NewTransportProtocolsInWirelessMulti-hop Ad-hocNetworkswlr/228S06/TransAdhocSlides.pdf ·...

Motivation The File Transfer Problem Analysis And Comparison The Fairness Problem Proposed Algorithms Extensions

New Transport Protocols In Wireless Multi-hop

Ad-hoc Networks

Jingyi Shaoshao14@eecs

EE228a, UC Berkeley

March 7, 2006


Outline

1 Motivation

2 The File Transfer Problem

3 Analysis And Comparison

4 The Fairness Problem

5 Proposed Algorithms

6 Extensions


Motivation

The exploding number of ideas of wireless applications.

Wireless channels are intrinsic unreliable due to multi-pathfading, and have limited spectrum resources.


Motivation



How do we build scalable efficient wireless networks?


Motivation




Cross-layer design


Motivation




Cross-layer designMulti-hop approach


Outline

1 Motivation





6 Extensions


Assumptions

Linear multi-hop network

n+1 stationary nodes; Node 0 wants to transfer a file to Noden via n hops; intermediate nodes do not have their own traffic.

Discretize time.

Unreliable links with binary states (two models).

If a link is on, the source of that link can transfer 1 packet tothe sink of that link perfectly in one time step.

Ignore node processing delay and link propagation delay.

File size: m-packets.

Let T be the time taken to transfer the file.


Assumptions

Linear multi-hop network

n+1 stationary nodes; Node 0 wants to transfer a file to Noden via n hops; intermediate nodes do not have their own traffic.

Discretize time.

Unreliable links with binary states (two models).

If a link is on, the source of that link can transfer 1 packet tothe sink of that link perfectly in one time step.

Ignore node processing delay and link propagation delay.

File size: m-packets.

Let T be the time taken to transfer the file.

Question: what is the expectation of T?


Link Models

Model 1 (iid):During each time step, assume each link is independent andidentically distributed (i.i.d.) with probability p being on, andprobability (1− p) being off, independent across time as well.


Link Models


Model 2 (Markov):During each time step, assume each link is independent fromthe other links, and the state of each link follows a simpleMarkov Chain with transition probability p01 from off to on,and transition probability p10 from on to off.


Link Models


Model 2 (Markov):During each time step, assume each link is independent fromthe other links, and the state of each link follows a simpleMarkov Chain with transition probability p01 from off to on,and transition probability p10 from on to off.

The stationary distribution is π0 =p10

p01+p10and π1 =

p01

p01+p10


Proposed Protocols

Protocol 1Intermediate nodes forwards packets only.Destination can not store partial file (i.e. if the path is brokenbefore the transfer is complete, the transfer needs to start overagain when the path comes back on).


Proposed Protocols


Protocol 2Intermediate nodes forwards packets only.Destination can store partial file (i.e. if the path is brokenbefore the transfer is complete, the transfer can resume fromwhere it left off when the path comes back on).


Proposed Protocols



Protocol 3Intermediate nodes can cache completed file, but not partialfile.Destination can not store partial file.


Proposed Protocols



Protocol 3Intermediate nodes can cache completed file, but not partialfile.Destination can not store partial file.

Protocol 4Both the intermediate nodes and the destination can storepartial file


Outline

1 Motivation





6 Extensions


Model iid, Protocol 1

The file transfer is complete when all the links are on for m

consecutive time steps.

Pr(all links on during 1 timestep) = pn

E [T ] =1− pnm

(1− pn)pnm

0 2 4 6 8 10 12 14 16 18 2010

0

102

104

106

108

1010

1012

1014

1016

1018

1020

Model 1, Protocol 1, n=10, p=0.8

Number of Packets

Exp

ect

ed

Tim

e



The file transfer is complete when there are total n times stepswhen all links are on. Thus, T has negative binomial distributionwith parameter (m, pn).

E [T ] = m(1

pn)

0 2 4 6 8 10 12 14 16 18 2010

0

101

102

103


Number of Packets

Exp

ect

ed

Tim

e

SimulatedTheoretical


Model iid, Protocol 3, Original Version

Observation: since intermediate nodes can not cache partial file,at the time step when Node i (0 < i ≤ n) completes receiving thefile, Link i (the link between Node (i − 1) and Node i) must havebeen on for the previous m time steps. At this time, if Link (i + 1)has also been on for the previous m time steps, Node (i + 1) willhave completed receiving the entire file. Node i can not receive thecomplete file unless all the previous nodes have received thecomplete file.

