Rahul Vaze and Srikanth Iyer
Breaking the Poisson infinite delay stranglehold
IISc, Bangalore
Wireless Network
PPP distributed nodes
Wireless Network
PPP distributed nodes
sij = hij`(dij)
sij
Wireless Network
PPP distributed nodes
sij = hij`(dij)
sijI =
X
k2�\{i}
hkj`(dkj)
I
Wireless Network
PPP distributed nodes
sij = hij`(dij)
sijI =
X
k2�\{i}
hkj`(dkj)
I
SINRij =sij
I +N
Wireless Network
PPP distributed nodes
sij = hij`(dij)
sijI =
X
k2�\{i}
hkj`(dkj)
SINRij > �
I
SINRij =sij
I +N
Wireless Network
PPP distributed nodes
sij = hij`(dij)
sijI =
X
k2�\{i}
hkj`(dkj)
SINRij > �
I
SINRij =sij
I +N
SINR is a random variable: multiple attempts required
PPP networkNegative result
- Expected delay to any node is infinite
[Baccelli, Blacszczyszyn, Mirsadeghi, Adv. App. Prob. 2010]
PPP networkNegative result
- Expected delay to any node is infinite
[Baccelli, Blacszczyszyn, Mirsadeghi, Adv. App. Prob. 2010]
T3
T2
T1
PPP networkNegative result
- Expected delay to any node is infinite
[Baccelli, Blacszczyszyn, Mirsadeghi, Adv. App. Prob. 2010]
T3
T2
T1
min delay has infinite expectation
Negative ResultEven in the absence of interference (AWGN is sufficient)
sijI
Negative ResultEven in the absence of interference (AWGN is sufficient)
sijI
Negative ResultEven in the absence of interference (AWGN is sufficient)
sij
Negative ResultEven in the absence of interference (AWGN is sufficient)
sij
Unbounded sized holes in PPP
More Bad News.
Let be the successful distance traveled by a taggedparticle from origin in time
More Bad News.
d(t)t
Let be the successful distance traveled by a taggedparticle from origin in time
More Bad News.
V = limt!1
d(t)
t
d(t)t
Then speed or information velocity is
Let be the successful distance traveled by a taggedparticle from origin in time
More Bad News.
V = limt!1
d(t)
t
d(t)t
Then speed or information velocity is
In a PPP network V = 0 [Baccelli, et al, 2010]
Remedy
• Add a regular square grid [Baccelli, et al, 2010]
Remedy
• Add a regular square grid [Baccelli, et al, 2010]
A more practical solution
sourcedestination
A more practical solution
sourcedestination
�
A more practical solution
sourcedestination
dij�
A more practical solution
sourcedestination
dij�
A more practical solution
sourcedestination
dij�
A more practical solution
sourcedestination
dij�
sij = hij`(dij) SINRij =sij
I +N
A more practical solution
Power Control : power with prob.
sourcedestination
Pi = c`(dij)�1 pi = MP�1
i
dij�
sij = hij`(dij) SINRij =sij
I +N
Transmit with higher power Less frequently
A more practical solution
Power Control : power with prob.
sourcedestination
Pi = c`(dij)�1 pi = MP�1
i
dij�
sij = hij`(dij) SINRij =sij
I +N
Results
• Expected delay to nearest neighbor is finite
• Information velocity is positive
Negative Result Primer
sij sij = Pihij`(dij)
Negative Result Primer
sij sij = Pihij`(dij)
Time to exit to any other nodeT
Negative Result Primer
sij sij = Pihij`(dij)
Drop the Interference SINRij =sij
Ij +NSNRij =
sijN
Time to exit to any other nodeT
Negative Result Primer
sij sij = Pihij`(dij)
Drop the Interference SINRij =sij
Ij +NSNRij =
sijN
Time to exit to any other nodeT
P(T > k) = P(SNRij(t) < �, 8 j 2 �, t = 1, . . . , k)
Negative Result Primer
sij sij = Pihij`(dij)
Drop the Interference SINRij =sij
Ij +NSNRij =
sijN
Time to exit to any other nodeT
Conditioned over location of nodes, SIRs are independent
