ELEC 7770 Advanced VLSI Design Spring 2008 Constraint Graph and Performance Optimization

1. ELEC 7770Advanced VLSI DesignSpring 2008Constraint Graph and Performance Optimization Vishwani D. Agrawal James J. Danaher Professor ECE Department, Auburn University Auburn, AL 36849 vagrawal@eng.auburn.edu http://www.eng.auburn.edu/~vagrawal/COURSE/E7770_Spr08/course.html ELEC 7770: Advanced VLSI Design (Agrawal)

2. Retiming Theorem • Given a network G(V, E, W) and a cycle time T, (r1, . . . ) is a feasible retiming if and only if: • ri – rj ≤ wij for all edges (vi,vj) ε E • ri – rj ≤ W(vi,vj) – 1 for all node-pairs vi, vj such that D(vi,vj) > T Where, W(vi,vj) is the minimum weight path between vi and vj D(vi,vj) is the maximum delay among all minimum weight paths between vi and vj ELEC 7770: Advanced VLSI Design (Agrawal)

3. Timing Optimization • Find the clock period (T) by path analysis. • Set clock period to T/2 and find a feasible retiming. • If feasible, further reduce the clock period to half. • If not feasible, increase clock period. • Do a binary search for optimum clock period. • Retime the circuit. ELEC 7770: Advanced VLSI Design (Agrawal)

4. Representing a Constraint ri – rj ≤ wij or rj ≥ ri – wij – wij rj ri ELEC 7770: Advanced VLSI Design (Agrawal)

5. Constraint Graph -6 r1 ≥ r0 + 3 r1 ≥ r2 + 1 r2 ≥ r0 + 1 r2 ≥ r1 – 1 r3 ≥ r1 + 1 r3 ≥ r2 + 4 r0 ≥ r3 – 6 r1 3 1 r0 r3 -1 1 1 4 r2 ELEC 7770: Advanced VLSI Design (Agrawal)

6. Feasibility Condition • A set of values for variables can be found if and only if the constraint graph has no positive cycles. • This is also the condition for the solvability of the longest path problem, which provides a solution to the set of constraints. ELEC 7770: Advanced VLSI Design (Agrawal)

7. Example: Infeasible Constraints x2 x1 ≥ x2 + 6 x2 ≥ x1 – 3 6 x2 ≥ x1 – 3 x1 x2 3 -3 Positive cycle mean no longest path can be found. x1 ≥ x2 + 6 x1 0 3 6 ELEC 7770: Advanced VLSI Design (Agrawal)

8. Solving a Constraint Set -6 r1 ≥ r0 + 3 r1 ≥ r2 + 1 r2 ≥ r0 + 1 r2 ≥ r1 – 1 r3 ≥ r1 + 1 r3 ≥ r2 + 4 r0 ≥ r3 – 6 r1 3 1 r0 r3 -1 1 Longest path from source r0: r0, r1, r2, r3 Path lengths: s0=0, s1=3, s2=2, s3=6 Solution: r0=0, r1=3, r2=2, r3=6 1 4 r2 ELEC 7770: Advanced VLSI Design (Agrawal)

9. The General Path Problem • Find the shortest (or longest) path in a graph from a source vertex to all other vertices. • Graph has vertices and directed edges: • Edge weights can be positive or negative • Graph can be cyclic • Single source vertex – a vertex with 0 in-degree (not a necessary condition) • Inconsistent problems • Negative weight cycles for shortest path • Positive weight cycles for longest path ELEC 7770: Advanced VLSI Design (Agrawal)

10. Dijkstra’s Shortest Path Algorithm • Greedy algorithm. • Applies to directed acyclic graphs (DAG) with positive edge weights. • Computational complexity O(|E| + |V| log |V|) ≤ O(n2) • References: • A. Aho, J. Hopcroft and J. Ullman, Data Structures and Algorithms, Reading, Massachusetts: Addison-Wesley, 1983. • T. Cormen, C. Leiserson and R. Rivest, Introduction to Algorithms, New York: McGraw-Hill, 1990. ELEC 7770: Advanced VLSI Design (Agrawal)

11. Dijkstra’s Shortest Path Algorithm Example 1 v1 w01=15 3 v0 v3 10 source 2 6 v2 si = path weight (v0, vi) Each step marks the path with smallest weight and updates the unmarked path weights. ELEC 7770: Advanced VLSI Design (Agrawal)

12. Dijkstra’s Shortest Path Algorithm Example 2 v1 w01=15 3 v0 v3 6 source 2 10 v2 si = path weight (v0, vi) Each step marks the path with smallest weight and updates the unmarked path weights. ELEC 7770: Advanced VLSI Design (Agrawal)

