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Lateness Models Contents 1. Lawler’s algorithm which gives an optimal schedule with the minimum cost h max when the jobs are subject to precedence relationship 1 | prec | h max 2. A branch-and-bound algorithm for the scheduling problems with the objective to minimise lateness

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- 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.

- Lawler’s Algorithm
- Backward algorithm which gives an optimal schedule for
- 1 | prec | hmaxhmax= max ( h1(C1), ... ,hn(Cn) )
- hjare 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 ofset J, set of jobs still to be scheduled
- J' JC set of jobs that can be scheduled immediately before set J (schedulable jobs)

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

3

5

10

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].

2

3

5

10

2

1

3

5

10

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.

or

Two schedules are optimal: 1, 2, 3 and 2, 1, 3

Lawler’sAlgorithm 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 scheduledimmediately before set J.

Step 3.

If JC = then Stop

else k = k - 1

go to Step 2

3

5

10

2

3

2

5

10

1

2

3

5

10

Example. What will happen in the previous example if theprecedence 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

2

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 orderof 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

S1

S2

Sn

. . .

S12

S13

. . .

. . .

- 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

S1 S2... Sn =

S1

S2

Sn

. . .

S12

S13

. . .

- We need some means for obtaining a lower bound on the cost for any particular solution (the task is to minimise the cost).

fboundf(x), xS1

fbound f(x), xS2

there is no need to exploreS2

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 = PSij

For all SijP do

if lower bound of f(Sij) > fbound then remove Sij from P

Step 3

If not termination condition then go to Step 2

1,*,*,*

2,*,*,*

n,*,*,*

. . .

1,2,*,*

1,3,*,*

. . .

. . .

Branch-and-bound algorithm for1 | rj | Lmax

- Solution space contains n! schedules (n is number of jobs).
- Total enumeration is not viable !

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 notbe processed before the release time of jk

that is:

J set of jobs not yet scheduled

t is time when jk-1 is completed

- Lower bound:
- Preemptive earliest due date (EDD) rule is optimal for 1 | rjprmp | 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.

2

12

3

7

2

1

0

5

9

2

1

2

0

3

7

9

Example.

- Non-preemptive schedules

L1=3

L2=6

Lmax=6

0

L1=5

L2=-1

Lmax=5

- Preemptive schedule obtained using EDD

L1=3

L2=3

Lmax=3

- the lowest Lmax !

*,*,*,*

L.B. = 5

L.B. = 7

1,*,*,*

2,*,*,*

3,*,*,*

4,*,*,*

job 1 could be processedbefore job 4

1,2,*,*

1,3,*,*

job 2 could be processedbefore job 3

L.B. = 5

L.B. = 6

1,3,4,2

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 2can be processedbefore 4 !

3,*,*,*

job 2can be processedbefore 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,

- 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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