Single-Machine Sequencing with
Independent Jobs
Chapter 2
Elements of Sequencing and Scheduling
by Kenneth R. Baker
Byung-Hyun Ha
R3
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
1
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
 Examples to be employed for 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
2
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 advance
C4. One machine is continuously available and never kept idle while work
is waiting
C5. Once an operation begins, it proceeds without interruption (i.e. no
preemption is allowed)
3
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
4
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
under 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
5
Preliminaries
 Information for evaluating schedule
 Output from applying schedule
 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
6
Preliminaries
 Due date related penalty functions
Lj
Tj
dj
Cj
dj
Cj
Uj
1
dj
Cj
7
Preliminaries
 Performance measures
 1-dimensional aggregated quantities about all jobs
• Commonly used for evaluation of schedules
 Functions of set of completion time in schedule, usually
• Z = f(C1, C2, ..., Cn)
 Examples
•
•
•
•
•
•
Total flowtime -- F = j=1n Fj
Total tardiness -- T = j=1n Tj
Maximum flowtime -- Fmax = max1jn {Fj}
Maximum tardiness -- Tmax = max1jn {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
8
Preliminaries
 Definition -- 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
increases
(b) Z'  Z  Cj'  Cj for some job j, where Z = f(C1, C2, ..., Cn) is value of a
measure 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
 The measures presented in the previous slide are all regular
• Then, non-regular measures?
9
Preliminaries
 Definition -- dominant set
 A set of schedules is dominant if it is sufficient to be considered for
finding solution
• Reduced search space from complete enumeration
 Verifying dominant set D of schedules for regular measures
1. Consider an arbitrary schedule S (which contains completion times Cj)
that is excluded from D
2. Show that there exists a schedule S' in D, in which Cj'  Cj for all j
3. Hence, Z'  Z for any regular measure, and so S' is at least as good as S
4. Therefore, in searching for an optimal schedule, it is sufficient to consider
only schedules in D
 Basic single-machine problems

Dominant sets for regular measures
 Set of schedules without idle time  C4 could be relaxed
 Set of permutation schedules (without preemption)  C5 could be relaxed
10
Preliminaries
 Suppose we consider regular measures
 Theorem 1
 In the basic single-machine problem, schedule without inserted idle time
constitute a dominant set.
 Theorem 2
 In the basic single-machine problem, schedules without preemption
constitute a dominant set.
 HOMEWORK #1
 Prove Theorem 2
11
Problem without Due Dates: Elementary Results
 Outline
 Flowtime and inventory
 Minimizing total flowtime
• F-problem -- min. F = j=1n Fj
• 1 || Cj
 Minimizing total flowtime: the weighed version
• Fw-problem -- min. Fw = j=1n wjFj
• 1 || wjCj
12
Flowtime and Inventory
 Two closely related objectives
 Low inventory levels -- minimizing average # of jobs in system
 Rapid turnaround of customers -- minimizing total flowtime
 Notations
 J(t) -- number of jobs in system at time t, J -- time average of J(t)
 Fmax = F[n] = p1 + p2 + ... + pn -- makespan
J(t)
J(t)
n
n–1
n–2
n
n–1
n–2
2
2
1
1
p[1]
p[2]
p[3]
...
p[n-1] p[n]
t
A = np[1] + (n – 1)p[2] + ... + 2p[n-1] + p[n]
p[1]

A = JFmax = F
p[2]

