Computer System Architecture MCQ - Pipeline and Vector Processing

A hard disk has 63 sectors per track, 10 platters each with 2 recording surfaces and 1000 cylinders. The address of a sector is given as a triple (c, h, s), where c is the cylinder number, h is the surface number and s is the sector number. Thus, the 0th sector is addressed as (0, 0, 0), the 1st sector as (0, 0, 1), and so on.

41 The address (400, 16, 29) corresponds to sector number :

Sector no = 400 * 20 * 63 + 16 * 63 + 29 = 505037.

42 The address of 1039^th sector is

For option
(a) sector no. = 0 * 20 * 63 + 15 * 63 + 31 = 976 sector.
(b) Sector no = 0 * 20 * 63 + 16 * 63 + 30 = 1038
(c) Sector no = 0 * 20 * 63 + 16 * 63 + 31 = 1039 sector
Hence answer is (c).

43 An application loads 100 libraries at start-up. Loading each library requires exactly one disk access. The seek time of the disk to a random location is given as 10 ms. Rotational speed of disk is 6000 rpm. If all 100 libraries are loaded from random locations on the disk, how long does it take to load all libraries? (The time to transfer data from the disk block once the head has been positioned at the start of the block may be neglected.)

6000 rotations ________ 60 sec
1 rotation _________ 10 ms
Rotational latency = 5 ms
Time for one disk access = 15 ms
Time to load all libraries = 15 * 100
= 1500 ms = 1.5 sec

44 Consider a hard disk with 16 recording surfaces (0-15) having 16384 cylinders (0-16383) and each cylinder contain 64 sectors (0-63). Data storage capacity in each sector is 512 bytes. Data are organized cylinder-wise and addressing format is . A file of size 42797 KB is stored in the disk and the starting location of the file is . What is the cylinder number of the last sector of the file, if it is stored in a contiguous manner?

Since starting on disk starts from <1200, 9, 40> so no of sectors left on 9^th surface is 24
so, on 9^th surface total storage of “12288” B is possible Now, a part from 9^thsurface, on cylinder no. 1200 only 6 surface is left.
To storage possible on these 6 surface are
= 6 * 2⁶ * 2⁹ ￫ storage on each sector.
ᛎ
= 196608 B
So total on cylinder no. 1200, storage possible
= 196608 + 12288 = 208896 B
So Since file size is 42797 kB and out of which 208896 B are stored on cylinder no. 1200. So we are left only with only 43615232 B.
Since in 1 cylinder, Storage Possible is
= 2⁴ * 2⁶ * 2⁹ B = 524288 B
So we need about
= 43615232B / 524288B
= 83.189 more cylinders.
Hence we’ll need the 1284^th Cylinder to completely store the file. C₀₂ after 1283^rd cylinder we will left with data which will need 189

45 A hard disk with a transfer rate of 10 M bytes/ second is constantly transferring data to memory using DMA. The processor runs at 600 MHz, and takes 300 and 900 clock cycles to initiate and complete DMA transfer respectively. If the size of the transfer is 20 Kbytes, what is the percentage of processor time consumed for the transfer operation?