Bus-Driven Floorplanning Hua Xiang*, Xiaoping Tang +, Martin D. F. Wong* * Univ. Of Illinois at Urbana-Champaign + Cadence Design Systems Inc.

Bus-Driven Floorplanning

Hua Xiang*, Xiaoping Tang+, Martin D. F. Wong*

* Univ. Of Illinois at Urbana-Champaign+ Cadence Design Systems Inc.

Floorplanning & Bus Planning Bus planning

An important issue for floorplanning in DSM Buses:

Different widths Go through several module blocks

The positions of module blocks heavily affect bus planning.

An Example

(a) A floorplan with 2 buses. (b) Neither bus can be assigned.

Bus-Driven Floorplanning (BDF) Given:

A set of rectangular macro blocks A set of buses

Objective: Decide positions of blocks and buses

Constraints: No overlap between any two blocks No overlap between any two horizontal (vertical)

buses Buses go through all of the related blocks Minimize chip area as well as the total bus area

Preliminary Sequence Pair

A pair of sequences of n elements representing a list of n blocks. Block position relationship

(… bi … bj … , … bi … bj …) (… bi … bj … , … bj … bi …) Bus representation < H/V , w , {b1, b2, … , bk} >

Feasible Buses

Feasible Horizontal Bus Bus < H , w , {b1, b2, … , bk} > is feasible

ymax – ymin ≥ w, where ymax = min {yi + hi | i =1, …, k} ymin = max {yi | i =1, …, k}

Necessary Condition for One Bus

(a) A horizontal bus

( … A … D … B … C … , … A … D … B … C … )

(b) A vertical bus( … A … C … B … , … B … C … A … )

Necessary Condition for One Bus

Theorem 1 (Block Ordering) Given:

A sequence pair (X, Y) A bus u = {b1, b2, …, bk}

If u is feasible The ordering of the k blocks should be either the same or reverse in X and Y. Same order Horizontal bus Reverse order Vertical bus

Bus Ordering between Two Buses Given two horizontal buses: u = {a1, … , am} and v = {b1, … , bn} Let Su = {a1, … , am}, Sv = {b1, … , bn}. S = Su∪Sv and L = |S| ( L ≤ m + n ) Given a sequence pair (X, Y) = (… c1 … c2 … … cL … , … d1 … d2 … … dL …) where ci ∈S and di ∈S Subsequence pair: (X ’, Y ’) = ( c1 c2 … cL , d1 d2 … dL ) Let p[ci] = i and q[di] = i (i = 1, … , L)

Bus Ordering between Two Buses

Case 1 ; bus u is above bus v Bus Constraint

][][, bqbpSb u ][][, bqbpSb u

Two buses: {A, B, C} {D, E, F}SubSequence pair (D A E B F C , A D B E C F)

Bus Ordering between Two Buses

Case 2 ; bus u is below bus v Bus Constraint

][][, bqbpSb u ][][, bqbpSb u

Two buses: {A, B, C} {B, D, E}SubSequence pair (A D B C E , D A B E C)

Bus Ordering between Two Buses

Case 3 (Bus Crossing) ; u and v can’t be assigned at the same

time

][][, bqbpSb u ][][, bqbpSb u

Two buses: {A, B} {C, D}SubSequence pair (A C D B , C A B D)

Bus Ordering between Two Buses

Case 4 No Bus Constraint

][][, bqbpSb u

Two buses: {A, B, C} {D, E}SubSequence pair (A D B E C , A D B E C)

Multiple Bus Ordering The ordering of buses in a BDF

solution cannot be a cycle For example:

Bus u is above bus v Bus v is above bus w Bus w is above bus u

Bus Ordering Constraint Graph Each bus is represented by a node. If one bus u is above bus v (Case 1 or 2), then add one edge (u, v). If two buses u and v are crossing (Case 3), then two edges (u, v) and (v, u) are added. If two buses have no bus constraint (Case 4),

then no edges are added.

Cycles in Constraint Graph A cycle in a bus ordering constraint grap

h At least one bus cannot be assigned Two kinds of cycles

A cycle caused by bus crossing (Case 3) A cycle caused by multiple buses

Cycles in Constraint Graph (cont)

A cycle caused by bus crossing

Two buses: {A, B} {C, D}SubSequence pair (A C D B , C A B D)

Cycles in Constraint Graph (cont)

A cycle caused by multiple busesThree buses: {A1, A2} {B1, B2} {C1, C2}Sequence pair ( A1 B1 B2 C1 C2 A2 , B1 A1 C1 B2 A2 C2 )

Cycles in Constraint Graph (cont)

Remove fewest nodes from a constraint graph The graph contains no cycles

It is proved to be an NP-Complete problem. A heuristic approach

If a node whose in-degree or out-degree is zero, then the node should not be removed. If all nodes whose in-degree and out-degree are

non-zero, remove the node with the max degree.

Node Removal in Constraint Graph

(a) A constraint graph with cycles. (b) Nodes a, e and d are good nodes and they are not

considered.

