To solve the TMP optimally using problem-specific structural properties, an IE method is developed. Unlike the MIP model, which relies on a general-purpose solver, the IE method explicitly explores the solution space by constructing feasible job sequences. Each node in the enumeration tree represents a partial sequence of selected jobs, and the method systematically extends these sequences while checking feasibility with respect to time window constraints. This framework serves as the basis for incorporating additional techniques, including midpoint-based ordering for guiding the search and dominance rules derived from the partial ordering (⇒) to prune the search tree, thereby improving computational efficiency.
This section presents an IE procedure using a depth-first search (DFS) scheme to identify an optimal solution. Let denote the level in the enumeration tree. A node at level represents a partial solution consisting of scheduled jobs. The sequence of these jobs is defined by the path from the root node to the current node. By appending a not-yet-selected job to the current partial solution, a new sequence is generated, which corresponds to a child node at level + 1 in the tree.
By developing heuristic sorting, dominance rules, and an LP relaxation of subproblems, each discussed in detail below, this study effectively reduces the number of nodes generated in the enumeration tree, thereby significantly decreasing computation time.
Lemma 1 indicates that an earlier available starting time provides greater flexibility for scheduling jobs, and therefore cannot lead to a worse optimal objective value under the same job set.
Lemma 2 indicates that enlarging the set of available jobs cannot reduce the optimal objective value. With more candidate jobs to choose from, the decision maker has greater flexibility, and therefore the optimal objective value cannot be worse than that obtained from a subset of jobs. Together, Lemmas 1 and 2 show that increasing scheduling flexibility, either through earlier start times or larger job sets, cannot reduce the optimal objective value.
Based on the assumption that a selected job must be fully processed within its time window, Dominance Rule 1 follows directly.
This rule eliminates jobs that cannot be feasibly scheduled after the current partial sequence. If the earliest possible start time of job , given by , already leads to a completion time beyond its due date, then job cannot be included in any feasible extension of the current solution. Therefore, such nodes can be safely pruned without losing optimality.
4.3.1. Methods Based on Ordering Relations
This subsection utilizes two ordering relations: a partial ordering for deriving dominance rules, and a midpoint-based ordering for guiding job sequencing.
Definition 1. (Relation operator ⇒) For two jobs , the relation holds if and only if and .
According to this definition, jobs means that the time window of job is no later than that of job ; that is, both the ready date and the due date of job are less than or equal to those of job .
Definition 2. (Equivalence under relation ⇒) For two jobs , they are considered equivalent under the relation if and only if and .
This definition identifies jobs with identical time windows. Such jobs are indistinguishable with respect to the relation , and therefore can be treated equivalently when sorting the job set. Such jobs have identical scheduling flexibility and do not impose any ordering preference under the relation .
Definition 3. (Equivalence notation ⇔) If two jobs are equivalent under the relation , they are denoted as .
Lemma 3. For any two jobs , the relation holds if and only if both and .
Proof. (if) Suppose and , then , , , and . Hence, we have and , which implies jobs and are equivalent under , i.e., .
(only if) If , then by Definition 2, and . Therefore, both and hold. □
This lemma shows that the equivalence relation corresponds to a mutual relation under .
Lemma 4. The relation defined on the set is a partial ordering.
Proof. A partial ordering has to satisfy reflexivity, transitivity, and antisymmetry. reflexivity: For any job , since and , we have . transitivity: For any jobs , if and , then and . Hence, and , which implies . anti-symmetry: For any jobs , if and , then and . Therefore, and are equivalent under ; i.e., . □
The job set , together with the relation , forms a partially ordered set (or poset) denoted as . Two jobs are called incomparable if neither nor holds. This occurs when one job has an earlier ready date but a later due date than the other, i.e., when ( and ) or ( and ). In such cases, neither job dominates the other under this relation. A poset becomes a total ordering if no pair of elements is incomparable. However, in the problem considered in this study, incomparable job pairs may exist. Therefore, the poset is not a total ordering. This partial ordering is used to support the development of dominance rules.
Definition 4. (Relation operator →) For two jobs , the relation holds if and only if .
This condition is equivalent to comparing the midpoints of the time windows. Specifically, it implies that , i.e., the midpoint of the time window of job is less than or equal to that of job .
Definition 5. (Equivalence under relation →) Two jobs are considered equivalent under the relation if and only if .
This condition is equivalent to , i.e., the midpoints of the time windows of jobs and are identical. Such jobs are indistinguishable under the ordering defined by and may be treated equivalently during sorting.
Definition 6. (Equivalence notation ↔) If two jobs are equivalent under the relation , they are denoted by .
This notation indicates that the two jobs have identical midpoint values of their time windows.
Lemma 5. For two jobs , the relation holds if and only if both and hold.
Proof. (if) Suppose . Then, by Definition 5, . This implies and , and therefore both and hold.
This lemma shows that equivalence under corresponds to mutual ordering based on midpoint comparison.
