CELL SELECTION IN Guy Grebla OFDMA WIRELESS NETWORKS Slides: Moshe - - PowerPoint PPT Presentation

JOINT SCHEDULING AND FAST CELL SELECTION IN OFDMA WIRELESS NETWORKS

Reuven Cohen Guy Grebla

Slides: Moshe Gabel

MOSHE GABEL 1

MODERN CELLULAR NETWORKS

One base station per cell Multiple users in cell Divide cell to 3 or 6 sectors, (one antenna per sectors) Antennas run by same BS

MOSHE GABEL

A1 A2 A3

2

GOAL: TRANSMIT PACKETS TO USERS

Choose antenna for each packet. But which antenna?

 Best signal to the user?  What if overloaded?

How to transmit?

 Frequencies, sub-bands, slots.  Various schemes available.

MOSHE GABEL

A1 A2 A3

3

WHICH? BEST SINR MAY NOT BE BEST CHOICE

Antenna A1 Antenna A2 Load: High Low User SINR: Excellent OK

MOSHE GABEL 4

A1 A2 A3

HOW? FRACTIONAL FREQUENCY REUSE MODEL

3 sectors  4 sub-bands per cell 𝐺0 used in all 3 sectors 𝐺1 used in sector 1 only 𝐺2 used in sector 2 only 𝐺2 used in sector 3 only Also: allow various modulation and coding schemes (MCS)

MOSHE GABEL 5

𝐺01 𝐺02 𝐺03 𝐺1 𝐺2 𝐺3

OFDMA SCHEDULING AREAS AND BLOCKS

𝐺01 𝐺02 𝐺03

MOSHE GABEL 6

𝐺0 blocks 𝐺1 blocks 𝐺2 blocks 𝐺3 blocks 1ms sub-frame

Limited blocks per sub-frame Must prioritize packets.

SCHEDULING AREAS AND BLOCKS

𝐺01 𝐺02 𝐺03

MOSHE GABEL 7

𝐺0 blocks 𝐺1 blocks 𝐺2 blocks 𝐺3 blocks 1ms subframe this packet takes 3 blocks in 𝐺02, i.e. antenna 2 packet of length 2 blocks, in 𝐺1, (antenna 1) Antenna 2 will transmit 4 packets in sub- band 𝐺2

MCS: MODULATION AND CODING SCHEMES

Each packet can be transmitted in different ways MCS + packet length  # needed blocks (“cost”) MCS + SINR  success probability (“gain”)

MOSHE GABEL 8

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 … … … …

OVERALL GOAL (REVISED)

Given: Packets with lengths Scheduling areas with capacities MCS options Success probabilities Goal: select MCS and areas to schedule packets.

MOSHE GABEL 9

A1 A2 A3

QUANTIFY: ASSIGN PROFIT, SIZE TO PACKETS

Profit A good scheduler optimizes throughput, fairness, QoS, revenue, etc. Combine to one metric. Multiply by success probability (expected profit) Depends on scheduling area! Size Number of scheduled blocks needed to transmit. Depends on packet length and selected MCS. Depends on scheduling area! Start easy! 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

MOSHE GABEL 12

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

P1: OFDMA JOINT SCHEDULING

Input Scheduling areas (𝐺01, 𝐺02, 𝐺03, 𝐺1, 𝐺2, 𝐺3) with capacities Packets to transmit, with feasible areas (SINR > 1) Size and profit for each packet in each area

 MCS pre-selected!

Output Choose area per packet Maximizing profit, subject to: One or zero areas per packet Do not exceed area capacity Avoid interference in 𝐺0𝑘 area

 Met by requiring SINR > 1

MOSHE GABEL 13

GENERALIZED ASSIGNMENT PROBLEM

𝑜 items 𝑛 bins with capacity 𝐶

𝑘

𝑞𝑗𝑘 = profit of item 𝑗 if assigned to bin 𝑘 𝑡𝑗𝑘 = size of item 𝑗 if assigned to bin 𝑘 Goal: assign items to bins Maximize profit Do not exceed capacity

MOSHE GABEL 14

1 2 3 1

3 1 5

2

1 1 1

3

5 15 25

4

25 15 5

1 2 3 1

1 1 1

2

2 3 3

3

2 3 4

4

1 2 3

𝑞 = 𝑡 =

𝐶 =(2,3,4)

GAP FORMALIZATION

Maximize

𝑗=1 𝑜 𝑘=1 𝑛

𝑞𝑗𝑘𝑦𝑗𝑘 Subject to

𝑗=1 𝑜

𝑡𝑗𝑘𝑦𝑗𝑘 ≤ 𝐶

𝑘 for 1 ≤ 𝑘 ≤ 𝑛 𝑘=1 𝑛

𝑦𝑗𝑘 ≤ 1 for 1 ≤ 𝑗 ≤ 𝑜 𝑦𝑗𝑘 ∈ 0,1 for all 𝑗, 𝑑, 𝑘

MOSHE GABEL 15

Maximize profit within capacity each item in one bin or none

IF WE ONLY HAD ONE BIN…

One bin with capacity B. Maximize total profit such that total size < B. Does this remind you of anything? 0-1 knapsack problem.

