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To determine the optimal placement and routing effort levels, a designer should follow this procedure: 1. Set both placement and routing effort levels to minimum. 2. Run place and route to determine if the worst case timing is in fact due to a sub-optimal layout and not due to excessive logic levels. 3. Increment placement effort until timing is met or max effort has been reached. 4. If at max placement effort timing is not met, begin incrementally increasing the routing effort. 5. If timing closure cannot be met, revisit the architecture of the design. 6. If a high routing effort is required to meet timing, it may be a good idea to revisit the architecture of the design to optimize for timing (see 1).

The con guration space obstacle (C-obstacle) is de ned as OC = {c C : L(c) O = } The free C-space is Cf = C OC The proof for the following theorem and its corollary are analogous to those for Theorem 631 Theorem 638 A C-obstacle is an open set Corollary 633 The free C-space Cf is a closed set The con guration space obstacle OC may have more than one component For convenience, we may call each component an obstacle Theorem 639 Let L(cs ) = Ls and L(ct ) = Lt If there exists a collision-free path (motion) between Ls and Lt in W, then there is a path pC Cf connecting cs and ct , and vice versa.

Logic replication occurs early in the placement process for structures that fan out to other logic elements that cannot (for any number of reasons) reside in the same proximity. The problem this addresses is illustrated in Figure 16.4. In this scenario, the output of D2 fans out to two structures that are relatively distant from one another. Regardless of where the driver is placed, the resultant route will be lengthy to one of the driven elements. To eliminate potentially long routing delays, logic duplication will replicate the driver as shown in Figure 16.5.

If there exists a motion between Ls and Lt in W, then there must be a path pJ Jf between two points js , jt Jf such that L(js ) = Ls and L( jt ) = Lt Then, pC = f (pJ ) Cf is a path between cs = f (js ) and ct = f (jt ) The other half of the theorem follows directly from the de nition of Cf ..

If Alistair was trading one-million-dollar positions, then the 120 pips had made him twelve thousand dollars If he had traded a five-million-dollar position, then he had made sixty thousand The numbers amazed me I decided that the long-term charts worked better for me, after all Harvey was right from the beginning, and I never should have looked at the shorter time frames, which only tempted me to jump into quick positions The short-term charts appealed to my greed The long-term charts required patience, rewarded analysis, and produced in one trade more profit than any of the short-term trades I d seen..

Figure 2.1 A planar two-link arm manipulator: l1 and l2 are links, with their respective endpoints a and b; J0 and J1 are two revolute joints; 1 and 2 are joint angles. Both links are of the same thickness 2R. The robot s base coincides with the joint J0 and is xed.

Logic duplication should be used only on critical path nets with multiple loads that cannot be physically localized.