JOINT SCHEDULING AND FAST Reuven Cohen CELL SELECTION IN Guy Grebla OFDMA WIRELESS NETWORKS Slides: Moshe Gabel MOSHE GABEL 1
MODERN CELLULAR NETWORKS One base station per cell Multiple users in cell A1 A2 Divide cell to 3 or 6 sectors, A3 (one antenna per sectors) Antennas run by same BS MOSHE GABEL 2
GOAL: TRANSMIT PACKETS TO USERS Choose antenna for each packet. But which antenna? A1 Best signal to the user? What if overloaded? A2 A3 How to transmit? Frequencies, sub-bands, slots. Various schemes available. MOSHE GABEL 3
WHICH? BEST SINR MAY NOT BE BEST CHOICE Antenna A1 Antenna A2 A1 Load: High Low A2 A3 User Excellent OK SINR: MOSHE GABEL 4
HOW? FRACTIONAL FREQUENCY REUSE MODEL 3 sectors 4 sub-bands per cell 𝐺1 𝐺0 used in all 3 sectors 𝐺0 1 𝐺1 used in sector 1 only 𝐺 2 used in sector 2 only 𝐺0 2 𝐺3 𝐺 2 used in sector 3 only 𝐺0 3 𝐺2 Also: allow various modulation and coding schemes (MCS) MOSHE GABEL 5
OFDMA SCHEDULING AREAS AND BLOCKS 1ms sub-frame 𝐺0 1 Limited 𝐺0 blocks 𝐺0 2 blocks per 𝐺0 3 sub-frame 𝐺 1 blocks Must 𝐺 2 blocks prioritize packets. 𝐺 3 blocks MOSHE GABEL 6
SCHEDULING AREAS AND BLOCKS 1ms subframe this packet takes 𝐺0 1 3 blocks in 𝐺0 2 , 𝐺0 blocks 𝐺0 2 i.e. antenna 2 𝐺0 3 packet of length 𝐺 1 blocks 2 blocks, in 𝐺 1, (antenna 1) 𝐺 2 blocks Antenna 2 will transmit 4 𝐺 3 blocks packets in sub- band 𝐺 2 MOSHE GABEL 7
MCS: MODULATION AND CODING SCHEMES Each packet can be transmitted in different ways MCS + packet length # needed blocks (“cost”) success probability MCS + SINR (“gain”) Modulation Code rate Bits/symbol Success probability at SNR=5dB 64-QAM 2/3 4 0.9 16-QAM 3/4 3 0.95 QPSK 1/2 1 0.999 … … … … MOSHE GABEL 8
OVERALL GOAL (REVISED) Given: Packets with lengths Scheduling areas with capacities A1 MCS options A2 Success probabilities A3 Goal: select MCS and areas to schedule packets. MOSHE GABEL 9
QUANTIFY: ASSIGN PROFIT, SIZE TO PACKETS Profit Size Number of scheduled blocks A good scheduler optimizes needed to transmit. throughput, fairness, QoS, revenue, etc. Depends on packet length and selected MCS. Combine to one metric. Depends on scheduling area! Multiply by success probability (expected profit) Start easy! Depends on scheduling area! Assume MCS of each packet. MOSHE GABEL 10
SELECTING DEFAULT MCS SCHEME Assume each packet has minimum needed success probability choose most efficient MCS that obtains it. Example: packet with min probability = 0.95 Modulation Code rate Bits per Symbol # Required Blocks Pr[success] 64-QAM 2/3 4 1 0.9 16-QAM 3/4 3 2 0.95 16-QAM 1/2 2 2 0.98 QPSK 1/2 1 4 0.999 MOSHE GABEL 12
