job shop scheduling

1. Job Shop Scheduling
1

2. Job shop environment:
• m machines, n jobs
• objective function
• Each job follows a predetermined route
• Routes are not necessarily the same for each job
• Machine can be visited once or more than once
(recirculation)
2

3. Job Shop Problem
Network Formulation
• Let us consider the example with 4 m/c & 3 jobs
• The route of the jobs as well as their processing times are given
below
Job m/c sequence Processing time
1 1- 2 -3 P11=10 P21=8 P31=4
2 2-1-4-3 P22=8 P12=3 P42=5 P32=6
3 1-2-4 P13=4 P23=7 P43=3
where Pij  job J processed on m/c i
3

4. Construction of the Network
1,1 2,1 3,1
0
0
S 2,2 1,2 4,2 3,2 T
0
1,3 2,3 4,3
4

5. Problem : Jm | | Cmax
Node (i, j) represents the operation of jth job on ith machine
• Pij processing time of job j on machine i
• G = (N, A∪B)
• A: Solid (Conjunctive) arcs represent the precedence
relationships between operation of a single job.
(i, j ) → ( k , j ) ∈ A
• Operation (i, j) precedes (k, j)
5

6. • B: Broken (Disjunctive) arcs represent the precedence
relationships between operation of a single machine.
• Disjunctive arcs B represent conflicts on machines.
• Two operations (i, j) and (i, l) are connected by two arcs going
in opposite direction.
• Two dummy nodes S and T representing source and sink.
• Arcs from S to all first operations of jobs.
• Arcs from all last operations of jobs to T
• A feasible schedule corresponds to a selection of at most one
(disjunctive) broken arcs from each such pair such that the
resulting directed graph is acyclic
6

7. How to construct a feasible schedule?
Select D - a subset of disjunctive arcs (one from each pair) such
that the resulting directed graph G(D) has no cycles.
Graph G(D) contains conjunctive arcs + D.
D represents a feasible schedule.
A cycle in the graph corresponds to a schedule that is infeasible.
7

8. 10 8
1,1 2,1 3,1
0 4
0 8 3 5 6
S 2,2 1,2 4,2 3,2 T
0 3
4 7
1,3 2,3 4,3
• The makespan of a feasible schedule is determined by the longest path
in G(D) from S to T.
• Minimise makespan: find a selection of disjunctive arcs that
minimises the length of the longest path (the critical path).
8

9. Selection
• A subset D ⊂ B is called a selection if it contains from
each pair of disjunctive arcs exactly one.
• A selection D is feasible if the resulting directed graph
G (D) = (N, A ∪ D) i.e. graph with conjunctive and selected
disjunctive arcs is acyclic.
9

10. Remarks
1. A feasible selection leads to a sequence in which
operations have to be processed on machines.
2. Each feasible selection leads to a feasible schedule.
10

11. Ex.
Machines – M1, M2, M3
Jobs J1 where (3, 1)  (2, 1)  (1, 1)
J2 where (1, 2)  (3, 2)
J3 where (2, 3)  (1, 3)  (3, 3)
Duration
P31 = 4 P21 = 2 P11 = 1
P12 = 3 P32 = 3
P23 = 2 P13 = 4 P33 = 1
11

12. Feasible Selection
3,1 2,1 1,1
u 1,2 3,2 v
2,3 1,3 3,3
Represents conjunctive arcs
Selection
12

13. M1 J3 J 1 J2
Corresponding 
M2 J1 J3
Schedule
M3 J1 J2 J3
5 10 15 20
Make Span Cmax = 20
13

14. Selection for given schedule
3,1 2,1 1,1
u 1,2 3,2 v
2,3 1,3 3,3
Selection
14

15. M1 J2 J1 J3
M2 J3 J1
M3 J1 J2 J3
5 10 12
Make Span Cmax =12
15

16. Disjunctive Programming Formulation
Minimizing Cmax
Subject to
y kj
− y ≥ Pij for all
ij (i, j ) → (k , j ) ∈ A
where yij denotes starting time of operation (i, j)
Cmax − y ≥ Pij for all (i, j ) ∈ N
ij
y
}
−y ≥ p
ij il il
for all (i, l) & (i, j)
or
i = 1, 2, …..,m
y il
− y ≥ Pij
ij
y ij
≥ 0 for all (i, j ) ∈N
16

