Information Technology Reference
In-Depth Information
subject to: for
i
=1
,
2
, ..., N,
x
i,
1
=max(
x
i−
1
,
2
,x
i−
1
,
6
)+
s
i,
1
(12)
x
i,
2
=
x
i,
1
+
s
i,
2
(13)
x
i,
3
=max(
x
i−
1
,
6
,x
i,
2
)+
s
i,
3
(14)
x
i,
4
=max(
x
i−
1
,
5
,x
i,
3
)+
s
i,
4
(15)
x
i,
5
=
x
i,
4
+
s
i,
5
(16)
x
i,
6
=max(
x
i−
1
,
5
,x
i,
3
)+
s
i,
6
(17)
s
min
i,j
≤
s
i,j
, j
∈{
1
,
2
,
3
,
4
,
5
,
6
}
,
(18)
where
s
min
i,j
is a lower bound of time required
s
i,j
. This lower bound
s
i,j
for task
τ
i,j
is requested by the supervisory controller from the lower-level controller.
Solving optimization problem (11)-(18) gives the completion time
x
i,j
associ-
ated with task
τ
i,j
. Both the start of task
τ
i,
1
and task
τ
i,
3
requires the synchro-
nization of task
τ
i,
2
(carried out by the QC) and task
τ
i,
6
(carried out by the
AGV), which means that task
τ
i,
2
and task
τ
i,
6
are finished together. We define
the variable
x
qc
,
agv
as the meeting time of the QC and the AGV. Similarly we
define
x
agv
,
asc
as the meeting time of the AGV and the ASC, at which both the
AGV and ASC start their following task. The meeting time
x
qc
,
agv
and
x
agv
,
asc
are determined as follows:
x
qc
,
agv
=max(
x
i−
1
,
6
,x
i,
2
)
(19)
x
agv
,
asc
=max(
x
i−
1
,
5
,x
i,
3
)
.
(20)
The operation time
T
i,j
allowed by the supervisory controller for carrying out
task
τ
i,j
based on the meeting time above. According to Figure 3,
T
i,
1
,T
i,
2
are
linked to the QC,
T
i,
3
,T
i,
6
are linked to the AGV and
T
i,
4
,T
i,
5
are linked to the
ASC. Considering there is no synchronization of different types of equipment
(here we only consider QC, AGV and ASC) between task
τ
i,
1
and task
τ
i,
2
and
between task
τ
i,
4
and task
τ
i,
5
, for simplicity we assume
T
i,
1
=
T
i,
2
for the QC
and
T
i,
4
=
T
i,
5
for the ASC. Therefore, the operation time allowed by the higher
level for carrying out the tasks can be computed as follows: for
i
=1
,
2
, .., N
T
i,
1
=
x
qc
,
agv
−
x
i−
1
qc
,
agv
(21)
2
T
i,
2
=
x
qc
,
agv
−
x
i−
1
qc
,
agv
(22)
2
T
i,
3
=
x
agv
,
asc
−
x
i−
1
qc
,
agv
(23)
T
i,
4
=
x
agv
,
asc
−
x
i−
1
agv
,
asc
(24)
2
T
i,
5
=
x
agv
,
asc
−
x
i−
1
agv
,
asc
(25)
2
T
i,
6
=
x
i
+1
x
agv
,
asc
.
−
(26)
qc
,
agv
