5.1. Sufficient Conditions for an Optimal Pair of Job Permutations
In the proofs of several claims, we use a notion of the main machine, which is introduced within the proof of the following theorem.
Theorem 7. Consider the following conditions in Inequalities (7) or (8): If one of the above conditions holds, then any pair of job permutations  is a single-element dominant set  for the problem  with set  of the given jobs.
 Proof.  Let the condition in Inequalities (
7) hold. Then, we consider an arbitrary pair of job permutations 
 with any fixed scenario 
 and show that this pair of job permutations 
 is optimal for the individual deterministic problem 
 with scenario 
p, i.e., 
.
Let  () denote a completion time of all jobs  (jobs ) on machine  (machine ) in the schedule , where  and . For the problem , the maximal completion time of the jobs in schedule  may be calculated as follows: .
Machine  (machine ) is called a main machine for the schedule  if equality  holds (equality  holds, respectively).
For schedule 
, the following equality holds:
          
          where 
 and 
 denote total idle times of machine 
 and machine 
 in the schedule 
, respectively. We next show that, if the condition in Inequalities (
7) holds, then machine 
 is a main machine for schedule 
 and machine 
 has no idle time, i.e., machine 
 is completely filled in the segment 
 for processing jobs from the set 
. At the initial time 
, machine 
 begins to process jobs from the set 
 without idle times until the time moment 
From the first inequality in (
7), we obtain the following relations:
          
Therefore, at the time moment , machine  begins to process jobs from the set  without idle times and we obtain the following equality:  where  and machine  has no idle time. We next show that machine  is a main machine for the schedule . To this end, we consider the following two possible cases.
(a) Let machine  have no idle time.
By summing Inequalities (
7), we obtain the following inequality:
          
Thus, the following relations hold:
          
Hence, machine  is a main machine for the schedule .
(b) Let machine  have an idle time.
An idle time of machine  is only possible if some job  from set  is processed on machine  at the time moment  when this job  could be processed on machine .
Obviously, after the time moment 
 when machine 
 completes all jobs from set 
, machine 
 can process some jobs from set 
 without an idle time. Therefore, the inequality 
 holds and we obtain the following relations:
          
We conclude that, in case 
(b), machine 
 is a main machine for the schedule 
. Thus, if the condition in Inequalities (
7) holds, then machine 
 is a main machine for the schedule 
 and machine 
 has no idle time, i.e., equality 
 holds and machine 
 is completely filled in the segment 
 with processing jobs from the set 
.
Thus, the pair of permutations  is optimal for scenario . Since scenario p was chosen arbitrarily in the set T, we conclude that the pair of job permutations  is a singleton  for the problem  with set  of the given jobs. As a pair of permutations  is an arbitrary pair of job permutations in the set S, any pair of job permutations  is a singleton  for the problem  with job set .
The case when the condition in Inequalities (
8) holds may be analyzed similarly via replacing machine 
 by machine 
 and vice versa. □
 If  conditions of Theorem 7 hold, then in the optimal pair of job permutations  existing for the problem , the orders of jobs from sets  and  may be chosen arbitrarily. Theorem 7 implies the following two corollaries.
Corollary 3. If  the following inequality holds: then set , where  is an arbitrary permutation in set , is a dominant set of schedules for the problem  with set  of the given jobs.
 Proof.  We consider an arbitrary vector  of the job durations and an arbitrary permutation  in the set . The set  contains at least one Johnson’s permutation  for the deterministic problem  with job set  and scenario  (the components of vector  are equal to the corresponding components of vector p). We consider a pair of job permutations  and show that it is an optimal pair of job permutations for the problem  with job set  and scenario p. Without loss of generality, both permutations  and  are ordered in increasing order of the indexes of their jobs.
Similar to the proof of Theorem 7, one can show that, if the condition in Inequalities (
9) holds, then machine 
 processes jobs without idle times and equality 
 holds, where the value of 
 cannot be reduced. If machine 
 has no idle time, we obtain equalities
          
On the other hand, an idle time of machine 
 is only possible if some job 
 from set 
 is processed on machine 
 at the time moment 
 when job 
 could be processed on machine 
. In such a case, the value of 
 is equal to the makespan 
 for the problem 
 with job set 
 and scenario 
. As the permutation 
 is a Johnson’s permutation, the value of 
 cannot be reduced and we obtain the following equalities:
          
