1. 1 HARQ Feedback Channel Index Document Number: IEEE C802.16m-09/0947 Date Submitted: 2009-04-27 Source: Yujian Zhangyujian.zhang@intel.comyujian.zhang@intel.com Hujun Yinhujun.yin@intel.comhujun.yin@intel.com Yi Hsuanyi.hsuan@intel.comyi.hsuan@intel.com Intel Corporation Re:Category: AWD-Comments / Area: Chapter 15.3.6 (DL-CTRL) “Comments on AWD 15.3.6 DL-CTRL” Venue:IEEE Session#61, Cairo Base Contribution: C802.16m-09_1018.doc (or later versions) Purpose: Discussion and Adoption by TGm Notice: This document does not represent the agreed views of the IEEE 802.16 Working Group or any of its subgroups. It represents only the views of the participants listed in the “Source(s)” field above. It is offered as a basis for discussion. It is not binding on the contributor(s), who reserve(s) the right to add, amend or withdraw material contained herein. Release: The contributor grants a free, irrevocable license to the IEEE to incorporate material contained in this contribution, and any modifications thereof, in the creation of an IEEE Standards publication; to copyright in the IEEE’s name any IEEE Standards publication even though it may include portions of this contribution; and at the IEEE’s sole discretion to permit others to reproduce in whole or in part the resulting IEEE Standards publication. The contributor also acknowledges and accepts that this contribution may be made public by IEEE 802.16. Patent Policy: The contributor is familiar with the IEEE-SA Patent Policy and Procedures: and.http://standards.ieee.org/guides/bylaws/sect6-7.html#6http://standards.ieee.org/guides/opman/sect6.html#6.3 Further information is located at and.http://standards.ieee.org/board/pat/pat-material.htmlhttp://standards.ieee.org/board/pat

2. 2 Overview

3. 3 Introduction How could MS determine the ACK/NACK resource in either DL and UL –Explicit signal of the ACK/NACK resource –From some transmission attributes of corresponding data or STID A-MAP Resource A-MAP Order Data Resource index MCS … Tradeoffs between –Minimizing the overall resource occupied by ACK/NACK –Avoid the collision of ACK/NACK resources Design constraints –HARQ –Persistent scheduling –Group scheduling –MU-MIMO

4. 4 802.16e Rev2 Persistent scheduling –Index for ACKCH transmitted in DL Field ACK Disable can be used to disable the usage of DL HARQ Bitmap. –Index for ACKCH transmitted in UL Explicitly specified in DL MAP. Non-persistent scheduling –Index for ACKCH transmitted in DL Implicitly by toggling AI_SN bit Or explicitly by HARQ ACK IE –ACK/NACK position is determined from the order of HARQ burst in UL-MAP. Thanks to joint coding, there is no waste in ACK/NACK resource. –Index for ACKCH transmitted in UL ACK/NACK position is determined from the order of HARQ burst in DL-MAP. Thanks to joint coding, there is no waste in ACK/NACK resource. Can be also explicitly specified in DL MAP for MIMO transmissions.

5. 5 Index for HARQ Feedback Channel in UL

6. 6 A-MAP Order Solution When A-MAP order can be inferred from non-user specific A- MAP, one solution is to link HARQ Feedback Channel index with A-MAP order. For persistent scheduling, HARQ Feedback Channel resource pool is separated from non-persistent scheduling.

7. 7 A-MAP Order Solution - Example

8. 8 Index for HARQ Feedback A-MAP in DL

9. 9 Semi-Explicit Solution Explicit solution indicates the full HF-A-MAP index. Semi-explicit solution indicates a partial HF-A-MAP index, which combines with other known parameters (e.g. resource index, pilot index) to form the full HF-A-MAP index. Assume same amount of HF-A-MAP allocated, following is the comparison between 3 approaches. SchemesOverhead in A-MAPScheduler Flexibility ExplicitHigh Semi-explicitMedium ImplicitNoneLow

10. 10 Semi-Explicit Solution – Enhancement (1) One enhancement is to add one field in non-user-specific A-MAP to indicate which parameters are used to derive HF-A-MAP index, abbreviated as HFIP (HF- A-MAP Index Parameter) later –Assume HFIP has N HFIP bits. –Parameters could be considered: Resource index STID MCS value For one particular subframe, there might be the case that for one particular parameter set, even the complete combinations of all HF-A-MAP index still cannot avoid the HF-A-MAP conflicts. Using different parameter set could remove the conflicts in most cases. –Assume Parameters used are independent. The probability that there is no collision when HFIP field is not present is P nc –The probability that there is no collision when N HFIP bit HFIP field used is 1-(1- P nc )^(2^N HFIP ). Even if P nc =0.6, the new probability is 0.9919 when N HFIP =2. Scheduling procedure by ABS –First, ABS determines scheduling decisions. –Then ABS tries to find one HFIP configuration, with which at least one combination of HF-A-MAP index values can provide collision-free HF-A-MAP allocation.

