1. Single-Machine Sequencing with Independent Jobs Chapter 2 Elements of Sequencing and Scheduling by Kenneth R. Baker Byung-Hyun Ha R4

2. 1 Outline  Introduction  Preliminaries  Problem without due dates: elementary results  Flowtime and inventory  Minimizing total flowtime  Minimizing total flowtime: the weighed version  Problem with due dates: elementary results  Lateness criteria  Minimizing the number of tardy jobs  Minimizing total tardiness  Due dates as decisions  Summary

3. 2 Introduction  Pure sequencing problem  Specialized scheduling problem in which an ordering of jobs completely determines a schedule  Single-machine problem  Simplest pure sequencing problem, but important  Reveals a variety of scheduling topics in a tractable model A context in which to investigate many performance measures and solution techniques A building block in the development of a comprehensive understanding of scheduling concepts  Examples to be employed for complex problems Single-machine problems as a part of complex ones Solving imbedded single-machine problems and incorporating the results into larger problems  Bottleneck analysis in multiple-operation processes Decisions using resources as aggregated one

4. 3 Introduction  Characteristics of basic single-machine problem C1. A set of n independent, single-operation jobs is available for processing simultaneously (at time zero) C2. Setup times for the jobs are independent of job sequence and are included in processing times C3. Job descriptors are deterministic and known in advance C4. One machine is continuously available and never kept idle while work is waiting C5. Once an operation begins, it proceeds without interruption (i.e. no preemption is allowed)

5. 4 Introduction  Permutation schedule  Sequence of n jobs  permutation of job indices 1, 2,..., n  Single-machine problem  pure sequencing problem A schedules can be completely specified by a permutation of integers  Total # of distinct schedule – n!  Indication of position in sequence using brackets  [5] = 2  5th job in sequence is job 2  d [1]  due date of 1st job in sequence

6. 5 Preliminaries  Job descriptors  Input to scheduling problem  Examples p j -- processing time Amount of processing required by job j  p j will generally include both processing time and facility setup time under C2 r j -- ready time Time at which job j is ready (or available) for processing r j can be regarded as arrival time, but r j = 0 under C1 d j -- due date Time at which processing of job j is due to be completed

7. 6 Preliminaries  Information for evaluating schedule  Output from applying schedule  Examples C j -- completion time Time at which processing of job j is finished F j -- flowtime Time job j spends in system: F j = C j – r j Response of system to individual demands for service (turnaround time) L j -- lateness Amount of time by which completion time of job j exceeds its due date: L j = C j – d j Conformity of schedule to given due date Negative value whenever job is completed early T j -- tardiness L j, if job j fail to meet its due date; 0, otherwise: T j = max(0, L j ) U j -- unit penalty U j = 1, if C j  d j ; 0, otherwise

8. 7 Preliminaries  Due date related penalty functions LjLj CjCj djdj TjTj CjCj djdj UjUj CjCj djdj 1

9. 8 Preliminaries  Performance measures  1-dimensional aggregated quantities about all jobs Commonly used for evaluation of schedules  Functions of set of completion time in schedule, usually Z = f(C 1, C 2,..., C n )  Examples Total flowtime -- F =  j=1 n F j Total tardiness -- T =  j=1 n T j Maximum flowtime -- F max = max 1  j  n {F j } Maximum tardiness -- T max = max 1  j  n {T j } Number of tardy jobs, or total unit penalty -- U =  j=1 n U j Mean flowtime, mean tardiness, proportion of tardy jobs  Problem description  F-problem -- minimization of total flowtime  T-problem  C max -problem

10. 9 Preliminaries  Definition -- regular measures  Performance measure Z is regular, if (a) scheduling object is to minimize Z, and (b) Z can increase only if at least one of the completion times in schedule increases (b) Z'  Z  C j '  C j for some job j, where Z = f(C 1, C 2,..., C n ) is value of a measure of schedule S and Z' = f(C 1 ', C 2 ',..., C n ') is value of the same measure of some different schedule S' (b  ) C j '  C j for all job j  Z'  Z  The measures presented in the previous slide are all regular Then, non-regular measures?