 Which is NP-hard

MOSHE GABEL 16

S=2, P=5 S=3, P=5 S=1, P=4 S=4, P=5 S=2, P=1 S=2, P=3 S=1, P=2

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 = 𝑞obtained + 𝑞residual
• b. Apply 𝒝 over 𝑞residual and update assignments
• 3. Result is combination of bin assignments

Local Ratio Theorem shows: 𝛽-approximation to knapsack  (𝛽 + 1)-approximation to GAP

MOSHE GABEL 19

• R. Cohen, L. Katzir, D. Raz. An Efficient Approximation for the Generalized

Assignment Problem. Information Processing Letters 100.4 (2006): 162-166.

LOCAL RATIO THEOREM

If 𝐺 set of constraints 𝑞(), 𝑞1(), 𝑞2() : profit functions and 𝑞() = 𝑞1() + 𝑞2() 𝑦 is 𝛽-approximation to (𝐺, 𝑞1) and (𝐺, 𝑞2) Then 𝑦 is 𝛽-approximation to (𝐺, 𝑞)

MOSHE GABEL 20

• 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

LOCAL RATIO THEOREM PROOF

Let 𝑦∗, 𝑦1

∗, 𝑦2 ∗ be optimal solutions to 𝐺, 𝑞 , 𝐺, 𝑞1 , 𝐺, 𝑞1 .

Then: 𝑞 𝑦 = 𝑞1 𝑦 + 𝑞2 𝑦 ≥ 𝛽 ⋅ 𝑞1 𝑦1

∗ + 𝛽 ⋅ 𝑞2 𝑦2 ∗

≥ 𝛽 𝑞1 𝑦1

∗ + 𝑞2 𝑦2

∗

≥ 𝛽 𝑞1 𝑦∗ + 𝑞2 𝑦∗ = 𝛽 ⋅ 𝑞 𝑦∗ ∎

MOSHE GABEL 21

• R. Cohen, L. Katzir, D. Raz. An Efficient Approximation for the Generalized

Assignment Problem. Information Processing Letters 100.4 (2006): 162-166.

GAP ALGORITHM

NextBin( 𝒌, 𝒒𝒌 )

• 1. Use 𝒝 to solve bin 𝑘 with profit 𝑞𝑘 :

𝑇

𝑘 ← 𝒝 𝑘, 𝑞𝑘

• 2. If 𝑘 = 𝑛 (last bin) return assignment: 𝑇

𝑘 ←

𝑇

𝑘.

• 3. Decompose profit function: 𝑞𝑘 = 𝑞𝑘

𝐵 + 𝑞𝑘 𝐶:

Set 𝑞𝑘

𝐵 𝑗, 𝑙 ← 𝑞𝑘 𝑗, 𝑙

if 𝑗 ∈ 𝑇

𝑘 or 𝑙 = 𝑘

• therwise

, set 𝑞𝑘

𝐶 ← 𝑞𝑘 − 𝑞𝑘 𝐵

• 4. Set 𝑞𝑘+1 ← 𝑞𝑘

𝐶 , but without all-zero column of bin 𝑘

• 5. Recursive call: 𝑇

𝑘+1 ←NextBin( 𝑘+1, 𝑞𝑘+1 )

• 6. Return 𝑇

𝑘 ← 𝑇 𝑘+1 plus items in

𝑇

𝑘 not assigned in 𝑇 𝑘+1 … 𝑇𝑛

MOSHE GABEL 22

• R. Cohen, L. Katzir, D. Raz. An Efficient Approximation

for the Generalized Assignment Problem. Information Processing Letters 100.4 (2006): 162-166.

EXAMPLE

MOSHE GABEL 24

1 2 3 1

3 1 5

2

1 1 1

3

5 15 25

4

25 15 5

1 2 3 1

1 1 1

2

2 3 3

3

2 3 4

4

1 2 3

𝑞 = 𝑡 =

𝐶 =(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.

EXAMPLE: ITERATION 1

MOSHE GABEL 25

1 2 3 1

3 1 5

2

1 1 1

3

5 15 25

4

25 15 5

1 2 3 1

1 1 1

2

2 3 3

3

2 3 4

4

1 2 3

𝑞1 = 𝑡 =

𝐶 =(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.

