Bandwidth Allocation in Networks with Multiple
InterferencesReuven Bar-Yehuda
Gleb PolevoyDror Rawitz
Technion
1
Multiple interference
2
(1 )ii
1 1 2 1 2 31 (1 ) (1 )(1 ) (1 )(1 )(1 )
1
1 2 3
1 1 2 1 2 3:1 (1 ) (1 ) (1 ) 1 ii
Additive we can approximate to For small interferences
Interval selection with multiple interference
3
Base stations B={1,2,…,i,…,n}Interferences i <1Users U={1,2,…,j,…,m}Times {1,2,…,t,…,f}User j has a set of time interval
requests from base station i: Rij={Iij1,…,Iijk,….}
Each request ijk has a profit Pijk >0
Optimization problem: Allocating subsets of time intervals with maximum profit s.t:
• At most one interval per user• All intervals satisfied by a base
station are independent.
Rijj
'' : ( ')
(1 )ijk ijk
iI S t Ii i t alloc i
t
2
1
i
n
Main result: 7-approximation
This is achieved by getting:
k+1- approximation for strong interferences
-approximation for weak interferences
For k=2 it gives: 3+4=7 (will be shown)
4
1 ki i
1 ki i
131k
Interval selection with multiple interference
Linearization & Normalization
We can transform:
To:
Where:
5
'' : ( ')
1ijk ijk
iI S t I i i t alloc i
w
'' : ( ')
(1 )ijk ijk
iI S t Ii i t alloc i
''
log(1 )log
iiw
Maximize
s.t.
II SP
Common time &One req/user One req/base station
( { }) 1I Sw S I
6
Riii
t
2
1
i
n
R11
R22
Rnn
t0
Maximize
s.t.
Bad news: NP-Hard (add width-less expensive box)
Good news: FPTAS (Dynamic programming approach)
Generalization to many base stations: the bipartite is a forest.
II SP
Open knapsack
( { }) 1I Sw S I
1Not feasible
( { }) 1w S I
7
8
1
2
1
1
2
j
i
j
m n
uu
Baseu
uu
Baseu
Base u
Use open knapsack constraintsat interval’s right endpoints
maxI
I SP
( { }) 1Iw S Right II S
s.t. S contains at most one interval from a user contains at most one interval per base stationIS RightI S
Same user
9
1
2
1
1
2
j
i
j
m n
uu
Baseu
uu
Baseu
Base u
Strong interferences: w > 1/k
Same user
Same time
Let Î be an interval that ends first; 1 if I in conflict with Î
For all intervals I define: p1 (I) = 0 else
For every feasible x: p1 ·x k+1 Every Î-maximal solution is k+1 approximation . For every Î-maximal x: p1 ·x 1
Î
Algorithm MaxIS( R, p )If R = Φ then return Φ ;If I S p(I) 0 then return MaxIS( R - {I}, p);Let Î R that ends first;
p(Î) if I in conflict with ÎI S define: p1 (I) =
0 elseIS = MaxIS( R, p- p1 ) ;If IS is Î-maximal then return IS else return IS {Î};
10
Strong interferences: w > 1/kThe k+1 approximation algorithm
11
1
2
1
1
2
j
i
j
m n
uu
Baseu
uu
Baseu
Base u
Weak interferences: w ≤ 1/k
Same user
Interference conflict
Let Î be an interval that ends first; 0 if I not in any conflict with Î
For all intervals I define: p1 (I) = 1-1/k else if I same base or same user as Î w(I) else if I in interference conflict with Î
For every feasible x: p1 ·x 3-2/k
Every Î-maximal is For every Î-maximal x: p1 ·x 1-1/k
Î Same base station
13 approximation1k
1/7-approximationR9
R8 w½ > R7 w > ½ w½ >
R6R5 w½ >
R4R3 w > ½ w½ >
R2R1 w > ½ w > ½ w½ >
Algorithm:GRAY = Find 1/3-approximation for gray (w>1/2) intervals;COLORED = Find 1/4-approximation for colored intervalsReturn the one with the larger profitAnalysis:If GRAY* 3/7OPT then GRAY 1/3(3/7OPT)=1/7OPT elseCOLORED* 4/7OPT thus COLORED 1/4(4/7OPT)=1/7OPT
Interval selection with multiple interference
13
Base stations B={1,2,…,i,…,n}Interferences i <1Users U={1,2,…,j,…,m}Times {1,2,…,t,…,f}User j has a set of time interval
requests from base station i: Rij={Iij1,…,Iijk,….}
Each request ijk has a profit Pijk >0
Optimization problem: Allocating subsets of time intervals with maximum profit s.t:
• At most one interval per user• All intervals satisfied by a base
station are independent.
Riji
j
11 : ( 1)
(1 )ijk ijk
iI S t Ii i t alloc i
t
Frequency allocation with multiple interference
14
Base stations B={1,2,…,i,…,n}Interferences i <1Users U={1,2,…,j,…,m}Frequencies {1,2,…,t,…,f}User j has a set of bandwidth
demands from base station i: Rij={dij1,…,dijk,….}
Each demand dijk has a profit pijk >0
Optimization problem: Allocating demands with maximum profit s.t:
• At most one demand satisfied per user
• All demands satisfied by a base station are independent.