17. • Some ordering must exists among operation of different job
that are processed on same machine.
• Solution procedures for Jm / Cmax are based either on
enumerative or heuristic.
• No standard solution procedure available that will work
satisfactory.
• Two popular heuristic algorithms: (i) schedule generation
algorithm. (ii) shifting bottleneck heuristic algorithm;
17

18. Algorithm for Non Delay Schedule Generation
PS - A partial schedule containing t scheduled operations
t
S - The set of schedulable operations at stage t corresponding
t
to a given PS t
σ J - The earliest time at which operation J ∈ St could be started
φ J - The earliest time at which operation J ∈ St could be
completed
18

19. σ j is determine by the completion time of the direct
predecessor of operation J and latest completion time on the
machine required by operation J
– The larger of these quantities is σ J
– The potential finish time φ =σ +t
J J J
where t is processing time of operation J
J
Here (i, j, k) represents job i operation J on machine k
19

20. Algorithm
Step 1 – Let t = 0 and PS t = {φ}
S t includes all operations with no predecessor.
Step 2 – Determine σ
*
= min {σ }
j and the machine
*
m
S
J∈ t
σ
*
on which could be realized
*
Step 3 – For each operation J ∈ S t that requires machine m
and for whichσ J ∈σ
*
create a new partial schedule
in PS t
σJ
which operation J is added to and started at
time
20

21. Step 4 – For each new partial schedule PS t +1 created in step 3,
update the data set as follows
(a) Remove operation J from S t
(b) Form S t +1 by adding the direct successor of operation J
to S t
(c) Increment t by 1
Step 5 – Return to step 2 for each PS t +1 created in step 3 and
continue in this fashion until all non delay schedules
have been generated.
 The quality of the solution obtained by the heuristic
mainly depends on the effectiveness of priority rules
which are used in them.
21

22. A Sample Set of Priority Rules
1. SPT – Select the job with min. processing time
2. FCFS – Select the operation that entered S t earliest
3. MWKR – (Most work remaining) – Select the operation
associated with the job having the most work remaining to be
processed
4. MOPNR – (Most operation remaining) – Select the
operation that has the largest number of successor operation
5. Random – Select the operation at random
22

23. Ex. Find the schedule using non delay schedule generation
heuristic with following primary rules
First level priority rule – MWKR (Most work remaining)
Second level priority rule – SPT
Third level priority rule – Random order
23

24. Processing time Routing
Operation Operation
1 2 3 1 2 3
Job Job
1 2 3 4 1 1 2 3
2 4 4 1 2 3 2 1
3 2 2 3 3 2 3 1
4 3 3 1 4 1 3 2
24

25. At t=0 PS 0 = {φ} Job Operation
S 0
= ( ,1,1), (2,1,3), (3,1,2 ), (4,1,1)}
{1
σ 111 = σ 213 = σ 312 = σ 411 = 0 M/c
σ = Min{
σ}
*
=0 Since σ
*
J
S
J∈ t
σ =σ
*
J for all J ∈ S 0
*Therefore, priority rule must be σ → Earliest time at
J
involved to select among all which operation J ∈ S t
four operation [MWKR] could be started
25

26. R1 = 9 R2 = 9 R3 = 7 R4 = 7
MWKR = 9 is not unique.
This is occurring for job 1 and job 2
Now, a tie breaking rules is needed
SPT is used as tie breaking rule
Now t111 < t213
26

27. This means PS1 consists of operation {(1, 1, 1)} started at
time 0
PS1 = {(1, 1, 1)}
f1 = 2, f2 = 0, f3 = 0
M/c1 (1, 1, 1)
M/c2
M/c3
2 4 6
27

28. S 1
= {(1,2,2 ), ( 2,1,3), (3,1,2 ), ( 4,1,1)}
σ = Min{ σ ,σ ,σ ,σ }
*
122 213 312 411
= Min{2,0,0,2}
=0
At this stage σ J =σ
*
for two operations in S 1. Thus priority
rule must be involved to choose between (2, 1, 3) and (3, 1, 2)
28

