1 / 38

# Bounding Variance and Expectation of Longest Path Lengths in DAGs Jeff Edmonds, York University - PowerPoint PPT Presentation

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

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'Bounding Variance and Expectation of Longest Path Lengths in DAGs Jeff Edmonds, York University' - jerold

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

Jeff Edmonds, York University

Supratik Chakraborty, IIT Bombay

Motivation DAGs

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

Contributions DAGs

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

Series Graphs DAGs

If G is a series graph,

XG = ∑i xi

Exp[xG] = ∑i Exp[xi]

0 ≤ Var[xG] ≤ (∑i √Var[xi] )2

s

t

Series Graphs DAGs

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

Series Graphs DAGs

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]

Maximizing Var [ X DAGsG ]

There are no distributions xi for which

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

Var[XG] >

Proof uses lots of calculus.

Theorem

Cakes Maximizing Var [ X DAGsG ]

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.

Cakes Maximizing Var [ X DAGsG ]

There exists “cake” distributions xi such that

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

Var[XG] =

Theorem

Proved

Lower Bound of Var[ X DAGsG ]

TheoremVar[xG] ≥ 0

Continuum Results

Theorem

Every Var[XG] in this range achievable.

Upper bound of Exp [ X DAGsG ]

XG

r

tp

0

1

For st-path p, tp is interval for which p is the longest path.

 p P tp = 1

Upper bound of Exp [ X DAGsG ]

XG

r

tp

0

1

ti

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

ti =  p itp

Upper bound of Exp [ X DAGsG ]

Xi

r

tp

0

1

ti

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

Upper bound of Exp [ X DAGsG ]

Xi

r

tp

0

1

ti

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

it can only contribute

Upper bound of Exp [ X DAGsG ]

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?