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Single-Machine Sequencing with Independent Jobs. Chapter 2 Elements of Sequencing and Scheduling by Kenneth R. Baker Byung-Hyun Ha. Outline. Introduction Preliminaries Problem without due dates: elementary results Flowtime and inventory Minimizing total flowtime - PowerPoint PPT Presentation
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Single-Machine Sequencing with Independent Jobs
Chapter 2
Elements of Sequencing and Schedulingby Kenneth R. Baker
Byung-Hyun Ha
2
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
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
A way to 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
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 advanceC4. One machine is continuously available and never kept idle while work is
waitingC5. Once an operation begins, it proceeds without interruption (i.e. no
preemption is allowed)
5
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
Preliminaries Job descriptors
Input to scheduling problem Examples
• pj -- processing time• Amount of processing required by job j pj will generally include both processing time and facility setup time und
er C2• rj -- ready time
• Time at which job j is ready (or available) for processing• rj can be regarded as arrival time, but rj = 0 under C1
• dj -- due date• Time at which processing of job j is due to be completed
7
Preliminaries Information for evaluating schedule
Output from scheduling Examples
• Cj -- completion time• Time at which processing of job j is finished
• Fj -- flowtime• Time job j spends in system; Fj = Cj – rj
• Response of system to individual demands for service (turnaround time)• Lj -- lateness
• Amount of time by which completion time of job j exceeds its due date; Lj = Cj – dj
• Conformity of schedule to given due date• Negative value whenever job is completed early
• Tj -- tardiness
• Lj if job j fail to meet its due date, 0 otherwise; Tj = max{0, Lj}
• Uj -- unit penalty
• Uj = 1, if Cj dj; 0, otherwise
8
Preliminaries Due date related penalty functions
Lj
Cjdj
Tj
Cjdj
Uj
Cjdj
1
9
Preliminaries Performance measures
1-dimensional aggregated quantities about all jobs• Used generally for evaluation of schedules
Functions of set of completion time in schedule, usually• Z = f(C1, C2, ..., Cn)
Examples• Total flowtime -- F = j=1
n Fj
• Total tardiness -- T = j=1n Tj
• Maximum flowtime -- Fmax = max1jn {Fj}
• Maximum tardiness -- Tmax = max1jn {Tj}
• Number of tardy jobs, or total unit penalty -- U = j=1n Uj
• Mean flowtime, mean tardiness, proportion of tardy jobs Problem description
F-problem -- minimization of total flowtime T-problem Cmax-problem
10
Preliminaries 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 increa
ses(b) Z' Z Cj' Cj for some job j, where Z = f(C1, C2, ..., Cn) is value of a meas
ure of schedule S and Z' = f(C1', C2', ..., Cn') is value of the same measure of some different schedule S'
(b) Cj' Cj for all job j Z' Z , where ... Measures that presented in the previous slide are all regular
11
Preliminaries Dominant set
A set of schedules is dominant if it is sufficient to be considered for finding solution
• Reduced search space from complete enumeration Verification of dominant set D of schedules for regular measures
1. Consider an arbitrary schedule S (which contains completion times Cj) that excluded from D
2. Show that there exists a schedule S' in D, in which Cj' Cj for all j3. Therefore Z' Z for any regular measure, and so S' is at least as good as S4. Hence, in searching for an optimal schedule, it is sufficient to consider only s
chedules in D Dominant sets for regular measures in basic single-machine problems
Set of permutation schedules (without preemption) C5 could be relaxed Set of schedules without idle time C4 could be relaxed
12
Preliminaries Theorem 1
In the basic single-machine problem, schedule without inserted idle time constitute a dominant set.
Proof sketch of Theorem 1 Suppose schedule S by which machine is idle during interval (a, b). There exists schedule S' such that Cj' = Cj, if Cj a; Cj' = Cj – (b – a), oth
erwise. Cj' Cj for all j. Hence, Z' Z for any regular measure. Therefore, schedules without idle time constitute a dominant set.
Theorem 2 In the basic single-machine problem, schedules without preemption cons
titute a dominant set.