29. By applying MWKR, we get R2 = 9 R3 = 7 since R2 > R3
(2, 1, 3) is added to PS1 to form
PS2 = {(1, 1, 1), (2, 1, 3)}
f1 = 2, f2 = 0, f3 = 4
M/c1 (1, 1, 1)
M/c2
M/c3 (2, 1, 3)
2 4 6
29

30. Now
S 2
= {(1,2,2 ), ( 2,2,2 ), ( 3,1,2 ), ( 4,1,1)}
σ = Min{σ 122 ,σ 222 ,σ 312 ,σ 411}
*
= Min{ 2,4,0,2}
=0 Operation 1 of job 2 is to
be completed in M/c 3
30

31. *The minimum is for σ 312 & it is unique. Add this to partial
schedule PS3
PS3 = {(1, 1, 1), (2, 1, 3), (3, 1, 2)}
f1 = 2, f2 = 2, f3 = 4
M/c1 (1, 1, 1)
M/c2 (3, 1, 2)
M/c3 (2, 1, 3)
2 4 6
31

32. S 3
= {(1,2,2 ), ( 2,2,2 ), ( 3,2,3), ( 4,1,1)}
σ = Min{σ 122 ,σ 222 ,σ 323 ,σ 411}
*
=Min{2,4,4,2}
=2
32

33. σ =σ
*
• At this stage J for two operation
• Thus priority rule must e involved to choose between (1, 2, 2)
and (4, 1, 1)
R1 = 7 R4 = 7
• MWKR is not unique.
• Now SPT is used as tie breaker & t122 = t411
• After it is resolved randomly in favor of (4, 1, 1)
PS4 = {(1, 1, 1), (2, 1, 3), (3, 1, 2), (4, 1, 1)}
f1 = 5, f2 = 2, f3 = 4
33

34. S4 = {(1, 2, 2), (2, 2, 2), (3, 2, 3), (4, 2, 3)}
σ 122
=2 σ 222
=4 σ 323
=4 σ 423
=5
Operation 1 of job 4 in
M/c 1 takes 5 minutes
σ = Min{
σ ,σ ,σ ,σ }
*
122 222 323 423
= Min{2,4,4,5}
=2
34

35. • Now (1, 2, 2) is added to PSt
• Thus PS5 = {(1, 1, 1), (2, 1, 3), (3, 1, 2), (4, 1, 1), (1, 2, 2)}
f1 = 5, f2 = 5, f3 = 4
Now S5 = {(1, 3, 3), (2, 2, 2), (3, 2, 3), (4, 2, 3)}
σ 133
=5 σ 222
=5 σ 323
=4 σ 423
=5
σ = Min{
σ ,σ ,σ ,σ }
*
133 222 323 423
= Min{5,5,4,5}
=4
35

36. • This minimum corresponds to (3, 2, 3) only partial schedule
PS6 = {(1, 1, 1), (2, 1, 3), (3, 1, 2), (4, 1, 1), (1, 2, 2), (3, 2, 3)}
• In this way we have to proceed to 3 × 4 = 12 stage to complete
the entire schedule
The final schedule is given as
P12 = {(1, 1, 1), (2, 1, 3), (3, 1, 2), (4, 1, 1), (1, 2, 2), (3, 2, 3)
(2, 2, 2), (2, 3, 1), (4, 2, 3), (3, 3, 1), (1, 3, 3), (4, 3, 2)}
36

37. M/c1 (1, 1, 1) (4, 1, 1) (2, 3, 1) (3, 3, 1)
2 5 9 10 13
M/c2 (3, 1, 2) (1, 2, 2) (2, 2, 2) (4, 3, 2)
2
M/c3 (2, 1, 3) (3, 2, 3) (4, 2, 3) (1, 3, 3)
37

38. Further Reading
1. Scheduling, Theory, Algorithms, and Systems, Michael Pinedo,
Prentice Hall, 1995, or new: Second Addition, 2002
Chapter 6
or
2. Operations Scheduling with Applications in Manufacturing
and Services, Michael Pinedo and Xiuli Chao, McGraw Hill, 2000
Chapter 5
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39. THANKS
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