P(T > k) = P(SNRij(t) < �, 8 j 2 �, t = 1, . . . , k)
Negative Result Primer
sij sij = Pihij`(dij)
Drop the Interference SINRij =sij
Ij +NSNRij =
sijN
Time to exit to any other nodeT
Conditioned over location of nodes, SIRs are independent
P(T > k) = P(SNRij(t) < �, 8 j 2 �, t = 1, . . . , k)
P (T > k|�) =Y
j2�
P (SINRij(t) < �)k
Negative Result Primer
sij sij = Pihij`(dij)
Drop the Interference SINRij =sij
Ij +NSNRij =
sijN
Time to exit to any other nodeT
Conditioned over location of nodes, SIRs are independent
After unconditioning P(T > k) >1
k
P(T > k) = P(SNRij(t) < �, 8 j 2 �, t = 1, . . . , k)
P (T > k|�) =Y
j2�
P (SINRij(t) < �)k
Issues with New Policy
n(y)
n(x)
y
x
z
n(z)
Choice of cones at any time are correlated
Analysis Ideas - Power Control
Analysis Ideas - Power Control
P(T > k) = P(SINRon(o)(t) < �, t = 1, . . . , k)
Analysis Ideas - Power Control
Earlier it was sufficient to condition over �
P(T > k) = P(SINRon(o)(t) < �, t = 1, . . . , k)
Analysis Ideas - Power Control
Earlier it was sufficient to condition over �
With correlated choice of cones across time slots
Not any more
P(T > k) = P(SINRon(o)(t) < �, t = 1, . . . , k)
Analysis Ideas - Power Control
Earlier it was sufficient to condition over �
With correlated choice of cones across time slots condition over sigma field generated by and choice of cones �Gk
Not any more
P(T > k) = P(SINRon(o)(t) < �, t = 1, . . . , k)
Analysis Ideas - Power Control
Earlier it was sufficient to condition over �
With correlated choice of cones across time slots condition over sigma field generated by and choice of cones �Gk
Not any more
P(T > k) = E
(kY
t=1
P(SINRon(o)(t) < �|G
k
)
)
P(T > k) = P(SINRon(o)(t) < �, t = 1, . . . , k)
on(o)
x n1(x)
Analysis Ideas - Power Control
P(T > k) = E
(kY
t=1
P(SINRon(o)(t) < �|G
k
)
)
dij
Pij
on(o)
x n1(x)
Analysis Ideas - Power Control
P(T > k) = E
(kY
t=1
P(SINRon(o)(t) < �|G
k
)
)
dij
Pij
on(o)
x
n2(x)
n1(x)
Analysis Ideas - Power Control
P(T > k) = E
(kY
t=1
P(SINRon(o)(t) < �|G
k
)
)
dij
Pij
on(o)
x
n2(x)
n1(x)Px
(t) changes
Analysis Ideas - Power Control
P(T > k) = E
(kY
t=1
P(SINRon(o)(t) < �|G
k
)
)
for o, cone is fixed for k slots
dij
Pij
on(o)
x
n2(x)
n1(x)Px
(t) changes
Analysis Ideas - Power Control
For any interferer z, use the choice of cone that maximizes the interference seen at n(o) at time t=1,…,k
P(T > k) = E
(kY
t=1
P(SINRon(o)(t) < �|G
k
)
)
for o, cone is fixed for k slots
Analysis Ideas - Power Control
E {T} =1X
k=0
P(T > k) P(T > k) = E
(kY
t=1
P(SINRon(o)(t) < �|G
k
)
)
Analysis Ideas - Power Control
Power Control : power with prob.Pi = c`(dij)�1 pi = MP�1
i
E {T} =1X
k=0
P(T > k) P(T > k) = E
(kY
t=1
P(SINRon(o)(t) < �|G
k
)
)
Analysis Ideas - Power Control
For any interferer z, the average interference is bounded
Power Control : power with prob.Pi = c`(dij)�1 pi = MP�1
i
E {T} =1X
k=0
P(T > k) P(T > k) = E
(kY
t=1
P(SINRon(o)(t) < �|G
k
)
)
Analysis Ideas - Power Control
For any interferer z, the average interference is bounded
Power Control : power with prob.Pi = c`(dij)�1 pi = MP�1
i
pzPz M
E {T} =1X
k=0
P(T > k) P(T > k) = E
(kY
t=1
P(SINRon(o)(t) < �|G
k
)
)
Analysis Ideas - Power Control
For any interferer z, the average interference is bounded
Power Control : power with prob.Pi = c`(dij)�1 pi = MP�1
i
pzPz MZ
`(|x|)dx < 1Thus, if
E {T} =1X
k=0
P(T > k) P(T > k) = E
(kY
t=1
P(SINRon(o)(t) < �|G
k
)
)
