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23年 4月 19日 1
VLSI Physical Design Automation
Prof. David Pan
Office: ACES 5.434
Lecture 18. Global Routing (II)
223年 4月 19日
Global Routing Approaches
• Sequential Approach (Rip-up and Re-route)– Maze Routing– Line probing– Shortest Path Based Algorithms– Steiner Tree Based Algorithms
• Concurrent Approach– Integer Programming
323年 4月 19日
Sequential Approach
Algorithm: 1. Graph modeling of the routing regions 2. For each net k:
2.1 Find a route r for net k on the graph. 2.2 For each edge e in r:
2.2.1 capacity(e) = capacity(e) - 1 2.2.2 if capacity(e) < 0 then cost(e) = cost(e)
We can use differentmethods to do this.
423年 4月 19日
Maze Routing for Multi-Terminal Nets
523年 4月 19日
Maze Routing on Weighted Graph
623年 4月 19日
Example of Weighted Graph
723年 4月 19日
Mikami & Tabuchi’s Algorithm
823年 4月 19日
Line Probing
• Keep two lists of line segments, slist and tlist, for the source and the target respectively.
• If a line segment from slist intersects with one from tlist, a route is found; else, new line segments are generated from the escape points.
S
Tslist
tlist
Intersection
923年 4月 19日
Line Probing
• We can use all the grid vertices on the line segments as escape points:
• Always find a path but may not be optimal.
S
T 0
0
1
1
1 1
1
1111
11
1 1
0
0
Escape point
Iteration number
1023年 4月 19日
Hightower’s Algorithm
1123年 4月 19日
Line Probing
• We can pick just one escape point from each line segment.
• May fail to find a path even if one exists.
S
T
1
1
0
0
1
0
0
Escape point
Iteration number
1223年 4月 19日
1323年 4月 19日
Comparison of Algorithms
1423年 4月 19日
BFS based Maze Routing (A*)
• Need to search whole space?– Guide the search to the goal explicitly
• A* search is faster if you need “good path”, not “perfect path”– Use priority queue– C(n) = F(n)+H(n)
• F(n) is a computed cost from source to current location.
• H(n) is a predicted cost from current location to target.
• If H(n)=0, it becomes maze routing!
– Optimal (shortest path) when H(n) <= H’(n) (no overestimation)• H’(n) is the exact cost
• H(n)=0 never overestimates!
1523年 4月 19日
Maze vs A* routing (I)
1623年 4月 19日
Maze vs A* routing (II)
1723年 4月 19日
Shortest Path Based Algorithms
• For 2-terminal nets only.• Use Dijkstra’s algorithm to find the shortest path
between the source s and the sink t of a net.• Different from Maze Routing:
– The graph need not be a rectangular grid.– The edges need not be of unit length.
1823年 4月 19日
Dijkstra’s Shortest Path Algorithm
• Label of vertices = Shortest distance from S.• Let P be the set of permanently labeled vertices.
• Initially, – P = Empty Set.– Label of S = 0, Label of all other vertices = infinity.
• While (T is not in P) do– Pick the vertex v with the min. label among all vertices not in P.– Add v to P.– Update the label for all neighbours of v.
1923年 4月 19日
Dijkstra’s Algorithm: Example
00
8888
8888
1010
5522
22 33 44 66
11
99
77SS
BB TT
AA CC
00
1010
55
8888
1010
5522
22 33 44 66
11
99
77SS
BB TT
AA CC
00
88
55
1414
77
1010
5522
22 33 44 66
11
99
77SS
BB TT
AA CC
00
88
55
99
77
1010
5522
22 33 44 66
11
99
77SS
BB TT
AA CC
00
88
55
99
77
1010
5522
22 33 44 66
11
99
77SS
BB TT
AA CC
00
88
55
1313
77
1010
5522
22 33 44 66
11
99
77SS
BB TT
AA CC
P (Permanently Labeled)P (Permanently Labeled) Min. Label VertexMin. Label Vertex
2023年 4月 19日
Steiner Tree Based Algorithms
• For multi-terminal nets.• Find Steiner tree instead of shortest path.• Construct a Steiner tree from the minimum spanning
trees (MST)
2123年 4月 19日
FLUTE Overview
• Solve Rectilinear Steiner minimal tree (RSMT) problem:– Given pin positions, find a rectilinear Steiner tree with minimum WL
• Basic idea:– LUT to handle small nets
– Net breaking technique to recursively break large nets
• Handling of small nets (with a few pins) is extremely well:– Optimal and extremely efficient for nets up to 7 pins
• So FLUTE is especially suitable for VLSI applications:– Over all 1.57 million nets in 18 IBM circuits [ISPD 98]
• Average error is 0.72%• Runtime faster than minimum spanning tree algorithm
– Open source available http://home.eng.iastate.edu/~cnchu/flute.htm
[courtesy of Prof. Chris Chu, Iowa State Univ.]
