Single Sink Edge Installation

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Single Sink Edge Installation

• Kunal Talwar
• UC Berkeley

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Problem Definition

• Given
• A graph G(V,E) , sink t
• Sources s1,s2,, sm
• k Discount types
• building cost per unit length sq
• routing cost per unit length dq
• Find cheapest installation to route a unit of
  demand from each source si

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Example

• Given sources
• and the underlying graph

• Find subgraph and cable type for each edge to
  minimize total cost

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Example

• Given sources
• and the underlying graph
• Find subgraph and cable type for each edge to
  minimize total cost

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Related Work

• Salman, et.al. 97 - A constant factor for single
  cable type case (LASTs)
• Awerbuch, Azar 98 - O(log n loglog n)-
  approximation (tree embedding)
• Meyerson, et.al. 00 - O(log n) (comb.)
• Garg et.al. 01 - O(k) (LP rounding)
• Guha et.al. 01 - O(1) (you just heard !)
• This work LP rounding - O(1).
• Goel, Estrin, 03 O(log n) oblivious to sq,dq

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Special case

• Only one cable type
• Extraspecial subcases
• If d0 ) build steiner tree.
• If s0 ) use shortest path tree.
• In general, want a
• Light Approximate Shortestpath Tree
• KRY 94 LASTs tree of cost at most 2 times a
  given connecting tree, with dT(root,v) at most 3
  times dG(root,v).

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Single cable type Algorithm

• Build a steiner tree.
• Convert to LAST !
• Analysis -
• OPT s(optimal steiner tree).
• OPT d åv dG(s,v).
• Cost 2s(steiner tree) d 3åv dG(s,v) 7OPT.

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About OPT

• Its a tree !
• As we travel from a source to sink
• Total traffic only increases
• .. So thicker and thicker cables
• We let the LP know the above

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Integer Program

• Variable zqe is 1 if edge e has discount type q
  installed
• Flow variable fjeq is 1 if flow from j uses
  discount type q on edge e
• Objective function
• cost of building the network

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Integer Program

• Variable zqe is 1 if edge e has discount type q
  installed
• Flow variable fjeq is 1 if flow from j uses
  discount type q on edge e
• Objective function
• cost of building the network
• cost of routing the demands

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Integer program

• Subject to
• Flow conservation
• Flow monotonicity
• Outflow 1
• Route on edge e only if edge built
• Integrality contraints

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Rounding the linear program

• Top down
• Use the linear program solution to guide the
  algorithm
• Use the linear program cost as the lower bound

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Algorithm outline

• Tk t
• For each discount type q (from highest to lowest)
• Identify what to connect in this stage
• Connect it to Tq1 with discount type q to get Tq
  

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Identifying what to connect

• For a fractional solution f, flow from vj
• travels some average number
  of edges on low discount types.

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Identifying what to connect

• For a fractional solution f, flow from vj
• travels some average number
  of edges on low discount types.
• Beyond that radius, fractional solution uses high
  discount types

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Identifying what to connect

• For a fractional solution f, flow from vj
• travels some average number
  of edges on low discount types.
• Beyond that radius, fractional solution uses high
  discount types
• Form balls of radius 2Rqj around node vj

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Identifying what to connect

• For a fractional solution f, flow from vj
• travels some average number
  of edges on low discount types.
• Beyond that radius, fractional solution uses high
  discount types
• Form balls of radius 2Rqj around node vj
• Select a set of non intersecting balls in
  increasing order of radii.

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Identifying what to connect

• For a fractional solution f, flow from vj
• travels some average number
  of edges on low discount types.
• Beyond that radius, fractional solution uses high
  discount types
• Form balls of radius 2Rqj around node vj
• Select a set of non intersecting balls in
  increasing order of radii.
• Each node vl has a buddy within distance 4Rql

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Identifying what to connect.
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Identifying what to connect.
vj
Less Than 4Rqj
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Connecting it

• Contract all selected balls
• Shrink Tq1
• Build a Steiner tree on the contracted nodes
• Convert to LAST
• Each selected vertex has a proxy in its ball at
  distance at most 2Rqj

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Connecting it .
j
i
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Connecting it .
j
i
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Connecting it .
j
i
Less than 2Rqj
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Connecting it .
j
Less Than 4Rqj
i
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Connecting it .
Each node (using its buddy) has someone in the
tree Tq within 6Rqj
Less Than 6Rqj
j
i
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Analysis Building Cost

• Key Lemma Let D(S) be the set of edges on the
  boundary of S B(vj,2Rqj), r 2 Sc. Let z,f be
  any feasible solution to the LP. Then

Proof
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Lemma proof
t
Flow f
Flow (1-f)
vj
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Lemma proof

• A total flow of 1 leaves S
• Flow crossing the boundary on low discounts
  reaches there on low discounts. (monotonicity)
• Suppose the high discounts built at D(S) lt 1/2
• Then gt half the flow travels at least 2Rqj on low
  discounts
• This flow itself contributes gt Rqj to Rqj.
  Contradiction.