Thus, the pair of job permutation  is optimal for the problem  with scenario . The optimal pair of job permutations for the problem  with scenario  belongs to the set . As vector p is an arbitrary vector in the set T, the set  contains an optimal pair of job permutations for all scenarios from set T. Due to Definition 4, the set  is a dominant set of schedules for the problem  with job set . □
 Corollary 4. Consider the following inequality: If  the above inequality holds, then set , where  is an arbitrary permutation in set , is a dominant set of schedules for the problem  with set  of the given jobs.
 This claim may be proven similar to Corollary 3. If the conditions of Corollary 3 (Corollary 4) hold, then the order for processing jobs from set  (set , respectively) in the optimal schedule  for the problem  may be arbitrary. Since the orders of jobs from the sets  and  are fixed in the optimal schedule (Remark 1), we need to determine only orders for processing jobs from set  (set , respectively). To do this, we will consider two uncertain problems  with job set  and with the machine route  and that with job set  and with the opposite machine route .
Lemma 2. If  is a set of permutations from the dominant set for the problem  with job set , then  is a dominant set for the problem  with job set .
 The proof of Lemma 2 and those for other statements in this section are given in 
Appendix A.
Lemma 3. Let  be a set of permutations from the dominant set for the problem  with job set , . Then,  is a dominant set for the problem  with job set .
 The proof of this claim is similar to that for Lemma 2 (see 
Appendix A).
Theorem 8. Let  be a set of permutations from the dominant set for the problem  with job set , and let  be a set of permutations from the dominant set for the problem  with job set . Then,  is a dominant set for the problem  with job set .
 Theorem 9. Let a pair of identical permutations  determine a single-element J-solution for the problem  with job set , and let a pair of identical permutations  determine a single-element J-solution for the problem  with job set . Then, the pairs of permutations  are a single-element dominant set  for the problem  with job set .
 The following claim follows directly from Theorem 9.
Corollary 5. If the conditions of Theorem 9 hold, then there exists a single pair of job permutations, which is an optimal pair of job permutations for the problem  with job set  and any scenario .
 Theorem 9 implies also the following corollary proven in 
Appendix A.
Corollary 6. If the conditions of Theorem 9 hold, then there exists a single pair of job permutations which is a J-solution for the problem  with job set .
 Note that the criterion for a single-element J-solution for the problem  is given in Theorem 2.
  5.2. Precedence Digraphs Determining a Minimal Dominant Set of Schedules
In 
Section 4.2, it is assumed that 
 and 
, i.e., 
. Based on the results presented in 
Section 4.2, we can determine a binary relation 
 for the problem 
 with job set 
 and a binary relation 
 for this problem with job set 
. For job set 
, the binary relation 
 determines the digraph 
 with the vertex set 
 and the arc set 
. For job set 
, the binary relation 
 determines the digraph 
 with the vertex set 
 and the arc set 
.
Let us consider the problem  with job set  and the corresponding digraph  (the same results for the problem  with job set  can be derived in a similar way).
Definition 5. Two jobs,  and , , are called conflict jobs if they are not in the relation , i.e.,  and .
 Due to Definitions 2 and 3, for the conflict jobs 
 and 
, 
, Inequalities (
4) and (
5) do not hold either for the case 
 with 
 or for the case 
 with 
.
Definition 6. The subset  is called a conflict set of jobs if, for any job , either relation  or relation  holds for each job  (provided that any proper subset of the set  does not possess such a property).
 From Definition 6, it follows that the conflict set 
 is a minimal set (with respect to the inclusion). Obviously, there may exist several conflict sets in the set 
. (A conflict set of the jobs 
 can be determined similarly.) Let the strict order 
 for the problem 
 with job set 
 be represented as follows:
        where all jobs between braces are conflict ones and each of these jobs is in relation 
 with any job located outside the brackets in Relation (
10). In such a case, an optimal order for processing jobs from the set 
 is determined as follows: 
.
Due to Theorem 5, we obtain that set  of the permutations generated by the digraph  contains an optimal Johnson’s permutation for each vector  of the durations of jobs from the set . Thus, due to Definition 1, the singleton , where , is a J-solution for the problem  with job set . Analogously, the singleton , where , is a J-solution for the problem  with job set . We can determine a dominant set of schedules for the problem  with job set  as follows: . The following theorems allow us to reduce a dominant set for the problem . We use the following notation: 
Theorem 10. Let the strict order  over set  be determined as follows: . Consider the following inequality: If  the above inequality holds, then set  with  is a dominant set of schedules for the problem  with job set .
 Proof.  We consider an arbitrary vector  of the job durations and an arbitrary permutation  from the set . The set  contains at least one optimal Johnson’s permutation  for the problem  with job set  and vector  of the job durations (components of this vector are equal to the corresponding components of the vector p).
We consider a pair of job permutations 
 and show that it is an optimal pair of job permutations for the problem 
 with set 
 of the jobs and scenario 
p. To this end, we show that the value of 
 cannot be reduced. Indeed, an idle time for machine 
 is only possible if some job 
 from the set 
 is processed on machine 
 at the same time when job 
 could be processed on machine 
. In such a case, 
 is equal to the makespan 
 for the problem 
 with job set 
 and vector 
 of the job durations. As permutation 
 is a Johnson’s permutation, the value of
          
