1. 7/15/2015 1 VLSI Placement Prof. Shiyan Hu shiyan@mtu.edu Office: EERC 731

2. 2 7/15/2015 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. –The total wire length is minimized. –Subject to area constraint or the area of the resulting block is minimized

3. 3 7/15/2015 Placement can Make A Difference Random Initial Placement Final Placement

4. 4 7/15/2015 Partitioning: Objective: Given a set of interconnected blocks, produce two sets that are of equal size, and such that the number of nets connecting the two sets is minimized.

5. 5 7/15/2015 FM Partitioning: Initial Random Placement After Cut 1 After Cut 2 list_of_sets = entire_chip; while(any_set_has_2_or_more_objects(list_of_sets)) { for_each_set_in(list_of_sets) { partition_it(); } /* each time through this loop the number of */ /* sets in the list doubles. */ }

6. 6 7/15/2015 FM Partitioning: -2 1 0 0 0 2 0 0 1 -2 - 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.

7. 7 7/15/2015 -2 1 0 0 0 2 0 0 1 -2 FM Partitioning:

8. 8 7/15/2015 -2 1 0 -2 0 0 1 -2

9. 9 7/15/2015 -2 1 0 -2 0 0 1 -2

10. 10 7/15/2015 -2 1 0 -2 0 0 1 -2

11. 11 7/15/2015 -2 1 0 -2 0 -2

12. 12 7/15/2015 -2 1 0 -2 0 -2

13. 13 7/15/2015 -2 1 0 -2 0 -2

14. 14 7/15/2015 -2 1 -2 0 1 -2

15. 15 7/15/2015 -2 1 -2 0 1 -2

16. 16 7/15/2015 -2 1 -2 0 1 -2

17. 17 7/15/2015 -2 -3 -2 0 1 -2

18. 18 7/15/2015 -2 - 1 -3 -2 0 1 -2

19. 19 7/15/2015 -2 - 1 -3 -2 0 1 -2

20. 20 7/15/2015 -2 - 1 -3 -2 -2

21. 21 7/15/2015 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

22. 22 7/15/2015 x2 x1 x=100 x=200 Example:

23. 23 7/15/2015 x2 x1 x=100x=200 Interpretation of matrices A and B: The diagonal values A[i,i] correspond to the number of connections to xi The off diagonal values A[i,j] are -1 if object i is connected to object j, 0 otherwise The values B[i] correspond to the sum of the locations of fixed objects connected to object i Example:

24. 24 7/15/2015 Quadratic Placement aGlobal optimization: solves a sequence of quadratic programming problems aPartitioning: enforces the non-overlap constraints

25. 25 7/15/2015 Solution of the Original QP

26. 26 7/15/2015 Partitioning Use FM to cut.

27. 27 7/15/2015 Perform the Global Optimization again with additional constraints that the center of gravities should be in the center of regions. Applying the Idea Recursively Center of Gravities

28. 28 7/15/2015 Process of Gordian (a) Global placement with 1 region(b) Global placement with 4 region(c) Final placements

29. 29 7/15/2015 Quadratic Techniques: Pros: - mathematically well behaved - efficient solution techniques Cons: - solution of Ax + B = 0 is not a legal placement, so generally some additional partitioning techniques are required. - solution of Ax + B = 0 is minimizes wirelength squared, not linear wire length.