1. Chapter 7: Job shops Sections 7.1 and 7.2 (skip section 7.3)
IOE/MFG 543
Chapter 7: Job shops
Sections 7.1 and 7.2
(skip section 7.3)

2. Job shop (Jm) m machines, n jobs
Each job has its own routing on the machines
The flow shop is a special case where all the jobs follow the same route (recall F3||Cmax is NP-hard)
If a job is to be processed on the same machine more than once => recirculation

3. Section 7.1. Disjunctive graph for Jm||Cmax
In Jm||Cmax there is no recirculation
All jobs start at a source node U and finish at the sink node V
Node (i,j) is the processing of job j on machine i
Let N denote the set of nodes
The arc from node (i,j) has weight pij
The arcs going out of the source node have zero weights
Let A denote the set of arcs

4. Disjunctive graph (2)
For each machine i connect all nodes of the form (i,j) and (i,k) with two arcs (one going in each direction)
The arcs for each machine form a clique, i.e., a sub-graph in which any two nodes are connected to one another
The two arcs are disjunctive, i.e., only one of them is utilized
If the operation (i,j) precedes operation (i,k) then only the arc from (i,j) to (i,k) and not the arc from (i,k) to (i,j) is utilized
Let B denote the set of disjunctive arcs

5. Using the graph to find a feasible schedule
Select one disjunctive arc from each pair of disjunctive arcs
If the selection results in an acyclic graph, i.e., contains no cycles, then the corresponding schedule is feasible
The makespan of the schedule is determined by the critical path method

6. Minimizing the makespan
To minimize the makespan we need to select the arcs that results in a schedule that minimizes the makespan
Disjunctive program
yij is the starting time of operation (i,j)
minimize
Cmax
subject to
ykj-yij≥pij
 (i,j)(k,j)
Cmax-yij≥ pij
 (i,j)
yij-yil≥pil or yil-yij≥pij
 (i,l) and (i,j) , i=1,…,m
yij≥0

7. Minimizing the makespan (2)
Solving this problem can be achieved by a branch and bound algorithm
The algorithm is computationally expensive even for a problem with a modest number of jobs and machines
Use heuristic methods instead

8. Section 7.2: The shifting bottleneck heuristic
Example 7.2.2
job
machine sequence
processing times 1
1, 2, 3
p11=10, p21=8, p31=4 2
2, 1, 4, 3
8, 3, 5, 6 3
1, 2, 4
4, 7, 3

9. Example 7.2.2: Disjunctive graph10 8
1,1
2,1
3,1 4 5 6 8 3 V U
2,2
1,2
4,2
3,2 3 4 7
1,3
2,3
4,3

10. Example 7.2.2: Iteration 1 M0=Set of machines already scheduled=
Only include the conjuctive arcs
Compute Cmax(M0)=22 10 8
1,1
2,1
3,1 4 5 6 8 3 V U
2,2
1,2
4,2
3,2 3 4 7
1,3
2,3
4,3

11. Ex. 7.2.2: Iter. 1 Which machine to sched. first?
Solve 1|rj|Lmax for each machine i
Release dates
rij=longest path from U to node (i,j)
Due dates
dij= Cmax(M0)
- longest path from (i,j) to V
+ pij
The machine with the largest minimum Lmax is the bottleneck machine

12. Ex. 7.2.2: Iter. 1. 1|rj|Lmax job j 1 2 3 p1j 10 4 r1j 8 d1j 11 128
d1j 11 12
Machine 1:
How do we solve this problem?
Branch and bound algorithm gives Lmax(1)=5
Machine 2:
Branch and bound algorithm gives Lmax(2)=5
job j 1 2 3
p2j 8 7
r2j 10 4
d2j 18 19

