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EE4271 VLSI Design, Fall 2013
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06/23/22 1 EE4271 VLSI Design, Fall 2013 VLSI Quadratic Placement
description
EE4271 VLSI Design, Fall 2013. VLSI Quadratic Placement. Problem formulation. Input: Blocks (standard cells and macros) B 1 , ... , B n Shapes and Pin Positions for each block B i Nets N 1 , ... , N m Output: Coordinates (x i , y i ) for block B i . - PowerPoint PPT Presentation
Transcript of EE4271 VLSI Design, Fall 2013
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EE4271 VLSI Design, Fall 2013
VLSI Quadratic Placement
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Problem formulation
• Input:– Blocks (standard cells and macros) B1, ... , Bn
– Shapes and Pin Positions for each block Bi
– Nets N1, ... , Nm
• Output:– Coordinates (xi , yi ) for block Bi.
– The total wirelength as an estimation of timing is minimized.
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Placement Can Make A Difference
Random InitialPlacement
FinalPlacement
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Partitioning:
Given a set of interconnected blocks, produce two sets thatare of equal size, and such that the number of nets connecting the two sets is minimized.
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FM Partitioning:
Initial Random Placement
After Cut 1
After Cut 2
The more crossings along the cut, the longer the total wirelength. The idea is to reduce the crossings.
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FM Partitioning:
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- each object is assigned a gain- objects are put into a sorted gain list- the object with the highest gain is selected and moved.- the moved object is "locked"- gains of "touched" objects are recomputed- gain lists are resorted
Object Gain: The amount of change in cut crossings that will occur if an object is moved from its current partition into the other partition
Moves are made based on object gain.
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FM Partitioning:
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Analytical Placement
• Write down the placement problem as an analytical mathematical problem
• Quadratic placement– Sum of squared wire length is quadratic in the cell
coordinates.– So the wirelength minimization problem can be formulated
as a quadratic program.– It can be proved that the quadratic program is convex,
hence polynomial time solvable
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Cost x1 1002 x1 x22 x2 2002
x1Cost 2x1 100 2x1 x2
x2Cost 2x1 x2 2x2 200
setting the partial derivatives = 0 we solve for the minimum Cost:
Ax + B = 0
= 04 2 2 4
x1x2
200 400
= 02 1 1 2
x1x2
100 200
x1=400/3 x2=500/3
x2x1
x=100
x=200Example:
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Quadratic Placement
Form quadratic placement problem
Perform partitioning to reduce overlap
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Solution of the Original QP
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Partitioning
• Use FM to cut.
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• Perform the quadratic placement on each region with additional constraints that the center of gravities should be in the center of regions.
Applying the Idea Recursively
Center of Gravities
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Process
(a) Global placement with 1 region (b) Global placement with 4 region (c) Final placements
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Summary
• Definition of VLSI quadratic placement problem• Partitioning technique
