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## PowerPoint Slideshow about ' Multi-Layer Channel Routing Complexity and Algorithms' - yosefu

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Chapter 4

A General Framework for Track Assignment in

Multi-Layer Channel Routing

- Presented By-
- Sumaya Kazary
- Std. ID: 0409052009

TAH Basics

- The Algorithm TAH assigns intervals to tracks from top to bottom in the presence of Vertical constraints.
- 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.

3

5

4

1

7

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I3

I1

I2

I7

I5

I4

I6

Some Notations & a few definitions- dmax→channel density.
- vmax→the length of the longest path in the VCG.
- idvi→ indegree of vertex vi in the VCG.
- dpvi→length of the longest path from a source to vi.
- htvi→ length of the longest path from a sink vertex to vi.
- spvi→span of a net ni.
- zdvi→ zonal density of a vertex v.

For the following channel

TOP: 3 1 3 0 0 5 6 0 3 0 0

BOTTOM: 1 2 4 2 4 1 5 7 0 7 6

The routing solution using TAH framework for VH routing is

3

1

3

0

0

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6

0

0

0

3

1

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4

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1

5

7

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I1

I2

I7

I5

I4

I6

0

3

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3

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- The Channel contains total 7 nets.
- Density of this channel is dmax = 4.

In the TAH algorithm ,for each iteration, certain no. of intervals are assigned in a track from top to bottom.

- In each iteration ,it computes a clique such that there is no overlapped interval is present & assign it to a track.
- At the beginning, it compares the channel density (dmax)and the longest path in the VCG(vmax).
- Two cases may arise:-
- dmax>vmax
- dmax≤vmax

3

5

4

1

7

2

Executing the Algorithm Steps:S1={3,6,7}- Iteration-1As, dmax=vmax=4. So case II

and S3={6} [As S3={vi| htvi=vmax}]

Step Efor vertex 6 < od6,1,sp6> is <1,1,4>.

So ,

Step Flimitvi=max(dmax,vmax)-htvi+1

wt:<f(t),ht,zd,sp>

limit3=4-3+1=2>t .So,f(1)3=(1-1+1)÷(2-1+1)=1/2.

limit6=4-4+1=1=t. So,f(1)6=1-1+1=1.

limit7=4-1+1=4>t. So,f(1)7=(1-1+1)÷(4-1+1)=1/4.

Now, ht3=3,ht6=4 ,ht7=1; zd3=4,zd6=3,zd7=3 and

sp3=8,sp6=4, sp7=2 So, 3:<1/2,3,4,8>

6:<1,4,3,4> 7: <1/4,1,3,2>

C1={6}.

So,C2={6}.

Step GIn step E, span usage (spvi) by a net was maximized, now it is checked with minimizing, so the weights are <odvi, 1, -spvi>.In this case,6:<1, 1, -4> .

So, again let’s pick C3 = {6}.

Step H similar to step F, so, C4 = {6}

So net 6 is assigned to track 1.

Now, HNCG & VCG are rearranged excluding node 6 and its adjacent edges.

5

4

1

7

2

Executing the Algorithm Steps:S1={3,5,7}- Iteration-2Now , dmax>vmax. So case I. and S2={1,2,3,4}

Step ANow, S1 S2={3}

for vertex 3 < cn3,1,sp3> is <1,1,8>.

So ,

Step Blimitvi=max(dmax,vmax)-htvi+1 wt:<f(t),zd,ht,sp>

limit3=4-3+1=2=t .So,f(2)3=2-2+1=1. zd3=4, ht3=3, sp3=8

limit5=4-3+1=2=t. So,f(2)5=2-2+1=1. zd5=3, ht5=3, sp5=1

limit7=4-1+1=4>t. So,f(2)7=(2-1+1)÷(4-1+1)=1/2. zd7=2, ht7=1, sp7=2

So, 3:<1,4,3,8>; 5:<1,3,3,1>7: <1/2,2,1,2>

C1={3}.

So,C2={3}.

Step CIn step A, span usage (spvi) by a net was maximized, now it is checked with minimizing, so the weights are <cnvi, 1, -spvi>.In this case,3:<1, 1, -8> .

So, again let’s pick C3 = {3}.

Step D Similar to step F, so, C4 = {3}

So net 3 is assigned to track 2.

Now, again HNCG & VCG are re-arranged excluding node 3 and its adjacent edges. And iteration 3 is proceed.

Time complexity

- For initial part ,time complexity--
- To compute the HNCG & G* is O(n+e).
- To compute the VCG is O(n).
- To compute the span of every interval is O(n).
- htvi and dpvi is computed by 1 scan of VCG needs O(n).

So,for initial part, it takes time O(n+e).

- Each iteration part takes O(n+e).
- If total track=k, then total time complexity of the algorithm is O(k(n+e)).

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