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스케줄 이론 Ch 8 Extensions on Basic Single Machine Problem

스케줄 이론 Ch 8 Extensions on Basic Single Machine Problem. 박진우 ( 서울대 ) 배준수 ( 전북대 ). 1. Introduction. Previous assumptions C1: n single operation jobs at time 0 C2: at most one job at a time C3: sequence independent set up time C4: job descriptor deterministic and known C5: no break down

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스케줄 이론 Ch 8 Extensions on Basic Single Machine Problem

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  1. 스케줄 이론 Ch 8 Extensions on Basic Single Machine Problem 박진우(서울대) 배준수(전북대)

  2. 1. Introduction • Previous assumptions • C1: n single operation jobs at time 0 • C2: at most one job at a time • C3: sequence independent set up time • C4: job descriptor deterministic and known • C5: no break down • C6: no idle(derived) • C7: no preemption(derived) • In this chapter, we consider relaxation of these assumptions • C1: different release dates • C1: precedence • C3: sequence dependent • If we have dynamic ready times, then the assumptions C6 and C7 need to be reconsidered.

  3. 2. Non-simultaneous arrivals • Dynamic version: different ready times • Inserted idle time (Condition 6) ? • Job preemption (Condition 7) ? • Example: total job tardiness • (a) Satisfies C6 and C7 • T1+T2=5 • (b) Inserted idle time allowed • T1+T2=2 • (c) Preemption allowed • T1+T2=1 1 2 5 7 2 1 1 3 8 1 2 1 1 3 7

  4. C[1] R[1] 2. Non-simultaneous arrivals • We can think of 2 modes of operations. • preemption-resume : schedule with no inserted idle time is a dominant set. • preemption-repeat : schedule with no preemption is a dominant set. • (a) Preempt-Resume • Most of the rules with 5 Assumptions can apply. • (With minor modifications in some case) • Decision Points : • ① Completion of a job • ② Arrival of a job. • (E.g.) min   : EDD rule at every decision points • At each job completion, apply EDD among the set of available jobs • At each job arrival, if newly available job has EDD than current processing job, then switch. • (E.g.) min     • At Completion time Decision Point : SPT • At Arrival Time Decision Point : SRPT(Shortest Remaining Processing Time)  Preempt

  5. 2. Non-simultaneous arrivals • (Terminology) • Dispatching Rule : Decisions are based only on the current status(myopic). • Look-ahead Rule : (Analogy)   • (cf) Planning, Routing, Dispatching, Expediting. • (b) Preempt-Repeat • More complicated but fortunately "Permutation Schedule" still dominates the solution set. • Only general methodologies are available up to this time (D.P., B&B, other combinatorial schemes). • Especially, B&B seems very much promising because "Preempt-Resume" solution can be used for bounds. • We shall assume that the preempt-repeat mode applies from now on unless otherwise specified.

  6. 2. Non-simultaneous arrivals • (e.g.)                      - Branching from the first job assignment • B & B: Preempt-resume schedule can be used as lower bounds for preempt-repeat cases. Solve for 3 job problem given above using B&B with jump tracking scheme.

  7. 2. Non-simultaneous arrivals • Min. Makespan(head-body-tail problem) • (rj, pj, qj) = (ready time, processing time, delivery time) = (first operation, second operation, third operation) = (head, body, tail) • Body: a bottleneck facility • Head, Tail: plentiful, parallel • NP hard • (Algorithm 8.1): The Largest Tail(LT) Procedure • Step1. Initially, let t = 0 • Step2. If there are no unscheduled jobs at time t, set t equal to the min. ready time among unscheduled jobs; otherwise proceed. • Step3. Find job j with the largest qj among unscheduled jobs available at time t. Schedule job j to begin at time t. • Step4. Increase t by pj. If all n jobs are scheduled, stop; otherwise, return to Step 2. • Makespan where job i initiates a block, job k is critical job • Optimality condition: qk qj for all jobs j from i to k(뒤의 dk≥dj와 같은 조건) • Min. Lmax문제와 Min. Makespan의 문제가 동일한 문제가 됨

  8. 2. Non-simultaneous arrivals • (Theorem 8.1) • (Proof of 8.1) Let qj= dmax - dj , then Lmax = max { Cj - dj } = max {Cj –( dmax - qj )} = max {Cj + qj } – dmax Consider the relaxed problem in which we eliminate all jobs except those from 1 to k in the final sequence(the Block). Next set the release dates of the remaining jobs equal to ri . This relaxed problem is essentially the basic single machine sequencing problem with ri serving as time 0. Its optimal Lmax is no larger than the optimal solution of the original problem. Original and relaxed problem have the same objective function value. Lmax is attained by EDD sequence. So it is the optimal.

