Lateness Models
Contents
1. Lawler’s algorithm which gives an optimal schedule
with the minimum cost hmax
when the jobs are subject to precedence relationship
1 | prec | hmax
2. A branch-and-bound algorithm for the scheduling problems
with the objective to minimise lateness
1 | rj | Lmax
Literature:
• Scheduling, Theory, Algorithms, and Systems, Michael Pinedo,
Prentice Hall, 1995, Chapter 3.2
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or new: Second Addition, 2002, Chapter 3.
Lawler’s Algorithm
• Backward algorithm which gives an optimal schedule for
1 | prec | hmaxhmax = max ( h1(C1), ... ,hn(Cn) )
hj are nondecreasing cost functions
Notation
makespan Cmax =  pj completion of the last job
J
set of jobs already scheduled


C

p
,
C
j
max 
they have to be processed during the time interval  max j
J

JC
J'  JC

complement of set J, set of jobs still to be scheduled
set of jobs that can be scheduled immediately
before set J (schedulable jobs)
2
Lawler’s Algorithm for 1 | | hmax
Step 1.
J=
JC = {1,...,n}
k=n
Step 2.
Let j* be such that







h j*   p j   minC h j   p j 
 jJ C  jJ
 jJ C 
Place j* in J in the k-th order position
Delete j* from JC
Step 3.
If JC =  then Stop
else k = k - 1
go to Step 2
3
Example (no precedence relationships between jobs)
jobs
pj
hj(Cj)
1
2
1+C1
2
3
1.2C2
3
5
10
J=
JC={1, 2, 3} jobs still to be scheduled
Cmax = 10
h1(10) = 11
h2(10) =12
h3(10) =10
Job 3 is scheduled last and has to be processed in [5, 10].
...
3
5
10
4
J = {3}
JC={1, 2} jobs still to be scheduled
Cmax = 5
h1(5) = 6
h2(5) = 6
Either job 1 or job 2 may be processed before job 3.
1
2
5
or
2
3
1
10
3
5
10
Two schedules are optimal: 1, 2, 3 and 2, 1, 3
5
Lawler’s Algorithm for 1 | prec | hmax
Step 1.
J = ,
JC = {1,...,n}
J' the set of all jobs with no successors
k=n
Step 2.
Let j* be such that







h j*   p j   min h j   p j 
 jJ C  jJ '  jJ C 
Place j* in J in the k-th order position
Delete j* from JC
Modify J' to represent the set of jobs which can be scheduled
immediately before set J.
Step 3.
If JC =  then Stop
else k = k - 1
go to Step 2
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Example. What will happen in the previous example if the
precedence 1  2 has to be taken into account?
J=
JC={1, 2, 3} still to be scheduled
J'={2, 3} have no successors
Cmax = 10
h2(10) = 12
h3(10) = 10
...
5
J = {3}
JC={1, 2} still to be scheduled
J'={2} can be scheduled immediately before J
Cmax = 5
h2(5) = 6
J = {3, 2}
JC={1}
h1(2) = 3
Optimal schedule: 1, 2, 3,
h = 10
J'={1}
3
2
2
1
3
5
2
2
10
10
3
5
10
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1 || Lmax is the special case of the 1 | prec | hmax
where hj = Cj - dj
algorithm results in the schedule that orders jobs in increasing order
of their due dates - earliest due date first rule (EDD)
1 | rj | Lmax
is NP hard , branch-and-bound is used
1 | rj , prec | Lmax
similar branch-and-bound
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Branch-and-bound algorithm
• Search space can grow very large as the number of variables
in the problem increases!
• Branch-and-bound is a heuristic that works on the idea of successive
partitioning of the search space.
S
S1
S12
S2
S13
...
Sn
S = S1 S2... Sn
S1 S2...  Sn = 
...
...
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• We need some means for obtaining a lower bound on the cost for
any particular solution (the task is to minimise the cost).
S
fbound f(x), xS1
S1
S12
S2
S13
...
...
Sn
fbound  f(x), xS2
there is no need to explore S2
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Branch-and-bound algorithm
Step 1
Initialise P =  Si (determine the partitions)
Initialise fbound
Step 2
Remove best partition Si from P
Reduce or subdivide Si into Sij
Update fbound
P = P  Sij
For all SijP do
if lower bound of f(Sij) > fbound then remove Sij from P
Step 3
If not termination condition then go to Step 2
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Branch-and-bound algorithm for 1 | rj | Lmax
 Solution space contains n! schedules (n is number of jobs).
Total enumeration is not viable !
*,*,*,*
1,*,*,*
1,2,*,*
1,3,*,*
2,*,*,*
...
n,*,*,*
...
...
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*,*,*,*
1,*,*,*
1,2,*,*
1,3,*,*
2,*,*,*
...
...
n,*,*,*
1st level
2nd level
...
Branching rule:
k-1 level, j1, ... , jk-1 are scheduled,
jk need to be considered if no job still to be scheduled can not
be processed before the release time of jk
that is:
r jk  min (max(t , rl )  pl )
lJ
J set of jobs not yet scheduled
t is time when jk-1 is completed
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Lower bound:
• Preemptive earliest due date (EDD) rule is optimal for
1 | rj prmp | Lmax
A preemptive schedule will have a maximum lateness
not greater than a non-preemtive schedule.
• If a preemptive EDD rule gives a nonpreemptive schedule then
all nodes with a larger lower bound can be disregarded.
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Example.
jobs
pj
rj
dj
1
4
3
4
2
5
0
6
• Non-preemptive schedules
1
2
3
0
2
0
12
7
1
5
9
L1=3
L2=6
Lmax=6
L1=5
L2=-1
Lmax=5
• Preemptive schedule obtained using EDD
L1=3
2
1
2
 the lowest Lmax !
L
=3
2
15
0
3
7
9
Lmax=3
Example
jobs
pj
rj
dj
1
4
0
8
2
2
1
12
4
5
5
10
3,*,*,*
4,*,*,*
*,*,*,*
L.B. = 7
L.B. = 5
1,*,*,*
1,2,*,*
3
6
3
11
2,*,*,*
1,3,*,*
L.B. = 5
L.B. = 6
job 1 could be processed
before job 4
job 2 could be processed
before job 3
1,3,4,2
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*, *, *, *
1,*,*,*
1 [0, 4]
3 [4, 5]
4 [5, 10]
3 [10, 15]
2 [15, 17]
1,2,*,*
1 [0, 4]
2 [4, 6]
4 [6, 11]
3 [11, 17]
L1=-4
L4=0
L3=4
L2=5
L1=-4
L2=-6
L4=1
L3=6
2,*,*,*
2 [1, 3]
1 [3, 7]
4 [7, 12]
3 [12, 18]
L2=-9
L1=-1
L4=2
L3=7
1,3,*,*
1 [0, 4]
3 [4, 10]
4 [10, 15]
2 [15, 17]
L1=-4
L3=-1
L4=5
L3=5
4,*,*,*
either job 1 or 2
can be processed
before 4 !
3,*,*,*
job 2
can be processed
before 3 !
Schedule: 1, 3, 4, 2,
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Summary
 1 | prec | hmax , hmax=max( h1(C1), ... ,hn(Cn) ), Lawler’s algorithm
 1 || Lmax
EDD rule
 1 | rj | Lmax
is NP hard , branch-and-bound is used
1 | rj , prec | Lmax similar branch-and-bound
 1 | rj, prmp | Lmax preemptive EDD rule
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