气象互助社 发表于 2024-4-10 17:15:53

WRF系列教程4:WRF并行核数如何确定?


使用WRF模式进行模拟,一个常涉及的问题就是应该使用多少核用于并行计算?并行计算,并不是核数越多越好,核数增加,则相应的通信成本也会增加。如下图所示,当核数提升到一定程度时,模式每个积分步长所用时间就难以继续下降。


模式并行scaling,图片来自WRF官网培训PPT,Dave Gill


WRF运行的并行核数如何确定?
mpirun -np? ./wrf.exe
并行的最小核数,取决于运行的个例需要多大内存
例如北美大陆1500×1500网格数的个例,常规的节点(每个节点36核)则最少需要4个节点,如果使用内存大的节点,则只需要2个节点。


CONUS模拟配置,图片来自WRF官网培训PPT,Dave Gill

对于一个不考虑内存的计算任务,并行和串行都能完成计算,用时长短的区别而已。但是分布式内存的并行方式(编译WRF时并行方式选择dm),与串行最大的区别是能提供更大的内存,承担规模更大的任务。举个例子,一个小区域即使模拟一万年,个人PC和超算都能完成,快慢的区别而已。但是模拟一个全球高分辨率的任务,个人PC提供的内存无法存储这么大的内容,而超算分布式内存并行的方式就可以将其分解存储在不同的节点中。

对WRF模拟而言,具体的意思就是分布式内存并行的方式提供的内存,能够满足模拟更大的区域以及更高的分辨率。内存不够,就会出现段错误。


并行的最大核数,取决于WRF模式内部的通信模板
当并行核数增加时,相应的通信成本会增加,WRF会将模拟区域剖分成一个个小块,每个核计算一小块,每个小块差分计算需要用到相邻小块的值,因此每步积分都需要通信获取相邻块的值。

WRF分解模拟区域时,每个小块的长和宽必须不小于10。因此运行模式时,使用过多的核也会报错。

例如北美大陆1500×1500网格数的个例,在X和Y方向分别最多可以分解为150份,也就是150×150小块,最多能使用150×150=22500核。


最小最大核数经验的计算方法
如果是多层嵌套,并行使用的最小核数应基于网格数最多的domain计算,
而并行最大核数应基于网格数最少的domain。例如,假设有2个domain,并且domain01是200x200网格数,而domain02是300x300网格数,可用来下述公式大致估计:
最大核数 =(200/25)*(200/25)= 64个处理器
最小核数 =(300/100)*( 300/100)= 9个处理器


推荐的核数
因为不同用户的模拟区域范围、分辨率以及嵌套层数都不一样,并且每个人所拥有的计算资源以及对时效性的要求都不一致,最终使用的并行核数其实难以有一个统一的推荐。简单粗暴的方法就是不同核数测试一下,得到一个合适的值。并行核数最好是节点核数的整数倍,比如一个节点是24核,那么可以采用24,48,64等整数倍。


参考
https://www2.mmm.ucar.edu/wrf/users/tutorial/presentation_pdfs/202101/gill_computation.pdf
https://www.cnblogs.com/jiangleads/p/12885651.html




文章来源于微信公众号:气海同途

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