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LI et al.: IMPROVED ABC ALGORITHM FOR SOLVING HFF WITH DYNAMIC OPERATION SKIPPING 3
B. Molten Iron Scheduling Problem
To make the problem closer to the industrial reality, we
consider minimization of the weighted sum of four penalty
values, i.e., the average sojourn time, the total earliness, the
total tardiness, and the skipping rate of the critical machines.
The sojourn time of a job is the duration between the com-
pletion time in the first stage and the starting time in the last
stage. Minimizing sojourn times leads to minimization of wait-
ing times, and thus, energy consumption can be reduced [14].
The skipping rate penalty is to compute the sum of the
skipping rate of the devices in the pre- and post-processing
stages.
Due to its complexity, we consider the molten iron schedul-
ing problem as an HFF-D problem in a static environ-
ment. The notations used in this paper are summarized as
follows.
i Index of the jobs, i = 1, 2,...,n.
k Index of the machines, k = 1, 2,...,m.
j Index of the stages, j = 1, 2,...,5.
n Total number of jobs.
m Total number of machines.
p
i,j
Processing time of job i at stage j.
E
j
Set of parallel machines at stage j.
m
j
Number of parallel machines at stage j.
J Set of n jobs, J ={J
1
, J
2
,...,J
n
}.
T
h,j
Transfer time from stage h to stage j.
[d
i
s
, d
i
e
] Due time window of job i.
w
1
Penalty coefficient for the average sojourn time.
w
2
Penalty coefficient for the earliness.
w
3
Penalty coefficient for the tardiness.
w
4
Penalty coefficient for the skipping rate.
b
i,j
Starting time of job i at stage j.
e
i,j
Completion time of job i at stage j.
i
Set of stages to be accessed by job i.
x
i,j,k
If job i is processed on machine k at stage j,
x
i,j,k
= 1; otherwise x
i,j,k
= 0.
z
i,h,j
If job i is to be processed at stage j immediately
after its completion at stage h, z
i,h,j
= 1; otherwise
z
i,h,j
= 0.
y
i,l,j
If job i is preceding job l to be processed at stage j,
y
i,l,j
= 1; if jobs l and i start at the same time at
stage j, y
i,l,j
= 1/2; otherwise y
i,l,j
= 0.
By using the above notations, we give the formulation of
this paper as follows:
min f = w
1
F
1
+ w
2
F
2
+ w
3
F
3
+ w
4
F
4
F
1
=
n
i=1
b
i,5
− e
i,1
n (1)
F
2
=
n
i=1
max
0, d
i
s
− b
i,5
(2)
F
3
=
n
i=1
max
0, b
i,5
− d
i
e
(3)
F
4
=
⎛
⎝
1 −
⎛
⎝
j=2,4
m
j
k=1
n
i=1
x
i,j,k
⎞
⎠
2n
⎞
⎠
× 100 (4)
j∈
i
k∈E
j
x
i,j,k
= 1, ∀i ∈ J (5)
b
i,j
−
b
i,h
+ p
i,h
+ T
h,j
· z
i,h,j
≥ 0, ∀i ∈ J, h, j ∈
i
(6)
y
i,l,j
+ y
l,i,j
= 1, ∀i, l ∈ J, j ∈
(
i
∩
l
)
(7)
b
l,j
−
b
i,j
+ p
i,j
+ U ·
3 − y
i,l,j
− x
i,j,k
− x
l,j,k
≥ 0,
∀i, l ∈ J, k ∈ E
j
, j ∈
(
i
∩
l
)
(8)
x
i,j,k
∈{0, 1}, ∀i ∈ J, j ∈{1, 2, 3, 4, 5}, k ∈ E
j
(9)
z
i,h,j
∈{0, 1}, ∀i ∈ J, h, j ∈{1, 2, 3, 4, 5} (10)
y
i,l,j
∈
0,
1
2
, 1
, ∀i, l ∈ J, j ∈{1, 2, 3, 4, 5}. (11)
The four functions (1)–(4) are to minimize the penalty
caused by the average sojourn time, the total earliness, the
total tardiness, and the skipping rate penalty. Constraints (5)
make sure that each operation can select one and only one
available machine in each stage which it must go through.
Constraints (6) guarantee that for two consecutive operations
of each job, the next one can be started only after the com-
pletion of the preceding one plus the transfer time between
the two stages. Constraints (7) and (8) make sure that pro-
cessing overlap of jobs on the same machine is not permitted.
Constraints (9)–(11) define the value ranges for the decision
variables.
C. Example Problem Instance
The following example will help in illustrating this complex
HFF-D problem. Consider an instance with five torpedoes and
stages. There are two blast furnaces in the first stage, two
preslag-pouring devices in the second stage, two dephospho-
rization or desulphurization devices in the third stage, two
post-slag-pouring devices in the fourth stage, and two pouring
devices in the last stage. In the five torpedoes, the second and
third torpedoes are of the special molten irons, while the others
are of the common molten iron. That is, n = 5, m = 10, E
1
=
{1, 2}, E
2
={1, 2}, E
3
={1, 2}, E
4
={1, 2}, E
5
={1, 2}.Let
w
1
= 1, w
2
= 1, w
3
= 0.5, and w
4
= 1. The set of special
iron ={2, 3} and the set of common iron ={1, 4, 5}.The
processing times p
ij
, the transfer times T
ij
, and the due date
window are given as follows:
p
ij
5×5
=
⎡
⎢
⎢
⎢
⎢
⎣
30
30
30
30
30
30
30
30
30
30
30
30
30
30
30
40
40
30
40
30
30
30
30
30
30
⎤
⎥
⎥
⎥
⎥
⎦
T
ij
5×5
=
⎡
⎢
⎢
⎢
⎢
⎣
0
10
20
30
40
0
0
15
30
45
0
0
0
10
20
0
0
0
0
10
0
0
0
0
0
⎤
⎥
⎥
⎥
⎥
⎦
d
i
s
d
i
e
5×5
=
150
170
180
200
200
220
200
220
220
230
.
The Gantt chart of a solution for the above problem instance
isshowninFig.2, where each operation is represented by
a rectangle labeled with the job number. For example, in the
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