HOMEWORK #1 Prove Theorem 2 RIGOROUSLY
13
Problem without Due Dates: Elementary Results Flowtime and inventory Minimizing total flowtime Minimizing total flowtime: the weighed version
14
Flowtime and Inventory
Two closely related objectives Rapid turnaround of customers -- minimizing total flowtime Low inventory levels -- minimizing average # of jobs in system
Notations J(t) -- number of jobs in system at time t, J -- time average of J(t) Fmax = p1 + p2 + ... + pn -- makespan
A = np[1] + (n – 1)p[2] + ... + 2p[n-1] + p[n] A = JFmax = F F = F[1] + F[2] + ... + F[n]
n
n – 2n – 1
21
J(t)
p[1] p[2] p[3] p[n-1] p[n]... t
n
n – 2n – 1
21
J(t)
p[1] p[2] p[3] p[n-1] p[n]... t
15
Flowtime and Inventory
Relationship between flowtime and inventory J = F/Fmax -- 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
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
Minimizing Total Flowtime
Equivalent problem to F-problem Minimizing the area under J(t) function Using shortest processing time (SPT) sequencing
• Sequencing jobs in nondecreasing order of processing times
n
n – 2n – 1
21
J(t)
p[1] p[2] p[3] p[n-1] p[n]... t
17
Minimizing Total Flowtime
Theorem 3 Total flowtime is minimized by Shortest Processing Time (SPT) sequenci
ng (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 th
at pi pj . Construct sequence S' in which job i and j are interchanged. Fi + Fj Fj' + Fi' implies that F F', which is a contradiction. Therefore, for all S, if S is optimal, then S is SPT sequence.
Si j
S'ij
FjFi
Fj'Fi'
...
... ...
...
18
Minimizing Total Flowtime
Proof by contradiction Method
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• Suppose not. That is, suppose there is a greatest integer N.• 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.
19
Minimizing Total Flowtime
Theorem 3, revisited Predicates
• O(S) -- S is optimal, SPT(S) -- S is 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. That is, suppose there is S such that not SPT(S) but O(S).• Then, S', F F' by .• In S, there exists a pair of adjacent jobs, i and j, with j following i, such that pi
pj .• Let S' be the same sequence with S except job i and j are interchanged.• F F' since Fi + Fj Fj' + Fi' .• So S is not optimal by , which is a contradiction.
20
Minimizing Total Flowtime
Proof of Theorem 3 by construction1. Begin with any non-SPT sequence2. Find a pair of adjacent jobs i and j, with j following i, such that pi pj .
3. Interchange jobs i and j in sequence4. Return to Step 2 iteratively, improving the performance measure each ti
me, until eventually the SPT sequence is constructed.
Another perspective of Theorem 3 Total flowtime as scalar product of two vectors
j=1n Fj = j=1
n i=1j 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
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
Minimizing Total Flowtime: Weighted Version
Flowtime and inventory Fw = j=1
n wjFj -- total weighted flowtime V(t) -- total value of inventory in the system at time t V -- time average of V(t) A = j=1
n p[j] i=jn w[i] = VFmax = Fw = j=1
n wjFj
• Sequence which minimizes one will minimize the other
wj
w[n]
V(t)
p[1] p[2] p[3] p[n-1] p[n]... t
w[n] + w[n-1]
23
Minimizing Total Flowtime: Weighted Version
Theorem 4 Total weighted flowtime is minimized by Shortest Weighted Processing T
ime (SWPT) sequencing (p[1]/w[1] p[2]/w[2] ... p[n]/w[n]).
HOMEWORK #2 Prove Theorem 4
24
Problem with Due Dates: Elementary Results Lateness criteria Minimizing the number of tardy jobs Minimizing total tardiness Due dates as decisions
25
Lateness Criteria Total lateness and maximum lateness
Assuming 3 jobs {1, 2, 3}• p1 = 5, p2 = 2, p3 = 3
• d1 = 6, d2 = 4, d3 = 8
• r1 = r2 = r3 = 0 Sequencing
• (2, 1, 3) Lateness (Cj – dj)
• 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, Lmax = 2
Theorem 5 Total lateness is minimized by SPT sequencing.
