12_singlewcompl

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Single Machine Deterministic Models

Jobs: J1,J2, ...,Jn

Assumptions:

The machine is always available throughout the scheduling period.

The machine cannot process more than one job at a time.

Each job must spend on the machine a prescribed length of time.


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!t

tJktS

k

at timeprocessedisobnoi0

at timeprocessedisobi)(

1

3

2

J1J2 J3 ...

S(t)

t


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Requirements that may restrict the feasibility of schedules:

precedence constraints

no preemptions release dates

deadlines

Whether some feasible schedule exist? NP hard

Objective function fis used to compare schedules.

f(S) < f(S') whenever schedule S is considered to be better than S'

problem of minimising f(S) over the set of feasible schedules.


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1. Completion Time Models

Due date related objectives:

2. Lateness Models

3. Tardiness Models

4. Sequence-Dependent Setup Problems


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Completion Time Models

Contents

1. An algorithm which gives an optimal schedule

with the minimum total weighted completion time

1 || wjCj

2. An algorithm which gives an optimal schedule

with the minimum total weighted completion timewhen the jobs are subject to precedence relationship

that take the form of chains

1 | chain | wjCj


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Literature:

Scheduling, Theory, Algorithms, andSystems, Michael Pinedo,Prentice Hall, 1995, or new: Second Addition,2002, Chapter 3.


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1 || wjCj

Theorem. The weighted shortest processing time first rule (WSPT) is

optimal for 1 || wjCj

WSPT: jobs are ordered in decreasing order ofwj/pj

The next follows trivially:

The problem 1 || Cj is solved by a sequence Swith jobs arranged innondecreasing order of processing times.


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Proof.By contradiction.

S is a schedule, not WSPT, that is optimal.

j and kare two adjacent jobs such that

k

k

j

j

p

w

p

w which implies that wjpk< wkpj

t t+ pj + pk

j kS:

t t+ pj + pk

k jS

......

... ...

S: (t+pj) wj + (t+pj+pk) wk= twj + pj wj + twk + pjwk + pkwkS: (t+pk) wk+ (t+pk+pj) wj = twk+ pkwk + twj + pkwj + pj wj

the completion time forS < completion time forS contradiction!


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1 | chain | wjCj

chain 1: 1 p 2p ... p k

chain 2: k+1 p k+2p ... p n

Lemma. If

!

!

!

!"

n

kjj

n

kjj

k

jj

k

jj

p

w

p

w

1

1

1

1

the chain of jobs 1,...,kprecedes the chain of jobs k+1,...,n.

Let l* satisfy

!

!

!

ee

!

!l

jj

l

j

j

kll

jj

l

j

j

pp1

1

1*

1

*

1 max

V factor of chain 1,...,k

l* is the job that determines the V factor of the chain


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Lemma. If job l* determines V (1,...,k) , then there exists an optimal

sequence that processes jobs 1,...,l* one after another without

interruption by jobs from other chains.

Algorithm

Whenever the machine is free, select among the remaining chains

the one with the highest V factor. Process this chain up to and includingthe job l* that determines its V factor.


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Example

chain 1: 1 p 2p 3 p 4

chain 2: 5p 6p 7

jobs 1 2 3 4 5 6 7

wj 6 18 12 8 8 17 18

pj 3 6 6 5 4 8 10

V factor ofchain 1 is determined by job 2: (6+18)/(3+6)=2.67V factor ofchain 2 is determined by job 6: (8+17)/(4+8)=2.08

chain 1 is selected: jobs 1,2

V factor of the remaining part ofchain 1 is determined by job 3:

12/6=2V factor ofchain 2 is determined by job 6: 2.08

chain 2 is selected: jobs 5,6


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V factor of the remaining part ofchain 1 is determined by job 3: 2

V factor of the remaining part ofchain 2 is determined by job 7:

18

/10=1.8

chain 1 is selected: job 3

V factor of the remaining part ofchain 1 is determined by job 4: 8/5=1.6

V factor of the remaining part ofchain 2 is determined by job 7: 1.8

chain 2 is selected: job 7

job 4 is scheduled last

the final schedule: 1,2,5,6, 3,7,4


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1 | prec | wjCj

Polynomial time algorithms for the more complex precedence

constraints than the simple chains are developed.

The problems with arbitrary precedence relation are NP hard.

1 | rj,prmp | w

jC

jpreemptive version of the WSPT rule

does not always lead to an optimal solution,

the problem is NP hard

1 | rj,prmp | Cj preemptive version of the SPT rule is optimal

1 | rj | Cj is NP hard


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Summary

1 || wjCj WSPT rule

1 | chain | wjCj a polynomial time algorithm is given

1 | prec | wjCj with arbitrary precedence relation is NP hard

1 | rj,prmp | wjCj the problem is NP hard

1 | rj,prmp | Cj preemptive version of the SPT rule is optimal

1 | rj | Cj is NP hard

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