2223年 4月 19日
Preliminary
• A net is a set of n pins• Degree of a net is the number of pins in it• Consider routing along Hanan grid
• Define edge lengths hi and vi:
2323年 4月 19日
Wirelength Vector (WV)
• WL can be expressed as a vector of the coefficients• Called Wirelength Vector
321
321
22
vvvhhhWL
321
321
32 vvvhhhWL
321
321 2vvvhhhWL
(1, 2, 1, 1, 1, 2) (1, 1, 1, 1, 2, 3) (1, 2, 1, 1, 1, 1)
• Observation: WL can be written as a linear combination of edge lengths with positive integral coefficients
2423年 4月 19日
Potentially Optimal WV (POWV)• To find optimal wirelength, can enumerate all WVs• However, most WVs can never produce optimal WL
– (1, 2, 1, 1, 1, 2) is redundant as it always produces a larger WL than (1, 2, 1, 1, 1, 1)
• Potentially Optimal Wirelength Vector (POWV) is a WV that may produce the optimal wirelength
(1, 2, 1, 1, 1, 2) (1, 2, 1, 1, 1, 1)
2523年 4月 19日
# of POWVs is Very Small• For any net,
– # of possible routing solutions is huge– # of WVs is much less– # of POWVs is very small
• For example, only 2 POWVs for the net below:
POWV(1,2,1,1,1,1)
POWV(1,1,1,1,2,1)
2623年 4月 19日
Sharing of POWVs Among Nets
• To find optimal WL, we can pre-compute all POWVs and store them in a lookup table
• However, there are infinite number of different nets
• We try to group together nets that can share the same set of POWVs
• For example, these two nets share the same set of POWVs:
2723年 4月 19日
Grouping by Vertical Sequence
• Define vertical sequence s1s2…sn to be the list of pin indexes sorted in y-coordinate
• Lemma: The set of all degree-n nets can be divided into n! groups according to the vertical sequence such that all nets in each group share the same set of POWVs
Vertical sequence= 3142
2823年 4月 19日
Steps in FLUTE WL Estimation1. Input a net 2. Find hi’s and vi’s
3 2 5262
3. Find vertical sequence
3142
3
2
14
4. Get POWVs from LUT
(1,2,1,1,1,1)
(1,1,1,1,2,1)
5. Find WL for each POWV and return the best
HPWL + h2 = 22 returnHPWL + v2 = 26
Remark:• One RSMT topology can also be pre-computed and stored for each
POWV• Impractical for high-degree nets (degree >= 9)
– Other technique to break down high-degree nets
2923年 4月 19日
Net Ordering
• In sequential approach, we need some net ordering.• A bad net ordering will increase the total wire length,
and may even prevent com-pletion of routing for some circuits which are indeed routable.
A
A
B
B
B first (Good order)
A
AB
B
A first(Bad order)
3023年 4月 19日
Criteria for Net Ordering
• Criticality of net - critical nets first.• Estimated wire length - short nets first since they are
less flexible.• Consider bounding rectangles (BR):
A
AB
B
B is in A’s BR
Which one should be routed firstand why? (Note that this rule ofthumb is not always applicable.)
3123年 4月 19日
Net Ordering (cont’d)
3223年 4月 19日
Rip-Up and Re-route
• It is impossible to get the optimal net ordering.• If some nets are failed to be routed, the rip-up and re-
route technique can be applied:
A
B
C
A
B
C
A
B
C
A
B
C
A
B
C
A
B
C
Cannot route C So rip-up Band route C first.
Finally route B.
3323年 4月 19日
Concurrent Approach
• Consider all the nets simultaneously.• Formulate as an integer program.• Given:
Lij = Total wire length of Tij
Ce = Capacity of edge e• Determine variable xij s.t. xij = 1 if Tij is used xij = 0 otherwise.
Tn1, Tn2, ... , Tnknnet n
::
::
T11, T12, ...... , T1k1net 1
Set of possible routing treesNets
3423年 4月 19日
Integer Program Formulation
jix
eCx
nix
xL
ij
eijTeji ij
ik
j ij
n
i
ik
j
ijij
,1or 0
edge allfor
,,1 allfor 1 s.t.
