In this section, we will return to the F (1, 2)|comp, s-batch|Cmax problem without the assumption of two fixed sequences. Without the assumption of fixed sequences here, the general problem of differentiation flowshop to minimize makespan, F (1, 2)||Cmax, was proven to be strongly NP-hard by a reduction from 3-partition by Herrmann and Lee (1992). From the existing result, it is obvious that F (1, 2)|comp, s-batch|Cmax is strongly NP-hard even without any batching concern.
The NP-hardness indicates that it is very unlikely to design a polynomial time algo-rithm for producing optimal solutions. Branch and bound algoalgo-rithms are one of the exact methods which is widely adopted for tackling hard optimization problems. Effective lower bounds, used to pruned off non-promising solutions, are crucial to the efficiency of branch and bound algorithms. In the following, we will develop a lower bound based upon the result of Theorem 3.1.
For any job set I ∪ J to be processed, we can derive another job set I′ ∪ J′ by rearranging the operations of the jobs to create a data set called ideal data set. In makespan minimization of a two-machine flowshop problem, a job is preferred to be processed first if its processing time on stage-one is shorter and its processing time on stage-two is longer. A data set is ideal if and only if the data set contains jobs where a job with a shorter processing time on stage-one has a longer processing time on stage-two. In the following, we create such an ideal data set I′∪ J′ from the given job set I ∪ J . For all k = 1, 2, . . . , n1, job Ik′ in I′ is defined by two parameters:
1. pIk′,1= pI(k),1, i.e., the k-th smallest element among pI1,1, pI2,1, . . . , pIn1,1. 2. pIk′,2= pI(k),2, i.e., the k-th largest element among pI1,2, pI2,2, . . . , pIn1,2.
In a similar way, job J′ in J′ is defined by pJ ,1 and pJ ,2 for all k = 1, 2, . . . , n2.
We index the jobs in I′ and J′ in non-decreasing order of their processing time on ma-chine M0 and denote the sequence of I′ (respectively, J′) as σI′ (respectively, σJ′).
Lemma 3.2. There is an optimal schedule of the instance I′ ∪ J′ where jobs of I′ are sequenced by σI′ and jobs of J′ are sequenced by σJ′.
Proof. Without loss of generality, we assume that the processing times of all opera-tions are distinct. Given a schedule of the instance S(I′ ∪ J′), for each pair of jobs
Ik′, Ik+1′ |k = 1, 2, . . . , n1− 1 in S(I′ ∪ J′), we swap their position if pI′k,1 > pIk+1,1′ . Note that if the condition is met, it implies pIk,2′ < pIk+1,2′ . The similar job-interchange technique is applied to job set J′. Since the total idle time on either dedicated machine will not increase after the job interchange, it is clear that the makespan will not increase in the derived schedule. Repeating the job interchange, if necessary, will finally lead to a schedule in which all jobs of I′ are sequenced by σI′ and jobs of J′ are sequenced by σJ′. We complete the proof.
Lemma 3.3. An optimal schedule of the job set I′∪ J′ can be found in O(n2) time.
Proof. To decide how to group jobs into batches, it can be easily done with the Algorithm WBwhich has the time complexity O(n2). Therefore, an optimal schedule of the job set I′∪ J′ can be found in O(n2) time and the lemma follows.
Lemma 3.4. The optimal makespan of the job set I′ ∪ J′ will be no greater than the optimal makespan of the original job set I ∪ J , i.e., Cmax∗ (I′∪ J′) ≤ Cmax∗ (I ∪ J ).
Proof. Consider the schedule S∗(I ∪ J ), i.e., the schedule of job set I ∪ J with optimal makespan or Cmax∗ (I ∪ J ). For each pair of the stage-one (respectively, stage-two) op-erations of jobs {(Ik, Ik+1) |k = 1, 2, . . . , n1 − 1}, we swap their positions if pIk,1 > pIk+1,1
(respectively, pIk,2 < pIk+1,2) and leave their stage-two (respectively, stage-one) opera-tions unaltered in their original posiopera-tions. In the similar way, we apply the operation-interchange technique to the operations of J on both stage-one and stage-two oper-ations. Clearly, the makespan will not increase after we change the positions of the
operations under the specific condition. Therefore, the derived schedule contains jobs of the job set I′ ∪ J′ and has a makespan that is no worse than the previous one in the same batch composition, i.e., Cmax(S(I′ ∪ J′)) ≤ Cmax∗ (I ∪ J ). The optimal makespan of job set I′ ∪ J′ has the minimum makespan among all possible schedules of job set I′ ∪ J′ or Cmax∗ (I′ ∪ J′) ≤ Cmax(S(I′ ∪ J′)). By transitivity, we can see that Cmax∗ (I′ ∪ J′) ≤ Cmax∗ (I ∪ J ) and complete the proof.
In summary, by rearranging the operations in the two given job sets, we will get two data sets, I′ and J′, which constitute an ideal data set. With the two fixed sequences of jobs, the scheduling problem is now equivalent to F (1, 2)|comp, fixed seq, s-batch|Cmax
which can be solved optimally by the algorithm in the previous section (by Theorem 3.1).
Thus we get the minimum makespan of the derived problem, we also obtain a lower bound for the original problem.
Theorem 3.2. A lower bound of F (1, 2)|comp, s-batch|Cmax can be found in O(n2).
Proof. The process of constructing I′ and J′ takes O(n log n) time because of the sorting operations. Interleaving two sequences takes O(n) time, and Algorithm WB takes O(n2) time to group jobs. Therefore, the overall time complexity for obtaining a lower bound of F (1, 2)|comp, s-batch|Cmax is O(n2).
An example is given in the following for illustrating the lower bound calculation for the problem F (1, 2)|comp, s-batch|Cmax. There are four jobs in two types to be scheduled:
type 1, I = {I1, I2} and type 2, J = {J2, J2}. The batch setup time is 1. The processing times of jobs are shown below.
Jobs I1 I2 J1 J2
stage-one pI1,1 = 2 pI2,1 = 5 pJ1,1 = 3 pJ2,1 = 4 stage-two pI1,2 = 3 pI2,2 = 4 pJ1,2 = 2 pJ2,2 = 6
An ideal job set I′∪ J′ can be derived from the given job set I ∪ J by the operation interchange technique. The processing times of jobs in I′∪ J′ are shown below.
Jobs I1′ I2′ J1′ J2′
stage-one pI1,1′ = 2 pI2,1′ = 5 pJ1,1′ = 3 pJ2,1′ = 4 stage-two pI1,2′ = 4 pI2,2′ = 3 pJ1,2′ = 6 pJ2,2′ = 2
Applying Algorithm HL, we can determine the sequence of jobs on machine M0. Let D = 0, we can get the due dates of all jobs as shown below.
The following is the recursive steps of Algorithm WB.
B(1) = min
B(2) = min
Therefore, the lower bound of the optimal makespan of job set I ∪ J is 19, the corre-sponding schedule is ((J1′), (I1′, I2′), (J2′)).