Bounding Variance and Expectation of Longest Path Lengths in DAGs Jeff Edmonds, York University

1. Bounding Variance and Expectation of Longest Path Lengths in DAGs Jeff Edmonds, York University Supratik Chakraborty, IIT Bombay

2. Motivation Statistical timing analysis of circuits Mean and std deviation of component delays provided by manufacturers Joint distributions of component delays difficult to obtain in practice

3. The Longest Path Problem Input: st-DAG G gives job precedence. For each edge i, xi is the time to complete job i Output: Time for all jobs to complete in parallel = length of longest st-path = Maxp i p xi = XG s x1 x3 x2 t Easy with Dynamic Programming

4. The Longest Path Problem Input: st-DAG G gives job precedence. For each edge i, xi is the time to complete job i Output: Time for all jobs to complete in parallel = length of longest st-path = Maxp i p xi = XG s Inter-dependent random variables x1 x3 x2 t Understand random variable XG

5. The Longest Path Problem Input: st-DAG G gives job precedence. For each edge i, xi is the time to complete job i Output: Time for all jobs to complete in parallel = length of longest st-path = Maxp i p xi = XG s Exp[Xi] & Var[Xi] x1 x3 x2 t Bound Exp[XG] & Var[XG]

6. The Longest Path Problem Input: XG = Max( x1+x2, x3 ) 4 5 0 1 s • Possible distributions : x1 x3 x2 t • Another possibility :

7. The Longest Path Problem Input: XG = Max( x1+x2, x3 ) Upper & Lower bounds s x1 x3 x2 t

8. Contributions Upper bounds of Exp[XG] and Var[XG] A spring “algorithm” for computing bounds Proof no distributions give higher values (skip) Cake distributions that achieve bounds Lower bounds of Exp[XG] and Var[XG] Continuum of values for Exp[XG] and Var[XG] Cake distributions that achieve any Exp[XG] and Var[XG] within range Special results for series-parallel graphs

9. Series Graphs If G is a series graph, XG = ∑i xi Exp[xG] = ∑i Exp[xi] 0 ≤ Var[xG] ≤ (∑i √Var[xi] )2 s t

10. Series Graphs If G is a parallel graph, XG = Maxi xi Maxi Exp[xi] ≤ Exp[xG] ≤ ? 0 ≤ Var[xG] ≤ ? s t

11. Representing Random Variables X 5 0 r 0 0.5 1 X : Two-valued random variable, prob 0.5 for each value

12. Representing Random Variables X 5 Z 0 r 0 0.5 1 X, Z : Two equivalent independent random variables.

13. Representing Random Variables X Y 5 0 r 0 0.5 1 X, Y : Two-valued random variables, prob 0.5 for each value X, Y have perfect negative correlation Exp( Max(x,y) ) = Exp(x) + Exp(y) Var( Max(x,y) ) = 0

14. Series Graphs If G is a parallel graph, XG = Maxi xi Maxi Exp[xi] ≤ Exp[xG] ≤ Min( ∑i Exp[xi], Maxi Exp[xi] + √∑i Var[xi] ) 0 ≤ Var[xG] ≤ ∑i Var[xi] s t

15. Series Parallel Graphs Theorem In a series-parallel graph, Rules for maximum variance applied recursively to obtain Max Var[XG]. Not so Max Exp[XG]

18. Maximizing Var [ XG ] There are no distributions xi for which Var[xi] = vi and Exp[xi] = mi Var[XG] > Proof uses lots of calculus. Theorem

19. Cakes Maximizing Var [ XG ] There exists “cake” distributions xi such that Var[xi] = vi and Exp[xi] = mi Var[XG] = Theorem

21. Cake Distribution s t Find a cake distribution for each edgewith correct Exp[xi] & Var[xi] to maximize Var[xG]

22. Cake Distribution s Exp[xi] t

23. Cake Distribution s t Var[xi] = ∑c (ε hc)2

24. Cake Distribution s t • Series graphs G: • XG ≈ x1 + x2 • Candle heights add • Want candle heights to be in same location

25. Cake Distribution s t • Parallel graphs G: • XG ≈ Max( x1 , x2 ) • Candle heights max • Want candle heights to be in different location

26. Cake Distribution s t A candle location for each st-path in G but in the end # candles ≈ # edges

27. Cake Distribution s t If edge i not in path p, candle for xi at location p has height 0 If candle is selected, then corresponding path pis the longest path

28. Cake Distribution s t “Springs” give give candle heights.

29. Cakes Maximizing Var [ XG ] There exists “cake” distributions xi such that Var[xi] = vi and Exp[xi] = mi Var[XG] = Theorem Proved

30. Lower Bound of Var[ XG ] TheoremVar[xG] ≥ 0 Continuum Results Theorem Every Var[XG] in this range achievable.

31. Lower bound of Exp [ XG ]

32. Upper bound of Exp [ XG ] XG r tp 0 1 For st-path p, tp is interval for which p is the longest path.  p P tp = 1

33. Upper bound of Exp [ XG ] XG r tp 0 1 ti For edge i, ti is interval for which i is in the longest path. ti =  p itp

34. Upper bound of Exp [ XG ] Xi r tp 0 1 ti If it can edge i contributes all of its mi =Exp[Xi] to Exp[XG]

35. Upper bound of Exp [ XG ] Xi r tp 0 1 ti But if vi = Var[Xi] is too small, it can only contribute

36. Upper bound of Exp [ XG ] XG r tp 0 1 ti

38. Conclusion & Future Work Tight analysis for upper bounds was achieved Cake distributions particularly important for achieving tight bounds A related question is that of finding tight bounds of mean and expectation of difference in longest paths to two given nodes in a DAG Spring algorithm involves solving non-linear constraints iteratively. Can an alternative algorithm be obtained?