Title Page An Application of Market Equilibrium in Distributed Load Balancing in Wireless Networking...

Title PageAn Application of Market Equilibrium in Distributed

Load Balancing in Wireless Networking

Algorithms and Economics of NetworksUW CSE-599m

Reference

Cell-Breathing in Wireless Networks, by Victor Bahl, MohammadTaghi Hajiaghayi, Kamal Jain,

Vahab Mirrokni, Lili Qiu, Amin Saberi

Wireless Devices

Wireless DevicesCell-phones, laptops with WiFi cardsReferred as clients or users interchangeably

Demand ConnectionsUniform for cell-phones (voice connection)Non-uniform for laptops (application dependent)

Access Points (APs)

Access PointsCell-towers, Wireless routers

CapacitiesTotal traffic they can serveInteger for Cell-towers

Variable Transmission PowerCapable of operating at various power levelsAssume levels are continuous real numbers

Clients to APs assignment

Assign clients to APs in an efficient wayNo over-loading of APsAssigning the maximum number of clientsSatisfying the maximum demand

One Heuristic Solution

A client connects to the AP with the best signal and the lightest loadRequires support both from AP and ClientsAPs have to communicate their current loadClients have WiFi cards from various vendors

running legacy softwareLimited benefit in practice

We would like …

A Client connects to the AP with the best received signal strength

An AP j transmitting at power level Pj then a client i at distance dij receives signal with strength

Pij = a.Pj.dij-α

where a and α are constants Captures various models of power attenuation

Cell Breathing Heuristic

An overloaded AP decreases its communication radius by decreasing power

A lightly loaded AP increases its communication radius by increasing power

Hopefully an equilibrium would be reached Will show that an equilibrium exist Can be computed in polynomial timeCan be reached by a tatonement process

Market Equilibrium – A distributed load balancing mechanism.

Demand = Supply No Production

Static SupplyAnalogous to Capacities of APs

PricesAnalogous to Powers at APs

Utilities Analogous to Received Signal Strength function

Analogousness is Inspirational

Our situation is analogous to Fisher setting with Linear Utilities

Fisher Setting Linear UtilitiesBuyers Goods

1 1 1 1, j jj

M u u x

, i i ij ijj

M u u x

, n n nj njj

M u u x

1q

jq

mq

ijx

ij iju x

Clients assignment to APsClients APs

1 1 1 1, max j jj

D u P x

, maxi i ij ijj

D u P x

, maxn n nj njj

D u P x

1C

jC

mC

ijx

ij ijP x

Analogousness is Inspirational

Our situation is analogous to Fisher setting with Linear Utilities

Get inspiration from various algorithms for the Fisher setting and develop algorithms for our setting

We do not know any reduction – in fact there are some key differences

Differences from the Market Equilibrium setting

Demand Price dependent in Market equilibrium setting Power independent in our setting

Is demand splittable? Yes for the Market equilibrium setting No for our setting

Under mild assumptions, market equilibrium clears both sides but our solution requires clearance on either one side Either all clients are served Or all APs are saturated

This also means two separate linear programs for these two separate cases

Three Approaches for Market Equilibrium

Convex Programming BasedEisenberg, Gale 1957

Primal-Dual BasedDevanur, Papadimitriou, Saberi, Vazirani 2004

Auction BasedGarg, Kapoor 2003

Three Approaches for Load Balancing

Linear ProgrammingMinimum weight complete matching

Primal-DualUses properties of bipartite graph matchingNo loop invariant!

AuctionUseful in dynamically changing situation

Another Application of Market Equilibria in Networking

Fleisher, Jain, Mahdian 2004 used market equilibrium inspiration to obtain Toll-Taxes in Multi-commodity Selfish Routing ProblemThis is essentially a distributed load balancing i.e.,

distributed congestion control problem

Linear Programming Based Solution

Create a complete bipartite graph One side is the set of all clients The other side is the set of all APs, conceptually each

AP is repeated as many times as its capacity The weight between client i and AP j is

wij = α.ln(dij) – ln(a) Find the minimum weight complete matching

Theorem

Minimum weight matching is supported by a power assignment to APs

Power assignment are the dual variables Two cases for the primal program

Solution can satisfy all clientsSolution can saturate all APs

Case 1 – Complete matching covers all clients

,

ijj A

minimize

subject to

i C 1

, 0

ij iji C j A

ij ji C

ij

w x

x

j A x C

i C j A x

Case 1 – Pick Dual Variables

,

ijj A

minimize

subject to

i C 1

, 0

ij iji C j A

i

ij j ji C

ij

w x

x

j A x C

i C j A x

Write Dual Program

maximize

subject to

,

0

i j ji C j A

i j ij

j

C

i C j A w

j A

Optimize the dual program

Choose Pj = eπj

Using the complementary slackness condition we will show that the minimum weight complete matching is supported by these power levels

Proof

Dual feasibility gives:

-λi ≥ πj – wij= ln(Pj) – α.ln(dij) + ln(a) = ln(a.Pj.dij-α)

Complementary slackness gives:

xij=1 implies -λi = ln(a.Pj.dij-α)

