스케줄 이론 Single Machine Independent Jobs - Part 2 Problems with due dates

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# 스케줄 이론 Single Machine Independent Jobs - Part 2 Problems with due dates - PowerPoint PPT Presentation

스케줄 이론 Single Machine Independent Jobs - Part 2 Problems with due dates. Proof. Proof of Theorem 2.3 SPT minimizes MFT. (cf) Proof by contradiction 과  Proof by construction

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## 스케줄 이론 Single Machine Independent Jobs - Part 2 Problems with due dates

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스케줄 이론

Single Machine Independent Jobs - Part 2

Problems with due dates

Proof
• Proof of Theorem 2.3 SPT minimizes MFT.

(cf) Proof by contradiction과  Proof by construction

• Start from any non-SPT sequence, locate two adjacent  jobs  "i" and "j"  violating SPT rule.  Interchange two jobs, and eventually we will get to SPT sequence which is optimal).
Single Machine Sequencing with Due Dates
• Theorem 2.5

Mean Lateness & Mean Waiting Time is minimized by SPT sequencing.

• Proof

Minimizing Flow Time is minimizing mean lateness or mean waiting time.

• (Think) SPT Rule is independent of Due Date. Lateness is related to Due Date.

Then why?              In fact,

Single Machine Sequencing with Due Dates
• Theorem 2.6

are minimized by EDD sequencing.

• Proof) By adjacent pair wise interchange.
• Theorem2.7 is maximized by Min. Slack Time Sequencing.
• Proof) Similar to Theorem 2.6
• What is the implication of this theorem?
• Now, suppose we have a schedule with ( perhaps sequenced by EDD rule). Then, is it possible to have schedules other than EDD which also satisfy due-date constraints and improve other measures of performance at the same time?
Single Machine Sequencing with Due Dates
• Theorem (Smith's Rule:1956, Mar., NRLQ)
• Proof
Single Machine Sequencing with Due Dates
• Example of Smith’s rule
• [5]=4, and candidates for [4] are jobs 1,2 but we choose job 2 since t2 > t1
• Candidates for [3] are jobs 1,3,5 but we choose job 1 since t1 > t5 > t3
• Candidates for [2] are jobs 3,5 but we choose job 5 since t5 > t3
• [1] =3, and Optimal Sequence is 3-5-1-2-4, with Mean Flow Time = 9.8 and Total Tardiness = 0
Proof
• Proof of Theorem 2.6

Consider a sequence S which is not in EDD sequence

⇒ There are some jobs i & j in S with j following i  and

Consider another sequence S' with i and j interchanged.

Set B i j Set A time

Single Machine Sequencing with Due Dates
• Costs related to scheduling (Benefits from good schedules)
• Inventory
• As inventory is related to Mean Flow Time, it may make us feel that the Processing time is shrinking. (Reducing Lead Time from the Customer's viewpoint)
• Facility
• Given Workload can be consumed by small # of facilities. (Overtime, additional shift, overload subcontract can be reduced )
• Lateness
• Construction project, etc. (It may be shown Explicitly)
• Not unreal problem (Inventory is acceptable, but missing due date is more critical)
• Are tardy jobs equally penalized no matter how late they are ?
• Minimization of number of tardy jobs may be used as a criterion for lateness.
Single Machine Sequencing with Due Dates
• Algorithm 2.1 (Moore & Hodgson's algorithm: Minimize ) Management Science, Sep., 1968
• Use the fact that "EDD minimizes ".
• 1. Set all jobs in EDD sequence and place all jobs in set E. Let the set .
• 2. If (there is no late job in E) then STOP. E is optimal. Otherwise, identify the first late job in E. Assume this is job [k] (k-th job in E).
• 3. Identify the job with maximum t(i.e. longest job) among the first k jobs in sequence. Remove this job from E and place it in L. Revise the completion times of the jobs in E and Goto 2.
• 의미: By EDD rule, we minimize .

