1. 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 or new: Second Addition, 2002, Chapter 3.

2. Lawler’s Algorithm
Backward algorithm which gives an optimal schedule for
1 | prec | hmax hmax = 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
they have to be processed during the time interval
JC complement of set J, set of jobs still to be scheduled
J'  JC set of jobs that can be scheduled immediately before set J (schedulable jobs)

3. Lawler’s Algorithm for 1 | | hmax
Step 1.
J = 
JC = {1,...,n}
k = n
Step 2.
Let j* be such that
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

4. Example (no precedence relationships between jobs)
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]. 10 5
... 3

5. 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. 10 5 3 2 1 or 10 5 3 1 2
Two schedules are optimal: 1, 2, 3 and 2, 1, 3

6. 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
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

7. 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
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} J'={1}
h1(2) = 3
Optimal schedule: 1, 2, 3,
hmax = 10 10 5
... 3 10 5 3 2 10 5 3 2 1 2

8. 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

9. 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 S2 Sn
S12
S13
. . .
S = S1 S2... Sn
S1 S2...  Sn = 

10. there is no need to explore S2
We need some means for obtaining a lower bound on the cost for any particular solution (the task is to minimise the cost). S S1 S2 Sn
S12
S13
. . .
fbound f(x), xS1
fbound  f(x), xS2
there is no need to explore S2

11. 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

12. Branch-and-bound algorithm for 1 | rj | Lmax
Solution space contains n! schedules (n is number of jobs).
Total enumeration is not viable !
*,*,*,*
1,*,*,*
2,*,*,*
n,*,*,*
1,2,*,*
1,3,*,*
. . .

13. k-1 level, j1, ... , jk-1 are scheduled,
*,*,*,*
1,*,*,*
2,*,*,*
n,*,*,*
1,2,*,*
1,3,*,*
. . .
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:
J set of jobs not yet scheduled
t is time when jk-1 is completed

14. 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.

15. Example.
Non-preemptive schedules
L1=3
L2=6
Lmax=6 1 2 3 7 12 2 1 5 9
L1=5
L2=-1
Lmax=5
Preemptive schedule obtained using EDD
L1=3
L2=3
Lmax=3 2 1 3 7 9
the lowest Lmax !

16. Example *,*,*,* L.B. = 5 L.B. = 7 1,*,*,* 2,*,*,* 3,*,*,* 4,*,*,*
job 1 could be processed before job 4
1,2,*,*
1,3,*,*
job 2 could be processed before job 3
L.B. = 5
L.B. = 6
1,3,4,2

17. *, *, *, *
1,*,*,*
1 [0, 4] L1=-4
3 [4, 5]
4 [5, 10] L4=0
3 [10, 15] L3=4
2 [15, 17] L2=5
2,*,*,*
2 [1, 3] L2=-9
1 [3, 7] L1=-1
4 [7, 12] L4=2
3 [12, 18] L3=7
4,*,*,*
either job 1 or 2 can be processed before 4 !
3,*,*,*
job 2 can be processed before 3 !
1,2,*,*
1 [0, 4] L1=-4
2 [4, 6] L2=-6
4 [6, 11] L4=1
3 [11, 17] L3=6
1,3,*,*
1 [0, 4] L1=-4
3 [4, 10] L3=-1
4 [10, 15] L4=5
2 [15, 17] L3=5
Schedule: 1, 3, 4, 2,

18. 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