The makespan for this problem is 51 units of time
.
10.10
CDS HEURISTIC
For large size problems, it would be difficult to get optimum solution in finite time, since the
flow shop scheduling is a combinatorial problem. This means the time complexity function of
flow shop problem is exponential in nature. Hence, we have to use efficient heuristics for large
size problems.
CDS (Campbell, Dudek and Smith) heuristic is one such heuristic used for flow shop scheduling.
The CDS heuristic corresponds to multistage use of Johnson’s rule applied to a new problem formed
from the original processing time.
At stage 1
t
1
j
1
=
t
j
i
and
t
1
j
2
=
t
jm
In other words, Johnson’s rule is applied to the first and
mth
operations and intermediate operations
are ignored.
At stage 2
T
2
j
1
=
t
j
i
+
t
j
2
and
t
2
j
2
=
t
jm
+
t
j m
–1
That is, Johnson’s rule is applied to the sum of the first two and the last two operation processing
times. In general at stage
i,
t
i
ji
=
ji
and
t
i
j
2
=
t
j
,
m

k
+1
For each stage
i
(i
= 1, 2, ...
m
– 1), the job order obtained is used to calculate a makespan for
the original problem. After
m
– 1, stages, the best makespan among the
m
– 1 schedule is identified.
(Some of the
m
– 1 sequences may be identical).
250
Operations Management
ILLUSTRATION 6:
Find the makespan using the CDS heuristic for the following flow shop
problem:
Job j
t
j1
t
j2
t
j3
t
j4
14378
23725
31247
43432
SOLUTION
Job (j)
M/c–1
t
j1
M/c–1 t
j2
148
235
317
432
The optimal sequence for the above problem is as shown below:
3214
The makespan calculation for the above schedule is shown below:
Processing time (in hour)
Job
M/c–1
M/c–2
M/c–3
M/c–4
In
Out
In
Out
In
Out
In
Out
3011337714
2
1
4
4
11
11
13
14
19
148111414212129
1
8
11
14
18
21
24
29
31
Makespan of this problem = 31.
Stage
2
Job (j)
M/c–1
M/c–2
T
j1
t
j2
1715
2107
3311
475
After applying Johnson's algorithm to the above problem, we get the sequence, 3124. The
makespan calculation is summarized in the following table:
Scheduling and Controlling Production Activities
251
Processing time (in hour)
Job
M/c–1
M/c–2
M/c–3
M/c–4
In
Out
In
Out
In
Out
In
Out
3011337714
1
1
5
5
8
8
15
15
23
2
5
8
8
15
15
17
23
28
4
8
11
15
19
19
22
28
30
The makes pan for the sequence 3124 is 30.
Stage
2
M/c–1
M/c–2
T
j1
T
j2
11418
21214
3713
4109
The application of Johnson’s algorithm to the above data yields the sequence 3214. The
determination of the corresponding makespan is shown below:
M/c–1
M/c–2
M/c–3
In
Out
In
Out
In
Out
30137714
2 141113 1419
14814212129
4
8
11
21
24
29
31
stage
sequence
Makespan
1
3214
31
2
3124
30
3
3214
31
The best sequence is 3124, which has the makespan of 30.
10.11
JOBSHOP PROBLEM
In jobshop problem, we assume that each job has
m
different operations. If some of the jobs are
having less than
m
operations, required number of dummy operations with zero process times is
assumed. By this assumption, the condition of equal number of operations for all the jobs is ensured.
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 Fall '17
 Krejewski
 The Land