256 Mbyte, 19 bits
256 Mbyte, 21 bits
512 Mbyte, 20 bits
64 GB, 28 bits
→ So the disk pack capacity = 16*128*256*512 bytes = 256 MB
→ To specify a sector we need the information about surface number, track number and sector number within a track.
→ Surface number needs 4 bits as there are 16 surfaces(24), track number needs 7 bits as there are 128 tracks(27) within a surface, within a track the sector number needs 8 bits as there are 256 sectors (28).
→ Total number bits needed to specify a particular sector = 4+7+8 = 19 bits.
Shortest seek-time first
First Come First Served
Tracks order is 36, 69, 167, 76, 42, 51, 126, 12 and 199
Track position is at 100 and moving towards 199.
sequence of disc access is 126, 167, 199, 12, 36, 42, 51, 69 and 76
From the above sequence , we can observe that disc moving from 100 to 126 from there to moving right side direction until last disc 199
From the last disc(199) to first disc(12) and from there to next disc(36) and so on.
C-SCAN sweeps the disk from end-to-end, but as soon it reaches one of the end tracks it then moves to the other end track without servicing any requesting location. As soon as it reaches the other end track it then starts servicing and grants requests headed to its direction
A Computer uses a memory unit with 256K word of 32 bits each. A binary instruction code is stored in one word of memory. The instruction has four parts: an indirect bit, an operation code and a register code part to specify one of 64 registers and an address part. How many bits are there in operation code, the register code part and the address part?
7, 7, 18
18, 7, 7
7, 6, 18
6, 7, 18
Now, the instruction is divided into four parts :
An indirect bit
Register code part : Since number of registers given as 64(26) so to identify each register uniquely 6-bits are needed.
Address part : 256K(218) word memory is mentioned so to identify each word uniquely 18-bits are needed.
Size of Operation code = Complete instruction size - (size of indirect bit + size of register code + size of address part)
Size of Operation code = 7-bits
→ Average access time is time to complete 1/2 revolution= 7.5milliseconds.
2 and 3
3 and 3
3 and 4
4 and 4
90-> 120-> 115-> 110-> 130-> 80-> 70-> 30-> 25-> 20
Change of direction(Total 3); 120->15; 110->130; 130->80
According to First Come First Serve:
90-> 120-> 30-> 70-> 115-> 130-> 110-> 80-> 20-> 25
Change of direction(Total 4); 120->30; 30->70; 130->110;20->25
remains the same
either remains constant or decreases
→ The density, in bits per linear inch, increases as we move from the outermost track to the innermost track (this same phenomenon is present in a phonograph record).
● When memory is full and process needs memory, inactive parts of process are put in swap space of disk.
For Avg. seek time:
Given that, time to move between successive tracks is 1ms.
Time to move from track1 to track1 = 0ms
Time to move from track1 to track2 = 1ms
Time to move from track1 to track3 = 2ms
⋮ Time to move from track1 to track500 = 499ms
∴ Avg. seek time = 0+1+2+...+499/500 = 249.5ms
Avg. rotational delay:
600 rotations in 60sec.
One rotation takes 60/600 s = 100ms ∴ Avg. rotational delay = 100/2 = 50ms
Data transfer time:
In one rotation, we can read data on one complete track
= 100 × 500 = 50,000B data is read in one complete rotation
One complete rotation takes 100ms.
100ms → 50,000B
250B → 100/50000 × 250 = 0.5ms
∴ Avg. time to transfer = 249.5 × 50 + 0.5 = 300ms
● Array occupies primary memory(RAM). So it also not correct.
● Records used to access data on secondary devices record by record sequentially.
Rotational latency : The amount of time to rotate the track when the read/write head comes to desired sector position. In simple disk, rotational latency is the time to rotate 1⁄2 disk to the access.
Here, we have to find number of sectors per track.
Total storage capacity / (Number of bytes per sector*number of sides*number of tracks per side)
=21504000 / (512*2*600)
→ Average seek time is the average of all possible seek times which technically is the time to do all possible seeks divided by the number of all possible seeks, but in practice it is determined by statistical methods or simply approximated as the time of a seek over one third of the number of tracks.
Rotational latency (sometimes called rotational delay or just latency :
It is the delay waiting for the rotation of the disk to bring the required disk sector under the read-write head. It depends on the rotational speed of a disk (or spindle motor), measured in revolutions per minute (RPM).
→ Maximum latency = 60/rpm
→ Average latency = 0.5*Latency Time
Transfer Time Transfer time is the time to transfer the data. It depends on the rotating speed of the disk and number of bytes to be transferred. Disk Access Time Disk Access Time = Seek Time + Rotational Latency + Transfer Time
a plate of data
a cylinder of data
a track of data
a block of data
100 or more
125 or more
150 or more
175 or more
→ For each such alignment, some sectors of the two disks may have the same color. Since each sector of the smaller disk will match the same color sector of the larger disk 100 times among all the 200 ways and there are 200 sectors in the smaller disk, the total number of matched color sectors among the 200 ways is 100 × 200 = 20,000.
→ Note that there are only 200 ways. Then there is at least one way that the number of matched color sectors is 20,000 / 200 = 100 or more.
high-speed direct access storage device.
low-speed direct access storage device.
medium-speed direct access storage device.
high-speed sequential access storage device.
Non-Uniform distribution of requests
arm starting or stopping inertia
higher capacity of tracks on the periphery of the plate
use of unfair arm scheduling policies
Maximum capacity of the disk = 100*10*10*128 bytes
= 10^4 * 128 bytes
= (128*10^4 / 1024) KB
= (10^4/ 8) = 1250 KB
Hence, Position of sector < 2000, 15, 55>
Consider the disk which has average seek time of 32 ns and rotational rate of 360 rpm (round per minute), each track of the disk has 512 sectors, each of size 512 bytes.
What is the time taken to read four continuous sectors? And what is the data transfer rate?
0.0443s and 936 Kbps
0.0843s and 1536 Kbps
0.1043s and 1736 kbps
0.0943s and 1636 kbps
Better disk throughout but poor disk space utilization
Better disk throughout and better disk space utilization
Poor disk throughput but better disk space utilization
Poor disk throughput and poor disk space utilization