Together they imply that i is connected to the AP with the strongest received signal strength

Case 2 – Complete matching saturates all APs

,

ijj A

minimize

subject to

i C 1

, 0

ij iji C j A

ij ji C

ij

w x

x

j A x C

i C j A x

Case 2 – The rest of the proof is similar

Optimizing Dual Program

Once the primal is optimized the dual can be optimized with the Dijkstra algorithm for the shortest path

Primal-Dual-Type Algorithm

Previous algorithm needs the input upfront

In practice, we need a tatonement process

The received signal strength formula does not work in case there are obstructions

A weaker assumption is that the received signal strength is directly proportional to the transmitted power – true even in the presence of obstructions

Cell-phones Cell-towers

Start with arbitrary non-zero powers

10

40

10

30

Powers and Received Signal Strength

10

40

10

30

8

8

4

7

RSS

Equality Edges

10

40

10

30

8

8

Max RSS

Equality Graph

Desirable APs for each Client

10

40

10

30

Maximum Matching

Maximum Matching, Deficiency = 1

10

40

10

30

Neighborhood Set

Neighborhood Set

10

40

10

30

S

T

Deficiency of a Set

Deficiency of S = Capacities on T - |S|

10

40

10

30

S

T

Simple Observation

Deficiency of a Set S ≤ Deficiency of the Maximum Matching

Maximum Deficiency over Sets ≤ Minimum Deficiency over Matching

Generalization of Hall’s Theorem

Maximum Deficiency over Sets = Minimum Deficiency over Matching

Maximum Deficiency over Sets = Deficiency of the Maximum Matching

Maximum Matching

Maximum Matching, Deficiency = 1

10

40

10

30

Most Deficient Sets

Two Most Deficient Sets

10

40

10

30

Smallest Most Deficient Set

Pick the smallest. Use Super-modularity!

10

40

10

30

S

Neighborhood Set

Neighborhood Set

10

40

10

30

S

T

Complement of the Neighborhood Set

Complement of the Neighborhood Set

10

40

10

30

S

Tc

Initialize r.

Initialize r = 1

10

40

10r

30r

S

Tc

About to start raising r.

Start Raising r

10

40

10r

30r

S

Tc

Equality edges about to be lost.

Equality edge which will be lost

10

40

10r

30r

S

Tc

Useless equality edges.

Did not belong to any maximum matching

10

40

10r

30r

S

Tc

Equality edges deleted.

Let it go

10

40

10r

30r

S

Tc

All other equality edges remain.

All other equality edges are preserved!

10

40

10r

30r

S

Tc

A new equality edge added

At some point a new equality appears. r =2

10

40

20

60

S

Tc

Subcase A – Deficiency Decreases

New equality edge gives an augmenting path

10

40

20

60

S

Tc

Subcase B – Deficiency does not decrease

New edge does not create any augmenting path

10

40

20

60

S

Tc

Smallest most deficient set increases

New S is a strict super set of old S!

10

40

20

60

S

Eventually Subcase A will happen

Eventually the size of the matching increases

10

40

20

60

S

Case 1 – Deficiency Reaches Zero

All Clients are served!

10

40

20

60

S

All APs are saturated

Or the algorithm will prove that none exist!

S

10

40

20

Unsplittable Demand

Solve the splittable case by solving the minimum weight matching linear program

Unsplittable Demand

,

ijj A

minimize

subject to

i C 1

, 0

ij iji C j A

i ij ji C

ij

w x

x

j A D x C

i C j A x

Unsplittable Demand

Solve the splittable case by solving the minimum weight matching linear program

In fact compute a basic feasible solution Assume that the number of clients is much

larger than the number of APs – a realistic assumption

Approximate Solution

All xij’s but a small number of xij’s are integral

Analysis of Basic Feasible Solution

,

ijj A

minimize

subject to

i C 1

, 0

ij iji C j A

i ij ji C

ij

w x

x

j A D x C

i C j A x

Approximate Solution

All xij’s but a small number of xij’s are integral Number of xij which are not integral is at most

the number of APs Most clients are served unsplittably Clients which are served splittably – do not

serve them The algorithm is still almost optimal

Discrete Power Levels

Over the shelf APs have only fixed number of discrete power levels

Equilibrium may not existIn fact it is NP-hard to test whether it exist or not

If every client has a deterministic tie breaking rule then we can compute the equilibrium – if exist under the tie breaking rule

Discrete Power Levels

Start with the maximum power levels for each AP

Take any overloaded AP and decrease its power level by one notch

If an equilibrium exist then it will be computed in time mk, where m is the number of APs and k is the number of power levels

This is a distributed tatonement process!

Proof.

Suppose Pj is an equilibrium power level for the jth AP.

Inductively prove that when j reaches the power level Pj then it will not be overloaded again.Here we use the deterministic tie breaking rule.

Conclusion.

Theory of market equilibrium is a good way of synchronizing independent entity’s to do distributed load balancing.

We simulated these algorithm. Observed meaningful results.

Thanks Kamal Jain for the main part of slides.