5 Conclusions and Directions for Future Work 4.2 Elmore Delay Model 4.1 Linear Delay Model

30 b 0 = pl(T new ; b), and q 0 = pl(T new ; v). We now construct ZST T 0 for S by cutting oo the subtree of T rooted at q and replacing it with T new minus the edge between q and z. Since t LD (T 0 ; q) = d(q; s i), it must be that t LD (T 0 ; q) t LD (T; q). If the strict inequality holds, we add extra wire between q and q 0 to enforce equality, and thereby retain zero skew. For convenience, we use e a 0 and e b 0 to represent the embeddings of edges e a and e b in T 0. We also use e q 0 to denote the partial edge between q 0 and q in T 0. Because the subtrees of T rooted at a and b were constructed according to DME, we have t LD (T; a) = t LD (T 0 ; a 0) and t LD (T; b) = t LD (T; b 0). Thus, because t LD (T 0 ; q) = t LD (T; q), it must be that je a j = je a 0 j + je q 0 j and je b j = je b 0 j + je q 0 j: (3) Because d(a; b) > d(ms(a); ms(b)) and d(a; b) > jt LD (a) ? t LD (b)j, d(a; b) is strictly greater than the merging cost between ms(a) and ms(b). Therefore, je a j > je a 0 j and je b j > je b 0 j: (4) Equations (3) and (4) imply that je q 0 j > 0, and thus je a j + je b j > je a 0 j + je b 0 j + je q 0 j: As a result, cost(T 0) < cost(T). 29 PA(q; ms) = < 2. Lemma 2 : Given a ZST T with topology G, let v be an internal node with children a and b. Suppose the subtrees of T rooted at a and b can be generated by the DME algorithm for some placement of v on ms(v), and also suppose that q = pl(T; v) 6 2 ms(v). Then a new ZST T 0 with the same topology can be constructed from T by moving the placement of v so that …