instability of fifo at arbitrarily low rates in the adversarial queuing model
DESCRIPTION
Instability of FIFO at Arbitrarily Low Rates in the Adversarial Queuing Model. Rajat Bhattacharjee Ashish Goel Stanford University. Instability of FIFO at Arbitrarily Low Rates in the Adversarial Queueing Model . IEEE Foundations of Computer Science (FOCS), 2003. - PowerPoint PPT PresentationTRANSCRIPT
Instability of FIFO at Arbitrarily Low Rates in the Adversarial Queuing Model
Rajat Bhattacharjee
Ashish GoelStanford
UniversityInstability of FIFO at Arbitrarily Low Rates in the Adversarial Queueing Model . IEEE Foundations of Computer Science (FOCS), 2003.
SIAM Journal on Computing 34(2): 318-332 (2004).
Overworked server
Betty
Server processes tasks at rate 1 Tasks are generated for the server at rate r What is the value of r such that the input quue of
the server would become unbounded (unstable)? Equivalently, what is the value of r s.t. there
would be a task, which would never be processed (unstable)?
r > 1
Overworked network
Is the network of servers stable at rate r < 1? Unstable at r > 0.85!!! [Andrews et al.]
Sample task
Adversarial Queuing Model
Borodin et al. [1996] Packets injected by an adversary instead
of a stochastic process Route given at the time of injection Each edge forwards at most one packet
in one time step Contention resolved by a protocol like
FIFO
Adversarial Queuing Model
Limitations on the adversary In any window of T time steps, a (w,r)
adversary can inject at most w+rT packets that need to traverse any edge in the network
w: burst size, r: injection rate (r<1) No identifiable hotspots in the system
Stability of protocols
Stability: bounded queue size and delay r-stable: stable against all (w,r)
adversaries Universally stable: r-stable for all r<1 Andrews et al. [1996]
Rings and DAGs are universally stable networks
Longest-in-system (global FIFO) and Shortest-in-system are universally stable protocols
FIFO is unstable at rate>0.85
Related Work
Tsaparas [1997]: Nearest-To-Go unstable at arbitrarily low rates
Gamarnik [1998]: Equivalence of Fluid Model and Adversarial Queuing Model
Andrews [2000]: Session-oriented model Goel [2001], Gamarnik[1999], Alvarez et al.
[2002]: Characterized universally stable networks
Bramson studied FIFO in stochastic models: Kelly-type networks Job-shop scheduling model
Stability of FIFO
Andrews et al. [1996]: unstable at rate > 0.85
Diaz et al. [2001]: 0.83 Koukopoulos et al. [2001]: 0.749 Lotker et al. [2002]: 0.5
Is FIFO stable below some threshold, or, is it unstable at arbitrarily low rates?
Our Result
FIFO is unstable at arbitrarily low injection rates in Adversarial Queuing Model
Size of the network is polynomial in 1/r Stability not possible even at rates which are
some inverse polylograthmic function of the network size (1/logc n)
Main idea: Construct a gadget which acts as a break Use gadget to create a network and flow
which is unstable at arbitrarily low rates
Proportional Share Property of FIFO
T = j r(j) T<=1: R(i) = r(i) T>1: R(i)= r(i)/T: we will use this a lot
Analysis of the flow
r(i) – the rate of arrival of packets which have traversed i of the k load edges
T 1+r at all times
Concatenation of gadgets
Output edges of the first gadget act as Input edges of the second A chain is a sequence of
concatenated gadgets More than one gadget
can be concatenated to a gadget
Induction: Phases
Beginning of a phase
s packetsin each input
queue
End of phases’>s packets
in each input queue
Subphases
At the beginning of subphase i,
there are si packets waiting
in the input queue of gadget i.These packets are chain traversing
for the rest of column A.
Next si time steps
In the next si time stepsrsi internal gadget
packetson each load edge andrsi/k chain traversing
packetson each input edge
are introduced
At the end of the phase
At the end of the phasethere are chain traversing packets
in each of the connectorswhich wish to traverse column B
Putting it all together
Parameters of the network Size of the ring: k Length of column: Length of connector:
Choose parameters such that: (1+r)k > 64 k3/r2, = 4k/r, = 2
Putting it all together
The number of packets in the column at the beginning of subphase i, si > s/2 s is the number of initial packets in each
input queue of column A Due to exponentially small “leak” from a
gadget Number of packets which survive each
connector per load edge is > rs/4k The number of connectors = 4k/r Hence, the number of packets in each of
the input queue of column B > s