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admin commented on issue admin/matmul-analysis#24

2026-09-04 03:43:39 +00:00

[P1] 切M/N(ASW) 的共享块重复读未按 L2 带宽计时；场景判定与分组预算 D 错误除以总 batch 数

???(commit 4843053): (1) ASW evaluate ?????????:A ??? n_cnt ? tile ?? ? GM ?? 1 ? + L2 ??? (n_cnt-1) ?(B ???? (m_cnt-1) ?),?? A/B ? l2_read_bytes ?????? 5.2TB/s ????; (2) ??????**? batch…

admin closed issue admin/matmul-analysis#24

2026-09-04 03:43:39 +00:00

[P1] 切M/N(ASW) 的共享块重复读未按 L2 带宽计时；场景判定与分组预算 D 错误除以总 batch 数

admin commented on issue admin/matmul-analysis#25

2026-09-04 03:43:39 +00:00

[P2] StreamK evaluate 使用假想 65536 元素 L0C 方形 tile，未按 plan 实际 single_core_m×n 计量（瘦长 case 搬入低估 ~16x）

???(commit 4843053):StreamK evaluate ? plan ?? tile ?????????--tile = single_m x single_n(????????),GM ?? = b*(mk+kn)dt(??? K ????;? batch ???????????? L2 ?),MMAD = 2bmn*k / ????,???…

admin closed issue admin/matmul-analysis#25

2026-09-04 03:43:39 +00:00

[P2] StreamK evaluate 使用假想 65536 元素 L0C 方形 tile，未按 plan 实际 single_core_m×n 计量（瘦长 case 搬入低估 ~16x）

admin pushed to main at admin/matmul-analysis

2026-09-04 03:43:26 +00:00

4843053ad3 Fix issues #23-#25 (+#26 标注): GM读写共享总线累加计时 / ASW-切M/N 共享块 GM首读1次+L2重复读(n-1)次、场景按单batch判定、分组预算不再除B / StreamK 按plan实际tile芯片口径评估 / 降核线性带宽假设文档标注

admin opened issue admin/matmul-analysis#23

2026-09-04 03:38:05 +00:00

[P1] GM 读写共享总线未建模：MTE2 读与 Fixpipe 直写 GM 并发时需按 (读+写)/1.6TB/s 累加计时

admin opened issue admin/matmul-analysis#24

2026-09-04 03:38:05 +00:00

[P1] 切M/N(ASW) 的共享块重复读未按 L2 带宽计时；场景判定与分组预算 D 错误除以总 batch 数

admin opened issue admin/matmul-analysis#25

2026-09-04 03:38:05 +00:00

[P2] StreamK evaluate 使用假想 65536 元素 L0C 方形 tile，未按 plan 实际 single_core_m×n 计量（瘦长 case 搬入低估 ~16x）

admin opened issue admin/matmul-analysis#26

2026-09-04 03:38:05 +00:00

[P3] 降核/低核数带宽线性假设需标注待标定（min_core_num=0.8C 处 90%+ 带宽利用率的 KB 锚点）

admin pushed to main at admin/matmul-analysis

2026-09-04 03:09:52 +00:00

f4b23d9f05 Expand cases_demo to 44 cases (B 4..128 x M 1..131072 x N/K 128..8192 incl 7168) + regenerate examples outputs

admin pushed to main at admin/matmul-analysis

2026-09-04 02:47:11 +00:00

5051e00fa0 T_cmd=0 仲裁改为策略优先 MergeBatch: K截断即优先(指令/命令次数省b0倍,未量化收益), L1绑定恒劣; 路由层以策略覆盖时延模型差额并显式标注

admin pushed to main at admin/matmul-analysis

2026-09-04 02:32:01 +00:00

a071d87831 Fix T_cmd=0 robustness: beats_iterbatch 阈值除零防护 (T_cmd<=0 判 MergeBatch 不胜出, 无命令节省) + 3 条回归测试

admin closed issue admin/matmul-analysis#20

2026-09-03 13:10:26 +00:00

[P3] #13/#15 的回归测试在第三轮被删除（修复代码仍在，失去保护）

admin commented on issue admin/matmul-analysis#21

2026-09-03 13:10:26 +00:00

[P3] 仓库卫生复发：9 个调试临时 csv 再次被提交入库（bug1*.csv/t*.csv/p2.csv），.gitignore 未覆盖

???(commit 9afe6ee):?? BMM_Theory ??? 9 ??????? csv(bug1*.csv/bug2.csv/bug_test.csv/p2.csv/t.csv/t_out.csv/t_p.csv);??? .gitignore ?? BMM_Theory ?? csv ??(bug*.csv/t*.csv/p2.csv ?,examples/ ????????)?

admin closed issue admin/matmul-analysis#21

2026-09-03 13:10:26 +00:00

[P3] 仓库卫生复发：9 个调试临时 csv 再次被提交入库（bug1*.csv/t*.csv/p2.csv），.gitignore 未覆盖

admin commented on issue admin/matmul-analysis#22

2026-09-03 13:10:26 +00:00

[P3] 文档与代码状态失同步：04_差异对照版本声明称"含 issue#11-#16 修复"与回退后的代码矛盾

???(commit 9afe6ee):docs/04_????bmmv3_vs_BMM_Theory.md ??? v1.1--??????"?? BMM_Theory main(? issue#11-#22 ??)"(? #17/#19 ????????????),?????????"????????"??;docs/02????/04_????.md ???????;docs/03…

admin closed issue admin/matmul-analysis#22

2026-09-03 13:10:26 +00:00

[P3] 文档与代码状态失同步：04_差异对照版本声明称"含 issue#11-#16 修复"与回退后的代码矛盾

admin closed issue admin/matmul-analysis#17

2026-09-03 13:10:25 +00:00

[P1] 第三轮整改回退了 #11/#12/#14 的修复：StreamK fixpipe 双重计账(55us)、K=1 空洞(49.4%)、advice 误导，需恢复或给出替代论证

admin commented on issue admin/matmul-analysis#18

2026-09-03 13:10:25 +00:00

[P2] 占位方案(used_core_num=0)可被 evaluate 判为可行并给出正常时延：recommend"无方案"与 evaluate"可行"矛盾

???(commit 9afe6ee):constraints.check_plan_constraints ?????? used_core_num<1 ??"??/?????, ????"(?????????),evaluate ??????? feasible=False + ?? violations;?????? test_issue18_placeholder_plan_infe…

admin closed issue admin/matmul-analysis#18

2026-09-03 13:10:25 +00:00

[P2] 占位方案(used_core_num=0)可被 evaluate 判为可行并给出正常时延：recommend"无方案"与 evaluate"可行"矛盾