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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 - PowerPoint PPT Presentation
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Bounding Variance and Expectation of Longest Path Lengths in DAGs
Jeff Edmonds, York UniversitySupratik 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,
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 = X
G
t
x1
x2
s
x3
Easy with Dynamic Programming
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 = X
G
t
x1
x2
s
x3
Inter-dependent random variables
Understand random variable XG
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 = X
G
t
x1
x2
s
x3
Exp[Xi] & Var[Xi]
Bound Exp[XG] & Var[XG]
The Longest Path Problem Input:
t
x1
x2
s
x3
G (x1, x
2, x
3) xG
Exp (2, 2, 4) ?
Var (1, 1, 0) ?
Prob (x1, x
2, x
3) xG
0.5 (1, 1, 4) 4
0.5 (3, 3, 4) 6
Possible distributions :
Prob (x1, x
2, x
3) xG
0.5 (1, 3, 4) 4
0.5 (3, 1, 4) 4
Another possibility :
51
40
XG = Max( x1+x2, x3 )
The Longest Path Problem Input:
t
x1
x2
s
x3
G (x1, x
2, x
3) xG
Exp (2, 2, 4) ?
Var (1, 1, 0) ?
Upper & Lower bounds
XG = Max( x1+x2, x3 )
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[XG] and Var[X
G]
Continuum of values for Exp[XG] 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,
XG = ∑i xi
Exp[xG] = ∑i Exp[xi]
0 ≤ Var[xG] ≤ (∑i √Var[xi] )2
t
s
Series Graphs
If G is a parallel graph,
XG = Maxi xi
Maxi Exp[xi] ≤ Exp[xG] ≤ ?
0 ≤ Var[xG] ≤ ?
t
s
Representing Random Variables
r0 1
0
5X
X : Two-valued random variable, prob 0.5 for each value
0.5
Representing Random Variables
r0 1
0
5X
0.5
X, Z : Two equivalent independent random variables.
Z
Representing Random Variables
r0 1
0
5X Y
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 GraphsIf 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]
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[XG]
Maximizing Var [ XG ]
There are no distributions xi for which Var[x
i] = v
i and Exp[x
i] = m
i
Var[XG] >
Proof uses lots of calculus.
Theorem
Cakes Maximizing Var [ XG ]
There exists “cake” distributions xi such that Var[x
i] = v
i and Exp[x
i] = m
i
Var[XG] =
Theorem
Cake Distribution
t
s
G (x1, x
2, ...) xG
Exp (2, 8, ....) ?
Var (3, 2, ....) ?
Find a cake distribution for each edgewith correct Exp[xi] & Var[xi]
to maximize Var[xG]
Cake Distribution
t
s
G (x1, x
2, ...) xG
Exp (2, 8, ....) ?
Var (3, 2, ....) ?
Exp[xi]
Cake Distribution
t
s
G (x1, x
2, ...) xG
Exp (2, 8, ....) ?
Var (3, 2, ....) ?
Var[xi] = ∑c (ε hc)2
Cake Distribution
t
s
G (x1, x
2, ...) xG
Exp (2, 8, ....) ?
Var (3, 2, ....) ?
Series graphs G: •XG ≈ x1 + x2
• Candle heights add•Want candle heights to be in same location
Cake Distribution
t
s
G (x1, x
2, ...) xG
Exp (2, 8, ....) ?
Var (3, 2, ....) ?
Parallel graphs G: •XG ≈ Max( x1 , x2 )• Candle heights max•Want candle heights to be in different location
Cake Distribution
t
s
A candle locationfor each st-path in G
G (x1, x
2, ...) xG
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 xi at location p
has height 0
G (x1, x
2, ...) xG
Exp (2, 8, ....) ?
Var (3, 2, ....) ?
If candle is selected,then corresponding path p
is the longest path
Cake Distribution
t
s
G (x1, x
2, ...) xG
Exp (2, 8, ....) ?
Var (3, 2, ....) ?
“Springs” give give candle heights.
Cakes Maximizing Var [ XG ]
There exists “cake” distributions xi such that Var[x
i] = v
i and Exp[x
i] = m
i
Var[XG] =
Theorem
Proved
Lower Bound of Var[ XG ]
TheoremVar[xG] ≥ 0
Continuum Results
Theorem
Every Var[XG] in this range achievable.
Lower bound of Exp [ XG ]
r0 1
XG
Upper bound of Exp [ XG ]
p
For st-path p, p is interval for
which p is the longest path.
p P
p = 1
r0 1
XG
Upper bound of Exp [ XG ]
p
i
For edge i, i is interval for which i
is in the longest path.
i =
p i
p
r0 1
Xi
Upper bound of Exp [ XG ]
p
i
If it can edge i contributes all of its
mi =Exp[Xi]
to Exp[XG
]
r0 1
Xi
Upper bound of Exp [ XG ]
p
i
But if vi = Var[Xi] is too small,
it can only contribute
r0 1
XG
Upper bound of Exp [ XG ]
p
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?