Let Ti be the time interval from when Node (i − 1) receives thecomplete file to Node i receives the complete file.


Model iid, Protocol 3, Original Version

T =n∑

i=1

Ti

where T1 is as in Protocol 1 for 1 link, and for i > 1, Ti is a mixedrandom variable: Ti = 0 with probability pm, and Ti has the samedistribution as T1 with probability (1− pm).

E [T ] =n∑

i=1

E [Ti ]

= E [T1] +n∑

i=2

E [Ti ]

= E [T1] + (n − 1)(1− pm)E [T1]

where E [T1] =1−pm

(1−p)pm


Model iid, Protocol 3, Improved Version

At the time Node i (0 < i ≤ n) just completes receiving the file, ifLink i has been on for the previous k (0 ≤ k ≤ m) time steps,Node i + 1 has received k packets. For i > 1,

Ti =

{

0, with probability pm

Ak , with probability (1− p)pk for 0 ≤ k < m

where Ak is a mixed R.V. with

E [Ak ] = (m − k)pm−k +m−k−1∑

j=0

(j + 1 + E [T1])pj(1− p)


Model iid, Protocol 3, Improved Version

E [Ti ] =

m−1∑

k=0

(1− p)pkE [Ak ], i > 1

E [T ] = E [T1] +n∑

i=2

E [Ti ]

= E [T1] + (n − 1)m−1∑

k=0

(1− p)pkE [Ak ]


Compare Two Versions of Protocol 3 For Model iid

Fix the number of links

0 2 4 6 8 10 12 14 16 18 2010

0

101

102

103

104


Number of Packets

Exp

ect

ed

Tim

e

Simulated ITheoretical ITheoretical O

Fix the number of packets

0 5 10 15 20 25 3010

1

102

103

104

Model 1, Protocol 3, m=10, p=0.8

Number of Links

Exp

ect

ed

Tim

e

Simulated ITheoretical ITheoretical O



Approach: represent the network as a Markov chain with statespace X ⊆ Zn. Each state x ∈ X is an n-dimensional vectorwhose ith component represents the number of packets Nodei has successfully received from Node (i − 1).Clearly, m ≥ x1 ≥ x2 ≥ ... ≥ xn−1 ≥ xn ≥ 0.




The number of states grows exponentially with n, but if weare patient, we can write down the exact chain with alltransition probabilities, and we can theoretically use first stepanalysis to compute the desired expectation.




The number of states grows exponentially with n, but if weare patient, we can write down the exact chain with alltransition probabilities, and we can theoretically use first stepanalysis to compute the desired expectation.

No closed form expression for theoretical analysis, yet.


Model iid, Protocol 4, Example

The Markov chain for n=2, m=4.


Model iid, Protocol 4, Another Approach

The network can be seen as a series of concatenated queues,where each intermediate node is a server of a queue, and thearrival process for Node i (0 < i < n) is the departure processof Node (i − 1). At each node, the processing time isgeometric with parameter p.




For simplicity, assume the packets of the file arrive at Node 0with geometric inter-arrival distribution with parameter q andq < p.(If Node 0 has the complete file at the beginning of thetransfer, then the departure process at Node 0 has geometricinter-arrival distribution with parameter p.)




For simplicity, assume the packets of the file arrive at Node 0with geometric inter-arrival distribution with parameter q andq < p.(If Node 0 has the complete file at the beginning of thetransfer, then the departure process at Node 0 has geometricinter-arrival distribution with parameter p.)

From queuing theory, we know a stationary distribution existsiff q < p, and in stationary distribution, the departure processat each node is a random process with geometric inter-arrivaldistribution with parameter q.


Model iid, Protocols Comparison


0 2 4 6 8 10 12 14 16 18 2010

0

101

102

103

104

Model 1, n=10, p=0.8

Number of Packets

Exp

ect

ed

Tim

e

P2P3 IP4 Sim


0 5 10 15 20 25 3010

1

102

103

104

Model 1, m=10, p=0.8

Number of Links

Exp

ect

ed

Tim

e

P2P3 IP4 Sim


Model Markov

The link states are not independent across time anymore.