11. 11 Semi-Explicit Solution – Enhancement (2) Example Solution with 2 bit HFIP: –Select one parameter as baseline, e.g. STID. Such parameter is always used for HF-A-MAP index (with N index bits) derivation. –HFIP indicates the combination of other parameters. Assume value signaled in HFIP is j, corresponding parameter set is M(j) = {m i } –Calculation HF-A-MAP index signaled in A-MAP is n. Number of HF-A-MAPs configured is N HF-A-MAP Associated HF-A-MAP index is (Σm i +n) mod N HF-A-MAP 2 bit HFIP j Parameter Set M(j) STIDResourceMCS 00 ■□□ 01 ■□■ 10 ■■□ 11 ■■■

12. 12 Example Assume –N HFIP=2, value of HFIP=2, means STID and Resource index are used to calculate HF-A-MAP index. –N HF-A-MAP = 16 –Index of Resource: 15 STID: 334 MCS: 12 –Index signaled in A-MAP n = 1. HF-A-MAP index is: –(15 + 334 + 1) mod 16 = 14

13. 13 Semi-Explicit Solution – Evaluation Assumptions Methodology –For each random draw of the parameters (resource index, STID, and MCS value), determine whether HF-A-MAP can be allocated without collision or not (one example algorithm is given in Backup). Assumption: –Resource allocation of different AMSs could be overlapping with each other (e.g. MU- MIMO). System Bandwidth (MHz)1020 N HFIP 2 N HF-A-MAP 1632 N index 0..(log 2 N HF-A-MAP -1) Resource index range0..470..95 MCS0..32 STID0..4095

14. 14 Semi-Explicit Solution –Results (1) Denote N alloc as the number of allocations. For original semi-explicit solution (without HFIP field), necessary N index depends on N HF-A-MAP –When N HF-A-MAP =16, N index =3 is sufficient –When N HF-A-MAP =32, N index =4 is sufficient Note: Theoretically, N index =0 is the case of birthday paradox (http://en.wikipedia.org/wiki/Birthday_paradox ), with the successful allocation probability ashttp://en.wikipedia.org/wiki/Birthday_paradox

15. 15 Semi-Explicit Solution –Results (2) Benefits of the enhanced solution by introducing HFIP field. –When N HF-A-MAP =16, N index =2 is sufficient when N alloc <=14. –When N HF-A-MAP =32, N index =3 is sufficient –In both scenarios, for each assignment A-MAP, the enhanced solution can save 1 bit compared with original semi-explicit solution, save 2 bit compared with explicit solution.

16. 16 Comparison (1) Total Overhead (in one subframe) required to indicate HF-A-MAP index in A-MAP includes non-user-specific A-MAP, assignment A-MAP and HF-A-MAP. –Assume Non-user specific A-MAP has code rate of 1/8, only consider the overhead of HFIP. HF-A-MAP has code rate of 1/8 Assignment A-MAP has average code rate of ¼ (with code rate from 1/8 to ½). For the overhead in assignment A-MAP, only consider the partial HF-A-MAP index. –Then 1 bit of non-user-specific A-MAP or HF-A-MAP is equivalent to 2 bits of assignment A-MAP. For fair comparison, for each N alloc, the required N HF-A-MAP is calculated for each option and total overhead is compared. –N HF-A-MAP is allocated in the granularity of 4. –For semi-explicit option (including both original and enhancement solutions), the criterion is that the percentage of successful allocation is larger than 99%.

17. 17 Comparison (2) From total overhead point of view, enhanced solution is better than both original solution and explicit solution. Therefore it is recommended to adopt the enhanced solution.

18. 18 Backup

19. 19 DL HF-A-MAP Resource Overhead Estimation DL HF-A-MAP resource overhead is related to two issues –The number of simultaneous UL transmissions per subframe –HF-A-MAP index scheme. Assumptions –10 MHz, TDD with 4:4 ratio –One packet needs 1.2 HF-A-MAP due to HARQ –20 ms VoIP frame length Estimation: –Number of HF-A-MAP’s per subframe N HF-A-MAP for persistent scheduling. VoIP capacity is >= 60 active users/MHz/sector (50% activity factor) as per SRD. N HF-A-MAP =60*10/2*50%*1.2/4/4=11.25

20. 20 Example Algorithm (1) To calculate the solution, i.e. find the combination of HF-A-MAP index values (indicated in assignment A-MAP), Hopcroft-Karp algorithm (http://en.wikipedia.org/wiki/Hopcroft– Karp_algorithm ) could be used. Note that other algorithms also exit, but current discussion is focused on this algorithm.http://en.wikipedia.org/wiki/Hopcroft– Karp_algorithm Assume the total number of HF-A-MAPs for allocation (excluding those already allocated, e.g. retransmission without A-MAP, persistent scheduling) is N alloc, construct a bipartitie graph (http://en.wikipedia.org/wiki/Bipartite_graph) (U,V,E) with two disjoint sets U and V, and edges Ehttp://en.wikipedia.org/wiki/Bipartite_graph –U contains all the HF-A-MAPs for allocation with size of N alloc –V contains all the legal absolute HF-A-MAP index values (with the range 0..N HF-A-MAP -1). “Legal” means that the value could be derived from at least one of the indicated HF-A-MAP index values. –E associates each value in U with 2^N index values in V. Hopcroft-Karp algorithm produces a maximum-cardinality matching of this bipartitie graph. –If the cardinality of the matching equals N alloc, the result is the solution to assign HF-A-MAP index values. –Otherwise, there is no solution; So ABS has to try another parameter set. Complexity –In the worse case, the complexity of the algorithm is O(m*sqrt(n)), where m is the number of edges in the graph and n is the number of vertices in the graph. In our case, m= N alloc *(2^N index ), while maximum n= N alloc +N HF-A-MAP, assume N alloc =N HF-A-MAP, the worst case complexity is shown below, which is rather trivial for ABS implementation.

21. 21 Example Algorithm (2) Support of BPSK/QPSK transmission of HF-A-MAP. –If BPSK transmission of HF-A-MAP is needed, one can run the previous algorithm as follows. –Suppose that only 1 BPSK transmission is needed, then ABS can first try to allocate HF-A-MAP index for this particular channel. There are maximum 2^N index values. The algorithm is as follows: For each value of HF-A-MAP index ABS apply previous algorithm (removing current channel from set U, and removing the two HF-A-MAPs (the QPSK pair) from set V). If the allocation is successful, end Otherwise, try next value. –The above algorithm can be easily generalized when there are more than 1 BPSK channels.