          cannot be reduced. In the beginning of the permutation 
, the jobs of set 
 are arranged in the Johnson’s order. Thus, if machine 
 has an idle time while processing these jobs, this idle time cannot be reduced. From Inequality (
11), it follows that machine 
 has no idle time while processing jobs from the conflict set.
In the end of the permutation , jobs of set  are arranged in Johnson’s order. Therefore, if machine  has an idle time while processing these jobs, this idle time cannot be reduced. Thus, the value of  cannot be reduced by changing the order of jobs in the conflict set.
We obtain the qualities  The pair of job permutations  is optimal for the problem  with scenario . Thus, set  contains an optimal pair of job permutations for the problem  with scenario . As vector p is an arbitrary vector in set T, set  contains an optimal pair of job permutations for each vector from set T. Due to Definition 4, set  is a dominant set of schedules for the problem  with job set . □
 Theorem 11. Let the partial strict order  over set  be determined as follows: . Consider the following inequality:If the above inequality holds for all  then the set , where , is a dominant set for the problem  with job set .  Proof.  We consider an arbitrary scenario  and a pair of job permutations  , where  is a Johnson’s permutation of the jobs from the set  with vector  of the job durations (components of this vector are equal to the corresponding components of vector p). We next show that this pair of job permutations  is optimal for the individual deterministic problem  with scenario p, i.e., .
If  conditions of Theorem 11 hold, then machine  processes jobs from the conflict set  without idle times. At the initial time , machine  begins to process jobs from the permutation  without idle times. Let a time moment  be as follows:  At the time moment , job  is ready for processing on machine .
On the other hand, at the time 
, machine 
 begins to process jobs from the set 
 without idle times and then jobs from the permutation 
. Let 
 denote the first time moment when machine 
 is ready for processing job 
. Obviously, the following inequality holds: 
 From the condition in Inequality (
12) with 
, we obtain inequality 
Therefore, the following relations hold:
          
Machine  processes job  without an idle time between job  and job .
Analogously, using , one can show that machine  processes jobs from the conflict set  without idle times between jobs  and , between jobs  and , and so on to between jobs  and . To end this proof, we have to show that the value of  cannot be reduced.
An idle time for machine  is only possible between some jobs from the set . However, the permutation  is a Johnson’s permutation of the jobs from the set  for the vector  of the job durations. Therefore, the value of  cannot be reduced. On the other hand, in the permutation , all jobs  and all jobs  are arranged in Johnson’s orders. Therefore, if machine  has an idle time while processing these jobs, this idle time cannot be reduced. It is clear that machine  has no idle time while processing jobs from the conflict set. Thus, the value of  cannot be reduced by changing the order of jobs from the conflict set. We obtain the equalities 
It is shown that the pair of job permutations  is optimal for the problem  with vector  of job durations. As vector p is an arbitrary one in set T, the set  contains an optimal pair of job permutations for each scenario from set T. Due to Definition 4, the set  is a dominant set of schedules for the problem  with job set . □
 The proof of the following theorem is given in 
Appendix A.
Theorem 12. Let the partial strict order  over set  have the form . If inequalitieshold for all indexes , then the set , where , is a dominant set of pairs of permutations for the problem  with job set .  Similarly, one can prove sufficient conditions for the existence of an optimal job permutation for the problem  with job set , when the partial strict order  on the set  has the following form: .
To apply Theorems 11 and 12, one can construct a job permutation that satisfies the strict order . Then, one can check the conditions of Theorems 11 and 12 for the constructed permutation. If  the set of jobs  is empty in the constructed permutation, one needs to check conditions of Theorem 12. If the set of jobs  is empty, one needs to check the conditions of Theorem 11. It is needed to construct only one permutation to check Theorem 11 and only one permutation to check Theorem 12.
  5.3. Two Illustrative Examples
Example 1. We consider the uncertain job-shop scheduling problem  with lower and upper bounds of the job durations given in Table 1.  These bounds determine the set T of possible scenarios. In Example 1, jobs , , and  have the machine route ; jobs , , and  have the machine route ; and job  (job , respectively) has to be processed only on machine  (on machine , respectively). Thus, , , , .
We check the conditions of Theorem 7 for a single pair of job permutations, which is optimal for all scenarios 
T. For the given jobs, the condition in Inequalities (
7) of Theorem 7 holds due to the following relations:
Due to Theorem 7, the order of jobs from the set  and the order of jobs from the set  may be arbitrary in the optimal pair of job permutations for the problem  under consideration. Thus, any pair of job permutations  is a single-element dominant set  for Example 1.
Example 2. Let us now consider the problem  with numerical input data given in Table 1 with the following two exceptions:  and .  We check the condition in Inequalities (
7) of Theorem 7 and obtain
        
