Channel Assignment in Cellular Networks

Ivan Stojmenovic

www.site.uottawa.ca/~ivan

Ivan@site.uottawa.ca

Overview

• Fixed channel assignment

• Multicoloring – co-channel interference

• General problem statement

• Genetic algorithms

• Results and details

• Fixed/dynamic channel and power assignment

Cell structure• Implements space division multiplex: base station

covers a certain transmission area (cell)

• Mobile users communicate only via the base station

• Advantages of cell structures:– higher capacity, higher number of users– less transmission power needed– more robust, decentralized– base station deals with interference locally

• Cell sizes from some 100 m in cities to, e.g., 35 km on the country side (GSM) - even more for higher frequencies

Cellular architecture

One low power transmitter per cell

Frequency reuse–limited spectrum

Cell splitting to increase capacityA

B

Reuse distance: minimum distance between two cells using same channel for satisfactory signal to noise ratio

Measured in # of cells in between

Problems– Propagation path loss for signal power: quadratic or higher in

distance – fixed network needed for the base stations– handover (changing from one cell to another) necessary– interference with other cells:

• Co-channel interference: Transmission on same frequency

• Adjacent channel interference:Transmission on close frequencies

Reuse pattern for reuse distance 2?

One frequency can be (re)used in all cells of the same color

Minimize number of frequencies=colors

Reuse distance 2 – reuse pattern

One frequency can be (re)used in all cells of the same color

Reuse pattern for reuse distance 3?

Reuse distance 3 – reuse pattern

Frequency planning I• Frequency reuse only with a certain

distance between the base stations• Standard model using 7 frequencies:

• Note pattern for repeating the same color: one north, two east-north

f4

f5

f1f3

f2

f6

f7

f3f2

f4

f5

f1

Fixed and Dynamic assignment

• Fixed frequency assignment: permanent– certain frequencies are assigned to a certain cell– problem: different traffic load in different cells

• Dynamic frequency assignment: temporary– base station chooses frequencies depending on the

frequencies already used in neighbor cells– more capacity in cells with more traffic– assignment can also be based on interference

measurements

3 cell clusterwith 3 sector antennas

f1f1 f1f2f3

f2

f3

f2

f3h1

h2

h3g1

g2

g3

h1h2

h3g1

g2g3

g1g2g3

Cell breathing• CDM systems: cell size depends on current load

• Additional traffic appears as noise to other users

• If the noise level is too high users drop out of cells

Multicoloring• Weight w(v) of cell v = # of requested frequencies

• Reuse distance r

• Minimize # channels used: NP hard problem

• Multi-coloring = multi-frequencing

• Channel= Frequency= ColorChannel= Frequency= Color

• HybridHybrid CA = combination fixed/dyn. frequencies

• Graph representation: weighted nodes, two nodes connected by edge iff their distance is < r

• same colors cannot be assigned to edge endpoints

Hexagon graphs: reuse distance 2

What is the graph for reuse distance 3?

Lower bounds for hexagonal graphsD= Maximum total weight on any cliqueLower bound on number of channels: D

D/3

D/2 D/6

D/2

D/2 D/2

D/2

D/2

D/2D/2D/2

D/2

000

Odd cycle bound: induced 9-cycle, each weight D/2

Channels needed in this cycle: 9D/2

Each channels can be used at most 4 times.

Needs 9/8D channels

Fixed allocations – reuse distance 2D= maximum number of channels in a node or 3-cycle

Red : 1, 4, 7, 10, …Green: 2, 5, 8, 11, … Blue: 3, 6, 9, 12, …

Total # channels: 3D Performance ratio: 3

Janssen, Kilakos, Marcotte ’95: D/2 red, blue and green each

D/2

D/2

D/2

Each node takes as many channels as needed from its own set

If necessary, RED borrow from GREEN BLUE borrow from RED GREEN borrow from BLUE

If a node has D/2+x channels, no neighbor has more than D/2-x channels

3D/2 channels used, performance ratio: 3/2

4/3 approximation for reuse distance 2• McDiarmid-Reed 97, Narayanan-Shende 97, Scabanel-Ubeda-Zerovnik 98

• Base color graph RED, GREEN, BLUE

• D/3 RED, GREEN, BLUE, PURPLE channels

• Each vertex uses at most D/3 channels from own set

• Certain ‘heavy’ vertices (>D/3 colors) borrow from ‘light’ neighbors: red from green, green from blue, blue from red

• Purple channels used if/when needed; at most one vertex in 3-cycle will need them (why?)

