i. floorplanning with fixed moduleslimsk.ece.gatech.edu/book/slides/pdf/ilp-floorplanning.pdf ·...

Practical Problems in VLSI Physical Design ILP Floorplanning (1/22)

Fixed modules only, no rotation allowedm1 (4,5), m2 (3,7), m3 (6,4), m4 (7,7)

I. Floorplanning with Fixed Modules

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ILP Formulation

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Non-Overlapping Constraints (cont)

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Additional Constraints

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SolutionsUsing GLPK we get the following solutions:

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Final FloorplanWhy the non-optimality?

Due to linear approximation of area objective (= y*)Chip width/height constraints also affectedIn fact, our ILP solution (y* = 12) is optimal under these conditions.

Practical Problems in VLSI Physical Design ILP Floorplanning (7/22)

II. Floorplanning with RotationFixed modules, rotation allowed

Fixed modules: m1 (4,5), m2 (3,7), m3 (6,4), m4 (7,7)Need 4 more binary variables for rotation: z1, z2, z3, z4

We use M = max{W,H} = 23

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ILP Formulation

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Non-Overlapping Constraints (cont)

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Non-Overlapping Constraints (cont)

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Additional Constraints

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SolutionsUsing GLPK we get the following solutions:

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III. Floorplanning with Flexible Modules2 Fixed modules:

m1 (4,5), m2 (3,7) (rotation allowed)

2 Flexible modules: m3: area = 24, aspect ratio [0.5, 2]m4: area = 49, aspect ratio [0.3, 2.5]

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Linear Approximation

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Linear Approximation (cont)

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Upper Bound of Chip Dimension

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Non-Overlap Constraint

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Non-Overlap Constraint (cont)

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More Constraints

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Solutions

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ComparisonFixed modules only = 12 × 12Rotation allowed = 11 × 11Flexible modules used = 10.46 × 10.32

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Approximation Error and OverlapDue to linear approximation

Approximated area of m3 = 3.46 × 5.2 = 17.99 (actually 24)Approximated area of m4 = 3.83 × 7.32 = 28.04 (actually 49)Real area of m3 = 3.46 × 6.94 = 24Real area of m4 = 3.83 × 12.79 = 49Floorplan area increases, overlap occurs