The Load Distance Balancing Problem

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The Load Distance Balancing Problem

• Eddie Bortnikov (Yahoo!)
• Samir Khuller (Maryland)
• Yishay Mansour (Google)
• Seffi Naor (Technion)

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The Load-Distance Balancing Problem

• Given n clients and k servers s1, s2,sk we need
  to assign each client to a server.
• Cost for client i assigned to server sj is as
  follows
• Cost(i) Distance(i,sj) Delay(j,Lj)
• Delay(j,Lj) is a FUNCTION of the number of
  clients Lj assigned to sj.
• OBJECTIVE Min Max Cost(i)

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An Example
s2
s1
s3
C
E
D
A
B

• Each server has its own delay function (can be
  arbitrary, we just assume its non-decreasing).
• Note that C is closer to s1, but prefers to
  attach to s2 since Dist(C,s2)Delay(s2,2)ltDist(C,s
  1)Delay(s1,3)
• Objective is to Minimize Max Cost for any client

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Related Work

• Lots of research on locating facilities. Here the
  facilities are all given we just have to compute
  assignment of clients to facilities.
• Notion of capacities has been used for various
  covering problems such as Vertex Cover, K
  Centers, Facility Location etc.

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Main Results

• The problem is NP-hard.
• We develop a polynomial time 2 approx.
• We show that the bound 2 cannot be improved to
  2-e unless NPP.
• With triangle inequality in the distance function
  the hardness reduces to 5/3-e.
• When all clients and servers are on a line its
  solvable in polynomial time.
• For Min Sum Cost(i), we can solve in polynomial
  time using Min-Weight Matching.

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NP-hardness by Exact Set Cover

• Given N elements and a collection S of K sets
  (each set has size m). Does there exist a subset
  S of S, such that each element belongs to
  exactly one set in S?
• In other words, we need to pick exactly N/m
  subsets from S, to cover each element once.

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Example of Exact Cover

• Here we have N16 elements and 9 sets (m4).
• The FOUR blue sets form an EXACT COVER, and we
  discard the FIVE orange sets.

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Reduction from Exact Cover

• Each element is a client. In addition we create a
  collection of M(K-N/m) dummy clients.
• Subset Sj in S corresponds to server sj.
• Dist(dummy,server)d1
• Dist(i,sj) d2 if i e Sj, o.w. 8
• d2 gtgt d1

d1
Dummy clients
sj
d2
i
Clients (elements)
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Reduction from Exact Cover

• Delay functions for servers are basically a step
  function.
• Delay(j,Lj) ?-d2, when load is at most m.
• Delay(j,Lj) ?-d1, when load exceeds m, but is at
  most M.

?-d1
?-d2
m
M
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Reduction from Exact Cover

• Suppose there is a solution to exact cover, then
  there is a solution to the LDB problem with delay
  at most ?.
• For each chosen subset Sj, the corresp. server sj
  gets m clients each at distance d2.Since the
  delay is ?-d2, the total cost is at most ?.
• For the remaining subsets, those dummies are all
  assigned to the remaining servers (K-N/m), each
  gets M dummies.
• The proof in the other direction requires some
  work!

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Proof (cont.)

• Each server can support at most M dummy clients
  if the total cost does not exceed ?, and no more
  than m real clients.
• Suppose a server supports both real and dummy
  clients then the total number of servers with
  real clients is k gt N/m.
• These serve at most (mk-N) dummy clients, while
  the rest can serve only M(k-k) dummy clients.
• Adding the two shows (some algebra needed) that
  we can only assign lt M(k-N/m) dummy clients if
  Mgtm.

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Hardness Results follow.

• We can set d1e and d2?-e. A solution to EXACT
  COVER exists if and only if a solution with cost
  ? exists for LDB.
• If there is no solution to EXACT COVER, then
  every solution to LDB has cost 2(?-e).
• However, here we do violate triangle inequality
  in the distance function.

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Hardness results with triangle inequality

• We need to set d1? ?, and d2 ?-e.
• With these parameters, the distance between a
  (real) client i and a server sj such that i is
  not in Sj, is at least 5/3?- e.
• A solution to EXACT COVER exists if and only if a
  solution with cost ? exists for LDB.
• If there is no solution to EXACT COVER, then
  every solution to LDB has cost 5/3?-e.

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Approximation Algorithm with factor 2

• Suppose a solution exists with maximum cost ?.
• For each server sj, we can compute an upper bound
  on the number of clients that can be served with
  a delay of at most ? (say Lj).
• For each client i we can compute the subset of
  servers that are within distance ? (Si).
• Now its just a flow problem to check if an
  assignment exists where each client i is assigned
  to a server from Si and each sj has load at most
  Lj.
• Minimizing ? gives a trivial 2 approximation.

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All servers and clients on a line

• Use dynamic programming!

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Minimizing the Sum of Costs

• We reduce this to min cost matching in a
  bipartite graph.
• Let G(X,Y,E) where nodes in X correspond to n
  clients and there are nk nodes in Y. We have n
  nodes corresp. to each server.
• We ask for a min cost matching to find a
  solution.

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Capacitated K Centers

• The related problem of choosing K facilities has
  been considered (each client should be assigned
  to a closeby facility and the load on the
  facility should not be too high) Khuller
  Sussman (K,L,5R) or ((2/c)K, cL, 2R) related to
  K-centers clustering.

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Conclusions

• Can we improve the 2 approximation when triangle
  inequality holds?
• Can we improve the 2 approximation when Delay
  functions satisfy specific properties? What is a
  natural delay function?
• Are there other special cases that can be solved
  in polynomial time?