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Earliness and Tardiness Penalties Chapter 5 Elements of Sequencing and Scheduling by Kenneth R. Baker Byung-Hyun Ha R1
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1 Outline Introduction Minimizing deviations from a common due date Four basic results Due date as decisions The restricted version Different earliness and tardiness penalties Quadratic penalties Job dependent penalties Distinct due dates Summary
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2 Introduction Until now Basic single-machine model with regular measures of performance, which are nondecreasing in job completion times Among regular measures, total tardiness criterion has been a standard way of measuring conformance to due dates The measure does not penalize jobs completed early Just-In-Time (JIT) production “Inventory is evil” Earliness, as well as tardiness, should be discouraged E/T criterion in basic single-machine model Earliness and tardiness E j = max{0, d j – C j } = (d j – C j ) + T j = max{0, C j – d j } = (C j – d j ) + Linear penalty function with unit earliness (tardiness) penalty j ( j ) f(S) = j=1 n ( j (d j – C j ) + + j (C j – d j ) + ) = j=1 n ( j E j + j T j ) Nonregular measure
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3 Introduction Variations in E/T criterion Decision variables Job sequence with due dates given Due dates and job sequence Setting due dates internally, as targets to guide the progress of shop floor activities Due dates Common due dates (d j = d) Several items constitute a single customer’s order Assembly environment where components should all be ready at the same time Distinct due dates Penalties Common penalties ( j = , j = ) Distinct penalties Role of penalty functions Guiding solutions toward the target of meeting all due date exactly Measuring suboptimal performance of nonideal schedules
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4 Minimizing Deviations from a Common Due Date Basic E/T problem Minimizing sum of absolute deviations of job completion times from common due date (d j = d, j = j = 1) f(S) = j=1 n |C j – d j | = j=1 n (E j + T j ) Due date can be in the middle of jobs? Tightness of due date d Restricted version vs. unrestricted version d d
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5 Basic E/T Problem, Unrestricted Theorem 1 In the basic E/T model, schedules without inserted idle time constitute a dominant set. Theorem 2 In the basic E/T model, jobs that complete on or before the due date can be sequenced in LPT order, while jobs that start late can be sequenced in SPT order. V-shaped schedule Exercise Prove Theorem 1 using proof by contradiction. Prove Theorem 2 using proof by contradiction.
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6 Basic E/T Problem, Unrestricted Theorem 3 In the basic E/T model, there is an optimal schedule in which some job completes exactly at the due date. Proof sketch of Theorem 3 (proof by contradiction) Suppose S is an optimal schedule where C i – p i d C i. Let b (a) denote the number of early (tardy) jobs in sequence. Case 1 (a b) Consider S' where S is shifted earlier by t = C i – d. Increase in earliness (decrease in lateness) penalty is b t (a t). Hence, f(S) f(S'), because a t b t. Case 2 (a b) Consider S' where S is shifted later by t = d – (C i – p i ). Decrease in earliness (increase in lateness) penalty is b t (a t). Hence, f(S) f(S'), because a t b t. Therefore, in either case a schedule with the property of the theorem is at least as good as S.
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7 Basic E/T Problem, Unrestricted Properties of optimal schedule by Theorem 1, 2, 3 Optimum is describable by a sequence of jobs and a start time of 1st job V-shaped schedule 2 n candidates instead of n! candidates Analysis on optimal schedule Notations A (B) -- set of jobs completing after (on or before) the due date a = |A|, b = |B| Ai (Bi) -- ith job in A (B) Earliness penalty for job Bi -- E Bi = p B(i+1) + p B(i+2) +... + p Bb Total penalty for the jobs in B C B = i=1 b E Bi = i=1 b (p B(i+1) + p B(i+2) +... + p Bb ) = 0p B1 + 1p B2 +... + (b – 2)p B(b–1) + (b – 1)p Bb. Total penalty for the jobs in A C A = ap A1 + (a – 1)p A2 +... + 2p A(a–1) + 1p Aa. f(S) = C A + C B minimized by assigning jobs regarding processing times
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8 Basic E/T Problem, Unrestricted Algorithm 1: Solving the Basic E/T Problem 1.Assign the longest job to set B. 2.Find the next two longest jobs. Assign one to B and one to A. 3.Repeat Step 2 until there are no jobs left, or until there is one job left, in which case assign this job to either A or B. Finally, order the jobs in B by LPT and the jobs in A by SPT. Exercise: solve basic E/T problem with jobs below and d = 24. Job j123456 pjpj 134679
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9 Basic E/T Problem, Unrestricted Algorithm 1* Considering secondary measure: minimum total completion time Same as Algorithm 1 except that, in Step 2, shorter job is assigned to B and, in Step 3, if n is even, assign the shortest job in A Theorem 4 In the basic E/T model, there is an optimal schedule in which the bth job in sequence completes at time d, where b is the smallest integer greater than or equal to n/2. Due date for unrestricted version Supposing jobs are indexed SPT order The problem is unrestricted for d , where = p n + p n–2 + p n–4 +... For unrestricted problem, Algorithm 1* will produce optimal schedule Exercise: When d = 18, is it unrestricted? When d = 17? Job j123456 pjpj 134679
