CPU_CLK_UNHALTED.THREAD事件中的CLOCK是时钟周期还是机器周期呢?@Peter Wang

[chou_m_]

我是一个新手。在收集内存访问事件时有一些疑惑，现有如下几个问题想请假大家。

1.对于CPU_CLK_UNHALTED.THREAD事件中的CLOCK指的是时钟周期还是机器周期？

2.如果CPU_CLK_UNHALTED.THREAD指的是机器周期数的话，而且从教科书上知道执行一条指令所花费的机器周期数至少应该是1，那通过CPU_CLK_UNHALTED.THREAD / INST_RETIRED.ANY计算CPI时，结果应该大于等于1啊，但是从我的测试结果来看是小于1的，这是为什么呢？

3.MEM_TRANS_RETIRED.LOAD_LATENCY_GT_4，这个事件描述是载入数据超过4个什么(时钟周期还是机器周期)的延时？类似的量纲，我该到哪儿去查呢？

4.在收集访存事件时，最后的统计结果出现了Total_Latency_MEM_UOPS_RETIRED.ALL_LOADS_PS这一个数据项，它的单位是什么呢？我的测试结果是182144273216，如果是纳秒的话，那应该是182秒多一点，但我的整个程序只运行了不到一分钟。我想知道这是为什么？

谢谢！

[Peter_W_Intel]

1. CPU Clocktick, it means there are N CPU cycles per second 

2. Consider there are pipelines in architecture to prefetch / decode / execute uops / retired simultaneously, so each instruction will be little than 1, if you don't run SSE/AVX instructions.

3. MEM_TRANS_RETIRED.LOAD_LATENCY_GT_4, loads with latency value 4 ; for example, if L1 cache misses, load has 4-6 cycle delay. See detail info from VTune(TM) Amplifier's helper.

4. Total_Latency_MEM_UOPS_RETIRED.ALL_LOADS_PS; accumulate latency (cycles) from all memory loads in your program

[chou_m_]

@Peter Wang。王先生，您好。麻烦想再向您确认一下。

1. CPU Clocktick, it means there are N CPU cycles per second 

2. Consider there are pipelines in architecture to prefetch / decode / execute uops / retired simultaneously, so each instruction will be little than 1, if you don't run SSE/AVX instructions.

3. MEM_TRANS_RETIRED.LOAD_LATENCY_GT_4, loads with latency value 4 ; for example, if L1 cache misses, load has 4-6 cycle delay. See detail info from VTune(TM) Amplifier's helper.

4. Total_Latency_MEM_UOPS_RETIRED.ALL_LOADS_PS; accumulate latency (cycles) from all memory loads in your program

对于您在3、4回答的cycle是不同于1中cycle，3、4中的cycle是指的是时钟周期对吗？是通过1/(2.4G)来计算的吧？（我的CPU是Intel(R) Xeon(R) CPU E5-2407 v2 @ 2.40GHz）。那对于4中，我的测试结果是182144273216，换成绝对时间是 182144273216/(2.4*10^9) = 75s，但是我的程序只运行了不到一分钟，访存延时应该不会超过1分钟吧？问题出在哪儿呢？

谢谢您的回答！

[chou_m_]

Peter Wang wrote:

1. CPU Clocktick, it means there are N CPU cycles per second 

2. Consider there are pipelines in architecture to prefetch / decode / execute uops / retired simultaneously, so each instruction will be little than 1, if you don't run SSE/AVX instructions.

3. MEM_TRANS_RETIRED.LOAD_LATENCY_GT_4, loads with latency value 4 ; for example, if L1 cache misses, load has 4-6 cycle delay. See detail info from VTune(TM) Amplifier's helper.

4. Total_Latency_MEM_UOPS_RETIRED.ALL_LOADS_PS; accumulate latency (cycles) from all memory loads in your program

王先生，您好。麻烦想再向您确认一下。

对于您在3、4回答的cycle是不同于1中cycle，3、4中的cycle是指的是时钟周期对吗？是通过1/(2.4G)来计算的吧？（我的CPU是Intel(R) Xeon(R) CPU E5-2407 v2 @ 2.40GHz）。那对于4中，我的测试结果是182144273216，换成绝对时间是 182144273216/(2.4*10^9) = 75s，但是我的程序只运行了不到一分钟，访存延时应该不会超过1分钟吧？问题出在哪儿呢？

谢谢您的回答！

[Peter_W_Intel]

@chu m

Total latency was accumulated from multiple cores, please consider *parallelism* - don't compare with elapsed time.
 