11. 10 Preliminaries  Definition -- dominant set  A set of schedules is dominant if it is sufficient to be considered for finding solution Reduced search space from complete enumeration  Verifying dominant set D of schedules for regular measures 1. Consider an arbitrary schedule S (which contains completion times C j ) that is excluded from D 2. Show that there exists a schedule S' in D, in which C j '  C j for all j 3. Hence, Z'  Z for any regular measure, and so S' is at least as good as S 4. Therefore, in searching for an optimal schedule, it is sufficient to consider only schedules in D  Basic single-machine problems  Dominant sets for regular measures  Set of schedules without idle time  C4 could be relaxed  Set of permutation schedules (without preemption)  C5 could be relaxed

12. 11 Preliminaries  Suppose we consider regular measures  Theorem 1  In the basic single-machine problem, schedule without inserted idle time constitute a dominant set.  Theorem 2  In the basic single-machine problem, schedules without preemption constitute a dominant set.  HOMEWORK #1  Prove Theorem 2

13. 12 Problem without Due Dates: Elementary Results  Outline  Flowtime and inventory  Minimizing total flowtime F-problem -- min. F =  j=1 n F j 1 ||  C j  Minimizing total flowtime: the weighed version F w -problem -- min. F w =  j=1 n w j F j 1 ||  w j C j

14. 13 Flowtime and Inventory  Two closely related objectives  Low inventory levels -- minimizing average # of jobs in system  Rapid turnaround of customers -- minimizing total flowtime  Notations  J(t) -- number of jobs in system at time t, J -- time average of J(t)  F max = F [n] = p 1 + p 2 +... + p n -- makespan A = np [1] + (n – 1)p [2] +... + 2p [n-1] + p [n]  A = JF max = F  F = F [1] + F [2] +... + F [n] n n – 2 n – 1 2 1 J(t)J(t) p [1] p [2] p [3] p [n-1] p[n]p[n]...t n n – 2 n – 1 2 1 J(t)J(t) p [1] p [2] p [3] p [n-1] p[n]p[n]...t flowtime-inventory relation

15. 14 Flowtime and Inventory  Relationship between flowtime and inventory  J = F/F max -- J is directly proportional to F  The relationship extends well beyond single-machine problem In dynamic environment where jobs arrive over time In infinite-horizon models where new work arrives continuously In probabilistic systems where process times are uncertain In situations where inventory costs may vary among jobs  Emphasis on F-problem and its generalization in theory  Because total flowtime actually encompasses a broader range of scheduling-related costs  Little’s law  L = W, where L -- long-term average number of customers in a stable system -- long-term average arrival rate W -- long-term average time a customer spends in the system

16. 15 Minimizing Total Flowtime  Minimizing the area under J(t) function  Equivalent problem to F-problem  Steepest slope to the left, the next steepest slope,...  Using shortest processing time (SPT) sequencing Sequencing jobs in nondecreasing order of processing times n n – 2 n – 1 2 1 J(t)J(t) p [1] p [2] p [3] p [n-1] p[n]p[n]...t

17. 16 Minimizing Total Flowtime  Theorem 3  Total flowtime is minimized by Shortest Processing Time (SPT) sequencing (p [1]  p [2] ...  p [n] ).  Proof sketch of Theorem 3 (proof by contradiction)  Suppose there is a sequence S that is optimal but not SPT sequence.  In S, there exists a pair of adjacent jobs, i and j, with j following i, such that p i  p j.  Construct sequence S' in which job i and j are interchanged.  F i + F j  F j ' + F i ' implies that F  F', which is a contradiction.  Therefore, for all S, if S is optimal, then S is SPT sequence. S ij S' ij CjCj CiCi Ci'Ci'Cj'Cj'... p(B)p(B)p(B)p(B)

