Job Shop Scheduling
1. Modelling Job Shop Problems
2. Shifting Bottleneck Heuristic
3. Branch & Bound
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Topic 1
Job Shop Scheduling Problems
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Job Shops

Have m machines and n jobs
Jm || Cmax

Each job visits some or all of the machines
–


Only once (or sometimes multiple times if recirculation is allowed)
Customer order of small batches
–
Wafer fabrication in semiconductor industry
–
Hospital
Very difficult to solve
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Job Shop Example

Constraints
–
Job follows a specified route of operations
(Conjunctive constraints)
–
One job at a time on each machine
(Disjunctive constraints)
Machine 1
Machine 2
(1,1)
(2,3)
Machine 3
Machine 4
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(1,2) (1,3)
(2,1)
(2,2)
(3,1)
(4,3)
(3,3)
(4,2)
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Definitions

A schedule is non-delay if no machine is kept idle (or free
resource is kept unused) when there is an operation available

A schedule is called active if -without delaying other operationsno operation can be completed earlier

For “regular” objectives: at least one optimal schedule is active
(but not necessarily non-delay)
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Non-active Schedule
Machine 1
(1,1)
Machine 2
(2,3)
Machine 3
(2,1)
(2,2)
(2,1)
(3,2)
0
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4
6
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6
Active Schedule, not Non-delay
Machine 1
(1,1)
Machine 2
(2,3)
Machine 3
(2,2)
(2,1)
(3,2)
0
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4
6
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7
Non-delay schedule
Machine 1
(1,1)
Machine 2
(2,3)
Machine 3
(2,1)
(2,2)
(3,2)
0
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4
6
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8
Graph Representation (J-on-N)

Each job follows a given route through the job shop
–
Picturing each job as a row of nodes: (i,j)=operation on machine i of job j
(Conjunctive arcs A)
Source
(1,1)
(2,1)
(1,2)
(2,2)
(2,3)
(1,3)
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(3,1)
(4,2)
(4,3)
Sink
(3,3)
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Graph Representation

To model the machine constraints, introduce the arc set B (...),
giving ‘a clique’ of bidirected arc-pairs on each machine
(Disjunctive arc-pairs B)
Source

(1,1)
(2,1)
(1,2)
(2,2)
(2,3)
(1,3)
(3,1)
(4,2)
(4,3)
Sink
(3,3)
Full Graph G(N, AB)
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Solving the Problem

Select one arc from each pair of disjunctive arcs: D out of B
Source

(1,1)
(2,1)
(1,2)
(2,2)
(2,3)
(1,3)
(3,1)
(4,2)
(4,3)
Sink
(3,3)
Longest path in G(D)=G(N, A  D) determines the makespan
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Feasibility of the Schedule

Are all selections feasible?
Source
(1,1)
(2,1)
(1,2)
(2,2)
(2,3)
(1,3)
(3,1)
(4,2)
(4,3)
Sink
(3,3)
Resulting graph G(D) should be acyclic
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Disjunctive Programming
Minimize Cmax
s.t.
Cmax  yij  pij
for all operat's (i, j )
ykj  yij  pij
for all (i, j )  (k , j )
// arcs in A
yij  yil  pil or yil  yij  pij
for all (i, j ) and (i, l )
// arcs in B
yij  0
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for all (i, j )
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Solution Methods


Exact solution
–
Branch & Bound
–
20 machines and 20 jobs
Dispatching rules (16+)
–

Shifting Bottleneck
Search heuristics
–
Tabu, SA, GA, etc.
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Topic 2
Shifting Bottleneck Heuristic for
Job Shop Scheduling Problems
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Shifting Bottleneck

Minimize makespan in a job shop

Let M denote the set of machines

Let M0  M be machines for which disjunctive arcs have been
selected

Basic idea:
–
Select a machine in M - M0 to be included in M0
–
Sequence the operations on this machine
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Example

Three jobs on four machines
Job j
1
2
3

Sequence
1, 2, 3
2, 1, 4, 3
1, 2, 4
Processing Times on Machine i
1
2
3
10
8
4
3
8
6
4
7
-
4
5
3
total
22
22
14
Processing job 1 (or 2) without delay requires 22
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Iteration 1
M0  
(1,1)
10
(2,1)
8
(3,1)
4
0
Source
0
(2,2)
8
(1,2)
3
(4,2)
5
(3,2)
0
(1,3)
4
(2,3)
7
6
Sink
3
(4,3)
Cmax ( M 0 )  22
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Selecting a Machine

Set up a non-preemptive single machine maximum lateness
(1||Lmax) problem for Machine 1:
Job j
p(1, j)
r(1, j)
d(1, j)
1
10
0
10
2
3
8
11
3
4
0
12
Optimum sequence is 1, 2, 3 with Lmax(1) = 5

