考虑一个以每分钟 15000 转 (RPM) 旋转且传输速率为 50 × 10 6字节/秒的典型磁盘。如果磁盘的平均寻道时间是平均旋转延迟的两倍，控制器的传输时间是磁盘传输时间的 10 倍，则读取或写入磁盘的 512 字节扇区的平均时间(以毫秒为单位)为 _____________
(一) 6.1答案：(一)
解释：
Disk latency = Seek Time + Rotation Time + Transfer Time + Controller Overhead
Seek Time? Depends no. tracks the arm moves and seek speed of disk
Rotation Time? depends on rotational speed and how far the sector is from the head
Transfer Time? depends on data rate (bandwidth) of disk (bit density) and the size of request
Disk latency = Seek Time + Rotation Time +
Transfer Time + Controller Overhead
Average Rotational Time = (0.5)/(15000 / 60) = 2 miliseconds
[On average half rotation is made]
It is given that the average seek time is twice the average rotational delay
So Avg. Seek Time = 2 * 2 = 4 miliseconds.
Transfer Time = 512 / (50 × 106 bytes/sec)
= 10.24 microseconds
Given that controller time is 10 times the average transfer time
Controller Overhead = 10 * 10.24 microseconds
= 0.1 miliseconds
Disk latency = Seek Time + Rotation Time +
Transfer Time + Controller Overhead
= 4 + 2 + 10.24 * 10-3 + 0.1 miliseconds
= 6.1 miliseconds
参考 http://cse.unl.edu/~jiang/cse430/Lecture%20Notes/reference-ppt-slides/Disk_Storage_Systems_2.ppt
这个问题的测验