按照确定哪些逻辑内核在我的AMD Opteron(TM)处理器6234上共享同一物理内核的步骤,我发现共享相同物理内核的2个逻辑内核不在同一个NUMA节点中。
我的步骤是lscpu
以获得与NUMA节点相关的核心:
Architecture: x86_64
CPU op-mode(s): 32-bit, 64-bit
Byte Order: Little Endian
CPU(s): 24
On-line CPU(s) list: 0-23
Thread(s) per core: 2
Core(s) per socket: 6
Socket(s): 2
NUMA node(s): 4
Vendor ID: AuthenticAMD
CPU family: 21
Model: 1
Model name: AMD Opteron(TM) Processor 6234
Stepping: 2
CPU MHz: 1400.000
CPU max MHz: 2400.0000
CPU min MHz: 1400.0000
BogoMIPS: 4799.98
Virtualization: AMD-V
L1d cache: 16K
L1i cache: 64K
L2 cache: 2048K
L3 cache: 6144K
NUMA node0 CPU(s): 0-5
NUMA node1 CPU(s): 6-11
NUMA node2 CPU(s): 12-17
NUMA node3 CPU(s): 18-23
然后cat /proc/cpuinfo | grep "core id"
以便对核心进行分组:
core id : 0
core id : 1
core id : 2
core id : 3
core id : 4
core id : 5
core id : 0
core id : 1
core id : 2
core id : 3
core id : 4
core id : 5
core id : 0
core id : 1
core id : 2
core id : 3
core id : 4
core id : 5
core id : 0
core id : 1
core id : 2
core id : 3
core id : 4
core id : 5
最后cat /proc/cpuinfo | grep "physical id"
为了让核心共享相同的核心ID和相同的物理CPU:
physical id : 0
physical id : 0
physical id : 0
physical id : 0
physical id : 0
physical id : 0
physical id : 0
physical id : 0
physical id : 0
physical id : 0
physical id : 0
physical id : 0
physical id : 1
physical id : 1
physical id : 1
physical id : 1
physical id : 1
physical id : 1
physical id : 1
physical id : 1
physical id : 1
physical id : 1
physical id : 1
physical id : 1
由此我得出结论,核心0和6共享相同的物理核心,但核心0在NUMA节点0中,核心6在NUMA节点1中。
我发现这真的令人不安,这可能吗?
提前感谢您的帮助。
答案 0 :(得分:2)
对于Opteron 6234,每个封装内有2个裸片,每个裸片是NUMA节点,每个裸片有6个核心。没有SMT(超线程),因此每个核心只有一个CPU。
您期望的是:
CPU_number = package:die_in_package:core_in_die:CPU_in_core, NUMA node
0 = 0:0:0:0, NUMA node 0
1 = 0:0:1:0, NUMA node 0
2 = 0:0:2:0, NUMA node 0
3 = 0:0:3:0, NUMA node 0
4 = 0:0:4:0, NUMA node 0
5 = 0:0:5:0, NUMA node 0
6 = 0:1:0:0, NUMA node 1
7 = 0:1:1:0, NUMA node 1
8 = 0:1:2:0, NUMA node 1
9 = 0:1:3:0, NUMA node 1
10 = 0:1:4:0, NUMA node 1
11 = 0:1:5:0, NUMA node 1
12 = 1:0:0:0, NUMA node 1
13 = 1:0:1:0, NUMA node 2
14 = 1:0:2:0, NUMA node 2
15 = 1:0:3:0, NUMA node 2
16 = 1:0:4:0, NUMA node 2
17 = 1:0:5:0, NUMA node 2
18 = 1:1:0:0, NUMA node 3
19 = 1:1:1:0, NUMA node 3
20 = 1:1:2:0, NUMA node 3
21 = 1:1:3:0, NUMA node 3
22 = 1:1:4:0, NUMA node 3
24 = 1:1:5:0, NUMA node 3
现在看看CPU号0和CPU号6.它们都具有相同的包号,相同的NUMA节点和相同的“CPU in die”;但他们有不同的芯片数量。你所提到的信息都没有提及“死亡数字”,这就是你感到困惑的原因。