Proof sketch of Theorem 5 L = j=1
n Lj = j=1n (Cj – dj) = j=1
n (Fj – dj) = j=1n Fj – j=1
n dj .
2 31
2 7 10
26
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 adjacient jobs i and j, where dj di but Ci Cj . max{Li, Lj} max{Li', Lj'} Lmax of S Lmax of S'
S i j
Cj
Ci
... ...
di
dj
S' j i
Cj'
... ...
di Ci'
dj
Li
Lj
Lj'
Li'
S i j
Cj
Ci
... ...
di
dj
S' j i
Cj'
... ...
di Ci'
dj
Li
Lj
Lj' Li'
S i j
Cj
Ci
... ...
dj
di
S' j i
Cj'
... ...
Ci' di
Lj
Li
Lj' Li'
dj
27
Lateness Criteria Another measure of urgency
Slack time -- sj = dj – t – pj
• 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 is maximi
zed by MST sequencing
28
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
i j k l m... ...
di
dj
dk
dl
dm
i j...
di dj
...
29
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
... ...
B A
Early jobs Late jobs
longest ... ...... [k-1]
d[k-1] d[k]
30
Minimizing the Number of Tardy Jobs Proof of Optimality of Algorithm 1
See Pinedo, 2008, p. 48.
Exercise Get optimal sequence which minimize U.
Weighted version of U-problem NP-hard, i.e., as difficult as, or even more difficult than, NP-Complete
Job j 1 2 3 4 5
pj
dj
12
78
69
410
312
31
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 problem
First simple and possible analysis Consider only two adjacent jobs, i and j, and figure out the appropriate or
der of decreasing total tardiness Let
• Tij = Ti(S) + Tj(S) = max{p(B) + pi – di, 0} + max{p(B) + pi + pj – dj, 0}
• Tji = Tj(S') + Ti(S') = max{p(B) + pj – dj, 0} + max{p(B) + pi + pj – di, 0}
i j... ...
B AS
j i... ...
B AS'
p(B) p(B)
32
Minimizing Total Tardiness Agreeability
Two set of parameters, uj and vj are agreeable if ui uj implies vi vj
Case 1: agreeable processing time and due dates (pi pj, di dj) Remember
• Tij = max{p(B) + pi – di, 0} + max{p(B) + pi + pj – dj, 0}
• Tji = max{p(B) + pj – dj, 0} + max{p(B) + pi + pj – di, 0}
Case 1.1: p(B) + pi di
• Tij = max{p(B) + pi + pj – dj, 0}
• Case 1.1.1: p(B) + pj dj , p(B) + pi + pj di
• Tji = max{p(B) + pi + pj – di, 0} max{p(B) + pi + pj – dj, 0} = Tij
• Case 1.1.2: p(B) + pj dj , p(B) + pi + pj di
• Tji = max{p(B) + pj – dj, 0} max{p(B) + pi + pj – dj, 0} = Tij
• Case 1.1.3: p(B) + pj dj , p(B) + pi + pj di
• Tij – Tji = (p(B) + pi + pj – dj) – (p(B) + pj – dj) – (p(B) + pi + pj – di) = –p(B) – pj + di –p(B) – pi + di 0
• Therefore, Case 1.1 yield Tij Tji , so it is preferable that job j precede job i.
33
Minimizing Total Tardiness Case 1: pi pj, di dj (cont’d)
Remember• Tij = max{p(B) + pi – di, 0} + max{p(B) + pi + pj – dj, 0}
• Tji = max{p(B) + pj – dj, 0} + max{p(B) + pi + pj – di, 0}
Case 1.2: p(B) + pi di
• Tij = p(B) + pi – dj + p(B) + pi + pj – dj
• Tji = max{p(B) + pj – dj, 0} + p(B) + pi + pj – di
• Tij – Tji = p(B) + pi – dj – max{p(B) + pj – dj, 0}
• Case 1.2.1: p(B) + pj dj
• Tij – Tji = p(B) + pi – dj p(B) + pi – di 0
• Case 1.2.2: p(B) + pj di
• Tij – Tji = p(B) + pi – dj – (p(B) + pj – dj) = pi – pj 0
• Therefore, Case 1.2 yield Tij Tji , so it is preferable that job j precede job i.