Lemma 6. The relation defined on the set is a total preorder.
Proof. The relation satisfies the following properties: Reflexivity: For any job , , hence . Transitivity: For jobs , if and , then and . Adding these two inequalities yields , which implies . Totality: For any two jobs , either or , hence either or (or both). □
The job set , together with the relation , forms a totally preordered set. Two jobs may be equivalent under this relation if they share the same midpoint. By grouping equivalent jobs into equivalence classes, the induced ordering among these classes becomes a total ordering.
Lemma 7. For any two jobs , if , then .
Proof. Since , we have and . Adding these two inequalities yields . Therefore, . □
This lemma shows that the partial ordering is consistent with the midpoint-based ordering .
Lemma 8. Suppose two jobs satisfy . Then, for any job , the following two statements hold.
- (1)
If , then .
- (2)
If , then .
Proof. Since we have .
- (1)
If , then we have .
Substituting for , we obtain , which implies .
- (2)
If , then we have .
Substituting for , we obtain , which implies . □
This lemma shows that jobs with identical midpoints are interchangeable under the relation .
Lemma 9. If no pair of jobs in satisfies the equivalence relation , then there exists a unique sequence of jobs sorted by the relation . Furthermore, if there exists a subset such that every pair of jobs in satisfies the relation , then the jobs in appear consecutively as a group in the sequence sorted by .
Proof. The result follows from Lemma 8 and the transitivity and totality properties of the relation . If no pair of jobs is equivalent under , then all jobs are strictly ordered by their midpoint values, yielding a unique sorted sequence. If a subset consists of jobs equivalent under , then by Lemma 8 these jobs are interchangeable under the relation , and therefore they must appear consecutively in any sequence sorted by . □
Under the condition that two jobs
satisfy
, there are ten possible cases for the relationships among
,
,
, and
. These cases are detailed in
Appendix A. Among them, Cases 1 through 6 also satisfy
, with Case 4 further satisfying
. In addition, Cases 8 and 10 satisfy the equivalence condition
.
Lemma 10. If and ( or ), then cannot hold.
Proof. Assume . Then . Suppose, for contradiction, holds. Then, and . Adding these inequalities gives . Combining this with , we obtain .
Since and , if either inequality were strict, then , which is impossible. Therefore, and . This contradicts the assumption that or . Therefore, cannot hold. □
This lemma indicates that a strict ordering under the midpoint-based relation prevents a reverse dominance relationship under .
Lemma 11. Consider two jobs such that and . Suppose is an optimal schedule in which job is scheduled immediately after job . Then, there exists a feasible schedule obtained by exchanging the positions of and , while keeping the positions of all other jobs in unchanged. Moreover, is also optimal.
Proof. Let , and , denote the start and completion times of jobs and , respectively, in schedule . Let , and , denote the start and completion times of jobs and , respectively, in schedule . Since is feasible, , , , and . Construct schedule by swapping the positions of jobs and . Then, , .
- (1)
Feasibility of job : Since = and , we have . Because , we have , hence . Also, since job is scheduled before job in , its completion time satisfies . Thus, job remains feasible.
- (2)
Feasibility of job : Since and implies , we have . Also, because job is scheduled after job , its start time satisfies . Thus, job remains feasible.
- (3)
Optimality: Since the set of selected jobs remains unchanged, the objective value is unchanged. Therefore, is also an optimal schedule. □
This lemma shows that when job is no later and no looser than job , scheduling before is always at least as good as the reverse order.
Lemma 12. If two jobs and satisfy , then any schedule in which job is scheduled immediately after job cannot be a unique optimal schedule.
Proof. Consider two cases.
: In this case, the time window of job lies entirely before . Therefore, it is impossible to schedule job before job . Hence, any schedule in which job is scheduled immediately after job is infeasible.
: In this case, the time windows overlap. By Lemma 11, any optimal schedule in which job is scheduled immediately after job can be transformed into another optimal schedule by exchanging the position of and . Therefore, such a schedule cannot be unique. □
Therefore, when , it is unnecessary to consider any schedule in which job is scheduled immediately after . Following Lemma 12, we establish the following dominance rule:
Dominance rule 2. If two jobs and satisfy , then any child node obtained by scheduling job immediately after job in the current partial solution can be fathomed. □
This rule follows from Lemma 12, since any such node either corresponds to an infeasible schedule or can be transformed into another node with an equivalent or better solution by exchanging the positions of jobs and .
Lemma 13. If every pair of jobs in is comparable under the relation , then there exists an optimal solution in which the sequence of selected jobs follows the nondecreasing order defined by . Consequently, given this sorted order, at any node representing a selected job, it is sufficient to branch only to jobs that appear later in the sequence.