18. 17 Minimizing Total Flowtime  Proof by contradiction  General procedure 1. Suppose the statement to be proved is false. That is, suppose that the negation of the statement is true. 2. Show that this supposition leads logically to a contradiction. 3. Conclude that the statement to be proved is true.  Exercise: prove that there is no greatest integer. 1.Suppose not. That is, suppose there is a greatest integer N. 2.Then, N  n for every integer n. Let M = N + 1. M is an integer since it is a sum of integers. M  N, since M = N + 1. So N is not the greatest integer, which is a contradiction. 3.Therefore, there is no greatest integer.

19. 18 Minimizing Total Flowtime  Theorem 3, revisited  Predicates O(S) -- S is optimal, SPT(S) -- S is in SPT sequence  Definitions  S, O(S)  (  S', F  F')   S,  O(S)  (  S', F  F') ---   Theorem 3  S, O(S)  SPT(S)  Negation of Theorem 3  (  S, O(S)  SPT(S))   (  S,  O(S)  SPT(S))   S, O(S)   SPT(S) ---   Proof of Theorem 3 by contradiction Suppose not. Then, by , there exists S such that O(S)   SPT(S). Then, since O(S),  S', F  F' by . Because  SPT(S), in S, there exists a pair of adjacent jobs, i and j, with j following i, such that p i  p j. Let S' be the same sequence with S except job i and j are interchanged. F  F' because F i + F j  F j ' + F i '. That is  S', F  F'. So  O(S) by , which is a contradiction to O(S).

20. 19 Minimizing Total Flowtime  Proof of Theorem 3 by construction 1.Begin with any non-SPT sequence. 2.Find a pair of adjacent jobs i and j, with j following i, such that p i  p j. 3.Interchange jobs i and j in sequence. 4.Return to Step 2 iteratively, improving the performance measure each time, until eventually the SPT sequence is constructed.  Another perspective of Theorem 3  Total flowtime as scalar product of two vectors  j=1 n F j =  j=1 n  i=1 j p [i] =  j=1 n (n – j + 1)p [j]  Solution of minimizing such a scalar product One sequence with nonincreasing, the other with nondecreasing  Since one is already nonincreasing, the other should be nondecreasing

21. 20 Minimizing Total Flowtime  Related properties with Theorem 3  SPT sequencing minimizes J as well as F  If waiting time of job j is defined as its time spent in system prior to the start of its processing, SPT minimizes total waiting time.  SPT minimizes the maximum waiting time

22. 21 Minimizing Total Flowtime: Weighted Version  Flowtime and inventory  F w =  j=1 n w j F j -- total weighted flowtime  V(t) -- total value of inventory in the system at time t V -- time average of V(t)  Generalized flowtime-inventory relation A =  j=1 n p [j]  i=j n w [i] = VF max = F w =  j=1 n w j F j Sequence which minimizes one will minimize the other wjwj w[n]w[n] V(t)V(t) p [1] p [2] p [3] p [n-1] p[n]p[n]...t w [n] + w [n-1]

23. 22 Minimizing Total Flowtime: Weighted Version  Theorem 4  Total weighted flowtime is minimized by Shortest Weighted Processing Time (SWPT) sequencing (p [1] /w [1]  p [2] /w [2] ... p [n] /w [n] ).  HOMEWORK #2  Prove Theorem 4

24. 23 Problem with Due Dates: Elementary Results  Outline  Lateness criteria L max -problem -- min. L max = max j=1,...,n {L j } 1 || L max +1 ||  L j, 1 || T max, 1 || –L min, 1 || T max (1),  C j (2)  Minimizing the number of tardy jobs U-problem -- min. U =  j=1 n U j 1 ||  U j +1 ||  w j U j  Minimizing total tardiness T-problem -- min. T =  j=1 n T j 1 ||  T j  Due dates as decisions