Theorem 8 For any pair of jobs, suppose that processing times and due dates are agreeabl
e. Then total tardiness (T) is minimized by SPT (or, equivalently, EDD) sequencing.
34
Minimizing Total Tardiness Case 2: parameters are not agreeable (pi pj, di dj)
Remember• Tij = max{p(B) + pi – di, 0} + max{p(B) + pi + pj – dj, 0}
• Tji = max{p(B) + pj – dj, 0} + max{p(B) + pi + pj – di, 0}
Case 2.1: p(B) + pi di
• Tji = max{p(B) + pi + pj – di, 0} max{p(B) + pi + pj – dj, 0} = Tij
• Therefore, Case 2.1 yield Tji Tij , so it is preferable that job i (with earlier due date) precede job j.
Case 2.2: p(B) + pi di
• Case 2.2.1: p(B) + pi + pj dj
• Tij = p(B) + pi – di
• Tji = p(B) + pi + pj – di Tij
• Therefore, Case 2.2.1 yields Tji Tij , so it preferable that job i (with earlier due date) precede job j.
35
Minimizing Total Tardiness Case 2: (pi pj, di dj) (cont’d)
Remember• Tij = max{p(B) + pi – di, 0} + max{p(B) + pi + pj – dj, 0}
• Tji = max{p(B) + pj – dj, 0} + max{p(B) + pi + pj – di, 0}
Case 2.2: p(B) + pi di (cont’d)• Case 2.2.2: p(B) + pi + pj dj
• Case 2.2.2.1: p(B) + pj dj
Tij = p(B) + pi – di + p(B) + pi + pj – dj
Tji = p(B) + pi + pj – di
Tij – Tji = p(B) + pi – dj
Therefore, Case 2.2.2.1 yields that it is preferable that job i precede job j unless p(B) + pi dj , where job j (shorter job) may precede job i.
• Case 2.2.2.2: p(B) + pj dj
Tij = p(B) + pi – di + p(B) + pi + pj – dj
Tji = p(B) + pj – dj + p(B) + pi + pj – di
Tij – Tji = pi – pj 0
Therefore, Case 2.2.2.2 yield Tij Tji , so it is preferable that job j (shorter job) precede job i.
36
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{pi, pj} max{di, dj},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 j 1 2 3
pj
dj
38
86
310
37
Minimizing Total Tardiness Modified due date (MDD)
dj' = max{dj, t + pj}
MDD priority rule If jobs i and j are the candidates to begin at time t, then the job with the earli
er modified due date should come first. MDD priority rule is consistent with Theorem 9.
HOMEWORK #3 Prove MDD priority rule is consistent with Theorem 9. Hint: divide into cases and conquer, e.g.,
Case 1: agreeable, i.e., pi pj, di dj :
t + max{pi, pj} = t + pj, max{di, dj} = dj
Case 1.1: t + pj dj job i comes first according to Theorem 9
dj' = max{dj, t + pj} = t + pj
Case 1.1.1: di t + pi
di' = max{di, t + pi} = di dj t + pj = dj' job i comes first from the rule
Case 1.1.2: ...
38
Minimizing Total Tardiness 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.
39
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 (rj) Selection of due date depends only on information about the job itself
Possible rules selecting due date CON -- constant flow allowance: dj = rj + SLK -- equal slack flow allowance: dj = rj + pj + TWK -- total work flow allowance: dj = rj + pj
Some results CON due dates are dominated by either SLK or TWK due dates TWK will tend to be the best rule most of time
40
Summary Basic single-machine model
Fundamental in the study of sequencing and scheduling
Scheduling objectives Solution strategies
Using simple sequencing rules, such as SPT and EDD Using more intricate construction for U-problem We could not solve T-problem