• |alloc(ijk)|= dijk
Riji
j
{1,2,... }: ( )
(1 )ijkt fijk t alloc ijk
t
Main result: 12-approximation
This is achieved by getting:
- approximation for strong interferences
-approximation for weak interferences
For k=2 it gives: 5+7=12
15
1 ki i
1 ki i
2 1k
Frequency allocation with multiple interference
251k
Thank you !
The Local-Ratio Technique: Basic definitions
Given a profit [penalty] vector p.
Maximize[Minimize] p·x Subject to: feasibility constraints F(x)
x is r-approximation if F(x) and p·x [] r · p·x*
An algorithm is r-approximation if for any p, Fit returns an r-approximation
17
The Local-Ratio Theorem:
x is an r-approximation with respect to p1
x is an r-approximation with respect to p- p1
x is an r-approximation with respect to p
Proof: (For maximization) p1 · x r × p1*
p2 · x r × p2*
p · x r × ( p1*+ p2*) r × ( p1 + p2 )*
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Special case: Optimization is 1-approximation
x is an optimum with respect to p1
x is an optimum with respect to p- p1
x is an optimum with respect to p
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A Local-Ratio Schema for Maximization[Minimization] problems:
Algorithm r-ApproxMax[Min]( Set, p )
If Set = Φ then return Φ ;If I Set p(I) 0 then return r-ApproxMax( Set-{I}, p ) ;[If I Set p(I)=0 then return {I} r-ApproxMin( Set-{I}, p ) ;]
Define “good” p1 ;
REC = r-ApproxMax[Min]( S, p- p1 ) ;If REC is not an r-approximation w.r.t. p1 then “fix it”;
return REC; 20
The Local-Ratio Theorem: Applications
Applications to some optimization algorithms (r = 1) : ( MST )Minimum Spanning Tree (Kruskal)
( SHORTEST-PATH )s-t Shortest Path (Dijkstra) )LONGEST-PATH( s-t DAG Longest Path (Can be done with dynamic programming)
)INTERVAL-IS( Independents-Set in Interval Graphs Usually done with dynamic programming ) )LONG-SEQ( Longest (weighted) monotone subsequence (Can be done with dynamic programming)
( MIN_CUT )Minimum Capacity s,t Cut (e.g. Ford, Dinitz) Applications to some 2-Approximation algorithms: (r = 2)
( VC )Minimum Vertex Cover (Bar-Yehuda and Even) ( FVS )Vertex Feedback Set (Becker and Geiger)
( GSF )Generalized Steiner Forest (Williamson, Goemans, Mihail, and Vazirani) ( Min 2SAT )Minimum Two-Satisfibility (Gusfield and Pitt)
( 2VIP )Two Variable Integer Programming (Bar-Yehuda and Rawitz) ( PVC )Partial Vertex Cover (Bar-Yehuda)
( GVC )Generalized Vertex Cover (Bar-Yehuda and Rawitz) Applications to some other Approximations :
( SC )Minimum Set Cover (Bar-Yehuda and Even) ( PSC )Partial Set Cover (Bar-Yehuda)
( MSP )Maximum Set Packing (Arkin and Hasin)
Applications Resource Allocation and Scheduling :.…
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Single request to Single base stationI19I18I17I16I15I14I12I12I11
Maximize s.t: For each instance I:
For each freq. t:
I
IxIp )(
}1,0{Ix
)()(:
1IetIsIIx
R1j = {I1j}j
22
Single base station: How to select P1 to get optimization?
I19I18I17I16I15I14I13I12I11
Î time
Let Î be an interval that ends first; 1 if I in conflict with Î
For all intervals I define: p1 (I) = 0 else
For every feasible x: p1 ·x 1 Every Î-
maximal is optimal. For every Î-maximal x: p1 ·x 1
P1=1
P1=1
P1=1
P1=1
P1=0
P1=0
P1=0
P1=0
P1=0
23
Single base station: An Optimization Algorithm
I19I18I17I16I15I14I13I12I11 Î
time
Algorithm MaxIS( S, p )If S = Φ then return Φ ;If I S p(I) 0 then return MaxIS( S - {I}, p);Let Î S that ends first;
p(Î) if I in conflict with ÎI S define: p1 (I) =
0 elseIS = MaxIS( S, p- p1 ) ;If IS is Î-maximal then return IS else return IS {Î};
P1=0
P1=0
P1=0
P1=0
P1=0
P1=P(Î )
P1=P(Î )
P1=P(Î )
P1=P(Î )
24
Single base station: Running Example
P)I1( = 5 -5
P)I4( = 9 -5 -4
P)I3( = 5 -5
P)I2( = 3 -5
P)I6( = 6 -4 -2
P)I5( = 3 -4
-5 -4 -2
25
Approximation for weak interferencesFA-Weak(R, p)
If R= return (, )Let be minimum in R
26
:Frequency allocation with multiple interference
1 2 1
2
ˆ1 ifˆDefine ( , , ) 2 elseif and define
2else
(1 )ˆ ˆˆ̂ ˆ ˆDefine { : ( , , ) 0}
( , ) ( , )ˆˆ̂return Augume
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n
e
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ak
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ijk
ijk ijk
i ip i j k d j j p p p
d wf
R ijk p i j k
S A R p
S A ijk
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