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
The PRE incidence matrix is::
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 
The POST incidence matrix is:
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 
The incidence matrix(W) is:
-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 
The discrete places are:  P D{ P1, P2 }
The continuous places are:  P C{ P3, P4 }

The discrete marking is: m0D  = [ 1, 0 ]
The continuous marking is: m0C  = [ 180.00, 0.00 ]

The discrete transitions are:  Td{ T 1, T 2 }
The continuous transitions are:  Tc{ T3, T4 }

The timing of the discrete transitions : Td  = { T1=90, T2=60 }
The flow rates of the continuous transitions : Tc  = { U3=3.00, U4=2.00 }

The conflicts :
1) Place P1(discrete) with transitions T1(d) T3(c) is a conflict of Case 3
2) Place P2(discrete) with transitions T2(d) T4(c) is a conflict of Case 3
The priorities of the conflicts and sharing ß(if exists):
1) level -1 level 1



2) level -1 level 1



ALGORITHM 6.1

 

ALGORITHM 6.1 - STEP 1: Initialization

ALGORITHM 6.1 - STEP 1.2

Initial marking m˜ C=[ 180.00 0.00]
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 4

ALGORITHM 6.1 - STEP 4

The SUBNET 1 has the priority 0
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 1

ALGORITHM 6.1 - STEP 5

We treat the priority level 0
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=0.00

ALGORITHM 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 10

ALGORITHM 6.1 - STEP 10


UPDATE TOS
There are no other future D2 events
Future C1 events:
The marking m3 becomes 0 at the moment 60.00

ALGORITHM 6.1 - STEP 11


Next time ts=60.00

ALGORITHM 6.1 - STEP 2


D2(60.00)-event(ø)

ALGORITHM 6.1 - STEP 3


D1(60.00)=ø then GO TO STEP 4

ALGORITHM 6.1 - STEP 4

The SUBNET 1 has the priority 0
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 1

ALGORITHM 6.1 - STEP 5

We treat the priority level 0
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=0.00

ALGORITHM 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 10

ALGORITHM 6.1 - STEP 10


UPDATE TOS
There are no other future D2 events
There are no other future C1 events

ALGORITHM 6.1 - STEP 11


Next time ts=90.00

ALGORITHM 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 4

ALGORITHM 6.1 - STEP 4

The SUBNET 1 has the priority 0
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 1

ALGORITHM 6.1 - STEP 5

We treat the priority level 0
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=2.00

ALGORITHM 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 10

ALGORITHM 6.1 - STEP 10


UPDATE TOS
There are no other future D2 events
Future C1 events:
The marking m4 becomes 0 at the moment 180.00

ALGORITHM 6.1 - STEP 11


Next time ts=150.00

ALGORITHM 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 4

ALGORITHM 6.1 - STEP 4

The SUBNET 1 has the priority 0
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 1

ALGORITHM 6.1 - STEP 5

We treat the priority level 0
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=0.00

ALGORITHM 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 10

ALGORITHM 6.1 - STEP 10


UPDATE TOS
There are no other future D2 events
Future C1 events:
The marking m3 becomes 0 at the moment 190.00

ALGORITHM 6.1 - STEP 11


Next time ts=190.00

ALGORITHM 6.1 - STEP 2


D2(190.00)-event(ø)

ALGORITHM 6.1 - STEP 3


D1(190.00)=ø then GO TO STEP 4

ALGORITHM 6.1 - STEP 4

The SUBNET 1 has the priority 0
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 1

ALGORITHM 6.1 - STEP 5

We treat the priority level 0
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=0.00

ALGORITHM 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 10

ALGORITHM 6.1 - STEP 10


UPDATE TOS
There are no other future D2 events
There are no other future C1 events

ALGORITHM 6.1 - STEP 11


Next time ts=240.00

ALGORITHM 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 4

ALGORITHM 6.1 - STEP 4

The SUBNET 1 has the priority 0
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 1

ALGORITHM 6.1 - STEP 5

We treat the priority level 0
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=2.00

ALGORITHM 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 10

ALGORITHM 6.1 - STEP 10


UPDATE TOS
There are no other future D2 events
Future C1 events:
The marking m4 becomes 0 at the moment 330.00

ALGORITHM 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