相同的物理核心不同的NUMA节点

时间:2018-01-31 16:18:22

标签: numa hyperthreading

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。

1 个答案:

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。