Thus, the condition of Inequalities (
7) does not hold for Example 2. We check the condition of Inequalities (
8) of Theorem 7 and obtain
        
However, we see that the condition of Equation (
8) does not hold:
From Equation (
14), it follows that the condition of Inequalities (
9) of Corollary 3 does not hold. On the other hand, due to Equation (
15), conditions of Corollary 4 hold. Thus, the order for processing jobs from set 
 in the optimal schedule 
 for the problem 
 may be arbitrary. One can fix permutation 
 with the increasing order of the indexes of their jobs: 
. Since the orders of jobs from the sets 
 and 
 are fixed in the optimal schedule (Remark 1), i.e., 
 and 
, we need to determine the order for processing jobs in set 
. To this end, we consider the problem 
 with job set 
. We see that conditions of Theorem 2 do not hold for the jobs in set 
 since 
, 
, and 
; however the following inequalities hold: 
 and 
.
We next construct the binary relation 
 over set 
 based on Definition 3 and Theorem 1. Due to checking Inequalities (
4) and (
5), we conclude that the inequality in Equation (
5) holds for the pair of jobs 
 and 
. We obtain the relation 
. Analogously, we obtain the relation 
. For the pair of jobs 
 and 
, neither Inequality (
4) nor Inequality (
5) hold. Therefore, the partial strict order 
 over set 
 has the following form: 
. The job set 
 is a conflict set of these jobs (Definition 6).
Let us check whether the sufficient conditions given in 
Section 5.2 hold.
We check the conditions of Theorem 10 for the jobs from set 
. For 
 and 
, we obtain the following equalities: 
 The condition of Theorem 10 does not hold since the following relations hold:
For checking the conditions of Theorem 11, we need to check both permutations of the jobs from set , which satisfy the partial strict order : , where  and .
We consider permutation 
. As in the previous case, 
, 
, 
, and we must consider two inequalities in the condition in Equaiton (
12) with 
 and 
. For 
, we obtain the following:
However, for 
, we obtain
        
Thus, the conditions of Theorem 11 do not hold for permutation .
We consider permutation 
, where 
 and 
. Again, we must test the two inequalities in Equation (
12), where either 
 or 
. For 
, we obtain
        
However, for 
, we obtain
        
Thus, the conditions of Theorem 11 do not hold for permutation .
Note that we do not check the conditions of Theorem 12 since the conflict set of jobs 
 is located at the end of the partial strict order 
. We conclude that none of the proven sufficient conditions are satisfied for a schedule optimality. Thus, there does not exist a pair of permutations of the jobs in set 
 which is optimal for any scenario 
. The  
J-solution 
 for Example 2 consists of the following two pairs of job permutations: 
, where
        
We next show that none of these two pairs of job permutations is optimal for all scenarios  using the following two scenarios:  and  For scenario , only pair of permutations  is optimal with  since . On the other hand, for scenario , only the pair of permutations  is optimal with  since .
Note that the whole set S of the semi-active schedules has the cardinality . Thus, for solving Example 2, one needs to consider only two pairs of job permutations  instead of 36 semi-active schedules.