Title: Example_page220
The number of places is: 4 |
The number of discrete places is: 2 |
The number of transitions is: 4 |
The number of discrete transitions is: 2 |
1.0  | 0.0  | 1.0  | 0.0  |
0.0  | 1.0  | 0.0  | 1.0  |
0.0  | 0.0  | 1.0  | 0.0  |
0.0  | 0.0  | 0.0  | 1.0  |
0.0  | 1.0  | 1.0  | 0.0  |
1.0  | 0.0  | 0.0  | 1.0  |
0.0  | 0.0  | 0.0  | 1.0  |
0.0  | 0.0  | 1.0  | 0.0  |
-1.0  | 1.0  | 0.0  | 0.0  |
1.0  | -1.0  | 0.0  | 0.0  |
0.0  | 0.0  | -1.0  | 1.0  |
0.0  | 0.0  | 1.0  | -1.0  |
ALGORITHM 6.1 - STEP 1: Initialization
Initial marking m˜ C=[ 180.00 0.00]ALGORITHM 6.1 - STEP 1.2
Initialization of the time-ordered sequence TOS: Let ts=t0
Possibly events:
C1(0)=D1(0)=D2(0)=ø
D1(90)={T1}ALGORITHM 6.1 - STEP 2
D2(0.00)-event(ø)ALGORITHM 6.1 - STEP 3
D1(0.00)=ø then GO TO STEP 4The SUBNET 1 has the priority 0ALGORITHM 6.1 - STEP 4
m(P3) = 180.00
m(P4) = 0.00
and the transitions:
T3 with maximal speed 3.00
T4 with maximal speed 0.00
This subnet does not have conflicts.
The PRE incidence matrix is:
1.0  0.0 
0.0  1.0 
The POST incidence matrix is:
0.0  1.0 
1.0  0.0 
The priority levels: 0
On the priority level 0 there are the subnets 1We treat the priority level 0ALGORITHM 6.1 - STEP 5
This priority level contains a single subnet and this is the subnet 1
We calculate the values Vj` using (6.16) in Section 6.2.1.1:
V`3=3.00*1.00=3.00
V`4=2.00*0.00=0.00ALGORITHM 6.1 - STEP 6
In the subnet 1 the instantaneous firing speeds are: v3=3.00 v4=0.00
All the priority levels have been treated, GO TO STEP 10ALGORITHM 6.1 - STEP 10
UPDATE TOS
There are no other future D2 events
Future C1 events:
The marking m3 becomes 0 at the moment 60.00ALGORITHM 6.1 - STEP 11
Next time ts=60.00ALGORITHM 6.1 - STEP 2
D2(60.00)-event(ø)ALGORITHM 6.1 - STEP 3
D1(60.00)=ø then GO TO STEP 4The SUBNET 1 has the priority 0ALGORITHM 6.1 - STEP 4
m(P3) = 0.00
m(P4) = 180.00
and the transitions:
T3 with maximal speed 3.00
T4 with maximal speed 0.00
This subnet does not have conflicts.
The PRE incidence matrix is:
1.0  0.0 
0.0  1.0 
The POST incidence matrix is:
0.0  1.0 
1.0  0.0 
The priority levels: 0
On the priority level 0 there are the subnets 1We treat the priority level 0ALGORITHM 6.1 - STEP 5
This priority level contains a single subnet and this is the subnet 1
We calculate the values Vj` using (6.16) in Section 6.2.1.1:
V`3=3.00*1.00=3.00
V`4=2.00*0.00=0.00ALGORITHM 6.1 - STEP 6
In the subnet 1 the instantaneous firing speeds are: v3=0.00 v4=0.00
All the priority levels have been treated, GO TO STEP 10ALGORITHM 6.1 - STEP 10
UPDATE TOS
There are no other future D2 events
There are no other future C1 eventsALGORITHM 6.1 - STEP 11
Next time ts=90.00ALGORITHM 6.1 - STEP 2
D2(90.00)-event(ø)ALGORITHM 6.1 - STEP 3
D1(90.00)={T1 } The firing of the transitions: T1
The discrete marking is m˜ D=(0 1 )
The continuous marking is m˜ C=(0.00 180.00 )ALGORITHM 6.1 - STEP 2
D2(90.00)-event(ø)ALGORITHM 6.1 - STEP 3
D1(90.00)=ø then GO TO STEP 4The SUBNET 1 has the priority 0ALGORITHM 6.1 - STEP 4
m(P3) = 0.00
m(P4) = 180.00
and the transitions:
T3 with maximal speed 0.00
T4 with maximal speed 2.00
This subnet does not have conflicts.