• If only one heavy vertex then how it borrows?

• max 2 nodes borrow (why?); G=D/3+x, B=D/3+y,

• green borrows from ?, blue from ?

• x+y<=D/3 (why?)

• In practice, reuse distances 3 or 4 may be used

Feder-Shende algorithm-reuse dist. 3

• Base color underlying graph with 7 colors

• Assign L channels to each color class

• Every node takes as many channels as it needs from its base color set

• Heavy node (>L colors) borrows any unused channels from its neighbors

• L=D/3 algorithm with performance ratio 7/3

• Reuse distance r perform. ratio 18r2/(3r2+20)

• 2: 2.25, 3: 3.44, 4: 4.23, 5: 4.73 (Narayanan)• k-colorable graph perf. ratio k/2 (Janssen-Kilakos 95)

Adjacent channel interferenceReceiver filter

f1 f3f2interference

Co-site constraint: channels in the same cell must be c0 apart

Adjacent-site constraint: channels assigned to neighboring cells must be c1 apart

Inter-site constraint: channels assigned to cells that are r cells apart must be cr apart

Lower bounds: co-site and adjacent-site

Gamst ’86

c0 max {w(u), w(v), w(x)}

c1 max{vC w(v) | C is a clique}

max {c0 w(u), (c0–c1)w(u)+ c1vC,vu w(v) | C is a clique containing u} when c0 2c1

u

v x

c0c1c0<2c1

Algorithm: interleaving channels of different color classes

3-colorable graphsDistance between channels = max(c0/3, c1)

Borrowing impossible

Distance between channels = max(c0/2, c1)

Borrowing possible

Borrowed channels = change colordynamic CA=online distributed CA

Channels with ongoing calls can(not) be borrowed = (non)recoloring

k-local algorithm: node changes channels based on weights within k cells

Desirable qualities of CA algorithms

• Minimize connection set-up time

• Conserve energy at mobile host

• Adapt to changing load distribution

• Fault tolerance

• Scalability

• Low computation and communication overhead

• Minimize handoffs

• Maximize number of calls that can be accepted concurrently

Research problem: several power levels at mobile hosts

• If mobile phone is ‘near’ base station, it may switch to lower power level

• Interference from other hosts increases

• Interference of that host to other node decreases

• Are there benefits of using two power levels?

• Fixed or dynamic channel and power assignment and multicoloring: simplest cases

• Fixed or dynamic channel and power assignment with co-site, adjacent-site and inter-site constraints: Genetic algorithms, simulated annealing, …

Genetic algorithms• Rechenberg 1960, Holland 1975 …

• Part of evolutionary computing in AI

• Solution to a problem is evolved (Darwin’s theory)

• Represent solutions as a chromosomes = search space

• Generate initial population of solutions (‘chromosomes’) at random or from other method

• REPEAT

• Evaluate the fitness f(x) of each chromosome x

• Perform crossover, mutation and generate new population, using f(x) in selecting probabilities

• UNTIL satisfactory solution found or timeout

Fixed channel assignment problem• INPUT: n = number of cells

Compatibility matrix C, C[i,j]= minimal channel separation between cells i and j, 1i,jnd[i] = number of channels demanded by cell i

• OUTPUT: S[i,k] = channel # of k-th call of cell i, 1kd[i]

• CONSTRAINTS: |S[i,k]-S[j,L]|C[i,j],1kd[i], 1Ld[j], (i,k)(j,L)

• GOAL: minimize m= max S[i,k] = # channels

• reducable to graph coloring problem NP-complete• GA solution space: m fixed, F[j,k]=0/1 if channel k

is not assigned/assigned to cell j, 1km, 1jn.

