1. Multi-Layer Channel Routing Complexity and Algorithm Rajat K. Pal

2. Chapter- 4 A Framework for Track Assignment Presented by: Abu Sayed Chowdhury 0409052008 Multi-Layer Routing (Extensions of the TAH Framework)

3. Pre-requisite Channel: A channel is a horizontal routing area with fixed pins on the top and bottom TAH Framework: Algorithm for computing reduced wire length channel routing solutions using a purely graph theoretic framework, TAH (Track Assignment Heuristic) that was designed for computing minimum area routing solutions

4. Pre-requisite (Contd.) Net Cells have rectangular shapes with terminals located on their border. A set of terminals that must be electronically connected together, constitute a net. Terminals of the same net are given the same integer level

5. C E A D B F C E A D B F VLSI Layout Pre-requisite (Contd.) Net

6. Channel Routing Terminology Upper boundary Lower boundary Tracks Terminals Via TrunksBranches Dogleg

7. Routing Layer Models HV model VH model HVH model VHV model Layer 1 Layer 2 Layer 3 Via 1 layer 2 layers 3 layers

8.  An example channel instance of ten nets  Intervals of the nets are placed in five different tracks  The length of the channel (i.e., the number of columns) is 15  Arrows indicate that the terminals to be connected, either on the top or at the bottom, to complete the required interconnection of all the nets present in the channel

9. TAH Basics The Algorithm TAH assigns intervals to tracks from top to bottom. In the first iteration the algorithm assigns a set of non-overlapping intervals to the top most track. Then it delete the nets corresponding to these intervals from the channel. In the second iteration it assigns a set of non- overlapping intervals to the second track. The iterative process continues till all the nets are assigned to tracks in the channel.

11. I2I2 I7I7 I5I5 I4I4 I1I1 I3I3 I6I6 014 5 1 670 49 10 0 23 5 2 68 9 8 70 9 35 I8I8 I9I9 I 10  The Channel contains total 10 nets  Density of this channel is ( ) is 5  Intervals of nets 1, 4, 5, 3, and 2 are overlapping to each other, and form a clique

12. I2I2 I7I7 I5I5 I4I4 I1I1 I3I3 I6I6 014 5 1 670 49 10 0 23 5 2 68 9 8 70 9 35 I8I8 I9I9 I 10 1 4 6 7 8 10 2 9 53 VCG

13. I4I4 014 5 1 670 49 10 0 23 5 2 68 9 8 70 9 35 I 10 1 6 7 8 2 9 53 VCG Track 1

14. I7I7 I1I1 014 5 1 670 49 10 0 23 5 2 68 9 8 70 9 35 6 8 2 9 53 VCG Track 2

15. I5I5 014 5 1 670 49 10 0 23 5 2 68 9 8 70 9 35 8 23 VCG Track 3 I6I6 I9I9

16. 014 5 1 670 49 10 0 23 5 2 68 9 8 70 9 35 3 VCG Track 4 I2I2 I8I8

17. 014 5 1 670 49 10 0 23 5 2 68 9 8 70 9 35 3 VCG Track 4 I2I2 I8I8

18. 014 5 1 670 49 10 0 23 5 2 68 9 8 70 9 35 Track 5 I3I3

19. Finally using TAH

20. A Two Layer No Dog-leg routing solution for Ex.2 using TAH

21. Thanks to All