RTL Design Flow - Computation Structures Group

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Gordian Placement <strong>Flow</strong> Global Optimization minimization of wire length module coordinates position constraints module coordinates Final Placement adoption of style dependent constraints Complexity space: O(m) time: Q( m 1.5 log 2 m) Final placement •standard cell •macro-cell &SOG Partitioning of the module set and dissection of the placement region Regions with ≤ k modules Data flow in the placement procedure GORDIAN Intuitive formulation Given a series of points x1, x2, x3, … xn and a connectivity matrix C describing the connections between them (If cij = 1 there is a connection between xi and xj) Find a location for each xj that minimizes the total sum of all spring tensions between each pair xi Problem has an obvious (trivial) solution – what is it? xj Gordian: A Quadratic Placement Approach • Global optimization: solves a sequence of quadratic programming problems • Partitioning: enforces the non-overlap constraints Improving the intuitive formulation To avoid the trivial solution add constraints: Hx=b These may be very natural - e.g. endpoints (pads) x1 xn To integrate the notion of ``critical nets’’ Add weights wij to nets xi wij xj wij -some springs have more tension should pull associated vertices closer
Modeling the Net’s Wire Length (x u ,y u ) y 2 2 ∑ [ ( x −x ) + ( y − y ) ] Lv = u←Mv uv v ( xuv = xu + ξ uv ; module u (x v ,y v ) uv v connection to other modules l vu v ( , ) vu vu η ξ pin vu net node x The length Lv of a net v is measured by the squared distances from its points to the net’s center ( ρ , ρ ) ' ' F v u y uv = y u + y ) vu Quadratic Optimization Problem E D A B C ( l ) M ⎡M ρ ⎢ = ⎢* ρ'⎢0 ⎢ M ⎣M Linearly constrained quadratic programming problem min{ Φ( x) = x x∈R m T s.t. A ) x = T C x + d x } ( l u( l ) ρ ) ( uρ , v A A B M * 0 M C M * 0 M D M 0 * M E M 0 * M F M 0 * M G M ⎤ ⎥ L⎥ L⎥ ⎥ M ⎦ Accounts for fixed modules Wire-length for movable modules Center-of-gravity constraints Problem is computationally tractable, and well behaved Commercial solvers available: mostek Toy Example: Kurt Keutzer Cost =(x 1 − 100) 2 +(x 1 − x 2 ) 2 +(x 2 − 200) 2 Cost = 2(x x 1 − 100)+2(x1 − x2 ) 1 Cost = x 2 − 2(x1 − x2 )+2(x2 − 200) setting the partial derivatives = 0 we solve for the minimum Cost: Ax + B = 0 4 −2 −2 4 x 1 x 2 + −200 −400 2 −1 −1 2 x1 x2 + −100 −200 x1=400/3 x2=500/3 x=100 x=200 = 0 = 0 x1 x2 D. Pan Global Optimization Using Quadratic Placement Quadratic placement clumps cells in center Partitioning divides cells into two regions Placement region is also divided into two regions New center-of-gravity constraints are added to the constraint matrix to be used on the next level of global optimization Global connectivity is still conserved 10
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RTL Design Flow - Computation Structures Group

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