𝑇1 = 𝑗1, 𝑗4

1 2 3 1

3 3 3

2

1

3

5

4

25 25 25

1 2 3 1

• 2

2

2

1 1

3

15 25

4

0 -10 20

𝑞1 = 𝑞1

𝐵

𝑞1

𝐶

profit gained from 𝑇1 residual profit

𝒒𝟑

EXAMPLE: ITERATION 2

MOSHE GABEL 26

2 3 1

• 2

2

2

1 1

3

15 25

4

• 10 -20

2 3 1

1 1

2

3 3

3

3 4

4

2 3

𝑞2 = 𝑡 =

𝐶 =(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.

𝑇2 = 𝑗3

2 3 1

• 2

2

1

3

15 15

4

• 10 0

2 3 1

2

2

1

3

10

4

0 -20

𝑞2 = 𝑞2

𝐵

𝑞2

𝐶

profit gained from 𝑇2 residual profit

𝒒𝟑

EXAMPLE: ITERATION 3

MOSHE GABEL 27

3 1

2

2

1

3

10

4

• 20

3 1

1

2

3

3

4

4

3

𝑞3 = 𝑡 =

𝐶 =(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.

𝑇3 = 𝑗3

3 1

• 2

2

1

3

15

4

• 10

3 1 2 3 4

𝑞3 = 𝑞3

𝐵

𝑞3

𝐶

profit gained from 𝑇3 residual profit

EXAMPLE OUTPUT

Final assignment: Bin 1 = {𝑗1, 𝑗4} Bin 2 = {} Bin 3 = 𝑗3 Notes: i3 ∈ 𝑇3 overrides earlier i3 ∈ 𝑇2 Result not optimal

MOSHE GABEL 28

1 2 3 1

3 1 5

2

1 1 1

3

5 15 25

4

25 15 5

1 2 3 1

1 1 1

2

2 3 3

3

2 3 4

4

1 2 3

𝑞 = 𝑡 =

𝐶 =(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.

PROOF SKETCH

Solution is feasible, but how good is it? Using induction and Local Ratio we can show: 𝑇

𝑘 is a 𝛽 + 1 -approximation to 𝑞𝑘 𝐵 and 𝑞𝑘 𝐶 and therefore 𝑞𝑘

MOSHE GABEL 29

𝑄 𝑄

1

𝑄

1 A

𝑄

1 𝐶

𝑄2 𝑄2

𝐵

𝑄2

𝐶

𝑄3 𝑄3

𝐵

𝑄3

𝐶

induction

INDUCTION BASE

Base is on the final recursive call, 𝑘 = 𝑛 Only one bin, one column in 𝑞𝑛. 𝒝 is 𝛽-approximation so 𝑇𝑛 is 𝛽-approximation  𝛽-approximation is also 𝛽 + 1 -approximation  𝑇𝑛 = 𝑇𝑛 Induction step will go backwards: from 𝑘 + 1 to 𝑘

MOSHE GABEL 30

INDUCTIVE STEP (1)

Assume 𝑇

𝑘+1 is 𝛽 + 1 -approximation to 𝑞𝑘+1, show for 𝑇 𝑘, 𝑞𝑘

Assignment 𝑇

𝑘 contains items from 𝑇 𝑘+1 plus items from

𝑇

𝑘

𝒒𝒌

𝑪 is identical to 𝒒𝒌+𝟐 except all-zero column for bin 𝑘.

 𝑇

𝑘+1 is 𝛽 + 1 -approximation to 𝑞𝑘 𝐶.

 𝑇

𝑘 is 𝛽 + 1 -approximation to 𝑞𝑘 𝐶

What about 𝑞𝑘

𝐵?

MOSHE GABEL 31

INDUCTIVE STEP (2)

𝑞𝑘

𝐵 has 3 components:

• 1. Column for bin 𝑘

 Identical to column in 𝑞𝑘

• 2. Rows for items selected in

𝑇

𝑘

 Identical to profit in column j 𝑞𝑘

• 3. Everything else is zeros.

 Cannot contribute to profit

MOSHE GABEL 32

1 2 3 1

3 3 3

2

1

3

5

4

25 25 25

𝑞1

𝐵

INDUCTIVE STEP (3)