Consider rj as earliest starting time and dj as latest finish time
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Selecting a Machine

Set up a non-preemptive single machine maximum lateness
(1||Lmax) problem for Machine 2:
Job j
p(2, j)
r(2, j)
d(2, j)
1
8
10
18
2
8
0
8
3
7
4
19
Optimum sequence is 2, 3, 1 with Lmax(2) = 5
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Selecting a Machine

Similarly,
Lmax (3)  4
Lmax (4)  0
 Either Machine 1 or Machine 2 is the bottleneck
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Iteration 2
M 0  {1}
(1,1)
0
Source
0
10
(2,1)
8
(3,1)
4
10
(2,2)
0
8
(1,2)
3
4
(1,3)
3
(2,3)
(4,2)
7
5
(3,2)
6
Sink
3
(4,3)
Cmax ({1})  Cmax ()  Lmax (1)  22  5  27
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Selecting a Machine

Set up a non-preemptive single machine maximum lateness
(1||Lmax) problem for Machine 2:
Old:
Job j
p(2, j)
r(2, j)
d(2, j)
1
8
10
18
2
8
0
8
3
7
4
19
New:
Job j
p(2, j)
r(2, j)
d(2, j)
1
8
10
23
2
8
0
10
3
7
17
24
Optimum sequence is 2, 1, 3 with Lmax(2) = 1
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Selecting a Machine

Set up a non-preemptive single machine maximum lateness
(1||Lmax) problem for Machine 3:
Job j
p(3, j)
r(3, j)
d(3, j)
1
4
18
27
2
6
18
27
Optimum sequence is 1, 2 with Lmax(3) = 1
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Selecting a Machine

Similarly,
Lmax (3)  1
Lmax (4)  0
 Either Machine 2 or Machine 3 is the bottleneck
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Iteration 3
M 0  {1, 2}
(1,1)
0
(2,1)
10
0
Source
10
8
0
(1,2)
3
(1,3)
4
(3,1)
4
8
8
(2,2)
8
3
(2,3)
(4,2)
7
5
(3,2)
6
Sink
3
(4,3)
Cmax ({1, 2})  Cmax ({1})  Lmax (2)  27  1  28
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Shifting Bottleneck Heuristic
1.
M0:=; G=(N, u, v, A); Cmax := longest path-length in G between u and v
2.
iM-M0 do
// solving a single machine problem
a. for all operations o=(i,j) on machine i do
- find longest paths u-o and o-v of lengths lu,o and lo,v, say,
- define release time rij = lu,o, due date dij = Cmax - (lo,v -pij)
b. for this single machine problem on i: determine L*max(i) with choices D*(i)
3.
4.
k := arg max iM0 L*max(i) ;
// decide for bottleneck machine k
Insert arcs D(k)=D*(k) in G and update Cmax
// Cmax may increase
iM0-{k} do
// Re-sequencing ‘decided’ machines
a. remove D(i) from G and recompute Cmax
b.
recompute D*(i) as in step 2 above
c.
insert D(i)= D*(i) in G and update Cmax
// Cmax may decrease
5. M0:= M0  {k}; if M0<M then (add delayed precedence and) return to step 2.
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Discussion

Procedure continues until all disjunctive arcs have been added

Watch out for cycles, because of other (scheduled) machines
–


delayed precedence constraints may be required, see example in book
Very effective
–
Relatively fast
–
Good solutions
–
More general Job Shop problems can be solved as well
‘Just a heuristic’
–
No guarantee of optimum
–
The subproblem Max. Lateness, 1 | rj , prec | Lmax , is NP-hard
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Discussion

Shifting bottleneck can be applied generally

Basic idea
–
Solve problem “one variable at a time”
–
Determine the “most important” variable
–
Find the best value of that variable
–
Move on to the “second most important” ….
–
Here we treat each machine as a variable
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Topic 3
Branch & Bound for
Job Shop Scheduling Problems
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Branch and Bound

Minimize makespan

Notation

–
operation (i,j) has duration pij
–
Let W denote operations whose predecessors have all been
scheduled
–
Let rij be the earliest possible starting time of (i,j) in W.
FIRST SUBJECT: How to branch to active schedules
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Branching
 In the optimal schedule each machine has some order for its
operations: let machine i has first (i, j1), then (i, j2) etc.
 In partial schedules, B&B subproblems at level k, it is decided for
[some] machines i, in ancestor B&B nodes, which operations are
(i,j1), (i,j2),..., (i, jk(i) ) [possibly k(i)=0 –if i has no decisions yet]
Ancestor choices
New choices:
for some machine i* decide
on a next operation
(i*, jk(i)+1)=(i*, j')
(i*, jk(i)+1)=(i*, j''')
(i*, jk(i)+1)=(i*, j'')
!Avoid partial schedules that are not active!
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Which operations can be next in active schedule?
  { first of unscheduled operations for each job}
t ( ) 