One can use another Markov chain to keep track of the linkstates, Y ⊆ 2n.

Given the Markov model for each link, the overall link statecan be determined (again, patience is required).


Model Markov




The Brute Force Approach

Keep track both of the node states and the link states (X ,Y ), andin theory, one can compute the desired expectations.


Model Markov




The Brute Force Approach

Keep track both of the node states and the link states (X ,Y ), andin theory, one can compute the desired expectations.The Smart Approach

To be discovered.


Models Comparison, Protocol 2


0 2 4 6 8 10 12 14 16 18 2010

0

101

102

103

Protocol 2, n=10, p=0.8, p01

=0.5, p10

=0.125

Number of Packets

Exp

ect

ed T

ime

Model 1Model 2


0 5 10 15 20 25 3010

1

102

103

104

Protocol 2, m=10, p=0.8, p01

=0.5, p10

=0.125

Number of Links

Exp

ect

ed T

ime

Model 1Model 2


Models Comparison, Protocol 3 Improved


0 2 4 6 8 10 12 14 16 18 2010

0

101

102

103

104

Protocol 3 I, n=10, p=0.8, p01

=0.5, p10

=0.125

Number of Packets

Exp

ect

ed T

ime

Model 1Model 2


0 5 10 15 20 25 3010

1

102

103

Protocol 3 I, m=10, p=0.8, p01

=0.5, p10

=0.125

Number of Links

Exp

ect

ed T

ime

Model 1Model 2


Models Comparison, Protocol 4


0 2 4 6 8 10 12 14 16 18 2010

0

101

102

Protocol 4, n=10, p=0.8, p01

=0.5, p10

=0.125

Number of Packets

Exp

ect

ed T

ime

Model 1Model 2


0 5 10 15 20 25 30

101.2

101.3

101.4

101.5

101.6

Protocol 4, m=10, p=0.8, p01

=0.5, p10

=0.125

Number of Links

Exp

ect

ed T

ime

Model 1Model 2


Outline

1 Motivation





6 Extensions


Notations and Definitions

Flow: an acyclic route or path (unique and fixed) from asource to the corresponding destination, i.e. a series ofconnected links. (Each node in the network can have multiplesources and multiple destinations.)

For Link l :

Capacity, Cl (fixed such as link in wired network for now).The allocated rate of Flow i , ali .The incoming rate of Flow i , cli .The actual rate (outgoing) of Flow i , oli .

The rate of Flow i : ri = minl oli .

Feasible rate vector for the network: an ordered list of all therates of all the flows in the network that satisfies the capacityconstraints of each link.


Def. Cont.

Max-min fair rate: let R be a feasible rate vector for thenetwork, R is said to be max-min fair if for any other feasiblerate vector R ′, if any component r ′i > ri , then there existsr ′j < rj and rj ≤ ri . (Informally, one can not increase any ratewithout reducing an already smaller or equal rate.)


Def. Cont.


Note: max-min rate fully utilizes network resources, i.e. @R ′

such that ∀i , r ′i ≥ ri and ∃j , r′

j > rj .


Def. Cont.


Note: max-min rate fully utilizes network resources, i.e. @R ′

such that ∀i , r ′i ≥ ri and ∃j , r′

j > rj .

In general, max-min rate does not maximize total throughput.


Localized Allocation Without Feedback

If there’s only one link, round-robin scheduling achieves max-minfair rate (equal rate).If the network is more complicated, does localized round-robinwithout feedback still achieve max-min rate?


Localized Allocation Without Feedback

If there’s only one link, round-robin scheduling achieves max-minfair rate (equal rate).If the network is more complicated, does localized round-robinwithout feedback still achieve max-min rate?No.Counter Example:


Outline

1 Motivation





6 Extensions


Assumptions

The flows do not change (quasi-static).

Each flow wants to have rate as high as possible.

Each node knows the incoming and outgoing links capacities.

Each node knows the incoming and outgoing flows.

Perfect feedback with arbitrary precision from the downstream nodes.

Discrete time slots with perfect synchronization.


Distributed Max-min Algorithm

Step 0 (initialization): equally allocate each link capacityamong all flows sharing the same link, i.e. ali =

Cl

nwhere

there are n flows sharing Link l .