Analysis Ideas - Power Control
For any interferer z, the average interference is bounded
Power Control : power with prob.Pi = c`(dij)�1 pi = MP�1
i
pzPz MZ
`(|x|)dx < 1Thus, if
using Campbell’s Theorem we can show that expected delay is finite
E {T} =1X
k=0
P(T > k) P(T > k) = E
(kY
t=1
P(SINRon(o)(t) < �|G
k
)
)
Information Velocity
Look at tagged particle at origin
X1 + C1
R2
θ−1
X−1
X2
φ
R1
R−2
R−1
X1
θ−2
θ1
θ2
X−2
C1
o
V = limt!1
d(t)
t
dest
Information Velocity
Look at tagged particle at origin
X1 + C1
R2
θ−1
X−1
X2
φ
R1
R−2
R−1
X1
θ−2
θ1
θ2
X−2
C1
o
V = limt!1
d(t)
t
dest
Information Velocity
Look at tagged particle at origin
X1 + C1
R2
θ−1
X−1
X2
φ
R1
R−2
R−1
X1
θ−2
θ1
θ2
X−2
C1
o
V = limt!1
d(t)
t
dest
progress
Information Velocity
Look at tagged particle at origin
X1 + C1
R2
θ−1
X−1
X2
φ
R1
R−2
R−1
X1
θ−2
θ1
θ2
X−2
C1
T0
o
V = limt!1
d(t)
t
dest
progress
Information Velocity
Look at tagged particle at origin
X1 + C1
R2
θ−1
X−1
X2
φ
R1
R−2
R−1
X1
θ−2
θ1
θ2
X−2
C1
T0
T1
o
V = limt!1
d(t)
t
dest
progress
Information Velocity
Look at tagged particle at origin
X1 + C1
R2
θ−1
X−1
X2
φ
R1
R−2
R−1
X1
θ−2
θ1
θ2
X−2
C1
T0
T1
o
V = limt!1
d(t)
t
dest
progress
⌫ =
E{R cos(✓)}E{Ti}
First Guess
Information Velocity
Look at tagged particle at origin
X1 + C1
R2
θ−1
X−1
X2
φ
R1
R−2
R−1
X1
θ−2
θ1
θ2
X−2
C1
o
Information Velocity
Look at tagged particle at origin
X1 + C1
R2
θ−1
X−1
X2
φ
R1
R−2
R−1
X1
θ−2
θ1
θ2
X−2
C1
T0
o
Information Velocity
Look at tagged particle at origin
X1 + C1
R2
θ−1
X−1
X2
φ
R1
R−2
R−1
X1
θ−2
θ1
θ2
X−2
C1
T0
T1
o
Information Velocity
Look at tagged particle at origin
X1 + C1
R2
θ−1
X−1
X2
φ
R1
R−2
R−1
X1
θ−2
θ1
θ2
X−2
C1
T0
T1
T0,T1 are not identically distributed
o
Information Velocity
Look at tagged particle at origin
X1 + C1
R2
θ−1
X−1
X2
φ
R1
R−2
R−1
X1
θ−2
θ1
θ2
X−2
C1
T0
T1
T0,T1 are not identically distributed
vacant
o
Information Velocity
X1 + C1
R2
θ−1
X−1
X2
φ
R1
R−2
R−1
X1
θ−2
θ1
θ2
X−2
C1
o
Information Velocity
Dominate the delay by adding an infinite chain of points
X1 + C1
R2
θ−1
X−1
X2
φ
R1
R−2
R−1
X1
θ−2
θ1
θ2
X−2
C1
o
Information Velocity
Dominate the delay by adding an infinite chain of points
X1 + C1
R2
θ−1
X−1
X2
φ
R1
R−2
R−1
X1
θ−2
θ1
θ2
X−2
C1
same distribution as R and ✓
o
Information Velocity
Dominate the delay by adding an infinite chain of points
X1 + C1
R2
θ−1
X−1
X2
φ
R1
R−2
R−1
X1
θ−2
θ1
θ2
X−2
C1
same distribution as R and ✓fill Poisson points
o
Information Velocity
Dominate the delay by adding an infinite chain of points
X1 + C1
R2
θ−1
X−1
X2
φ
R1
R−2
R−1
X1
θ−2
θ1
θ2
X−2
C1
same distribution as R and ✓fill Poisson points
T̂0
T̂1
o
Information Velocity
Dominate the delay by adding an infinite chain of points
X1 + C1
R2
θ−1
X−1
X2
φ
R1
R−2
R−1
X1
θ−2
θ1
θ2
X−2
C1
same distribution as R and ✓fill Poisson points
T̂0, T̂1, ... are identically distributed
T̂0
T̂1
o
Information Velocity
Information Velocity
ˆT0, ˆT1, ... are stationary
similar to before E{T̂i} < 1
Information Velocity
ˆT0, ˆT1, ... are stationary
Birkoff’s Ergodic Theorem limn!1
1
n
n�1X
k=0
T̂k = T̂
similar to before
T̂where is a rv with mean
E{T̂i} < 1
E{T̂i}
Information Velocity
ˆT0, ˆT1, ... are stationary
v � lim
t!1
PN(t)k=1 Rk cos(✓k)PN(t)+1
k=1ˆTk
=
E[R cos(✓)]ˆT
Birkoff’s Ergodic Theorem
effective distance progress
limn!1
1
n
n�1X
k=0
T̂k = T̂
similar to before
T̂where is a rv with mean
⌫
E{T̂i} < 1
E{T̂i}