flowtime-inventory relation
p[3]
...
p[n-1] p[n]
t
F = F[1] + F[2] + ... + F[n]
13
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
In situations where inventory costs may vary among jobs
 Emphasis on F-problem and its generalization in theory
 Because total flowtime actually encompasses a broader range of
scheduling-related costs
 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
14
Minimizing Total Flowtime
 Minimizing the area under J(t) function
 Equivalent problem to F-problem
 Steepest slope to the left, the next steepest slope, ...
 Using shortest processing time (SPT) sequencing
• Sequencing jobs in nondecreasing order of processing times
J(t)
n
n–1
n–2
2
1
p[1]
p[2]
p[3]
...
p[n-1] p[n]
t
15
Minimizing Total Flowtime
 Theorem 3
 Total flowtime is minimized by Shortest Processing Time (SPT)
sequencing (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
that 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.
S
S'
...
p(B)
i
...
j
Ci
Cj
...
p(B)
j
...
i
Cj'
Ci'
16
Minimizing Total Flowtime
 Proof by contradiction
 General procedure
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.
1. Suppose not. That is, suppose there is a greatest integer N.
2. 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.
3. Therefore, there is no greatest integer.
17
Minimizing Total Flowtime
 Theorem 3, revisited
 Predicates
• O(S) -- S is optimal, SPT(S) -- S is in 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. Then, by , there exists S such that O(S)  SPT(S).
• Then, since O(S), S', F  F' by .
• Because SPT(S), 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' because Fi + Fj  Fj' + Fi' . That is S', F  F'.
• So O(S) by , which is a contradiction to O(S).
18
Minimizing Total Flowtime
 Proof of Theorem 3 by construction
1.
2.
3.
4.
Begin with any non-SPT sequence.
Find a pair of adjacent jobs i and j, with j following i, such that pi  pj .
Interchange jobs i and j in sequence.
Return to Step 2 iteratively, improving the performance measure each
time, until eventually the SPT sequence is constructed.
 Another perspective of Theorem 3
 Total flowtime as scalar product of two vectors
• j=1n Fj = j=1n i=1j p[i] = j=1n (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
19
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
20
Minimizing Total Flowtime: Weighted Version
 Flowtime and inventory
 Fw = j=1n wjFj -- total weighted flowtime
 V(t) -- total value of inventory in the system at time t
• V -- time average of V(t)
 Generalized flowtime-inventory relation
• A = j=1n p[j] i=jn w[i] = VFmax = Fw = j=1n wjFj
• Sequence which minimizes one will minimize the other
V(t)
wj
w[n] + w[n-1]
w[n]
p[1]
p[2]
p[3]
...
p[n-1] p[n]
t
21
Minimizing Total Flowtime: Weighted Version
 Theorem 4
 Total weighted flowtime is minimized by Shortest Weighted Processing
Time (SWPT) sequencing (p[1]/w[1]  p[2]/w[2]  ... p[n]/w[n]).
 HOMEWORK #2
 Prove Theorem 4
22
Problem with Due Dates: Elementary Results
 Outline
 Lateness criteria
• Lmax-problem -- min. Lmax = maxj=1,...,n {Lj}
• 1 || Lmax
+ 1 || Lj , 1 || Tmax , 1 || Tmax(1) , Cj(2)
 Minimizing the number of tardy jobs
• U-problem -- min. U = j=1n Uj
• 1 || Uj
+ 1 || wjUj
 Minimizing total tardiness
• T-problem -- min. T = j=1n Tj
• 1 || Tj
 Due dates as decisions
23
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)
2
1
2
3
7
10
 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=1n Lj = j=1n (Cj – dj) = j=1n (Fj – dj) = j=1n Fj – j=1n dj
24
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 adjacent jobs i and j, where dj  di but Ci  Cj .
 max{Li, Lj} = Lj  max{Li', Lj'}  Lmax  Lmax'
S
S'
...
p(B)
i
...
j
Ci
Cj
...
p(B)
j
...
i
Ci'
Cj'
25
Lateness Criteria
 Another measure of urgency
 Slack time at time t -- sj = dj – t – pj
• Among jobs with identical due dates, 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 (Lmin) is
maximized by MST sequencing
 Lmin is not a regular measure  restricted to schedules without inserted
idle time
26
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
...
i
j
k
l
...
m
dk
dj
dl
di
dm
 EDD may not be optimal if there are more than one tardy jobs
...
i
j
di
...
dj
27
Minimizing the Number of Tardy Jobs
 Optimal sequence form of U-problem
 Set B of early jobs and set A of late jobs
B
A
...
...
Early jobs
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.
...
longest
...
[k-1]
d[k-1]
[k]
d[k]
...
...
28
Minimizing the Number of Tardy Jobs
 Proof of optimality of Algorithm 1
 See p. 48 of Pinedo, 2008
 Exercise
 Get optimal sequence which minimize U.
Job j
1
2
3
4
5
pj
dj
1
2
7
8
6
9
4
10
3
12
 Weighted version of U-problem
 NP-hard, i.e., as difficult as, or even more difficult than, NP-Complete
29
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
 Adjacent-jobs analysis -- simple and possible analysis
 Consider only two adjacent jobs, i and j, and figure out the appropriate
order 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}
S
B
A
...
...
i
p(B)
j
B
S'
A
...
j
p(B)
i
...
30
Minimizing Total Tardiness
 Adjacent-jobs analysis (cont’d)
 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)
• Case 1.1: p(B) + pi  di
• It is preferable that job j precede job i.
• Case 1.2: p(B) + pi  di
• It is preferable that job j precede job i.
 Theorem 8
• For any pair of jobs, suppose that processing times and due dates are
agreeable. Then total tardiness (T) is minimized by SPT (or, equivalently,
EDD) sequencing.
31
Minimizing Total Tardiness
 Adjacent-jobs analysis (cont’d)
 Case 2: parameters are not agreeable (pi  pj, di  dj)
• Case 2.1: p(B) + pi  di
• 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
It preferable that job i (with earlier due date) precede job j.
• Case 2.2.2: p(B) + pi + pj  dj
» Case 2.2.2.1: p(B) + pj  dj
It is preferable that job i precede job j unless p(B) + pi  dj , in
which case job j (shorter job) may precede job i.
» Case 2.2.2.2: p(B) + pj  dj
It is preferable that job j (shorter job) precede job i.
32
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
3
8
8
6
3
10
33
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
earlier modified due date should come first.
 MDD priority rule is consistent with Theorem 9.
 Other 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.
 ...
34
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
35
Summary
 Basic single-machine model
 Fundamental in the study of sequencing and scheduling
 Scheduling objectives
 Flowtime-related, lateness-related, ...
 Problems




1 || Cj , 1 || wjCj
1 || Lj , 1 || Lmax , 1 || Tmax , 1 || Tmax(1) , Cj(2)
1 || Uj , 1 || wjUj
1 || Tj
 Solution strategies
 Using simple sequencing rules, such as SPT and EDD
 Using more intricate construction for U-problem
 We could not solve T-problem
36