25. 24 Lateness Criteria  Total lateness and maximum lateness  Assuming 3 jobs {1, 2, 3} p 1 = 5, p 2 = 2, p 3 = 3 d 1 = 6, d 2 = 4, d 3 = 8 r 1 = r 2 = r 3 = 0  Sequencing (2, 1, 3)  Lateness (C j – d j ) C [1] = 2, C [2] = 7, C [3] = 10, d [1] = 4, d [2] = 6, d [3] = 8 L [1] = 2 – 4 = –2, L [2] = 7 – 6 = 1, L [3] = 10 – 8 = 2 L = 1, L max = 2  Theorem 5  Total lateness is minimized by SPT sequencing.  Proof sketch of Theorem 5  L =  j=1 n L j =  j=1 n (C j – d j ) =  j=1 n (F j – d j ) =  j=1 n F j –  j=1 n d j 231 2710

26. 25 Lateness Criteria  Theorem 6  Maximum lateness and maximum tardiness are minimized by Earliest Due Date (EDD) sequencing (d [1]  d [2] ...  d [n] ).  Proof sketch of Theorem 6  Similar to proof of Theorem 3, we employ adjacent pairwise interchange of two adjacent jobs i and j, where d j  d i but C i  C j.  max{L i, L j } = L j  max{L i ', L j '}  L max  L max ' S ij S' ij CjCj CiCi Cj'Cj'Ci'Ci'... p(B)p(B)p(B)p(B)

27. 26 Lateness Criteria  Another measure of urgency  Slack time at time t -- s j = d j – t – p j Among jobs with identical due dates, longest job is most urgent  Minimum slack time (MST) sequencing d [1] – p [1]  d [2] – p [2] ...  d [n] – p [n]  Theorem 7  Among schedules with no idle time, the minimum jobs lateness (L min ) is maximized by MST sequencing  L min is not a regular measure  restricted to schedules without inserted idle time

28. 27 Minimizing the Number of Tardy Jobs  U-problem and EDD  EDD is optimal sequence if there is no tardy job (U = 0)  EDD is also optimal if there is only one tardy job  EDD may not be optimal if there are more than one tardy jobs ijklm... didi djdj dkdk dldl dmdm ij didi djdj

29. 28  Optimal sequence form of U-problem  Set B of early jobs and set A of late jobs  Algorithm 1 (Minimizing U) 1.Index jobs using EDD order and place all jobs in B. A = . 2.If no jobs in B are late, stop. Otherwise, identify the first late job j [k] in B. 3.Identify the longest job among the first k jobs in sequence. Remove this job from B and place it in A. Return to Step 2. Minimizing the Number of Tardy Jobs... BA Early jobsLate jobs longest[k][k]... [k-1] d [k-1] d[k]d[k]...

30. 29 Minimizing the Number of Tardy Jobs  Proof of optimality of Algorithm 1  See p. 48 of Pinedo, 2008  Exercise  Get optimal sequence which minimize U.  Weighted version of U-problem -- U w -problem  NP-hard, i.e., as difficult as, or even more difficult than, NP-Complete 0-1 knapsack problem reduces to U w -problem  0-1 knapsack problem  U w -problem Job j12345 pjdjpjdj 1212 7878 6969 4 10 3 12

31. 30 Minimizing Total Tardiness  Minimization of total tardiness  Quantification of qualitative goal of meeting job due dates  Time-dependent penalties on late jobs, no benefits from completing jobs early  Difficulties Tardiness is not a linear function of completion time NP-hard  Adjacent-jobs analysis -- simple and possible analysis  Consider only two adjacent jobs, i and j, and figure out the appropriate order of decreasing total tardiness  Let T ij = T i (S) + T j (S) = max{p(B) + p i – d i, 0} + max{p(B) + p i + p j – d j, 0} T ji = T j (S') + T i (S') = max{p(B) + p j – d j, 0} + max{p(B) + p i + p j – d i, 0} ij... BA S ji BA S' p(B)p(B)p(B)p(B)

32. 31 Minimizing Total Tardiness  Adjacent-jobs analysis (cont’d)  Agreeability Two set of parameters, u j and v j are agreeable if u i  u j implies v i  v j  Case 1: agreeable processing time and due dates (p i  p j, d i  d j ) Case 1.1: p(B) + p i  d i It is preferable that job j precede job i. Case 1.2: p(B) + p i  d i It is preferable that job j precede job i.  Theorem 8 For any pair of jobs, suppose that processing times and due dates are agreeable. Then total tardiness (T) is minimized by SPT (or, equivalently, EDD) sequencing.