Node Removal in Constraint Graph

(b) Nodes a, e and d are good nodes and they are not

considered.

(c) Node c has the max degree and it is removed

from the graph.

Node Removal in Constraint Graph

(a) A constraint graph with cycles. (d) Nodes c and i are removed to break

cycles.

BDF Algorithm

Simulated Annealing Perturbation (Move) Cost Function Evaluation Algorithm

Perturbation (Move)

Swap Exchange two blocks in the first sequence Exchange two blocks in the second

sequence Take constant time

Rotation Exchange the width and height of a block No change to the sequence pair Take constant time

Cost Function Objective

Fit in all buses Minimize the chip area Minimize the total bus area

Cost Function

C : the chip areaB : the total bus areaM : the number of unassigned buses

MBCCost

Evaluation Algorithm

Transform a sequence pair to a BDF solution How to detect infeasible buses ?

Violate block ordering Discard the bus Bus nodes form cycles in a constraint graph Node Remov

al How to decide block positions ?

Longest Common Subsequence Computation (LCS) How to assign buses ?

Assign a bus if its related blocks have been processed Related blocks may have to move up/left to meet bus align

ment requirement

Bus Assignment

Buses are processed from bottom to top one by one

Buses are ordered according to bus constraint graph

Bus u ={A, B, C} is above v ={E, B, G}. v should be assigned before u.

Bus Assignment (cont) Block alignment for one bus <H, w, {b1,

…, bk}>

ymax = max {yi | i = 1, …, k}

yi = max (yi, ymax + w – hi)

Bus_Overlap

(a) Two buses overlap (b) No overlap between two buses

(b) Blocks appear interlaced.

(A D B E C, A D B E C)

Bus Assignment (cont) Cases for Bus_Overlap

Two buses have ordering constraint (Case 1 or 2), no Bus_Overlap

Two buses have no ordering constraint (Case 4), there are 3 cases

(a) Two buses share at least one block.

(c) Blocks appear non-interlaced.

(A B C D E, A B C D E)Bus_Overlap may happen for (a) and (b)

Evaluation Algorithm

Evaluation_BDF (seq_pair, buses){ 1. Feasible_Bus_Checking_Orientation

// Remove buses which cannot be assigned due to block ordering or// cycles in constraint graphs. At the same time, decide bus

orientation

2. Bus_Ordering//Sort buses according to below-above/left-right relationship

3. Modified_lcs_computation//Calculate the positions of blocks and buses

4. Cost_Calculation 5. Return cost

}

Soft Block Adjustment

Blocks on longest common subsequence paths decide the chip size.

Each time, one soft block on a longest common subsequence path is selected, and its width and height are adjusted.

Simulated Annealing is applied again to find a compact floorplan.

Soft Block Adjustment (cont) An example

Blocks B, D, E are on a critical path. E is selected and adjusted. The chip size is

reduced.

Experimental Results

Hardware: workstation (2.4GHz) with 1G memory Software: C++ Test files

MCNC benchmarks Industry designs Bus grid test files

Experimental Results (cont)

(a) MCNC benchmarks

File Blocks BusesPacking Results

Soft Block Adjustment

Time(s)Deadspac

eTime(s)

Deadspace

apte 9 5 11 4.11% 12 (+1) 0.72%

xeorx 10 6 12 3.88% 13 (+1) 0.95%

hp 11 14 28 5.02% 28 (+0) 0.62%

ami33-1 33 8 61 6.02% 62 (+1) 0.94%

ami33-2 33 18 81 6.10% 86 (+5) 1.27%

ami49-1 49 9 98 5.42%101 (+3)

0.85%

ami49-2 49 12 278 6.09%281 (+3)

0.84%

ami49-3 49 15 265 7.40%268 (+3)

1.09%

Experimental Results (cont)Ami49-2 (after soft-adjustment)

49 blocks, 12 busesBuses

{0, 5, 9, 12, 18}{1, 10, 21, 25}{2, 28, 33}{3, 19, 22, 26, 29, 34}{4, 23, 27}{5, 35, 30, 6}{32, 31, 17}{11, 14, 15, 32, 33}{12, 8, 14}{44, 43, 7}{0, 3}{2, 47}

Experimental Results (cont)

(b) Industry designs

File cad1 cad2

Blocks 40 57

Buses 13 16

Time (s) 209 191

Deadspace 4.40% 5.16%

cad2 BDF Solution

Experimental Results (cont)

(c) Bus grid test files

File grid4 grid5 grid6 grid7

Blocks 16 25 36 49

Buses 8 10 12 14

Time (s) 1 23 103 150

Deadspace

0% 0% 0% 0%

Optimal BDF Solution for grid7

Conclusion

Analyze the relationship between bus ordering and sequence pair representation

Propose an algorithm for simultaneous floorplanning and bus planning using simulated annealing

Experimental results demonstrate its effectiveness and efficiency