• Optimization: Minimize number of interferences and satisfy demand

Our problem representation and solution space

• Each row F[j,k], 1km, is a combination of d[j] out of m elements (# of 1’s is = d[j])

• Cost function to minimize: C(F)= A+B

• A= total number of co-site constraint violations

• B= total number of adjacent and inter-site violations= parameter; C(F)=0 for optimal solution

• Initial population: generate restricted combinations:

• generate random combination of d[j] X’s and m-(c0+1)d[j] 0’s; replace each X by 100..0 (c0 0’s); shift circularly by random number in [0,c0]

Mutation• Each row=cell is mutated separately• Combinations in bit representation: x 1’s out of m bits• Mutation with equal probability for each bit: choose one out

of x 1’s and one out of m-x 0’s at random, swap: Ngo-Li ‘98

• Mutation with different probability for each bit: b[i]= # of conflicts of i-th selected channel with other channels in this and other cellsp[i]=b[i]/(b[1]+…+b[x])Repeat for 0’s: # of conflicts if that channel turned on

• Choosing bit with given probability: Generate at random r, 0 r 1, and choose i, p[1]+…p[i-1] r <p[1]+…+p[i]

Crossover• Regular GA crossover:

1011000110 1001111000 0101111000 0111000110

• Ngo-Li ’98: A and B two parents, each row separately, preserve # of 1’s in each row: push 10 and 01 columns in stack if top same;

pop for exchange if top different1011000110 1001101000 0101111000 0111010110

• Problem: # of swaps varies

New crossover• t= number of desired swaps in a row

• Mark positions in two combinations that differ

• let s 10’s and s 01’s are found

• Choose t out of s 10 at random and 01

• Choose t out of s 01 at random and 10

• Example: 1011000110 1001010010 0101111000 0111110110

s=4 t=2 $^$ ^^^$$ # **# # **# offspring

selected columns

Crossover needs further study• Problem: independent changes in each row=cell will

destroy good channel assignments of parents

• Two good solutions may have nothing in common• Try experiments with mutation only

(may be crossover has even negative impact !?)

• Evaluate impact of each column change by cost function and apply weighted probabilities for column selections

• Best value for t as function of s? t=s/2? Small t?

Combinatorial evolution strategy• Sandalidis, Stavroulakis and Rodriguez-Tellez ’98

• Generate individuals and evaluate them by f• Select best individual indiv; indiv1=indiv; counter=0; t=0;

• REPEAT t=t+1• IF counter=max-count THEN apply increased mutation rate

(destabilize to escape local minimum)

• Generate individuals from indiv1 and evaluate them by f

• Select best individual indiv2

• IF indiv2 better than indiv1 THEN {counter=0; indiv=indiv2} ELSE {counter=counter+1; indiv1=indiv2}

• UNTIL termination

• Applied for fixed, dynamic and hybrid CA

CES for dynamic channel assignment• n=49 cells, m=49 channels, call arrives at cell k• F[j,i]=0/1 if channel i is not assigned/assigned to

cell j, 1im, 1jn: current channel assignment for ongoing calls

• Reassignment of all ongoing calls at cell k (channel for each call may change) to accommodate new call

• V[k,i] = new channel assignment for cell k• CES minimizes energy function that includes: interference of new

assignment, reusing channels used in nearby cells, reusing channels according to base coloring scheme, and number of reassignments

• Centralized controller

• CES for Hybrid CA and for borrowing CA in FCA

Simple heuristics for FCA• Borndorfer, Eisenblatter, Grotschel, Martin ’98

(4240 total demand, m=75 channels, Germany)

• DSATUR: key[i]= # acceptable channels remained in cell i, cost[i,j]= total interference in cell i if channel j is selected

• Initialize key[i]= m; cost[i,j]=0; i,j

• WHILE cells with unsatisfied demand exist DO {

• Extract cell i with unsatisfied demand and minimum key[i];

• Let j be available channel which minimizes cost[i,j];

• Update cost[x,y] x,y by adding interference (i,j)

• Update key[x] x, reduce demand at cell i }

Hill climbing heuristic for FCA• Borndorfer, Eisenblatter, Grotschel, Martin ’98

• Two channel assignments are neighbors if one can be obtained from the other by replacing one channel by another in one of cells.

• PASS procedure for assignment A={(cell,channel)}:• Sort all (i,j)A by their interference in decreasing order

• FOR each (i,j)A in the order DO• Replace (i,j) by (i,j’) if later has same or lower interference

• Hill climbing for FCA: initialize A; A’=A

• REPEAT

• A=A’; A’= PASS(A)

• UNTIL A’=A or interference(A’)interference(A)