The PRE incidence matrix is:
1.0  0.0 
0.0  1.0 
The POST incidence matrix is:
0.0  1.0 
1.0  0.0 
The priority levels: 0
On the priority level 0 there are the subnets 1We treat the priority level 0ALGORITHM 6.1 - STEP 5
This priority level contains a single subnet and this is the subnet 1
We calculate the values Vj` using (6.16) in Section 6.2.1.1:
V`3=3.00*0.00=0.00
V`4=2.00*1.00=2.00ALGORITHM 6.1 - STEP 6
In the subnet 1 the instantaneous firing speeds are: v3=0.00 v4=2.00
All the priority levels have been treated, GO TO STEP 10ALGORITHM 6.1 - STEP 10
UPDATE TOS
There are no other future D2 events
Future C1 events:
The marking m4 becomes 0 at the moment 180.00ALGORITHM 6.1 - STEP 11
Next time ts=150.00ALGORITHM 6.1 - STEP 2
D2(150.00)-event(ø)ALGORITHM 6.1 - STEP 3
D1(150.00)={T2 } The firing of the transitions: T2
The discrete marking is m˜ D=(1 0 )
The continuous marking is m˜ C=(120.00 60.00 )ALGORITHM 6.1 - STEP 2
D2(150.00)-event(ø)ALGORITHM 6.1 - STEP 3
D1(150.00)=ø then GO TO STEP 4The SUBNET 1 has the priority 0ALGORITHM 6.1 - STEP 4
m(P3) = 120.00
m(P4) = 60.00
and the transitions:
T3 with maximal speed 3.00
T4 with maximal speed 0.00
This subnet does not have conflicts.
The PRE incidence matrix is:
1.0  0.0 
0.0  1.0 
The POST incidence matrix is:
0.0  1.0 
1.0  0.0 
The priority levels: 0
On the priority level 0 there are the subnets 1We treat the priority level 0ALGORITHM 6.1 - STEP 5
This priority level contains a single subnet and this is the subnet 1
We calculate the values Vj` using (6.16) in Section 6.2.1.1:
V`3=3.00*1.00=3.00
V`4=2.00*0.00=0.00ALGORITHM 6.1 - STEP 6
In the subnet 1 the instantaneous firing speeds are: v3=3.00 v4=0.00
All the priority levels have been treated, GO TO STEP 10ALGORITHM 6.1 - STEP 10
UPDATE TOS
There are no other future D2 events
Future C1 events:
The marking m3 becomes 0 at the moment 190.00ALGORITHM 6.1 - STEP 11
Next time ts=190.00ALGORITHM 6.1 - STEP 2
D2(190.00)-event(ø)ALGORITHM 6.1 - STEP 3
D1(190.00)=ø then GO TO STEP 4The SUBNET 1 has the priority 0ALGORITHM 6.1 - STEP 4
m(P3) = 0.00
m(P4) = 180.00
and the transitions:
T3 with maximal speed 3.00
T4 with maximal speed 0.00
This subnet does not have conflicts.
