i os造句

• Number of physical i os that are performed by this task
  此任务执行的物理i / o数。
• Number of pending i os that are waiting to be completed
  等待完成的挂起i / o数。
• Total byte count of i os that are performed by this task
  此任务执行的总i / o字节数。
• Average byte count of i os that are performed by this task
  此任务执行的平均i / o字节数。
• The number of physical i os that are performed by this worker
  此工作线程所执行的物理i / o数。
• Average number of bytes for physical i os for this worker
  此工作线程的物理i / o的平均字节数。
• To analyze statistics of physical i os on an index or heap partition
  对数据库中的指定对象具有control权限
• Total number of bytes for all pending physical i os for this worker
  此工作线程的所有挂起的物理i / o的字节总数。
• Buffer latch contention can indicate several issues , including hot pages and slow i os
  缓冲区闩锁争用可以指示出现了几个问题，包括热页和缓慢i / o 。
• Often it is the first user to invoke a transaction that incurs the cost of i os for reading in the indexes and data pages
  通常，第一个调用事务的用户将承受读取索引和数据页的成本。
• Uses an algorithm to avoid redundant statistics creation , which reduces the number of i os incurred during tuning
  使用算法以避免生成重复的统计信息，从而减少优化期间发生的i / o数量。
• For relatively small but heavily updated tables , a sufficiently large separate buffer pool may be able to eliminate both read and write i os
  对于相对较小但更新频繁的表，通过一个足够大的单独的缓冲池，也许可以同时减少读和写的i / o 。
• Each scheduler has a list of pending i os that are checked to determine whether they have been completed every time there is a context switch
  每个计划程序都有一个挂起i / o的列表，通过检查该列表，可以确定每次有上下文切换时它们是否已经完成。
• If it is higher than that , one striping unit needs more than one i o to be issued so that it may de - parallel disk i os in large disk arrays
  如果大于该尺寸，一个分段单元就需要不止一次地发出i / o ，以便它可以在大型磁盘阵列中近似地并行化磁盘i / o 。
• Disk i os are often the biggest factor influencing response time , but the underlying performance problems are easier to understand by looking at the getpage requests
  磁盘i / o常常是影响响应时间的最大因素，但是通过查看getpage ( gp )请求，更容易理解底层的性能问题。
• If the striping size is less than the maximum i o unit size , it may waste some of the i o capacity , causing a given number of i os to take more time to complete
  如果分段大小小于最大i / o单元的大小，就可能浪费部分i / o能力，导致要花费更多时间来完成给定数目的i / o 。
• This is accomplished by increasing vpsize as long as the number of dasd i os keeps decreasing until the cost of paging outweighs the benefit of the reduced i o
  为实现这一点，要使dasd i / o的数量持续下降，并不断增加vpsize ，直到换页的成本超出了通过减少i / o所带来的好处为止。
• As the technology has advanced , the tradeoff of reducing i os by adding more indexes or adding columns to existing indexes at the expense of extra disk space has become more and more attractive over the years
  随着技术的发展，通过增加更多的索引（或添加列到已有的索引中）来减少i / o ，以及由此消耗的额外磁盘空间，这几年两者之间的权衡已经变得越来越有吸引力。