#jobs early late

Single Machine Sequencing with Due Dates
• In the set E, we have at most one late job. If only one job is late, we can improve the schedule by choosing the job with the longest processing as the late job.
• Proof of Moore & Hodgson's algorithm
• The final schedule consists of .
• For set E, we optimize the schedule by EDD. (There is no late job)
• For set L, if we have m tardy jobs, processing time can be saved by rejecting it from E.
• An optimal schedule rejects at least m jobs as .
Single Machine Sequencing with Due Dates
• Minimizing Mean Tardiness
• A heuristic algorithm (called Wilkerson-Irwin Algorithm, 1971, AIIE Trans).
• Notation
• i precedes j ⇔ iⓟj.
• Among two adjacent jobs "i" and "j"  in a sequence where , on what due date conditions should interchange be allowed ?
Single Machine Sequencing with Due Dates
• Preliminary to Wilkerson-Irwin Algorithm.
• Case 1. dj is earlier than di
• 1.1. Job i is within due if jⓟi ⇒ jⓟi is admissible.
• 1.2. Job i is late even if iⓟj ⇒ jⓟi
• Case 2. di is earlier than dj
• 2.1 Job i is within due if iⓟj ⇒ iⓟj
• 2.2 Job i is late even if iⓟj
• 2.2.1 Job j is within due if iⓟj ⇒ iⓟj
• 2.2.2 Job j is within due if jⓟi , but Job j is late if iⓟj
• ⇒ iⓟj if Ci ≤ dj
• jⓟi if Ci ＞ dj
• 2.2.3 Job j is late if jⓟi ⇒ jⓟi
• 즉 EDD is applied first except when di < dj, di < tB + ti and  {dj < tB + ti  or dj < tB + tj}
• Or if ti ≥ tj  and di ＜ dj , we may switch to SPT rule from EDD( Switch "i" and  "j" ) when tB + max{ti, tj} ＞ max{di, dj}
Single Machine Sequencing with Due Dates
• Mystery (Fallacy of Heuristics)
• di ＜ dj, di ＜ tB + ti  and dj ＜ tB + ti  --- ①
• di ＜ dj, di ＜ tB + ti  and dj ＜ tB + tj  --- ②
• Assume that ti ≥ tj
• For ①, di ＜ tB + ti 는 redundant (di ＜ dj ) .
• But for ②, di ＜ tB + ti and  dj ＜ tB + tj  are different conditions.
• Of course, since the smaller quantity (tB + tj) is larger than the larger(dj) ⇒ tB + max{ti, tj} ＞ max{di, dj} (But the reverse(<=) does not hold)
• e.g.) Let tB = 0, ti=10, tj=1, di=3, dj=4. Then tB + max{ti, tj} ＞ max{di, dj}⇒ 10 > 4 ∴ holds
• But    di ＜ dj    3＜4   holds
• di ＜ tB + ti  3＜10  holds
•         dj ＜ tB + tj  4＜1  does not hold
• Two conditions are not equivalent.  So W-I Algorithm is a heuristic algorithm.
Single Machine Sequencing with Due Dates
• Wilkerson-Irwin Algorithm (A Heuristic Algorithm)

Step1

Step2

Single Machine Sequencing with Due Dates
• Wilkerson-Irwin Algorithm (A Heuristic Algorithm)

Step3

Step4

Single Machine Sequencing with Due Dates
• Example
• EDD order : 1-2-7-5-3-4-6-81. Compare 1 & 2.  t1＜ t2    S={1},  β=2,  γ=7  Fα=42
• 2. Check if    Fα + max{tβ, tγ} ≤ max{dβ, dγ} or if tβ ≤ tγ  42 + max{115, 95} ≤ max{283, 447}    ∴ S={1-2}, β =7, γ=5,  Fα=42+115 =157
• 3. tβ=95 ≤ tγ = 117    ∴ S={1-2-7}, β=5, γ=3,  Fα=157+95=252
Single Machine Sequencing with Due Dates
• 4. 252 + max{117,111} ≤ max{452,616}   ∴ S={1-2-7-5}, β=3, γ=4,  Fα=252+117=369
• 5.   369 + max{111,96} ≤ max{616,622}   ∴ S={1-2-7-5-3}, β=4, γ=6,  Fα=369+111=480
• 6.    480 + max{96,93} ≤ max{622,624}   ∴ S={1-2-7-5-3-4}, β=6, γ=8,  Fα=480+96=576
• 7.     S = { 1-2-7-5-3-4-6-8 }      j     1     2    7    5    3    4    6     8    Cj    42  157  252  369  480  576  669  767      dj 260  283  447  452  616  622  624   749
• (Discussion)  Heuristic algorithm is simple but gives sub-optimal solution.
Single Machine Sequencing with Due Dates
• (Definition) For each tardy job in the final sequence, we define TARDINESS INTERVAL
• (Theorem 2.8 )  W-I Algorithm minimizes if there is no overlapping tardiness interval (W-I gives Optimum in limited cases).
• (Discussion) Above theorem is only a sufficient condition. Even with overlapping tardiness intervals, a sequence could be optimal.
Single Machine Sequencing with Due Dates
• (Theorem 2.9) If EDD sequence produces no more than one tardy job, it yields min .
• (Proof)
• (Theorem 2.10) If dj = some constant for all j, then is minimized by SPT rule.
• (Proof) Make areas(the sum of tardiness) convex.
Single Machine Sequencing with Due Dates
• (Theorem 2.11) If it is impossible for any job to be on time in any sequence, then is minimized by SPT sequencing.
• (Proof)
• (Theorem 2.12) If SPT sequencing yields no jobs on time, then it minimizes .
• (Discussion)  This theorem is a Corollary of Theorem 2.11.
• Mean lateness = Mean Tardiness (in this case, is minimized by SPT)
Due Dates as Decisions
• Suppose we may set due dates of the incoming order:
• CON: Constant flow allowance, dj = rj + γ
• SLK: Equal slack flow allowance, dj = rj + pj +γ
• TWK: Total work flow allowance, dj = rj + α pj
• If possible tighter due date would be better: short flow allowance, better chance of getting order, etc.
• Minimizing the sum of due dates with the constraint

that Cj ≤ dj , for all j

• Sum of due dates can be minimized by ( ) Sequencing.
• CON is always dominated by other rules(simulation study reveals that TWK tends to be the rule most of the time.