13. Dijkstra’s Algorithm, G(V, E, W) s0(1) = 0 initialize source for ( i = 1 to n ) initialize path weights, n=|V| –1 si(1) = w0i repeat { Select an unmarked vertex vq such that sq is minimal Mark vq foreach ( unmarked vertex vi ) si = min { si, sq + wqi } } until (all vertices are marked) ELEC 7770: Advanced VLSI Design (Agrawal)

14. Try Dijkstra’s Algorithm for Your Graph http://www.dgp.toronto.edu/people/JamesStewart/270/9798s/Laffra/DijkstraApplet.html ELEC 7770: Advanced VLSI Design (Agrawal)

15. Dijkstra’s Longest Path Algorithm v1 Either change min to max Or change all positive weights to negatives w01=15 3 v0 v3 10 source 2 6 v2 v1 w01= -15 -3 v0 v3 -10 source -2 -6 v2 si = path length (v0, vi) ELEC 7770: Advanced VLSI Design (Agrawal)

16. Dijkstra’s Alg. Does Not Work for Cycles, Mixed Weights -2 v1 w01=15 3 v0 v3 5 source 2 4 v2 si = path weight (v0, vi) Algorithm stops because all vertices are marked. But, there exists a v0 to v3 path of length 5 ELEC 7770: Advanced VLSI Design (Agrawal)

17. Bellman’s Equations – Shortest Path vj vk wki wji For all vertices: si = min (sq + wqi) vq ε pred(vi) vi vm wmi wni vn sq = minimum path weight between source and vq ELEC 7770: Advanced VLSI Design (Agrawal)

18. Bellman-Ford Algorithm, G(V, E, W) Bellman-Ford { s0(1) = 0 initialize source for ( i = 1 to n ) initialize path weights, n = |V| – 1 si(1) = w0i for ( j = 1 to n ) n iterations for ( i = 1 to n ) si(j+1) = min { si(j), sk(j) + wkj } vk ε pred(vi) } if ( si(j+1) == si(j) i ) return (true) } return (false) Complexity = O(|V||E|) ≤ O(n3) ELEC 7770: Advanced VLSI Design (Agrawal)

19. Bellman-Ford Shortest Path n = 3 v1 w01=15 3 v0 v3 10 source 2 6 v2 si = path weight (v0, vi) ELEC 7770: Advanced VLSI Design (Agrawal)

20. Bellman-Ford Longest Path Reverse the sign of weights and solve shortest path problem. (Alternative: keep original weights and change min operator in algorithm to max.) n = 3 (shortest path) Weights reversed v1 w01= -15 -3 v0 v3 -10 source -2 -6 v2 si = path weight (v0, vi) ELEC 7770: Advanced VLSI Design (Agrawal)

21. Bellman’s Equations – Longest Path vj vk wki wji For all vertices: si = max (sq + wqi) vq ε pred(vi) vi vm wmi wni vn sq = maximum path weight between source and vq ELEC 7770: Advanced VLSI Design (Agrawal)

22. Bellman-Ford for Cycles, Neg. Weights n = 3 (shortest path) -2 v1 w01=15 3 v0 v3 5 source 2 4 v2 si = path weight (v0, vi) This was incorrect with Dijkstra’s shortest path algorithm ELEC 7770: Advanced VLSI Design (Agrawal)

23. Bellman-Ford for Negative Cycle n = 3 (shortest path) 2 v1 w01=15 -3 v0 v3 5 source 2 4 v2 si = path weight (v0, vi) Values not stabilized after n iterations. Inconsistent problem: negative cycle. ELEC 7770: Advanced VLSI Design (Agrawal)

24. Retiming Example FF a b c 10 5 5 Delay ELEC 7770: Advanced VLSI Design (Agrawal)

25. Retiming Graph FF a b c 10 5 5 1 0 0 h 0 a 10 b 5 1 c 5 Critical path = 15 It is the longest path consisting only of zero weight edges. ELEC 7770: Advanced VLSI Design (Agrawal)

26. Feasibility Constraints FF a b c 10 5 5 1 0 0 h 0 a 10 b 5 1 c 5 rh – ra ≤ 0 ra – rb ≤ 0 rb – rc ≤ 1 rc – rh ≤ 1 ri – rj ≤ wij  edges i → j Retiming should not cause negative edge weights. ELEC 7770: Advanced VLSI Design (Agrawal)