Proof. Since every pair of jobs is comparable under , the relation defines a total ordering on . Consider any optimal solution. If the sequence of selected jobs violates this ordering, then there exists at least one adjacent pair of jobs , such that but is scheduled before . By Lemma 11, such a pair can be exchanged without violating feasibility or optimality. Repeatedly applying such exchanges transforms the sequence into one that follows the order defined by without changing the objective value. Hence, there exists an optimal solution consistent with the sorted order.
Therefore, when constructing solutions, it is sufficient to branch only to jobs that appear later in the sorted sequence. □
This lemma shows that, under a total ordering, the search space can be restricted to sequences consistent with the ordering without loss of optimality.
4.3.2. Heuristic Sorting of Jobs
As discussed in Lemma 13 and Dominance Rule 2, the relation offers a strong structural property for sequencing jobs when searching for an optimal solution. Specifically, if , it is only necessary to consider scheduling job after job , rather than the reverse. However, the poset is not a total ordering; that is, not all pairs of jobs can be compared by . As a result, it may not be possible to derive a complete sorting of all jobs based solely on . In contrast, the relation provides a complete ordering of jobs based on the midpoints of their time windows. Therefore, the relation is used for dominance-based pruning, while the relation is used for heuristic ordering.
According to Lemma 10, sorting by does not contradict the ordering defined by when a job pair is comparable under . Among the ten possible cases describing the relationship between jobs and where holds, six cases (Cases 1 through 6) also satisfy . For these six cases, it is unnecessary to consider solutions in which job is scheduled immediately after job .
For a job pair and that satisfies but not , scheduling job before job is not guaranteed to yield an optimal solution. However, the relation implies that , which indicates that the interval (corresponding to scheduling job before job ) is wider than the interval (which results from scheduling job before job ). Therefore, when both jobs are selected under the condition , scheduling before provides more slack and flexibility than the reverse order. Consequently, sorting jobs according to the relation offers a useful heuristic for guiding the search process.
Moreover, since a depth-first search (DFS) strategy is employed in the enumeration tree, following the branching sequence defined by the sorted list increases the likelihood of generating a high-quality solution early. If the first solution generated by the sorted sequence is feasible and not eliminated by any dominance rule, it serves as a strong initial incumbent solution in the search for optimality.
4.3.3. Additional Dominance Rules
In addition to the dominance rules presented in the previous subsection, this subsection introduces further rules to reduce branching in the enumerative search tree. These additional rules exploit the structural properties of the scheduling problem and the ordering relations defined in this study to identify and eliminate dominated or redundant branches during the search process.
Figure 3 illustrates that job
can be inserted between the last scheduled job in
and job
without delaying the earliest completion time of job
j, denoted by
.
Lemma 14. Given a partial solution , consider the subproblem . If there exist two jobs such that , then the following inequality holds: .
Proof. The condition implies that job
can be inserted between the last scheduled job in
and job
without delaying the earliest completion time of job
, denoted by
(see
Figure 3). Consider two cases.
Job is selected in an optimal solution of . In this case, job is originally scheduled after job without affecting , we can exchange its position to occur before . This increases the total weight accumulated before by , while potentially creating additional space after for other jobs. Therefore, .
Suppose job is not selected in the optimal solution of . Since job can be inserted before job without affecting , including does not reduce the optimal value. Therefore,
Combining both cases and noting that , we obtain . □
This lemma shows that inserting job before job , when feasible without delaying the earliest completion time of job , cannot worsen the objective value and may improve it.
Dominance rule 3. Consider a subproblem of a partial solution . If there exist two jobs such that , then the child node corresponding to the partial schedule can be fathomed. □
This rule follows from Lemma 14. Since inserting job before job does not delay the earliest completion time of job and yields a solution that is at least as good as (and potentially better than) scheduling immediately after . Thus, branching directly from to can be skipped.
Lemma 15. Consider a subproblem of a partial solution . If there exist two jobs such that , , , and , then the following inequality holds: .
Proof. The two cases of adding job
or job
after
are illustrated in
Figure 4 and
Figure 5, respectively.
The first two conditions imply mutual exclusion:
If job is scheduled immediately after , then job cannot be scheduled afterward.
Similarly, if job is scheduled after , then job cannot be scheduled afterward.
Thus, for both subproblems and , the remaining job set is identical and equals . Let , . By assumption, . By Lemma 1, this implies . Since , we obtain . □
This lemma shows that when selecting one job that excludes the other, the job with the earlier completion time and no smaller weight dominates the alternative.
Following Lemma 15, we have Dominance rule 4. This rule follows from Lemma 15, since selecting job excludes job , and job yields an objective value that is at least as good as that obtained by selecting job .
Dominance rule 4. Consider a node representing solution and its subproblem . If there exist two jobs such that , , , and , then the child node corresponding to solution can be fathomed. □
If only one of the two jobs and can be scheduled after partial solution due to timing constraints, and job has a weight no smaller than that of job , and completes no later than job , then selecting job dominates selecting job . Therefore, branching to is unnecessary. That is, any solution obtained from node cannot outperform a corresponding solution obtained from node .