(1) /96 ■Letr(1)=0,r(2)=1,r(3)=0,r(4)=0 D (1) (2) 2D w,(3→2)=1w(3→2)+r(2)-r(3)=0+1-0=1 2 3 w,(4→2)=w(4→2)+r(2)-r(4)=0+1-0=1 w,(2→1)=w(2→1)+r(1)-r(2)=1+0-1=0 (2) (1) A retiming solution is feasible only if w-(e)>0 holds for all edges. (1) (2) 2D How to determine the retiming value,r(V),will be 2 3 discussed in section 4.4.2.And the method of solving systems of inequalities is used,which will be given in section 4.3. D (2) 2021年2月 52021年2月 5  Let r(1)=0, r(2)=1, r(3)=0, r(4)=0  A retiming solution is feasible only if wr (e)≥0 holds for all edges.  How to determine the retiming value, r(V), will be discussed in section 4.4.2. And the method of solving systems of inequalities is used, which will be given in section 4.3. 1 4 2 3 D D 2D (1) (1) (2) (2) D w (3  2)  w(3  2)  r(2)  r(3)  0 1 0 1 r wr (4  2)  w(4  2)  r(2)  r(4)  0 1 0 1 1 4 2 3 D D 2D (1) (1) (2) (2) wr (2 1)  w(2 1)  r(1)  r(2) 1 0 1 0