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

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Bounding Variance and Expectation of Longest Path Lengths in DAGs

Jeff Edmonds, York University

Supratik Chakraborty, IIT Bombay

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

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

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

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]

Input:

XG = Max( x1+x2, x3 )

4

5

0

1

s

- Possible distributions :

x1

x3

x2

t

- Another possibility :

Input:

XG = Max( x1+x2, x3 )

Upper & Lower bounds

s

x1

x3

x2

t

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

If G is a series graph,

XG = ∑i xi

Exp[xG] = ∑i Exp[xi]

0 ≤ Var[xG] ≤ (∑i √Var[xi] )2

s

t

If G is a parallel graph,

XG = Maxi xi

Maxi Exp[xi] ≤ Exp[xG] ≤ ?

0 ≤ Var[xG] ≤ ?

s

t

X

5

0

r

0

0.5

1

X : Two-valued random variable, prob 0.5 for each value

X

5

Z

0

r

0

0.5

1

X, Z : Two equivalent independent 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

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

Theorem

In a series-parallel graph,

Rules for maximum variance applied recursively to obtain Max Var[XG].

Not so Max Exp[XG]

There are no distributions xi for which

Var[xi] = vi and Exp[xi] = mi

Var[XG] >

Proof uses lots of calculus.

Theorem

There exists “cake” distributions xi such that

Var[xi] = vi and Exp[xi] = mi

Var[XG] =

Theorem

s

t

Find a cake distribution for each edgewith correct Exp[xi] & Var[xi]

to maximize Var[xG]

s

Exp[xi]

t

s

t

Var[xi] = ∑c (ε hc)2

s

t

- Series graphs G:
- XG ≈ x1 + x2
- Candle heights add
- Want candle heights to be in same location

s

t

- Parallel graphs G:
- XG ≈ Max( x1 , x2 )
- Candle heights max
- Want candle heights to be in different location

s

t

A candle location

for each st-path in G

but in the end

# candles ≈ # edges

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

s

t

“Springs” give

give candle heights.

There exists “cake” distributions xi such that

Var[xi] = vi and Exp[xi] = mi

Var[XG] =

Theorem

Proved

TheoremVar[xG] ≥ 0

Theorem

Every Var[XG] in this range achievable.

Lower 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

XG

r

tp

0

1

ti

For edge i, ti is interval for which i is in the longest path.

ti = p itp

Xi

r

tp

0

1

ti

If it can edge i contributes all of its mi =Exp[Xi] to Exp[XG]

Xi

r

tp

0

1

ti

But if vi = Var[Xi] is too small,

it can only contribute

XG

r

tp

0

1

ti

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?