  9. (Theorem 8.2) In the dynamic problem, suppose that the non-delay implementation of the EDD rule yields a sequence of jobs in EDD order.  Then this non-delay schedule is optimal. • (Theorem 8.3) In the dynamic problem, if the ready times and due dates are agreeable, then the non-delay implementation of EDD is optimal. • (Theorem 8.4) In the dynamic F problem, if the release dates and processing times are agreeable, then the non-delay implementation of SPT is optimal.

  10. 2. Non-simultaneous arrivals • (Theorem 8.5) In the dynamic T problem, if the ready times, processing times and due dates are all agreeable, then the non-delay implementation of MDD is optimal.( dj '= max{ dj , t+Pj } • Algorithm Minimize (dynamic version) • 1. Set all jobs in ERD(Earliest Release Date) sequence and place all jobs in set B.    Let set A = . • 2. If there is no late job in B then STOP. B is optimal.   Otherwise, identify the first late job in B. Assume this is job [k](k-th job in B). • 3. Remove one job from B so that the latest completion time among the first (k-1) jobs will be minimized. Place it in A and Goto step 2.

  11. 3. Technological constraints(Precedence constraints) • (Notation) • means that job i is the direct predecessor of job j. (i  j) • means that job i is the predecessor of job j. • Assume         • (a) Extensions of EDD rules (Min. ) • (Theorem for Modified EDD) • Transitivity holds for modified EDD rule • Recall that EDD minimizes • (Proof)

  12. Proof • (Proof of the Theorem for Modified EDD)

  13. 3. Technological constraints(Precedence constraints) • (Definition: Lawler's Problem) • Consider a measure, where Ci is Completion time of Job i and is Non-decreasing cost function of completion time under general precedence constraints.  And a problem • So  Minimize Lmaxor Tmax belongs to this category. • (Lawler's Theorem)  • (Proof)

  14. Proof • (Proof of Lawler's theorem)

  15. 3. Technological constraints(Precedence constraints) • (Lawler’s Algorithm: Minimize Lmax, Tmax under precedence constraints) • Let Jk be the last job under theorem. • Then the value of the objective function is max(A, Jk) where A is a permutation of the other (n-1) jobs. • The maximum cost of this sequence is the larger of , the cost of completing Jk last, and the maximum cost of completing the jobs in A. • An optimal sequence can be found by making both these terms as small as possible. Jk has been chosen so that is the minimum for all the jobs that could be processed last. • So to construct an optimal sequence, our task is simply to choose A so that the maximum cost of  completing its jobs is as small as possible.  • We have a new problem with the same form as our original, but with (n-1) jobs instead of n.  Now minimize A again. • ∴ ① Find the last job in n jobs using the theorem. Call it Jk. • ② Reduce the problem by eliminating Jk and go to ① if unscheduled job remains.

  16. 3. Technological constraints(Precedence constraints) • (Example)

  17. 3. Technological constraints(Precedence constraints) • (b) Extensions of "SPT" rules • (b-1) Problems with simple precedence constraints • (Definition) Job String (≈≠ Job Chain) • A set of jobs that must be processed in a fixed order without interruption. • (Namely, a job group where sequencing decision is already made) • (e.g.) Jobs sharing same set-ups being grouped into one job string. • The sequencing problem for job strings is one of sequencing these special job sets • (Notation)

  18. 3. Technological constraints(Precedence constraints) • There are 2 cases • (Case 1) Min. Mean String Flow Time = where s : # of strings, Fk : Flow time for string k • The string may be treated as “Jobs”, yielding an optimal sequence characterized by • (Case 2) Min. Mean Job Flow Time =     • SPT for mean string processing time.

  19. 3. Technological constraints(Precedence constraints) • (b-2) Problems with more general precedence constraints • Feasible subset refers to a set of jobs which can be processed once the lead job is complete without having to wait for the completion of jobs outside the set.

  20. 3. Technological constraints(Precedence constraints) • Precedence structure  Tree • Assembly Tree • Branching Tree 5 4 1 2 6 9 2 5 1 3 7 3 12 10 6 10 4 8 11 7 11 12 8 9

  21. 3. Technological constraints(Precedence constraints) • In a branching tree case, a feasible subset of job j, Sj must satisfy. • Example • job 1   {①} {① ②} {① ② ⑤} {① ② ⑥} {① ③} {① ③ ⑦} {① ② ③} {① ② ③ ⑤ ⑥ ⑦} • job 2   {②} {② ⑤} {② ⑥} • job 3   {③} {③ ⑦} • (Notation) Uj • Union of all feasible subsets of job j (job j and all of its successors)

  22. 3. Technological constraints(Precedence constraints) • (Theorem)   is minimized by the Algorithm for Branching Tree (ABT). • (Notation) • (Algorithm for Branching Tree : ABT)  Minimizing    in a Branching Tree. • Regard each feasible subset as a job string. • Step 1. For each job j find the minimum of weighted average for all possible job strings with job j as a lead job. And call it . • Step 2. Sequence the jobs with non-decreasing order of considering precedence requirement. Repeat Step 2 until the sequence is complete. • (cf) a ‘job by job’ procedure vs a ‘set by set’ procedure

  23. 3. Technological constraints(Precedence constraints) • Let    : largest feasible subset of job j on which   is attained. •  Once a job string with min is selected, then there is no reason to split it in the middle. If so selected, then above algorithm minimizes by the Theorem which was about   job string   ). •  We need to prove that there is no advantage by splitting the chain thus selected. • (But how ?)

