1. Channel Assignment in Cellular Networks Ivan Stojmenovic www.site.uottawa.ca/~ivan Ivan@site.uottawa.ca

2. Overview Fixed channel assignment Multicoloring – co-channel interference General problem statement Genetic algorithms Results and details Fixed/dynamic channel and power assignment

3. 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

4. Cellular architecture One low power transmitter per cell Frequency reuse–limited spectrum Cell splitting to increase capacity A B Reuse distance: minimum distance between two cells using same channel for satisfactory signal to noise ratio Measured in # of cells in between

5. 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

6. Reuse pattern for reuse distance 2? One frequency can be (re)used in all cells of the same color Minimize number of frequencies=colors

7. Reuse distance 2 – reuse pattern One frequency can be (re)used in all cells of the same color

8. Reuse pattern for reuse distance 3?

9. Reuse distance 3 – reuse pattern

10. 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 f4f4 f5f5 f1f1 f3f3 f2f2 f6f6 f7f7 f3f3 f2f2 f4f4 f5f5 f1f1

11. 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

12. 3 cell cluster with 3 sector antennas f1f1 f1f1 f1f1 f2f2 f3f3 f2f2 f3f3 f2f2 f3f3 h1h1 h2h2 h3h3 g1g1 g2g2 g3g3 h1h1 h2h2 h3h3 g1g1 g2g2 g3g3 g1g1 g2g2 g3g3

13. 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

14. 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

15. Hexagon graphs: reuse distance 2 What is the graph for reuse distance 3?

16. Lower bounds for hexagonal graphs D= Maximum total weight on any clique Lower bound on number of channels: D D/3 D/2D/6 D/2 0 00 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

17. Fixed allocations – reuse distance 2 D= 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: 3DPerformance ratio: 3 Janssen, Kilakos, Marcotte ’95: D/2 red, blue and green each 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

18. 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

19. 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 18r 2 /(3r 2 +20) 2: 2.25, 3: 3.44, 4: 4.23, 5: 4.73 (Narayanan) k-colorable graph  perf. ratio k/2 (Janssen-Kilakos 95)

20. Adjacent channel interference Receiver filter f1 f3f2 interference Co-site constraint: channels in the same cell must be  c 0 apart Adjacent-site constraint: channels assigned to neighboring cells must be  c 1 apart Inter-site constraint: channels assigned to cells that are r cells apart must be  c r apart

21. Lower bounds: co-site and adjacent-site Gamst ’86 c 0 max {w(u), w(v), w(x)} c 1 max{  v  C w(v) | C is a clique} max {c 0 w(u), (c 0 –c 1 )w(u)+ c 1  v  C,v  u w(v) | C is a clique containing u} when c 0  2c 1 u v x c0c0 c1c1 c 0 <2c 1 Algorithm: interleaving channels of different color classes

22. 3-colorable graphs Distance between channels = max(c 0 /3, c 1 ) Borrowing impossible Distance between channels = max(c 0 /2, c 1 ) Borrowing possible Borrowed channels = change color  dynamic 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

23. 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

24. 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, …

25. 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

26. Fixed channel assignment problem INPUT: n = number of cells Compatibility matrix C, C[i,j]= minimal channel separation between cells i and j, 1  i,j  n d[i] = number of channels demanded by cell i OUTPUT: S[i,k] = channel # of k-th call of cell i, 1  k  d[i] CONSTRAINTS: |S[i,k]-S[j,L]|  C[i,j],1  k  d[i], 1  L  d[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, 1  k  m, 1  j  n. Optimization: Minimize number of interferences and satisfy demand

27. Our problem representation and solution space Each row F[j,k], 1  k  m, 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- (c 0 +1)d[j] 0’s; replace each X by 100..0 (c 0 0’s); shift circularly  by random number in [0,c 0 ]

28. 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 cells p[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]

29. 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 different 1011000110  1001101000 0101111000  0111010110 Problem: # of swaps varies

30. 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

31. 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?

32. 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

33. 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, 1  i  m, 1  j  n: 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

34. 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 }

35. 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)