10.01.2013 Views

ComputerAided_Design_Engineering_amp_Manufactur.pdf

ComputerAided_Design_Engineering_amp_Manufactur.pdf

ComputerAided_Design_Engineering_amp_Manufactur.pdf

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

FIGURE 8.18 An ex<strong>amp</strong>le of the arc-ratio ARR.2. PN1 consists of all solid lines and holes. There are no paths in<br />

PN1 from t1 to t2. The NP from t1 to t2� must be accompanied by the NP from t3� to t4. There are two paths from<br />

f<br />

t1 to t4 in PN2; both � 2/1.<br />

The following theorems show that the ratio of maximum markings between two places equals their least<br />

marking ratio.<br />

Theorem 8: If pi pk in a synthesized net, then �M1 and M2 , where M1( pi)�0, M1( pk) � 0,<br />

M2( pk)�0, and M2( pi) � 0 such that<br />

Proof: Assume that P1 and P2 contain pi and pk , respectively, and start from the same place, ps, which<br />

has w tokens. If all these w tokens flow to and , respectively, then<br />

M1( pi) � ( w)<br />

( Ris) , M1( pk) � 0 and M2( pk) � ( w)<br />

( Rks) , M2( pi) � 0, respectively.<br />

We thus obtain<br />

2<br />

R 12<br />

m<br />

p3<br />

2<br />

p1<br />

t1 t1'<br />

4<br />

p2<br />

t2<br />

t3<br />

p4 2<br />

p4'<br />

t4<br />

The following theorems can be proved in a similar way.<br />

Theorem 9: If pi�p k in a synthesized net, � M such that M ( pi) � 0, M( pk) � 0, then<br />

p6<br />

p7<br />

entity 1 entity 2<br />

M1( pi) m<br />

---------------- � [ Rik]�. M2( pk) p i<br />

p k<br />

M1( pi) m<br />

---------------- � [ Rik]�. M2( pk) M( pi) -------------- m<br />

M( pk) �<br />

[ Rik]�. m<br />

4<br />

4<br />

p1'<br />

p2'<br />

t2'<br />

p3'<br />

t3'<br />

t4'

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!