title page an application of market equilibrium in distributed load balancing in wireless networking...
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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
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8
8
4
7
RSS
Equality Edges
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40
10
30
8
8
Max RSS
Equality Graph
Desirable APs for each Client
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30
Maximum Matching
Maximum Matching, Deficiency = 1
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40
10
30
Neighborhood Set
Neighborhood Set
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30
S
T
Deficiency of a Set
Deficiency of S = Capacities on T - |S|
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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
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30
Most Deficient Sets
Two Most Deficient Sets
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Smallest Most Deficient Set
Pick the smallest. Use Super-modularity!
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S
Neighborhood Set
Neighborhood Set
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S
T
Complement of the Neighborhood Set
Complement of the Neighborhood Set
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S
Tc
Initialize r.
Initialize r = 1
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10r
30r
S
Tc
About to start raising r.
Start Raising r
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40
10r
30r
S
Tc
Equality edges about to be lost.
Equality edge which will be lost
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40
10r
30r
S
Tc
Useless equality edges.
Did not belong to any maximum matching
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40
10r
30r
S
Tc
Equality edges deleted.
Let it go
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10r
30r
S
Tc
All other equality edges remain.
All other equality edges are preserved!
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10r
30r
S
Tc
A new equality edge added
At some point a new equality appears. r =2
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S
Tc
Subcase A – Deficiency Decreases
New equality edge gives an augmenting path
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S
Tc
Subcase B – Deficiency does not decrease
New edge does not create any augmenting path
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60
S
Tc
Smallest most deficient set increases
New S is a strict super set of old S!
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20
60
S
Eventually Subcase A will happen
Eventually the size of the matching increases
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20
60
S
Case 1 – Deficiency Reaches Zero
All Clients are served!
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20
60
S
All APs are saturated
Or the algorithm will prove that none exist!
S
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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.