The PRE incidence matrix is:
1.0  0.0 
0.0  1.0 
The POST incidence matrix is:
0.0  1.0 
1.0  0.0 
The priority levels: 0
On the priority level 0 there are the subnets 1We treat the priority level 0ALGORITHM 6.1 - STEP 5
This priority level contains a single subnet and this is the subnet 1
We calculate the values Vj` using (6.16) in Section 6.2.1.1:
V`3=3.00*1.00=3.00
V`4=2.00*0.00=0.00ALGORITHM 6.1 - STEP 6
In the subnet 1 the instantaneous firing speeds are: v3=0.00 v4=0.00
All the priority levels have been treated, GO TO STEP 10ALGORITHM 6.1 - STEP 10
UPDATE TOS
There are no other future D2 events
There are no other future C1 eventsALGORITHM 6.1 - STEP 11
Next time ts=240.00ALGORITHM 6.1 - STEP 2
D2(240.00)-event(ø)ALGORITHM 6.1 - STEP 3
D1(240.00)={T1 } The firing of the transitions: T1
The discrete marking is m˜ D=(0 1 )
The continuous marking is m˜ C=(0.00 180.00 )ALGORITHM 6.1 - STEP 2
D2(240.00)-event(ø)ALGORITHM 6.1 - STEP 3
D1(240.00)=ø then GO TO STEP 4The SUBNET 1 has the priority 0ALGORITHM 6.1 - STEP 4
m(P3) = 0.00
m(P4) = 180.00
and the transitions:
T3 with maximal speed 0.00
T4 with maximal speed 2.00
This subnet does not have conflicts.
The PRE incidence matrix is:
1.0  0.0 
0.0  1.0 
The POST incidence matrix is:
0.0  1.0 
1.0  0.0 
The priority levels: 0
On the priority level 0 there are the subnets 1We treat the priority level 0ALGORITHM 6.1 - STEP 5
This priority level contains a single subnet and this is the subnet 1
We calculate the values Vj` using (6.16) in Section 6.2.1.1:
V`3=3.00*0.00=0.00
V`4=2.00*1.00=2.00ALGORITHM 6.1 - STEP 6
In the subnet 1 the instantaneous firing speeds are: v3=0.00 v4=2.00
All the priority levels have been treated, GO TO STEP 10ALGORITHM 6.1 - STEP 10
UPDATE TOS
There are no other future D2 events
Future C1 events:
The marking m4 becomes 0 at the moment 330.00ALGORITHM 6.1 - STEP 11
Next time ts=300.00
t=0 mC,bef = ( 180 , 0 )
mC,aft = ( 180 , 0 )
Residual times = ( 90.00 , ø )
IB-state 1
Discrete markink mD = ( 1 , 0 )
Enabling vector eD = ( 1 , 0 )
Speed vector v = ( 3.00 , 0.00 )
t=60.00 we have the events:
m3=0
mC,bef = ( 0.00 , 180.00 )
mC,aft = ( 0.00 , 180.00 )
Residual times = ( 30.00 , ø )
IB-state 2
Discrete markink mD = ( 1 , 0 )
Enabling vector eD = ( 1 , 0 )
Speed vector v = ( 0.00 , 0.00 )
t=90.00 we have the events:
firing of [ (T1)1 ]
mC,bef = ( 0.00 , 180.00 )
mC,aft = ( 0.00 , 180.00 )
Residual times = ( ø , 60.00 )
IB-state 3
Discrete markink mD = ( 0 , 1 )
Enabling vector eD = ( 0 , 1 )
Speed vector v = ( 0.00 , 2.00 )
t=150.00 we have the events:
firing of [ (T2)1 ]
mC,bef = ( 120.00 , 60.00 )
mC,aft = ( 120.00 , 60.00 )
Residual times = ( 90.00 , ø )
IB-state 4
Discrete markink mD = ( 1 , 0 )
Enabling vector eD = ( 1 , 0 )
Speed vector v = ( 3.00 , 0.00 )
t=190.00 we have the events:
m3=0
mC,bef = ( 0.00 , 180.00 )
mC,aft = ( 0.00 , 180.00 )
Residual times = ( 50.00 , ø )
IB-state 5
Discrete markink mD = ( 1 , 0 )
Enabling vector eD = ( 1 , 0 )
Speed vector v = ( 0.00 , 0.00 )
t=240.00 we have the events:
firing of [ (T1)1 ]
mC,bef = ( 0.00 , 180.00 )
mC,aft = ( 0.00 , 180.00 )
IB-state 6 is identical with IB-state 3