Min.
s.t. ,
1
1 1
3523年 4月 19日
Concurrent Approach: Example
1,21,2
2,32,3
1,31,3
1,21,2
CCaa= = CCbb= = CCcc= = CCdd= 2= 2
bb
aa
cc
dd
Possible trees:Possible trees:
net 1:net 1: 22 33 33
net 2:net 2: 22 33 33
net 3:net 3: 22 22
1122
3322
11331122
SolutionSolution
1122
3322
11331122
;,1or 0
;1
;1
;1
s.t.
22332332 Min.
3231
232221
131211
3231232221131211
jix
xx
xxx
xxx
xxxxxxxx
ij
What are the constraintsfor edge capacity?
aCxxxxx 3123211312
3623年 4月 19日
Integer Programming Approach
Standard techniques to solve IP. No net ordering. Give global optimum. Can be extremely slow, especially for large problems. To make it faster, a fewer choices of routing trees for
each net can be used. May make the problem infeasible or give a bad solution.
Determining a good set of choices of routing trees is a hard problem by itself.
23年 4月 19日 37
Hierarchical Approach to Speed Up Integer Programming Formulation
For Global Routing
M. Burstein and R. Pelavin, “Hierarchical Wire Routing”,IEEE TCAD, vol. CAD-2,
pages 223-234, Oct. 1983.
3823年 4月 19日
Hierarchical Approach
• Large Integer Programs are difficult to solve.• Hierarchical Approach reduces global routing to routing
problems on a 2x2 grid.• Decompose recursively in a top-down fashion.• Those 2x2 routing problems can be solved optimally by
integer programming formulation.
3923年 4月 19日
Hierachical Approach: Example• Solving a 2xn routing problem hierarchically.
Level 1Level 1
Level 2Level 2
Level 3Level 3
Solution:Solution:
4023年 4月 19日
Types of 2x2 Routing Problems
Type 1Type 1 Type 7Type 7
Type 2Type 2
Type 3Type 3
Type 4Type 4
Type 5Type 5
Type 6Type 6
Type 8Type 8
Type 9Type 9
Type 10Type 10
Type 11Type 11
4123年 4月 19日
Objective Function of 2x2 Routing
Possible Routing Trees: Possible Routing Trees: TT1111, T, T1212, T, T2121, T, T2222,....., T,....., T11,111,1,..., T,..., T11,411,4
# of nets of each type: n# of nets of each type: n11, ..., n, ..., n1111
Determine xDetermine xijij: # of type-i nets using T: # of type-i nets using T ijij for routing. for routing. yyii: # of type-i nets that fails to route.: # of type-i nets that fails to route.
. minimize Want to
11,,111
1
i i
ij iji
y
inxy
4223年 4月 19日
Constraints of 2x2 Routing
CCaa
CCbb
CCcc
CCdd
BBabab BBbcbc
BBcdcdBBdada
aabb
ccdd
dij
ij
cij
ij
bij
ij
aij
ij
Cx
Cx
Cx
Cx
Tdji
Tcji
Tbji
Taji
s.t. ,
s.t. ,
s.t. ,
s.t. ,
daij
ij
cdij
ij
bcij
ij
abij
ij
Bx
Bx
Bx
Bx
daTji
cdTji
bcTji
abTji
regionin bend a has s.t. ,
regionin bend a has s.t. ,
regionin bend a has s.t. ,
regionin bend a has s.t. ,
Constraints on Edge Capacity:Constraints on Edge Capacity:
Constraints on # of Bends in a Region:Constraints on # of Bends in a Region:
4323年 4月 19日
Pop Quiz
• If you two nets, one with 2 pins, the other with 4 pins with a zero capacity edge– What is going to be the result?
Type 1Type 1
. minimize Want to
11,,111
1
i i
ij iji
y
inxy
Type 11Type 11
C=0C=1
4423年 4月 19日
ILP Formulation of 2x2 Routing
• Only 39 variables (28 xij and 11 yi) and19 constraints (plus 38 non-negative constrains).
• Problems of this size are usually not too difficult to solve.
jiyxinxy
y
iij
ij iji
i i
,0,011,,1 s.t.
Min.11
1
dij
ij
cij
ij
bij
ij
aij
ij
Cx
Cx
Cx
Cx
Tdji
Tcji
Tbji
Taji
s.t. ,
s.t. ,
s.t. ,
s.t. ,
daij
ij
cdij
ij
bcij
ij
abij
ij
Bx
Bx
Bx
Bx
daTji
cdTji
bcTji
abTji
regionin bend a has s.t. ,
regionin bend a has s.t. ,
regionin bend a has s.t. ,
regionin bend a has s.t. ,
