Single-Machine Sequencing with Independent Jobs

transcript

Chapter 2

Elements of Sequencing and Schedulingby Kenneth R. Baker

Byung-Hyun Ha

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

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

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)

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

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

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

Preliminaries Due date related penalty functions

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

• 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

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

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

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

Problem without Due Dates: Elementary Results Flowtime and inventory Minimizing total flowtime Minimizing total flowtime: the weighed version

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 – 2n – 1

p[1] p[2] p[3] p[n-1] p[n]... t

n – 2n – 1

p[1] p[2] p[3] p[n-1] p[n]... t

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

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 – 2n – 1

p[1] p[2] p[3] p[n-1] p[n]... t

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.

Fj'Fi'

... ...

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.

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.

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

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

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

p[1] p[2] p[3] p[n-1] p[n]... t

w[n] + w[n-1]

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

Problem with Due Dates: Elementary Results Lateness criteria Minimizing the number of tardy jobs Minimizing total tardiness Due dates as decisions

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 7 10

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' j i

... ...

di Ci'

... ...

S' j i

... ...

di Ci'

Lj' Li'

... ...

S' j i

... ...

Ci' di

Lj' Li'

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

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... ...

i j...

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

... ...

Early jobs Late jobs

longest ... ...... [k-1]

d[k-1] d[k]

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

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... ...

j i... ...

p(B) p(B)

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

• Tji = max{p(B) + pj – dj, 0} max{p(B) + pi + pj – dj, 0} = Tij

• 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.

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}

• 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.

Minimizing Total Tardiness Case 2: parameters are not agreeable (pi pj, di dj)

• 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 2.2.1: 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.

Minimizing Total Tardiness Case 2: (pi pj, di dj) (cont’d)

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.

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

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: ...

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.

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

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

Single-Machine Sequencing with Independent Jobs

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