Question:

Consider a disk with the following specifications: rotation speed of 6000 RPM, average seek time of 5 milliseconds, 500 sectors/track, 512-byte sectors. A file has content stored in 3000 sectors located randomly on the disk. Assuming average rotational latency, the total time (in seconds, rounded off to 2 decimal places) to read the entire file from the disk is ............

Show Hint

To calculate disk access time, sum up the seek time, rotational latency, and data transfer time.
Updated On: Jan 23, 2025
Hide Solution
collegedunia
Verified By Collegedunia

Solution and Explanation

The total time to read the file consists of: Seek time: Average seek time is \(5 \, \text{ms}\). Rotational latency: For a rotation speed of \(6000 \, \text{RPM}\), the time for one rotation is: \[ \text{Time per rotation} = \frac{60 \, \text{sec}}{6000} = 0.01 \, \text{sec}. \] The average rotational latency is half the time per rotation: \[ \text{Average rotational latency} = \frac{0.01}{2} = 0.005 \, \text{sec} = 5 \, \text{ms}. \] Data transfer time per sector: Each track has \(500 \, \text{sectors}\). The time to transfer one sector is: \[ \text{Time per sector} = \frac{\text{Time per rotation}}{\text{Sectors per track}} = \frac{0.01}{500} = 0.00002 \, \text{sec} = 20 \, \mu\text{s}. \] Total time per sector: Total time to access one sector is: \[ \text{Time per sector} = \text{Seek time} + \text{Rotational latency} + \text{Transfer time}. \] \[ \text{Time per sector} = 5 \, \text{ms} + 5 \, \text{ms} + 20 \, \mu\text{s} = 10.02 \, \text{ms}. \] Total time for 3000 sectors: \[ \text{Total time} = 3000 \times 10.02 \, \text{ms} = 30060 \, \text{ms} = 30.06 \, \text{sec}. \] Final Answer: \[ \boxed{29.5 \, \text{to} \, 30.5 \, \text{sec}} \]
Was this answer helpful?
0
0

Top Questions on Lexical analysis

View More Questions

Questions Asked in GATE CS exam

View More Questions