13. Ex. 7.2.2: Iter. 1. 1|rj|Lmax job j 1 2 3 p3j 4 6 - r3j 18 16 d3j 22
Machine 3:
Only 2 schedules! Lmax(3)=4
Machine 4:
Only 2 schedules… Lmax(4)=0
job j 1 2 3
p4j - 5
r4j 11
d4j 16 22

14. Ex. 7.2.2: Iter. 1 Which machine to sched. first? (2)
Machines 1 and 2 both have a minimum Lmax of 5
We can choose either machine so let’s pick machine 1 => M0={1} 10 8
1,1
2,1
3,1 4 10 5 6 8 3 V U
2,2
1,2
4,2
3,2 3 3 4 7
1,3
2,3
4,3

15. Ex. 7.2.2: Iteration 2 Compute Cmax(M0)=27 10 8 1,1 2,1 3,1 4 10 5 6 85 6 8 3 V U
2,2
1,2
4,2
3,2 3 3 4 7
1,3
2,3
4,3

16. Ex. 7.2.2: Iteration 2. 1|rj|Lmax
Machine 2
B&b algorithm gives Lmax(2)=1 w/sequence 2-1-3
Machine 3
Lmax(3)=1 w/sequences 1-2 and 2-1
Machine 4
Lmax(4)=0 w/sequence 2-3
Add machine 2 to M0 so M0={1,2}
job j 1 2 3
p2j 8 7
r2j 10 17
d2j 23 24
job j 1 2 3
p4j - 5
r4j 13 24
d4j 21 27

17. Ex. 7.2.2: Iteration 2 Resequencing10 8
Should we re-sequence machine 1 to decrease Cmax(M0)?
1-2-3 gives Lmax(1)=0
which is optimal since lateness of job 1 increases if it is delayed
Do not resequence machine 1
1,1
2,1
3,1 4 8 5 6 8 3 V U
2,2
1,2
4,2
3,2 8 3 4 7
1,3
2,3
4,3
job j 1 2 3
p1j 10 4
r1j 8
d1j 17 18

18. Ex. 7.2.2: Iteration 3 Compute Cmax(M0)=28 10 8 1,1 2,1 3,1 4 10 8 8 510 8 8 5 6 8 3 V U
2,2
1,2
4,2
3,2 3 3 4 7
1,3
2,3
4,3

19. Ex. 7.2.2: Iteration 3. 1|rj|Lmax
Machine 3
Lmax(3)=0
Machine 4
Lmax(4)=0 w/sequence 2-3
job j 1 2 3
p3j 4 6 -
r3j 18
d3j 28
job j 1 2 3
p4j - 5
r4j 13 25
d4j 22 28

20. Ex. 7.2.2: Iteration 3 Conclusion
Sequencing machines 3 and 4 does not increase the makespan of Cmax(M0)=28
No need to try to resequence 10 8
1,1
2,1
3,1 4 4 10 8 8 5 6 8 3 V U
2,2
1,2
4,2
3,2 5 3 3 4 7
1,3
2,3
4,3

21. Ex : Gantt chart 17 8 10 13 18 22 25 28 1 2 3 2 1 3 1 2 2 3

22. Algorithm 7.2.1: The shifting bottleneck heuristic
Initialization. Set M0=. Graph G contains only the conjunctive arcs
Solve a maximum lateness problem for each machine that has not been sequenced.
Sequence the machine that has the maximum minimum maximum lateness
Add it to M0 and add the disjunctive arcs to G for that machine
Resequence all machines in M0 one at a time, except the last added machine. Modify the disjunctive constraints accordingly
Stop if all machines have been sequenced, otherwise go to step 2.

23. An implementation issue with the heuristic
Recall: The disjunctive constraints must be selected such that the resulting graph is acyclic
In the SBH heuristic a cycle can be created
This can be prevented by adding precedence constraints (arcs) having a delay (a positive weight)

24. Example 7.2.3. Delayed precedence constraints
job
machine sequence
processing times 1
1, 2
p11=1, p21=1 2
2,1
1, 1 3 4