  24. Proof • Proof of optimality of the algorithm ABT( Alg. for Branching Tree)

  25. Proof ------------------- ①

  26. Proof

  27. Proof

  28. 3. Technological constraints(Precedence constraints) • (Definition) Antithetical • A pair of scheduling procedures are called antithetical if they produce opposite sequences in an n job 1 M/C problem. • 즉, R & R’ are antithetical if the job that is assigned to position i by R is assigned to position (n-i+1) by R’. • The solution procedures are applicable for tree structured precedence relationships but are not applicable for general kind of precedence relationships.

  29. 3. Technological constraints(Precedence constraints) • (c) Min   with General Precedence Structure • (Algorithm 4.2) Sidney's Algorithm (1975, O.R.) • Step 1. Among all initial sets, find the largest initial set, S*, that minimizes   . • Step 2. Add the jobs in S* to the sequence. • Step 3. Delete jobs in S* from consideration and go to Step 1. • Example • (Initial Sets) S {1,2,3,4} • If i ∈ S and k < i then k ∈ S. • But {1,3,4} is not an initial set  since 2 precedes 4.

  30. 4. Sequence dependent set up times • Frequently occurs in process type industries. • (e.g.) Paint Blending Process • Cleaning is necessary between job changes. • (e.g.) Job Shop: Different set-ups • Fixtures, Jigs, Tools.

  31. 4. Sequence dependent set up times • Make Span : The length of time required to complete all jobs • If we add final state (idle state), then it becomes • Now the problem becomes the famous (notorious) traveling salesman problem(TSP). • 4 Ways to solve TSP • ① D.P. • ② B.B. • ③ ILP Formulation • ④ Heuristic

  32. 4. Sequence dependent set up times • ① D.P. • (e.g.) Recall the general rule for D.P. • (a) Define optimal value function F(arguments) - G(J) in the previous model • (b) Develop a recurrence relation involving F(arg) • (c) Give boundary condition • Now, F(i, J) = the shortest path from i through set J to i0, the initial visiting city. • Final Solution is F(i0, K) where K is a set containing all cities other than i0.

  33. 4. Sequence dependent set up times • ② B & B • Most efficient solution algorithm known up to this point. • Follow the example of the text. • Rules: (a) tree, (b) branching policy, (c) formula for lower bounds, (d) terminal rule • (Reasoning) - Little, Murty, Sweeny and Karel(1963) • The volume of the solution is the sum of n elements from the distance matrix S specified by        . • Ingenuity lies in the reduction scheme (Select one element from each row and each column ⇒ lower bound) • Need to carry along the reduction matrix but it is not a big burden.

  34. 4. Sequence dependent set up times • ③ Heuristics • "Closest Unvisited city" • For n  20, Solutions are within 10% of optimum  if Sij's were reasonable.

  35. 4. Sequence dependent set up times • ④ Integer Linear Program • Then without returning to the original city, • (1) Prevents tours visiting less than n cities. • If only k (where k<n) cities were visited, then set of above conditions yields contradiction. • (2) Interpret ui as the position relative to some origin city.

  36. 4. Sequence dependent set up times • After then • Minimizing Total Costs in One-Machine Scheduling: Rinnooy Kan, Lageweg, Lenstra (Netherlands), 1975, Sep.-Oct., O.R. • Improved B & B. on Weighted Tardiness Problem using  Dominance Property. • Heuristic Procedures for the Single M/C Problem to Minimize Maximum Lateness: Larson, Dessouky, 1978,  AIIE Trans. • A simulation study for various Heuristic Rules for dynamic job ready time( ). • Minimizing the Range of Lateness on a Single M/C: Gupta & Sen, J of O.R.S. , 1984 • B & B for  • say,  Minimize the SPREAD of Lateness. • Min. Weighted Absolute Deviation in Single M/C Scheduling, Fry, Armstrong, Blackstone, IIE Trans, 1987. • Heuristics ( Sequencing then Insert Idle times Optimally ) • A Heuristic Scheduling Policy for Multi-Item, Single-M/C Production System with Time Varying, Stochastic Demands,  Leachman & Gascon, M.S., 1988 • Heuristic (by modifying the Target Cycle) • The Single M/C Early/Tardy Problem, Ow & Morton, M.S., 1989 • 2  Dispatch Rules.

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