P1: OFDMA JOINT SCHEDULING Input Output Scheduling areas Choose area per packet ( 𝐺0 1 , 𝐺0 2 , 𝐺0 3 , 𝐺1 , 𝐺2 , 𝐺3 ) Maximizing profit, subject to: with capacities One or zero areas per packet Packets to transmit, with feasible areas (SINR > 1) Do not exceed area capacity Size and profit for each Avoid interference in 𝐺0 𝑘 area packet in each area Met by requiring SINR > 1 MCS pre-selected! MOSHE GABEL 13
GENERALIZED ASSIGNMENT PROBLEM 1 2 3 𝑜 items 𝐶 = (2,3,4) 3 1 5 1 𝑛 bins with capacity 𝐶 𝑘 1 1 1 2 𝑞 = 𝑞 𝑗𝑘 = profit of item 𝑗 if assigned to bin 𝑘 5 15 25 3 25 15 5 𝑡 𝑗𝑘 = size of item 𝑗 if assigned to bin 𝑘 4 1 2 3 1 1 1 1 Goal : assign items to bins 2 3 3 2 Maximize profit 𝑡 = 2 3 4 3 Do not exceed capacity 1 2 3 4 MOSHE GABEL 14
GAP FORMALIZATION Maximize 𝑜 𝑛 𝑞 𝑗𝑘 𝑦 𝑗𝑘 Maximize profit 𝑗=1 𝑘=1 Subject to 𝑜 𝑡 𝑗𝑘 𝑦 𝑗𝑘 ≤ 𝐶 𝑘 for 1 ≤ 𝑘 ≤ 𝑛 within capacity 𝑗=1 𝑛 𝑦 𝑗𝑘 ≤ 1 for 1 ≤ 𝑗 ≤ 𝑜 each item in one 𝑘=1 bin or none 𝑦 𝑗𝑘 ∈ 0,1 for all 𝑗, 𝑑, 𝑘 MOSHE GABEL 15
IF WE ONLY HAD ONE BIN… One bin with capacity B. S=1, P=2 S=2, P=5 Maximize total profit such that total size < B. S=2, P=1 S=3, P=5 Does this remind you of anything? S=1, P=4 S=4, P=5 0-1 knapsack problem. S=2, P=3 Which is NP-hard MOSHE GABEL 16
SOLVING KNAPSACK Optimal solution in pseudo-polynomial time Using dynamic programming. Requires positive integer weights. Simple greedy algorithm for 2-approximation Sort by profit / size, insert from highest to lowest. (1+ ε )-approximation in polynomial time (FPTAS) By rounding profits and using dynamic programming. MOSHE GABEL 17
REMINDER Given value of the optimal solution OPT and 𝛽 > 1 , an 𝛃 -approximation returns a solution 𝑄 that is 𝑄 ≥ OPT 𝛽 (alternatively: OPT ≤ 𝛽𝑄 ) MOSHE GABEL 18
SOLVING GAP: SKETCH 1. = an 𝛽 -approximation algorithm for knapsack 2. Iterate over 𝑛 bins: a. In each iteration, separate profit to 𝑞 = 𝑞 obtained + 𝑞 residual b. Apply over 𝑞 residual and update assignments 3. Result is combination of bin assignments Local Ratio Theorem shows: 𝛽 -approximation to knapsack (𝛽 + 1) -approximation to GAP R. Cohen, L. Katzir, D. Raz. An Efficient Approximation for the Generalized Assignment Problem . Information Processing Letters 100.4 (2006): 162-166. MOSHE GABEL 19
LOCAL RATIO THEOREM If 𝐺 set of constraints 𝑞(), 𝑞 1 (), 𝑞 2 () : profit functions and 𝑞() = 𝑞 1 () + 𝑞 2 () 𝑦 is 𝛽 -approximation to (𝐺, 𝑞 1 ) and (𝐺, 𝑞 2 ) Then 𝑦 is 𝛽 -approximation to (𝐺, 𝑞) R. Bar-Yehuda and S. Even. A local-ratio theorem for approximating the weighted vertex cover problem . Annals of Discrete Mathematics, 25:27 – 45, 1985 MOSHE GABEL 20
LOCAL RATIO THEOREM PROOF ∗ be optimal solutions to 𝐺, 𝑞 , 𝐺, 𝑞 1 , 𝐺, 𝑞 1 . ∗ , 𝑦 2 Let 𝑦 ∗ , 𝑦 1 Then: 𝑞 𝑦 = 𝑞 1 𝑦 + 𝑞 2 𝑦 ∗ + 𝛽 ⋅ 𝑞 2 𝑦 2 ∗ ≥ 𝛽 ⋅ 𝑞 1 𝑦 1 ∗ + 𝑞 2 𝑦 2 ≥ 𝛽 𝑞 1 𝑦 1 ∗ ≥ 𝛽 𝑞 1 𝑦 ∗ + 𝑞 2 𝑦 ∗ = 𝛽 ⋅ 𝑞 𝑦 ∗ ∎ R. Cohen, L. Katzir, D. Raz. An Efficient Approximation for the Generalized Assignment Problem . Information Processing Letters 100.4 (2006): 162-166. MOSHE GABEL 21
