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

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

The Longest Path Problem  Input:  st-DAG G gives job precedence.  For each edge i, x i is the time to complete job i  Output:  Time for all jobs to complete in parallel = length of longest st-path = Max p  i  p x i = X G t x1x1 x2x2 s x3x3 Easy with Dynamic Programming

The Longest Path Problem  Input:  st-DAG G gives job precedence.  For each edge i, x i is the time to complete job i  Output:  Time for all jobs to complete in parallel = length of longest st-path = Max p  i  p x i = X G t x1x1 x2x2 s x3x3 Inter-dependent random variables Understand random variable X G

The Longest Path Problem  Input:  st-DAG G gives job precedence.  For each edge i, x i is the time to complete job i  Output:  Time for all jobs to complete in parallel = length of longest st-path = Max p  i  p x i = X G t x1x1 x2x2 s x3x3 Exp[X i ] & Var[X i ] Bound Exp[X G ] & Var[X G ]

The Longest Path Problem  Input: t x1x1 x2x2 s x3x3 G(x 1, x 2, x 3 ) x G Exp(2, 2, 4) ? Var(1, 1, 0) ? Prob(x 1, x 2, x 3 ) x G 0.5(1, 1, 4) 4 0.5(3, 3, 4) 6  Possible distributions : Prob(x 1, x 2, x 3 ) x G 0.5(1, 3, 4) 4 0.5(3, 1, 4) 4  Another possibility : X G = Max( x 1 +x 2, x 3 )

The Longest Path Problem  Input: t x1x1 x2x2 s x3x3 G(x 1, x 2, x 3 ) x G Exp(2, 2, 4) ? Var(1, 1, 0) ? Upper & Lower bounds X G = Max( x 1 +x 2, x 3 )

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

Series Graphs If G is a series graph, X G = ∑ i x i Exp[x G ] = ∑ i Exp[x i ] 0 ≤ Var[x G ] ≤ (∑ i √Var[x i ] ) 2 t s

Series Graphs If G is a parallel graph, X G = Max i x i Max i Exp[x i ] ≤ Exp[x G ] ≤ ? 0 ≤ Var[x G ] ≤ ? t s

Representing Random Variables r X X : Two-valued random variable, prob 0.5 for each value 0.5

Representing Random Variables r X 0.5 X, Z : Two equivalent independent random variables. Z

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

Series Graphs If G is a parallel graph, X G = Max i x i Max i Exp[x i ] ≤ Exp[x G ] ≤ Min( ∑ i Exp[x i ], Max i Exp[x i ] + √∑ i Var[x i ] ) 0 ≤ Var[x G ] ≤ ∑ i Var[x i ] t s

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

Maximizing Var [ X G ] There are no distributions x i for which  Var[x i ] = v i and Exp[x i ] = m i  Var[X G ] > Proof uses lots of calculus. Theorem

Cakes Maximizing Var [ X G ] There exists “cake” distributions x i such that  Var[x i ] = v i and Exp[x i ] = m i  Var[X G ] = Theorem

Cake Distribution t s G(x 1, x 2,...) x G Exp(2, 8,....) ? Var(3, 2,....) ? Find a cake distribution for each edge with correct Exp[x i ] & Var[x i ] to maximize Var[x G ]

Cake Distribution t s G(x 1, x 2,...) x G Exp(2, 8,....) ? Var(3, 2,....) ? Exp[x i ]

Cake Distribution t s G(x 1, x 2,...) x G Exp(2, 8,....) ? Var(3, 2,....) ? Var[x i ] = ∑ c (ε h c ) 2

Cake Distribution t s G(x 1, x 2,...) x G Exp(2, 8,....) ? Var(3, 2,....) ? Series graphs G: X G ≈ x 1 + x 2 Candle heights add Want candle heights to be in same location

Cake Distribution t s G(x 1, x 2,...) x G Exp(2, 8,....) ? Var(3, 2,....) ? Parallel graphs G: X G ≈ Max( x 1, x 2 ) Candle heights max Want candle heights to be in different location

Cake Distribution t s A candle location for each st-path in G G(x 1, x 2,...) x G Exp(2, 8,....) ? Var(3, 2,....) ? but in the end # candles ≈ # edges

Cake Distribution t s If edge i not in path p, candle for x i at location p has height 0 G(x 1, x 2,...) x G Exp(2, 8,....) ? Var(3, 2,....) ? If candle is selected, then corresponding path p is the longest path

Cake Distribution t s G(x 1, x 2,...) x G Exp(2, 8,....) ? Var(3, 2,....) ? “Springs” give give candle heights.

Cakes Maximizing Var [ X G ] There exists “cake” distributions x i such that  Var[x i ] = v i and Exp[x i ] = m i  Var[X G ] = Theorem Proved

Lower Bound of Var[ X G ] Theorem Var[x G ] ≥ 0 Continuum Results Theorem Every Var[X G ] in this range achievable.

Lower bound of Exp [ X G ]

r 01 XGXG Upper bound of Exp [ X G ] pp For st-path p,  p is interval for which p is the longest path.  p  P  p = 1

r 01 XGXG Upper bound of Exp [ X G ] pp ii For edge i,  i is interval for which i is in the longest path.  i =  p  i  p

r 01 XiXi Upper bound of Exp [ X G ] pp ii If it can edge i contributes all of its m i =Exp[X i ] to Exp[X G ]

r 01 XiXi Upper bound of Exp [ X G ] pp ii But if v i = Var[X i ] is too small, it can only contribute

r 01 XGXG Upper bound of Exp [ X G ] pp ii

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?