Step 1: each node feeds back the allocated rate to the upstream nodes for each flow through the node at the end ofeach time slot.

Step 2: update allocated rates at each node at the beginningof each time slot according to some fixed rules.

Go back to Step 1 and repeat.


Distributed Max-min Algorithm

Step 0 (initialization): equally allocate each link capacityamong all flows sharing the same link, i.e. ali =

Cl

nwhere

there are n flows sharing Link l .

Step 1: each node feeds back the allocated rate to the upstream nodes for each flow through the node at the end ofeach time slot.

Step 2: update allocated rates at each node at the beginningof each time slot according to some fixed rules.

Go back to Step 1 and repeat.

Claim: the above algorithm achieves max-min rate in steady state.


Allocated Rate Update Rule

Suppose Flow i goes through nodes (n − 1), n, and (n + 1) onlinks (l − 1), l , and (l +1). a(l+1)i is fed back to Node n, and ali isfed back to Node (n − 1).

At Node n:

If a(l+1)i ≤ ali , then ali = a(l+1)i ;

If a(l+1)i > ali , then increase ali up to a(l+1)i if possible, i.e.reverse ”waterfill” of the flow rates that need to be increased.

If Node (n + 1) is the destination of Flow i , then seta(l+1)i =∞.


Reverse Waterfill Illustration


Reverse Waterfill Illustration

The actual outgoing rate, oli , usually equals to the allocated rate,ali , except when the up-stream link allocation is smaller or in thecase where the allocated rates do not fill up the current linkcapacity. In the latter case, oli equals to

ali if ali < maxi ali ;

ali + el if ali = maxi ali , where el is chosen to meet the linkcapacity.


Example

Allocated Rates Table


Proof of Claim

Intuition: the actual flow rate ≥ the minimum of the initialallocated rates, which was fair in some sense.Assume steady state. Let i be a flow. Let L be the set of links i

goes through. For a link l ∈ L:

oli : the steady state outgoing rate of Flow i .

ali : the steady state allocated rate of Flow i .

aoli : initial allocated rate of Flow i .

Let ri = minl∈L oli be the flow rate of Flow i .Want to show: for any r ′i > ri , there exists j such that r ′j < rj andrj ≤ ri .Note: ri ≥ minl∈L ao

li ⇒ Two Cases.


Case 1: ri = minl∈L oli = minl∈L aoli

Order the links according to the natural ordering, i.e. l > k iffFlow i goes through Link k before Link l .Let k = maxl∈L(argminl∈L ao

li ).

It is easy to see that increasing ri leads to increasing oki , whichleads to decreasing okj = oki = ri . Further, note okj = rj .Therefore, ri is max-min fair.


Case 2: ri = minl∈L oli > minl∈L aoli

ri = oli for some l ∈ L.Subcase 1: ∃l , oli = ali .Let k = maxl∈L(argminl∈L ali ). Similar argument as Case 1.

Subcase 2: Suppose @l , oli = ali ,⇒ ∀l , oli > ali .Let k = maxl∈L(argminl∈L oli ).The sink of Link k is the destination of Flow i , and oki = aki .Contradiction.


Outline

1 Motivation





6 Extensions


Can One Combine Flows?

No, unless the flows have the same source, same destination, andthe same route.


Finite Precision Feedback

1-Bit Feedback

Goal: make ali to track a(l+1)i

If c(l+1)i ≤ a(l+1)i , Node (n + 1) feedsback +1.

If c(l+1)i > a(l+1)i , Node (n + 1) feedsback −1.

If Node n receives +1, increase ali by 1 unit.

If Node n receives −1, decrease ali by 1 unit (AIMD?).


Finite Precision Feedback

1-Bit Feedback

Goal: make ali to track a(l+1)i

If c(l+1)i ≤ a(l+1)i , Node (n + 1) feedsback +1.

If c(l+1)i > a(l+1)i , Node (n + 1) feedsback −1.

If Node n receives +1, increase ali by 1 unit.

If Node n receives −1, decrease ali by 1 unit (AIMD?).

Stability? Converge to the previous algorithm?


Variable Link Capacity

Known link capacity distribution

Two states with Bernoulli.Many states.How to define link rate?

Time scale of the changes

Slow (ignore transient state, and only consider steady state).Fast.

Simulation

What network topology?What link capacity distribution?

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