Single Sink Edge Installation Kunal Talwar UC Berkeley.
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Transcript of Single Sink Edge Installation Kunal Talwar UC Berkeley.
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Single Sink Edge Installation
Kunal TalwarUC 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 q
“routing” cost per unit length q
Find cheapest installation to route a unit of demand from each source si
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Example
Given sources and the underlying graph
t
Find subgraph and cable type for each edge to minimize total cost
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t
Example
Given sources and the underlying graphFind subgraph and cable type for each edge to minimize total cost
t
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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 q,q
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Special case
Only one cable typeExtraspecial subcases: If ) build steiner tree. If ) 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 ¸ (optimal steiner tree).OPT ¸ v dG(s,v).
Cost = 2(steiner tree) + 3v dG(s,v) · 7OPT.
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About OPT…
It’s 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 installedFlow variable fj
e;q is 1 if flow from j uses discount type q on edge eObjective function – cost of building the network –
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Integer Program
Variable zqe is 1 if edge e has
discount type q installedFlow variable fj
e;q is 1 if flow from j uses discount type q on edge eObjective 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
Linear
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Rounding the linear program
Top downUse the linear program solution to guide the algorithmUse 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 Tq+1 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 typesForm balls of radius Rq
j 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 typesForm balls of radius Rq
j 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 typesForm balls of radius Rq
j around node vj
Select a set of non intersecting balls in increasing order of radii.Each node vl has a buddy within distance 4Rq
l
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Identifying what to connect….
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Identifying what to connect….
LessThan4Rq
j
vj
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Connecting it…
Contract all selected ballsShrink Tq+1
Build a Steiner tree on the contracted nodesConvert to LASTEach selected vertex has a proxy in its ball at distance at most Rq
j
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Connecting it ….
j
i
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Connecting it ….
j
i
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Connecting it ….
Less than Rqj
j
i
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Connecting it ….
Less ThanRq
j
j
i
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Connecting it ….
LessThanRq
jj
i
Each node (using its buddy) has someone in the tree Tq within Rq
j
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Analysis: Building Cost
Key Lemma : Let (S) be the set of edges on the boundary of S ¶ B(vj,Rq
j), 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:
t
Flow f
Flow (1-f)
vj
S
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 (S) < 1/2
Then > half the flow travels at least 2Rq
j on low discounts
This flow itself contributes > Rqj
to Rqj. Contradiction.
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Lemma proved…
Lemma : Let (S) be the set of edges on the boundary of S ¶ B(vj,Rq
j), r 2 Sc. Let z,f be any feasible solution to the LP. Then
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Lemma proved…
Lemma : Let (S) be the set of edges on the boundary of S ¶ B(vj,Rq
j), 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 (S) be the set of edges on the boundary of S ¶ B(vj,Rq
j), r 2 Sc. Let z,f be any feasible solution to the LP. Then
Recall that Steiner tree LP is:
i.e. 2p¸ q zqe is a feasible fractional Steiner
tree.
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Building Cost (contd…)
Steiner tree LP has gap · 2Hence our steiner tree cost is no more than 2q times e p¸ q zp
e.
Thus the LASTq is no more than twice this.Let OPTb
q = OPT’s building cost for discount type q.Then LASTq cost is 8p¸ q (q/p)OPTb
p
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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 q+1 ¸ 2 q
q+1 · (1/2) q
Can be done with a factor 2 change in cost
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Building Cost (contd…)
p ¸ 2p-qq
LASTq cost is 8p¸ q (q/p)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 v=u0,u1,…uk=t uq- uq+1 uses discount type q.
v’s routing cost = q q dT(uq,uq+1)
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Routing costs…
u3= t
u0=vu1
u2 Proxy(v)2 Length of this= 2 d(u1,u2)
Routing cost on
discount type 2
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Routing costs…
u3= t
u0=vu1
u2 Proxy(v)2 Length of this= 2 d(u1,u2)· 3 2 d(u1,Proxy(v))
coz we built a LAST !
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Routing costs…
u3= t
u0=vu1
u2 Proxy(v)2 Length of this= 2 d(u1,u2)· 3 2 d(u1,Proxy(v))· 3 2(d(u1,v) + d(v,Proxy(v)))
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Routing costs…
u3= t
u0=vu1
u2 Proxy(v)2 Length of this= 2 d(u1,u2)· 3 2 d(u1,Proxy(v))· 3 2(d(u1,v) + d(v,Proxy(v)))
bounded by 6R2v
~ what fractional sol. pays
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Routing costs…
u3= t
u0=vu1
u2 Proxy(v)
already paid 1 ¸ 2 2 for this
2 Length of this= 2 d(u1,u2)· 3 2 d(u1,Proxy(v))· 3 2(d(u1,v) + d(v,Proxy(v)))
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Routing costs…
u3= t
u0=vu1
u2 Proxy(v)2 Length of this= 2 d(u1,u2)· 3 2 d(u1,Proxy(v))· 3 2(d(u1,v) + d(v,Proxy(v)))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 typesLAST + selection of balls ensures routing cost is not too highScaling crucial in both cases !
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Conclusions
Get a constant factor approximation algorithmThe natural LP has a constant integrality gap.
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Conclusions
Get a constant factor approximation algorithmThe natural LP has a constant integrality gap.
Open problems:More reasonableh constantsThe 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?