min
rij  pij  ri* j 0  pi* j 0
(i, j ) 
// first finish
 '(i*)  {(i*, j ) : operations on i * with release ri* j  t ()}
Machine i*
If active,
a schedule can have
(i*, j)  ' as
the next i*-operation,
but not (i*,k)  '
(i*, j)
(i*, j0)
0 t
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(i(i*,k)
*, k )   '
(i*, j0)
ri* j t() ri* k
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' is a subset of  , why?
Suppose (i*, j) in ' is not part of :
i.e., some other operation, say (i1, j) is first for job j.
(i1, j) is by definition in  and to be scheduled before (i*, j)
Gant-Chart of operations of j:
and of j0:
Then:
ri* j0 + pi* j0 = t() <= ri1 j + pi1 j <= ri* j,
(i*, j)
(i1, j)
(i*, j0)
t()
showing that (i*,j) is cannot be part of  '  {(i*, j) on i*| ri* j  t ()}
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Generating Active Schedules

Step 1. (Initialize)
–


Let  contain the first operations of each job; rij = 0 for all (i,j)   ;
Step 2. (Machine selection)
–
Compute of current partial schedule: time t () 
–
i* = associated machine
min rij  pij 
(i, j )  
Step 3. (Branching) // extends current schedule in all active ways
ri* j  t ()
–
Let ’ consist of operations (i*, j) on machine i* with
–
For each (i*, j) in ’ extend current partial schedule with (i*, j) next on i*
–
For each thus generated, extended partial schedule:
a. delete (i*, j) from , insert immediate follower (k, j) in ,
b. return with this schedule, as current, to step 2.
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Branching Tree
ancestor choices
Some machine i* decides
on next operation
(i*, jk(i)+1)=(i*, j')
(i*, jk(i)+1)=(i*, j''')
(i*, jk(i)+1)=(i*, j'')
Generating active schedules, that is
not all choices, only operations (i*, j' ) of '
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On Example
0
Source
(1,1)
0 (2,2) 8
0
10
(1,2)
4
(1,3)
(2,1)
3
(2,3)
8
(4,2)
7
(3,1)
4
5
6
(3,2)
3
(4,3)
Level 1
root
  {(1,1), (2, 2), (1,3)}
t ()  min{0  10, 0  8, 0  4}  4
i*  1
'  {(1,1), (1,3)}
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Sink
(1,1)
22
(1,3)
Branch at level 1 for possibilities
’ of (1, j1) at i* =1
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Level 1: when (1,1) is first on machine 1
(1,1)
0
Source
0
10
(2,2)
8
10
(2,1)
(3,1)
4
10
(1,2)
3
0
(1,3)
8
4
(2,3)
(4,2)
7
5
(3,2)
6
Sink
3
(4,3)
Cmax >= 24
Disjunctive Arcs
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Level 1: when (1,3) is first on machine 1
(1,1)
10
(2,1)
8
(3,1)
4
0
Source
0
(2,2)
0
8
(1,2)
3
(4,2)
5
(3,2)
4
4
4
(1,3)
(2,3)
7
6
Sink
3
(4,3)
Cmax >= 26
Disjunctive Arcs
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Branching Tree
Start, no disjunctive arcs
LB=22
(1,1) scheduled first
on machine 1
(1,3) scheduled first
on machine 1
LB = 24
LB = 26
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Next branching: at subproblem (1,1)
(1,1)
10
(2,1)
8
(3,1)
4
0
10
8 10
6
3
5
0
Source
(2,2)
(1,2)
(4,2)
(3,2)
0
3
7
4
(1,3)
(2,3)
(4,3)
Level 2
root
  {(2,2), (2,1), (1,3)}
t ()  min{0  8, 10  8, 10  4}  8
Sink
22
(1,1)
(1,3)
i*  2
'  {(2,2)}
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(2,2)
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Level 2: Extending to (2,2) as first on machine 2
As yet no new disjunctive Arcs
(1,1)
0
Source
0
8
8
(2,2)
0
10
10
(1,2)
(2,1)
3
8
(4,2)
(3,1)
5
4
(3,2)
6
Sink
8
(1,3)
4
(2,3)
7
3
(4,3)
Still LB=24
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Branching Tree
LB=22, no disjunctive arcs
(1,1) scheduled first
on machine 1
LB = 24
(1,3) scheduled first
on machine 1
(1,1) first on M1 and
(2,2) first on M2
LB = 26
LB = 24
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Lower Bounds

Lower bounds
–
Length of critical path in G(D’).

–
Linear programming relaxation.

–
Very quick but not very tight
Less quick but tighter
Maximum Lmax over all machines

Like in Shifting-Bottleneck Heuristic

slowest (m NP-hard problems), but tightest !
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