27. Constraint Graph FF a b c 10 5 5 -1 0 0 rh 0 ra 10 rb 5 -1 rc 5 rh – ra ≤ 0 ra – rb ≤ 0 rb – rc ≤ 1 rc – rh ≤ 1 ri – rj ≤ wij  edges i → j Retiming should not cause negative edge weights. Observation: Constraint graph has the same structure as the original retiming graph, with signs of weights reversed. Vertex labels are the retiming integer variables. ELEC 7770: Advanced VLSI Design (Agrawal)

28. Max Delay for Min Weight Paths 1 0 0 h 0 a 10 b 5 1 c 5 T = 15 W(b,c) = 1 D(b,c) = 10 W(b,h) = 2 D(b,h) = 10 W(b,a) = 2 D(b,a) = 20 W(c,h) = 1 D(c,h) = 5 W(c,a) = 1 D(c,a) = 15 W(c,b) = 1 D(c,b) = 20 W(h,a) = 0 D(h,a) = 10 W(h,b) = 0 D(h,b) = 15 W(h,c) = 1 D(h,c) = 20 W(a,b) = 0 D(a,b) = 15 W(a,c) = 1 D(a,c) = 20 W(a,h) = 2 D(a,h) = 20 ELEC 7770: Advanced VLSI Design (Agrawal)

29. Timing Optimization, T = 7.5? -1 Constraint graph (feasibility) 0 0 rh 0 ra 10 rb 5 -1 rc 5 ri – rj ≤ W(I,j) – 1  paths (i,j) such that D(i,j) > 7.5 W(b,c) = 1 D(b,c) = 10 W(b,h) = 2 D(b,h) = 10 W(b,a) = 2 D(b,a) = 20 W(c,h) = 1 D(c,h) = 5 W(c,a) = 1 D(c,a) = 15 W(c,b) = 1 D(c,b) = 20 W(h,a) = 0 D(h,a) = 10 W(h,b) = 0 D(h,b) = 15 W(h,c) = 1 D(h,c) = 20 W(a,b) = 0 D(a,b) = 15 W(a,c) = 1 D(a,c) = 20 W(a,h) = 2 D(a,h) = 20 ELEC 7770: Advanced VLSI Design (Agrawal)

30. Timing Optimization, T = 7.5? Positive cycle No solution -1 0 1 1 0 1 0 0 rh 0 ra 10 rb 5 -1 rc 5 -1 -1 0 0 -1 0 W(b,c) = 1 D(b,c) = 10 W(b,h) = 2 D(b,h) = 10 W(b,a) = 2 D(b,a) = 20 W(c,h) = 1 D(c,h) = 5 W(c,a) = 1 D(c,a) = 15 W(c,b) = 1 D(c,b) = 20 W(h,a) = 0 D(h,a) = 10 W(h,b) = 0 D(h,b) = 15 W(h,c) = 1 D(h,c) = 20 W(a,b) = 0 D(a,b) = 15 W(a,c) = 1 D(a,c) = 20 W(a,h) = 2 D(a,h) = 20 ELEC 7770: Advanced VLSI Design (Agrawal)

31. Timing Optimization, T = 11.25? -1 rh = 0 rb = 1 rc = 0 ra = 0 0 1 1 0 0 0 rh 0 ra 10 rb 5 -1 rc 5 -1 -1 0 0 W(b,c) = 1 D(b,c) = 10 W(b,h) = 2 D(b,h) = 10 W(b,a) = 2 D(b,a) = 20 W(c,h) = 1 D(c,h) = 5 W(c,a) = 1 D(c,a) = 15 W(c,b) = 1 D(c,b) = 20 W(h,a) = 0 D(h,a) = 10 W(h,b) = 0 D(h,b) = 15 W(h,c) = 1 D(h,c) = 20 W(a,b) = 0 D(a,b) = 15 W(a,c) = 1 D(a,c) = 20 W(a,h) = 2 D(a,h) = 20 ELEC 7770: Advanced VLSI Design (Agrawal)

32. Retiming Graph FF a b c 10 5 5 1 0 0 h 0 a 10 b 5 1 c 5 1 0 rc = 0 rh = 0 ra = 0 rb = 1 wij_retimed = wij + rj – ri ELEC 7770: Advanced VLSI Design (Agrawal)

33. Retimed Circuit FF a c b 10 5 5 Logic optimization will remove these. 1 1 0 0 h 0 a 10 b 5 c 5 rc = 0 rh = 0 ra = 0 rb = 1 Critical Path = 10 ELEC 7770: Advanced VLSI Design (Agrawal)