GAP ALGORITHM R. Cohen, L. Katzir, D. Raz. An Efficient Approximation for the Generalized Assignment Problem . Information Processing Letters 100.4 (2006): 162-166. NextBin( 𝒌 , 𝒒 𝒌 ) 1. Use to solve bin 𝑘 with profit 𝑞 𝑘 : 𝑇 𝑘 ← 𝑘, 𝑞 𝑘 2. If 𝑘 = 𝑛 (last bin) return assignment: 𝑇 𝑘 ← 𝑇 𝑘 . 𝐵 + 𝑞 𝑘 𝐶 : 3. Decompose profit function: 𝑞 𝑘 = 𝑞 𝑘 𝐵 𝑗, 𝑙 ← 𝑞 𝑘 𝑗, 𝑙 if 𝑗 ∈ 𝑇 𝑘 or 𝑙 = 𝑘 𝐶 ← 𝑞 𝑘 − 𝑞 𝑘 𝐵 Set 𝑞 𝑘 , set 𝑞 𝑘 0 otherwise 𝐶 , but without all-zero column of bin 𝑘 4. Set 𝑞 𝑘+1 ← 𝑞 𝑘 5. Recursive call: 𝑇 𝑘+1 ← NextBin( 𝑘 +1, 𝑞 𝑘+1 ) 6. Return 𝑇 𝑘 ← 𝑇 𝑇 𝑘 not assigned in 𝑇 𝑘+1 … 𝑇 𝑛 𝑘+1 plus items in MOSHE GABEL 22
EXAMPLE 1 2 3 3 1 5 1 1 1 1 2 𝑞 = 5 15 25 3 25 15 5 4 1 2 3 1 1 1 1 2 3 3 2 𝑡 = 2 3 4 3 𝐶 = (2,3,4) 1 2 3 4 R. Cohen, L. Katzir, D. Raz. An Efficient Approximation for the Generalized Assignment Problem . Information Processing Letters 100.4 (2006): 162-166. MOSHE GABEL 24
EXAMPLE: ITERATION 1 1 2 3 𝑇 1 = 𝑗 1 , 𝑗 4 3 1 5 1 𝐶 𝐵 𝑞 1 𝑞 1 1 1 1 2 𝑞 1 = 1 2 3 1 2 3 5 15 25 3 3 3 3 0 -2 2 1 1 25 15 5 4 1 0 0 0 1 1 2 2 𝑞 1 = 𝒒 𝟑 1 2 3 5 0 0 0 15 25 3 3 1 1 1 1 25 25 25 0 -10 20 4 4 2 3 3 2 𝑡 = profit gained 2 3 4 3 residual profit 𝑇 1 from 𝐶 = (2,3,4) 1 2 3 4 R. Cohen, L. Katzir, D. Raz. An Efficient Approximation for the Generalized Assignment Problem . Information Processing Letters 100.4 (2006): 162-166. MOSHE GABEL 25
EXAMPLE: ITERATION 2 2 3 𝑇 2 = 𝑗 3 -2 2 1 𝐶 𝐵 𝑞 2 𝑞 2 1 1 2 𝑞 2 = 2 3 2 3 15 25 3 -2 0 0 2 1 1 -10 -20 4 1 0 0 1 2 2 𝑞 2 = 𝒒 𝟑 2 3 15 15 0 10 3 3 1 1 1 -10 0 0 -20 4 4 3 3 2 𝑡 = profit gained 3 4 3 residual profit from 𝑇 2 𝐶 = (2,3,4) 2 3 4 R. Cohen, L. Katzir, D. Raz. An Efficient Approximation for the Generalized Assignment Problem . Information Processing Letters 100.4 (2006): 162-166. MOSHE GABEL 26
EXAMPLE: ITERATION 3 3 𝑇 3 = 𝑗 3 2 1 𝐶 𝐵 𝑞 3 𝑞 3 1 2 𝑞 3 = 3 3 10 3 -2 0 1 1 -20 4 1 0 2 2 𝑞 3 = 3 15 0 3 3 1 1 -10 0 4 4 3 2 𝑡 = profit gained 4 3 residual profit from 𝑇 3 𝐶 = (2,3,4) 3 4 R. Cohen, L. Katzir, D. Raz. An Efficient Approximation for the Generalized Assignment Problem . Information Processing Letters 100.4 (2006): 162-166. MOSHE GABEL 27
EXAMPLE OUTPUT 1 2 3 Final assignment: 3 1 5 1 Bin 1 = {𝑗 1 , 𝑗 4 } 1 1 1 2 𝑞 = 5 15 25 3 Bin 2 = {} 25 15 5 4 Bin 3 = 𝑗 3 1 2 3 1 1 1 1 Notes: 2 3 3 2 𝑡 = i 3 ∈ 𝑇 3 overrides earlier i 3 ∈ 𝑇 2 2 3 4 3 Result not optimal 𝐶 = (2,3,4) 1 2 3 4 R. Cohen, L. Katzir, D. Raz. An Efficient Approximation for the Generalized Assignment Problem . Information Processing Letters 100.4 (2006): 162-166. MOSHE GABEL 28
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