一、參數解釋

合適的HPL.dat參數設置才能夠正常運行以及達到較好的性能。

HPLinpack benchmark input file Innovative Computing Laboratory, University of Tennessee HPL.out output file name (if any) 8 device out (6=stdout,7=stderr,file) 1 # of problems sizes (N) 80000 Ns 1 # of NBs 1024 NBs 0 PMAP process mapping (0=Row-,1=Column-major) 1 # of process grids (P x Q) 1 Ps 1 Qs 16.0 threshold 1 # of panel fact 1 PFACTs (0=left, 1=Crout, 2=Right) 1 # of recursive stopping criterium 4 NBMINs (>= 1) 1 # of panels in recursion 2 NDIVs 1 # of recursive panel fact. 1 RFACTs (0=left, 1=Crout, 2=Right) 1 # of broadcast 0 BCASTs (0=1rg,1=1rM,2=2rg,3=2rM,4=Lng,5=LnM) 1 # of lookahead depth 2 DEPTHs (>=0) 2 SWAP (0=bin-exch,1=long,2=mix) 64 swapping threshold 0 L1 in (0=transposed,1=no-transposed) form 0 U in (0=transposed,1=no-transposed) form 1 Equilibration (0=no,1=yes) 8 memory alignment in double (> 0)

1、第1、2行為注釋說明行，不需要作修改

2、第3行說明如果輸出文件的話，文件的名字

3、第4行說明輸出結果文件的形式，為“6”時，測試結果輸出至標準輸出（stdout），為“7”時，測試結果輸出至標準錯誤輸出（stderr），為其它值時，測試結果輸出至第3行所指定的文件中

4、第5行說明求解問題（矩陣）的個數，也就是第6行要設置的參數的個數

5、第6行要設置矩陣的階，參數值要與第5行的數值相等。網上大多數都說N的值為N×N×8＝系統總內存×80％最優

6、第7行說明求解問題（矩陣）時采用的分塊方式的種數，也就是第8行要設置的參數的個數

7、第8行說明每一種分塊的大小。為提高數據的局部性，從而提高整體性能，HPL采用分塊矩陣的算法。NB值的選擇主要是通過實際測試得到最優值。

8、第9行是選擇處理器陣列是按列的排列方式還是按行的排列方式。

9、第10-12行說明二維處理器網格（P×Q）。二維處理器網格（P×Q）的要遵循以下幾個要求：P×Q＝進程數。這是HPL的硬性規定。

10、其他值采取默認即可。

二、單個節點上執行

命令：

./xhpl

結果：

================================================================================ HPLinpack 2.3 -- High-Performance Linpack benchmark -- December 2, 2018 Written by A. Petitet and R. Clint Whaley, Innovative Computing Laboratory, UTK Modified by Piotr Luszczek, Innovative Computing Laboratory, UTK Modified by Julien Langou, University of Colorado Denver ================================================================================An explanation of the input/output parameters follows: T/V : Wall time / encoded variant. N : The order of the coefficient matrix A. NB : The partitioning blocking factor. P : The number of process rows. Q : The number of process columns. Time : Time in seconds to solve the linear system. Gflops : Rate of execution for solving the linear system.The following parameter values will be used:N : 80000 NB : 1024 PMAP : Row-major process mapping P : 1 Q : 1 PFACT : Crout NBMIN : 4 NDIV : 2 RFACT : Crout BCAST : 1ring DEPTH : 2 SWAP : Mix (threshold = 64) L1 : transposed form U : transposed form EQUIL : yes ALIGN : 8 double precision words--------------------------------------------------------------------------------- The matrix A is randomly generated for each test. - The following scaled residual check will be computed:||Ax-b||_oo / ( eps * ( || x ||_oo * || A ||_oo + || b ||_oo ) * N ) - The relative machine precision (eps) is taken to be 1.110223e-16 - Computational tests pass if scaled residuals are less than 16.0================================================================================ T/V N NB P Q Time Gflops -------------------------------------------------------------------------------- WR20C2C4 80000 1024 1 1 729.77 4.6774e+02 HPL_pdgesv() start time Fri Jul 17 09:24:43 2020HPL_pdgesv() end time Fri Jul 17 09:36:53 2020-------------------------------------------------------------------------------- ||Ax-b||_oo/(eps*(||A||_oo*||x||_oo+||b||_oo)*N)= 2.16389188e-03 ...... PASSED ================================================================================Finished 1 tests with the following results:1 tests completed and passed residual checks,0 tests completed and failed residual checks,0 tests skipped because of illegal input values. --------------------------------------------------------------------------------End of Tests. ================================================================================

三、多個節點執行

命令：

第一種方式： mpirun -np N xhpl N為進程數第二種方式： mpirun -p4pg <p4file> xhpl 需要自己編寫配置文件，p4file指定每個進程在哪個節點運行

?

總結

以上是生活随笔為你收集整理的Linux下Linpack测试CPU性能的相关参数配置以及执行命令的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。