33. 32 Minimizing Total Tardiness  Adjacent-jobs analysis (cont’d)  Case 2: parameters are not agreeable (p i  p j, d i  d j ) Case 2.1: p(B) + p i  d i It is preferable that job i (with earlier due date) precede job j. Case 2.2: p(B) + p i  d i Case 2.2.1: p(B) + p i + p j  d j It preferable that job i (with earlier due date) precede job j. Case 2.2.2: p(B) + p i + p j  d j »Case 2.2.2.1: p(B) + p j  d j It is preferable that job i precede job j unless p(B) + p i  d j, in which case job j (shorter job) may precede job i. »Case 2.2.2.2: p(B) + p j  d j It is preferable that job j (shorter job) precede job i.

34. 33 Minimizing Total Tardiness  Theorem 9  In case of T-problem, if jobs i and j are the candidates to begin at time t, then the job with earlier due date should come first, except if t + max{p i, p j }  max{d i, d j }, in which case the shorter job should come first.  Combined results from Case 1 and Case 2  Review of Theorem 9  Outcome depends on time t when candidate jobs are compared  Not sufficient but necessary condition of T-problem It does not tell whether job i and j should come early or late in schedule  Consider following T-problem Check schedule (2, 1, 3) Check optimal schedule (1, 3, 2) Job j123 pjdjpjdj 3838 8686 3 10

35. 34 Minimizing Total Tardiness  Modified due date (MDD) at time t  d j ' = max{d j, t + p j }  MDD priority rule  If jobs i and j are the candidates to begin at time t, then the job with the earlier modified due date should come first.  MDD priority rule is consistent with Theorem 9.  Other specialized results for T-problem  If EDD sequence produces no more than one tardy job, it is optimal.  If all jobs have the same due dates, T is minimized by SPT sequencing.  If it is impossible for any job to be on time in any sequence, T is minimized by SPT sequencing.  If SPT sequencing yields no jobs on time, it minimizes T. ...

36. 35 Due Date as Decisions  Due date as a matter of negotiation  In practice Reasonable models would introduce much more complexity  A simple step, here Treating due date as a decision variable, subject to some constraints  Environment  Due date can be selected at job’s arrival time (r j )  Selection of due date depends only on information about the job itself  Due date decision problem  Decision variables -- d j  Objective Min. D =  j=1 n d j  Constraints C j  d j for all j  With full knowledge over all jobs, SPT works for optimal solution

37. 36 Due Date as Decisions  Due date decision problem (cont’d)  Possible rules selecting due date, considering only a job CON -- constant flow allowance: d j = r j +  SLK -- equal slack flow allowance: d j = r j + p j +  TWK -- total work flow allowance: d j = r j +  p j  Some results (Baker & Bertrand, 1981) CON due dates are dominated by either SLK or TWK due dates TWK will tend to be the best rule most of time

38. 37 Summary  Basic single-machine model  Fundamental in the study of sequencing and scheduling  Scheduling objectives  Flowtime-related, lateness-related,...  Problems  1 ||  C j, 1 ||  w j C j  1 ||  L j, 1 || L max, 1 || T max, 1 || –L min, 1 || T max (1),  C j (2)  1 ||  U j, 1 ||  w j U j  1 ||  T j  Solution strategies  Using simple sequencing rules, such as SPT and EDD  Using more intricate construction for U-problem  We could not solve T-problem  Due date decision problem