## Secondary-Storage

 Question 1

Consider a disk pack with a seek time of 4 milliseconds and rotational speed of 10000 rotations per minute (RPM). It has 600 sectors per track and each sector can store 512 bytes of data. Consider a file stored in the disk. The file contains 2000 sectors. Assume that every sector access necessitates a seek, and the average rotational latency for accessing each sector is half of the time for one complete rotation. The total time (in milliseconds) needed to read the entire file is _________.

 A 14020 B 14021 C 14022 D 14023
Computer-Organization       Secondary-Storage       GATE 2015 [Set-1]
Question 1 Explanation:
Given
Seek time = 4ms
60s→ 10000 rotations ∴ Rotational latency =1/2×6ms = 3ms
1track → 600sectors
⇒6ms ← 600sectors (1 rotation means 600 sectors (or)1 track)
1 track → 6ms/600 = 0.01ms
2000sector → 2000(0.01) = 20ms
∴total time needed to read the entire file is
= 2000(4+3) + 20
= 8000 + 6000+20
= 14020ms
 Question 2

Consider a typical disk that rotates at 15000 rotations per minute (RPM) and has a transfer rate of 50×106 bytes/sec. If the average seek time of the disk is twice the average rotational delay and the controller’s transfer time is 10 times the disk transfer time, the average time (in milliseconds) to read or write a 512-byte sector of the disk is _____________.

 A 6 B 6.1 C 6.2 D 6.3
Algorithms       Secondary-Storage       GATE 2015 [Set-2]
Question 2 Explanation:
15000 rotations → 60 sec
1 rotation → 60 /15000 sec = 4 ms
∴ Average rotational delay = 1/2 × 4 ms = 2 ms
Average seek time = 2 × Average rotational delay
= 2 × 2 ms = 4 ms
Time to transfer 512 byte,
50 × 106B ------- 1s
512 B ------- 512B/ 50×106B/s = 0.01 ms
∴ Controllers transfer time
= 10 × Disk transfer time
= 10 × 0.01 ms
= 0.01 ms
∴ Avg. time = (4 + 2 + 0.1 + 0.01)ms
= 6.11 ms
There are 2 questions to complete.

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