[chou_m_]

Peter Wang wrote:

@chu m

Total latency was accumulated from multiple cores, please consider *parallelism* - don't compare with elapsed time.
 

王先生，您好。我的程序是单进程，没有使用到多核啊，怎么会这样？

[Peter_W_Intel]

Please attach your zipped vtune's result. Also you can submit a ticket to https://premier.intel.com with data.

[chou_m_]

Peter Wang wrote:

Please attach your zipped vtune's result. Also you can submit a ticket to https://premier.intel.com with data.

王先生，您好。我将vtune的测试结果上传上来了，劳烦您帮忙看看。您后边给的网页打不开啊。:(

我的执行命令是：amplxe-cl -c memory-access -knob analyze-mem-objects=true -data-limit=0 -d 60 -- ./test_demo

501869

[Peter_W_Intel]

Thanks for your result file. See summary report:

CPU_CLK_UNHALTED.THREAD  38,904,000,000

MEM_TRANS_RETIRED.LOAD_LATENCY_GT_4  403,612,108

Total_Latency_MEM_TRANS_RETIRED.LOAD_LATENCY_GT_4 4,158,524,752

That was why Average Load latency =  Total_Latency_MEM_TRANS_RETIRED.LOAD_LATENCY_GT_4 / MEM_TRANS_RETIRED.LOAD_LATENCY_GT_4 = 10

Your CPU frequency is 2.4Ghz, CPU time (serial code, mostly) = CPU_CLK_UHHALED.THREAD counts / 2.4G = 16.2s

Time spent in Load = Total_Latency_MEM_TRANS_RETIRED.LOAD_LATENCY_GT / 2.4G = 1.84s

Load Memory bound = Load time / CPU time = (1.84 / 16.2) * 100% = 11.4%

[chou_m_]

Peter Wang wrote:

Thanks for your result file. See summary report:

CPU_CLK_UNHALTED.THREAD  38,904,000,000

MEM_TRANS_RETIRED.LOAD_LATENCY_GT_4  403,612,108

Total_Latency_MEM_TRANS_RETIRED.LOAD_LATENCY_GT_4 4,158,524,752

That was why Average Load latency =  Total_Latency_MEM_TRANS_RETIRED.LOAD_LATENCY_GT_4 / MEM_TRANS_RETIRED.LOAD_LATENCY_GT_4 = 10

Your CPU frequency is 2.4Ghz, CPU time (serial code, mostly) = CPU_CLK_UHHALED.THREAD counts / 2.4G = 16.2s

Time spent in Load = Total_Latency_MEM_TRANS_RETIRED.LOAD_LATENCY_GT / 2.4G = 1.84s

Load Memory bound = Load time / CPU time = (1.84 / 16.2) * 100% = 11.4%

王先生，您好。感谢您能耐心的回答我的问题。

但我还是有一些疑问想请教您，:)。

1.我注意到，您在计算Time spend in Load时，用的是Total_Latency_MEM_TRANS_RETIRED.LOAD_LATENCY_GT _4（4,158,524,752），我想知道为什么不使用Total_Latency_MEM_UOPS_RETIRED.ALL_LOADS_PS （429,400,644,100）？

2.上述两个参数之间有100倍的差距，这说明什么问题呢？

3.Total_Latency_MEM_TRANS_RETIRED.LOAD_LATENCY_GT_4是不是专门指的是从主存Load的时间，而Total_Latency_MEM_UOPS_RETIRED.ALL_LOADS_PS是不是指的是全部的Load时间（包括访问1、2、3级缓存及主存）？

4.就从上边的测试结果来看，有没有可能通过提高1、2、3缓存的命中率来大幅改善程序的性能？

问题有点多,:)，谢谢您的回答。

[Peter_W_Intel]

You need to use (trust) the formula that VTune uses, see "Memory Usage viewpoint" in bottom-up report, metric named "Average Latency (cycles)", move mouse to that item, says "..This metric shows average load of latency in cycles...Formula: Total_Latency_MEM_TRANS_RETIRED.LOAD_LATENCY_GT_4 /  MEM_TRANS_RETIRED.LOAD_LATENCY_GT "