𝒝 is 𝛽-approximation  𝑃𝑄𝑈

1 for component (1) is at most 𝛽 ⋅ 𝑞𝑘 𝐵

𝑇

𝑘

𝑞𝑘

𝐵 𝑗, 𝑙 is the same for all bins 𝑙

 𝑃𝑄𝑈2 for component (2) is at most 𝑞𝑘

𝐵

𝑇

𝑘

By definition: items in 𝑇

𝑘 are subset of items in 𝑇 𝑘

𝑞1

𝐵 𝑇 𝑘 ≥ 𝑞1 𝐵

𝑇

𝑘

𝑃𝑄𝑈

1 + 𝑃𝑄𝑈2 = 𝑃𝑄𝑈 ≤ 𝛽 + 1 ⋅ 𝑞1 𝐵 𝑇 𝑘

MOSHE GABEL 33

1 2 3 1

3 3 3

2

1

3

5

4

25 25 25

𝑞1

𝐵

FINISHING UP

We saw that for every 𝑘, 𝑇

𝑘 is 𝛽 + 1 -approximation to 𝑞𝑘 𝐵 and 𝑞𝑘 𝐶

By Local Ratio Theorem, 𝑇

𝑘 is 𝛽 + 1 -approximation to 𝑞𝑘 = 𝑞𝑘 𝐵 + 𝑞𝑘 𝐶

Thus we have 𝛽 + 1 -approximation for GAP ∎

MOSHE GABEL 34

P2: JOINT SCHEDULING WITH MCS SELECTION

Input Scheduling areas (𝐺01, 𝐺02, 𝐺03, 𝐺1, 𝐺2, 𝐺3) with capacities Packets to transmit, with feasible area and MCS combinations Size and profit for each packet in each area and MCS Output Choose area + MCS per packet Maximizing profit, subject to: One or zero areas per packet Do not exceed area capacity Avoid interference

 Met by requiring SINR > 1

MOSHE GABEL 35

MULTIPLE CHOICE GAP (MC-GAP)

New NP-hard problem combines GAP with multiple choice (MCKP) 𝑜 items 𝑛 bins with capacity 𝐶

𝑘

ℓ configurations 𝒒𝒋𝒅𝒌 = profit of item 𝑗 if assigned to bin 𝑘 with configuration 𝒅 𝒕𝒋𝒅𝒌 = size of item 𝑗 if assigned to bin 𝑘 with configuration 𝒅 Goal: assign (item, configuration) pairs to bins Maximize profit Do not exceed capacity

MOSHE GABEL 36

MC-GAP

Maximize

𝑗=1 𝑜 𝑑=1 ℓ 𝑘=1 𝑛

𝑞𝑗𝑑𝑘𝑦𝑗𝑑𝑘 Subject to

𝑗=1 𝑜 𝑑=1 ℓ

𝑡𝑗𝑑𝑘𝑦𝑗𝑑𝑘 ≤ 𝐶

𝑘 for 1 ≤ 𝑘 ≤ 𝑛 𝑑=1 ℓ 𝑘=1 𝑛

𝑦𝑗𝑑𝑘 ≤ 1 for 1 ≤ 𝑗 ≤ 𝑜 𝑦𝑗𝑘 ∈ 0,1 for all 𝑗, 𝑑, 𝑘

MOSHE GABEL 37

Maximize profit within capacity max one bin and configuration for each item

SOLVING MC-GAP

Same technique as GAP: each column is MCKP Given alg 𝒝 for multi choice knapsack… …repeat it for each bin with residual profit. 𝛽-approximation to MCKP  𝛽 + 1 -approximation to MC-GAP Even proof is the same.

MOSHE GABEL 38

EVALUATION VIA SIMULATION STUDY

Simulate middle cell of a small network Average 100 random instances

 (per point on figure)

One pending packet for each user

 Profit = success probability (optimize throughput)

40 blocks in each 𝐺0𝑘, 20 blocks in each of F1,F2,F3

MOSHE GABEL 39

SIMULATION STUDY: NETWORK PARAMETERS

Based on LTE specifications and interference model proposed in [Tabia, Gondran, Baala, 2011].

40

SIMULATION STUDY: ALGORITHMS

Separate benefit of joint scheduling and dynamic MCS selection Use optimal pseudo-polynomial solver for MCKP  gives 2-approximation for GAP/MC-GAP Compare to greedy “water-filling” algorithm

MOSHE GABEL 41

Intra-sector Inter-sector Default MCS (GAP)

Alg-3 Alg-1

Dynamic MCS (MC-GAP)

Alg-4 Alg-2

EVALUATION: GENERAL RESULTS

Run time below 1ms for over 200 users

 Modest hardware: 1 core Linux VM with 1GB of memory.

Obtains approximately 95% profit of optimal profit Verified against exponential search (15 packet instances)

MOSHE GABEL 42

UNIFORM USER DISTRIBUTION

Same number of users per sector no benefit from inter- sector scheduling better than water-filling Gain increases with load Load =

# waiting packets # blocks in frame

MOSHE GABEL 43

SKEWED USER DISTRIBUTION

User 20 times more likely to be in sector 1

MOSHE GABEL 44

CONCLUSIONS: RESULTS IN THIS WORK

Joint OFDMA scheduling = GAP Joint scheduling with MCS selection = GAP + MCKP  A new NP-hard problem  With efficient approximation Simulation study shows practical and substantial benefits

 4%-6% below optimal scheduling, < 1ms computation.  Almost x2 improvement in throughput

MOSHE GABEL 45

THANK YOU FOR LISTENING

Questions?

MOSHE GABEL 46

BS A1 A2 A3