S
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Lemma proved

• Lemma Let D(S) be the set of edges on the
  boundary of S B(vj,2Rqj), r 2 Sc. Let z,f be
  any feasible solution to the LP. Then

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Lemma proved

• Lemma Let D(S) be the set of edges on the
  boundary of S B(vj,2Rqj), r 2 Sc. Let z,f be
  any feasible solution to the LP. Then

• Recall that Steiner tree LP is

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Lemma proved

• Lemma Let D(S) be the set of edges on the
  boundary of S B(vj,2Rqj), r 2 Sc. Let z,f be
  any feasible solution to the LP. Then

• Recall that Steiner tree LP is
• i.e. 2åp q zqe is a feasible fractional Steiner
  tree.

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Building Cost (contd)

• Steiner tree LP has gap 2
• Hence our steiner tree cost is no more than 2sq
  times åe åp q zpe.
• Thus the LASTq is no more than twice this.
• Let OPTbq OPTs building cost for discount type
  q.
• Then LASTq cost is 8åp q (sq/sp)OPTbp

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Scaling

• We prune the discount
• types in the beginning to be
• sure that they are all different enough !
• More formally, ensure
• sq1 2 sq
• dq1 (1/2) dq
• Can be done with a factor 2 change in cost

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Building Cost (contd)

• sp 2p-qsq
• LASTq cost is 8åp q (sq/sp)OPTbp
• Now things work out fine !
• We get
• Total building cost
• 16 OPTb

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Routing costs

• Path from v to t uses increasingly higher
  discount type. Let the path be vu0,u1,ukt
• uq- uq1 uses discount type q.
• vs routing cost åq dq dT(uq,uq1)

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Routing costs
u3 t
Routing cost on discount type 2
u2
Proxy(v)
d2 Length of this d2 d(u1,u2)
u0v
u1
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Routing costs
u3 t
u2
Proxy(v)
d2 Length of this d2 d(u1,u2) 3 d2
d(u1,Proxy(v))
u0v
u1
coz we built a LAST !
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Routing costs
u3 t
u2
Proxy(v)
d2 Length of this d2 d(u1,u2) 3 d2
d(u1,Proxy(v)) 3 d2(d(u1,v) d(v,Proxy(v)))
u0v
u1
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Routing costs
u3 t
u2
Proxy(v)
d2 Length of this d2 d(u1,u2) 3 d2
d(u1,Proxy(v)) 3 d2(d(u1,v) d(v,Proxy(v)))
u0v
u1
bounded by 6R2v what fractional sol. pays
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Routing costs
u3 t
u2
Proxy(v)
d2 Length of this d2 d(u1,u2) 3 d2
d(u1,Proxy(v)) 3 d2(d(u1,v) d(v,Proxy(v)))
u0v
u1
already paid d1 2 d2 for this
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Routing costs
u3 t
u2
Proxy(v)
d2 Length of this d2 d(u1,u2) 3 d2
d(u1,Proxy(v)) 3 d2(d(u1,v) d(v,Proxy(v)))
u0v
u1
Total routing cost v pays in sol is O(what v pays
in LP)
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Hence.

• Theorem Algorithm described has cost within a
  constant factor of the LP optimum.
• Recap
• LP tells us how far from vj to go before LP can
  pay for building high discount types
• LAST selection of balls ensures routing cost is
  not too high
• Scaling crucial in both cases !

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Conclusions

• Get a constant factor approximation algorithm
• The natural LP has a constant integrality gap.

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Conclusions

• Get a constant factor approximation algorithm
• The natural LP has a constant integrality gap.
• Open problems
• More reasonableh constants
• The general buy-at-bulk problem
• Combinatorial lower bounds off by log
• LP might be the right approach

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• Paper available at
• http//www.cs.berkeley.edu/kunal/acads/bb.ps
• Questions?