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10 Basic E/T Problem, Unrestricted Due dates as decision One way of finding an optimal solution Set d = and utilize algorithm 1* optimal total penalty f(S) common due date d
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11 Restricted Version Basic E/T problem, restricted (d ) Optimal solution may contain a straddling job Theorem 1 and 2 hold, but Theorem 3 does not V-shaped schedules still constitute a dominant set Should optimal schedule start at time zero always? Three jobs with p 1 = 1, p 2 = 1, p 3 = 10, and d = 5 Optimal schedule, in which either the schedule starts at time zero, or some job completes exactly at the due date NP-hardness A dynamic programming technique (Hall et al., 1991) Solving problems with several hundreds of jobs
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12 Restricted Version An effective heuristic: S-A heuristic (Sundararaghavan and Ahmed, 1984) Assuming p 1 p 2 ... p n. 1.Let L = d and R = i=1 n p i – d. Let k = 1. 2.If L R, assign job k to the first available position in sequence and decrease L by p k. Otherwise, assign job k to the last available position in sequence and decrease R by p k. 3.If k n, increase k by 1 and go to Step 2. Otherwise, stop. Exercise Find good sequence for the jobs below with d = 90. Job j123456 pjpj 11011485053
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13 Restricted Version Adjustment of start time Delay of start time leads to reduction in total penalty, when e n/2 where e is number of jobs that finish before due date Schedule 6-3-2-1-4-5 of jobs below with d = 90 Job j123456 pjpj 11011485053
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14 Different Earliness and Tardiness Penalties A generalization of basic model Minimize f(S) = j=1 n ( E j + T j ) where -- holding cost (endogenous), -- tardiness penalty (exogenous) Properties of optimal solution Theorem 1, 2, and 3 hold Components of objective function C B = 0 p B1 + 1 p B2 +... + (b – 2) p B(b–1) + (b – 1) p Bb. C A = a p A1 + (a – 1) p A2 +... + 2 p A(a–1) + 1 p Aa. Algorithm 2: E/T with different earliness and tardiness penalties 1.Initially, sets B and A are empty, and jobs are in LPT order. 2.If |B| (1 + |A|), then assign the next job to B; otherwise, assign the next job to A. 3.Repeat Step 2 until all jobs have been scheduled. Exercise: consider jobs below with = 5, = 2, and d = 24. Job j123456 pjpj 134679
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15 Different Earliness and Tardiness Penalties Generalization of Theorem 4 In the basic E/T model with earliness penalty and tardiness penalty , there is an optimal schedule in which the bth job in the sequence completes at time d, where b is the smallest integer greater than or equal to n /( + ). Criterion for unrestricted version = p B1 + p B2 +... + p B(b–1) + p Bb Condition for delaying start of schedule e n /( + ) Effectiveness of modified S-A heuristic Tested by randomly generated problems = Problem SizeAverage ErrorNo. of OptimaAverage ErrorNo. of Optima n = 8 n = 10 n = 12 n = 15 0.40% 0.24% 0.26% 0.32% 10 9 4 1.52% 0.84% 0.66% 0.07% 5 7 10
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16 Quadratic Penalties Avoiding large deviations from due date Minimize f(S) = j=1 n (C j – d) 2 = j=1 n (E j 2 + T j 2 ) Due date d as decision variable d = = j=1 n C j /n Quadratic E/T problem, unrestricted f(S) = j=1 n (C j – ) 2 Problem of minimizing variance of completion times, but not easily solvable A heuristic solution (Vani and Raghavachari, 1987) Neighborhood search using pairwise interchanges
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17 Job Dependent Penalties Permitting each job to have its own penalties f(S) = j=1 n ( j E j + j T j ) NP-hardness A dynamic programming technique (Hall and Posner, 1991) Solving problems with hundreds of jobs in modest run times Generalization of Theorem 1–4 1.There is no inserted idle time. 2.Jobs that complete on or before the due date can be sequenced in non- increasing order of the ratio p j / j, and jobs that start late can be sequenced in non-decreasing order of the ratio p j / j. 3.One job completes at time d. 4.In an optimal schedule the bth job in sequence completes at time d, where b is the smallest integer satisfying the inequality i B ( j + j ) j=1 n j
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18 Distinct Due Dates Different due dates in job set f(S) = j=1 n ( j (d j – C j ) + + j (C j – d j ) + ) = j=1 n ( j E j + j T j ) NP-hardness T-problem reduces to this problem A solution technique Decomposing into two subproblems Finding a good job sequence Scheduling inserted idle time Solvable in polynomial time Refer to p. 74 of Pinedo, 2009 A neighborhood search (Armstrong and Blackstone, 1987) A branch-and-bound procedure (Darby-Dowman and Armstrong, 1986)
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19 Summary Earliness/tardiness problem From JIT concepts Nonregular performance measure Properties Optimum is describable by a sequence of jobs and a start time of 1st job V-shaped schedule 2 n candidates instead of n! candidates Restricted vs. unrestricted versions Difficulties in finding good schedules with tight due date Extended models Job-dependent penalty and due dates ...
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