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This chapter discusses the theoretical frameworks and algorithms for minimizing tardy jobs and total tardiness in single machine scheduling problems. It introduces various algorithms, such as the EDD rule and dynamic programming approaches, that help manage job scheduling effectively. Additionally, it delves into the classification of jobs into tardy and on-time categories, providing examples and computational complexities associated with these scheduling models. Understanding these principles aids managers in optimizing job scheduling to improve efficiency and reduce delays.
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IOE/MFG 543 Chapter 3: Single machine models (Sections 3.3-3.5)
Section 3.3: Number of tardy jobs 1||S Uj • Number of tardy jobs • Often used as a benchmark for managers (or % of on-time jobs) • Some jobs may have to wait really long
Characterization of an optimal schedule for 1||S Uj • Split jobs into two sets J=Jobs that are completed on time Jd=Jobs that are tardy • The jobs in J are scheduled according to an EDD rule • The order in which the jobs in Jd are completed is immaterial (since they are already tardy)
Algorithm 3.3.1 for 1||S Uj • Set J=, Jd=, Jc ={1,…,n} (set of jobs not yet considered for scheduling) • Determine job j* in Jc which has the EDD, i.e, dj*=min{dj : jJc} Add j* to J Delete j* from Jc Go to 3
Algorithm 3.3.1 for 1||S Uj (2) • If due date of job j* is met, i.e., if SjJ pj ≤ dj* go to 4, otherwise let k* be the job in J which has the longest processing time, i.e., pk*=maxjJ {pj} Delete k* from J Add k* to Jd • If Jc= STOP otherwise go to 2.
Algorithm 3.3.1 for 1||S Uj (3) • Computational complexity O(nlog(n)) • If it is implemented efficiently! • Theorem 3.3.2 • Algorithm 3.3.1 yields an optimal schedule for 1||S Uj • Proof: By induction
Example 3.3.3 • Use Algorithm 3.3.1 to determine the schedule that minimizes S Uj • How many jobs are tardy?
Total weighted number of tardy jobs 1||S wjUj • NP-hard • No polynomial time algorithm exists • If all jobs have the same due dates • i.e., d=dj for all jobs • Knapsack problem=> pseudopolynomial • Why not use WSPT? • Example 3.3.4 • What happens when WSPT is used? • What about 2-3-1?
Section 3.4: The total tardiness 1||S Tj • A more practical performance measure than the number of tardy jobs • May schedule the tardy jobs to minimize total tardiness • The problem is NP-hard in the ordinary sense • A pseudopolynomial dynamic programming algorithm exists
Dynamic programming • See Appendix B • Dynamic programming is an efficient sequential method for total enumeration • For 1||S Tj we can use Lemmas 3.4.1. and 3.4.3. to eliminate a number of schedules from consideration
Lemma 3.4.1. • If pj≤pk and dj≤dk,then there exists an optimal sequence in which job j is scheduled before job k • Proof: The result is fairly obvious, so we omit the proof (Good exercise!) • Consequences: • We can eliminate from consideration all sequences that do not satisfy this condition
Lemma 3.4.2. • For some job k let C'k be the latest completion time of job in any optimal sequence S' • Consider two sets of due dates: d1, …, dn and d1, …,dk-1,max(dk,C'k), dk+1, dn • S' is optimal for the first set and S'' is optimal for the second set • Lemma: Any sequence that is optimal for the second set of due dates is optimal for the first set as well • Proof: Skip
Enumerate the jobs by increasing due dates • Assume d1≤d2≤…≤dn • Let job k be such that pk=max(p1,…,pn) • Lemma 3.4.1=> there is an optimal sequence such that jobs 1,…,k-1 are scheduled before job k • The n-k jobs k+1,…,n are scheduled either before or after job k
Lemma 3.4.3 • There exists an integer d, 0≤d≤n-k, such that there is an optimal sequence S in which job k is preceded by all jobs j with j≤k+d and followed by all jobs j with j>k+d • Effectively reduces the number of schedules to be considered • Proof: Skip
Consequences of Lemma 3.4.3. • There is an optimal sequence that processes the jobs in this order: • jobs 1,2,…,k-1,k+1,…,k+d in some order • job k • jobs k+d+1, k+d+2, …, k+n in some order • How do we determine this sequence? • Algorithm 3.4.4.
Notation for Algorithm 3.4.4. • Ck(d)=Sj≤k+dpj is the completion time of job k • J(j,l,k) is the set of jobs in {j,j+1,…,l} that have a processing time less than or equal to job k and excludes job k itself • V(J(j,l,k),t) is the total tardiness of the jobs in J(j,l,k) under an optimal sequence assuming that this set starts processing at time t • k' is the job in J(j,l,k) with the largest processing time, i.e., pk'=max{pj : j J(j,l,k)}
Algorithm 3.4.4. for 1||S Tj • Initial conditions V(Ø,t)=0 V({j},t)=max(0,t+pj-dj) • Recursive relation V(J(j,l,k),t)=mind{V(J(j,k'+d,k'),t) +max(0,Ck'(d)-dk')+V(J(k'+d+1,l,k'),Ck'(d))} • Optimal value function: V({1,…,n},0)
The complexity of Algorithm 3.4.4. • At most n3 subsets J(j,l,k) • At most Spj time points • =>at most n3Spj recursive equations • Each equation requires at most O(n) operations • Overall complexity is O(n4Spj) => pseudopolynomial
Example 3.4.5. • Given the following data, use Algorithm 3.4.4. to solve the problem 1||S Tj
Section 3.5.The total tardiness 1||S wjTj • Thm. 3.5.2: The problem 1||S wjTj is strongly NP-hard • 3 partition reduces to 1||S wjTj • A dominance result exists. Lemma 3.5.1: • If there are two jobs j and k with dj≤dk , pj≤pk and wj≥wk then there is an optimal sequence in which job j appears before job k. • Branch and bound for 1||S wjTj • Branch: Start by scheduling the last jobs • Bound: Solve a transportation problem
