考虑一个寻道时间为 4 毫秒且转速为 10000 转/分钟 (RPM) 的磁盘组。每个磁道有 600 个扇区，每个扇区可以存储 512 字节的数据。考虑存储在磁盘中的文件。该文件包含 2000 个扇区。假设每个扇区访问都需要一次寻道，并且访问每个扇区的平均旋转延迟是一个完整旋转的一半时间。读取整个文件所需的总时间(以毫秒为单位)是_________。
(一) 14020
(二) 14000
(三) 25030
(四) 15000答案：(一)
解释：
Seek time (given) = 4ms
RPM = 10000 rotation in 1 min [60 sec]
So, 1 rotation will be =60/10000 =6ms [rotation speed]
Rotation latency= 1/2 * 6ms=3ms
# To access a file,
total time includes =seek time + rot. latency +transfer time
TO calc. transfer time, find transfer rate
Transfer rate = bytes on track /rotation speed
so, transfer rate = 600*512/6ms =51200 B/ms
transfer time= total bytes to be transferred/ transfer rate
so, Transfer time =2000*512/51200 = 20ms
Given as each sector requires seek tim + rot. latency
= 4ms+3ms =7ms
Total 2000 sector takes = 2000*7 ms =14000 ms
To read entire file ,total time = 14000 + 20(transfer time)
= 14020 ms
这个问题的测验