IOE/MFG 543
Chapter 5: Parallel machine
models
(Sections 5.3-5.6)
1
Section 5.3: Total completion
time Pm||S Cj



On a single machine
j p
Cj=Sk=1
(k)
where p(j) is the processing time of
the jth job processed on the machine
Then
SCj=np(1)+(n-1)p(2)+...+p(n)
=> Shortest Processing Time first
rule minimizes SCj
2
Total completion time
Pm||S Cj (2)
The same argument can be
extended to parallel machines
 Theorem 5.3.1

– The SPT rule is optimal for Pm||S Cj

In fact, a number of optimal
schedules exist
3
Weighted total completion
time Pm||S wjCj



WSPT rule is not always optimal
In practice it usually does pretty well
Worst case bound
S wjCj(WSPT) 1
< (1+√2) ≈ 1.2
S wjCj(OPT) 2
4
Other completion time
models

Precedence constraints Pm|prec|SCj
– Strongly NP-hard

Non-identical machines Rm||SCj
– Can be formulated as an integer program
– The solution of the corresponding linear
program gives an optimal schedule
=> Can be solved in polynomial time
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Section 5.4: Preemptions
Pm|prmp|S Cj

The SPT non-preemptive rule is still
optimal
– A special case of a more general result
for Qm|prmp|S Cj

Recall: vi is the speed on machine i in
Qm models
– pj can be interpreted as the required
work to complete job j (processing time
= pj/vi)
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Qm|prmp|S Cj

SRPT-FM rule:
– Process the jobs such that the job with the
shortest remaining processing time is put on the
fastest machine. The job with the second
shortest remaining processing time is put on the
second fastest machine, etc.
– Whenever the fastest machine completes a job,
all the remaining jobs are moved up on the
machines

Theorem 5.4.2
– The SRPT-FM rule is optimal for Qm|prmp|S Cj
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Example 5.4.3
Qm|prmp|S Cj



4 machines
machine i
1
2
3
4
vi
4
2
2
1
3
4
7 jobs
job j
1
2
5
6
7
pj
8 16 34 40 45 46 61
Use the SRPT-FM to solve Qm|prmp|S Cj
8
Section 5.5
Due date related objectives

Single machine problems that can be
solved “easily”
– e.g., 1|prec|Lmax, 1|rj,prmp|Lmax


Single machine tardiness problems
(1||STj) are NP-hard
If all due dates are 0
– Then Pm||Lmax is equivalent to Pm||Cmax
=> Pm||Lmax is NP-hard
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Qm|prmp|Lmax


One of few parallel machine problems
that can be solved in polynomial time
First question: Is there a schedule
such that Lmax ≤ z ?
– This can be formulated (backwards) as
Qm|rj,prmp|Cmax
– Let dj=dj+z be a hard deadline
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Qm|prmp|Lmax (2)

Solve the problem backwards
– Find the job k with the latest deadline
– Let rk=0 and rj=dk-dj

Solve Qm|rj,prmp|Cmax by applying
LRPT-FM
– The backward schedule is optimal for
Qm|prmp|Lmax
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Example 5.5.1
P2|prmp|Lmax

4 jobs

Let z=0

job j
pj
dj
1
3
4
2
3
5
job j
pj
rj
1
3
5
2
3
4
3
3
8
3
3
1
4
8
9
4
8
0
Solve P2|rj,prmp|Cmax by LRPT rule
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Section 5.6 Discussion


Parallel machine models are much
harder than single machine models!
Later in the course we will study
heuristics that can be used to obtain
“good” schedules
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