), where π1 and π2 are subsequences of jobs that precede job u and follow job v in permutation π, respectively. Then, π = (π1 , v, u, π2 ). We present the proof assuming that both sequences π1 and π2 are non-empty; otherwise, the corresponding part of the proof can be skipped.
Let us start again with a polynomially solvable problem 1|| w j C j . Considering its enhancement with several parallel machines, Bruno et al. (1974) show that problem P2|| w j C j with two identical parallel machines is NP-hard in the ordinary sense, while if the weights are equal, there is a polynomial-time algorithm to minimize C j on any number of unrelated parallel machines (see also Sect. 2). Using similar reasoning, a fairly full description of the complexity of the scheduling problems, including the “minimum hard,” “maximum easy,” and open problems, can be derived.
In Sect. 2, a priority rule is derived for problem 1| | w j C j of minimizing the weighted sum of the completion times on a single machine. In Sect. 3, the obtained results are extended to problems Pm| | C j and Qm| | C j of minimizing total completion time on parallel (identical or uniform) machines. A. Strusevich and K. 1 Minimizing a Linear Form Given two arrays a = (a1 , a2 , . . , an ) and b = (b1 , b2 , . . , bn ), and two arbitrary permutations π = (π(1), π(2), . . , π(n)) and σ = (σ(1), σ(2), .
f3-6
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