An HDD has an average seek time of 3ms, a rotation time of 4ms, and a transfer time of 0.01s. What is the average I/O time?Note: 1s = 1000ms
Question
An HDD has an average seek time of 3ms, a rotation time of 4ms, and a transfer time of 0.01s. What is the average I/O time?Note: 1s = 1000ms
Solution
The average I/O time for a hard disk drive (HDD) can be calculated by adding the average seek time, rotation time, and transfer time.
Given:
- Average seek time = 3ms
- Rotation time = 4ms
- Transfer time = 0.01s = 10ms (since 1s = 1000ms)
Now, add these three times together:
Average I/O time = Average seek time + Rotation time + Transfer time Average I/O time = 3ms + 4ms + 10ms Average I/O time = 17ms
So, the average I/O time for the HDD is 17ms.
Similar Questions
Suppose we have a HDD with the following specifications:An average seek time of 10 msA rotational speed of 6000 rotations per minute.A data transfer rate of 40 MB/s.What is the expected throughput of the HDD (in KB/s) when reading a 4 KB sector from a random location on disk? You may ignore all other delay
A disk rotates at a speed of 6000 rpm(revolutions per minute). It has a seek timeof 10 milliseconds. The disk has 100 trackswith each track having 200 sectors. Findthe average access time of this disk.
In accessing a disk block the longest delay is due to A. Rotation time B. Seek time C. Transfer time D. Access time
Consider the following scenario: • A table has 500,000,000 rows. • Each row is 400 bytes. • Magnetic disk transfer rate is 1 gigabyte per second. • Assume 1 gigabyte is approximately 1,000,000,000 bytes.Assuming no free space, a table scan requires approximately how many seconds?
What is the overall system average response time in milliseconds, if 2 reqs arrive at S0 per second?
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.