Disk Scheduling & Storage : MCQs & PYQs with Explanation

Option C

To calculate the maximum disk size, we use the formula:
Disk Size = Cylinders × Heads × Sectors per Track × Bytes per Sector.
Here, the important detail is that each head has two tracks, so the effective number of heads becomes 16*2 = 32. This increases the total storage capacity accordingly.
Substituting the values, we get, 1024*32*63*512 = 1,056,964,608 bytes. This represents the total number of bytes that can be stored on the disk based on the given configuration.
Now converting bytes into megabytes using decimal notation (1 MB = 106 bytes), we get approximately 1056 MB.

Option C

Disk scheduling algorithms manage the order in which disk I/O requests are processed. Common methods are FCFS (FIFO), SSTF, SCAN, LOOK, etc. The SCAN (also called Elevator algorithm) which moves the disk arm across the disk servicing requests in one direction.
In Batch Processing, jobs are processed in the order they arrive or according to some simple sequence. A common scheduling method is FIFO (First In First Out), where the first job to arrive is the first to be processed.
Time-sharing systems use Round Robin Scheduling to give each process a fair share of CPU.
Interrupts are handled in the reverse order of occurrence, often following a LIFO (Last In First Out) scheme where the most recent interrupt is serviced first.

Option B

In disk storage, data is organized into structures like tracks, sectors, and cylinders. The Sector is the smallest unit of data that can be physically read from or written to a disk.

• A block is a logical unit used by the operating system and may consist of one or more sectors. 
• A track is a circular path on the disk surface.
• A cylinder is a collection of tracks aligned vertically across platters.

Option A

Disk scheduling algorithms decide the order in which I/O requests are serviced to minimize disk head movement. Two common algorithms are:
SSTF (Shortest Seek Time First):

• Selects the request closest to the current head position.
• Reduces total head movement but can cause starvation for far-away requests.

SCAN (Elevator Algorithm):

• The disk head moves in one direction, servicing all requests until it reaches the end, then reverses direction.
• Provides more uniform wait times and avoids starvation.

Option D

RAID (Redundant Array of Independent Disks) is a storage technology that combines multiple physical disks into a single logical unit to improve performance and reliability (fault tolerance). Key characteristics of RAID include:

• Data Striping: Distributing data across multiple disks to improve read/write performance.
• Redundancy/Parity: Using extra disk space to store parity or mirrored data for fault tolerance.
• Logical View: The OS treats the RAID array as a single logical drive, even though it consists of multiple physical drives.

Option D

In RAID 5, data and parity information are distributed across all drives. The usable storage capacity is calculated using the formula:
RAID 5 Capacity = (N-1)*Size of Smallest Drive

• N is total number of drives.

In the given question, there are a total of 10 drives (Nine 1 TB drives and one 500 GB drive), and since RAID 5 always considers the smallest disk size, all drives are effectively treated as 500 GB each for capacity calculation.

• (10-1)*500 GB = 9*500 GB = 4500 GB = 4.5 TB

So,

• RAID 5 always considers the smallest disk size.
• One disk space is used for parity.
• Remaining disks store actual data.
   

Option A

Different Operating System techniques are used for different management tasks such as CPU scheduling, disk scheduling, page replacement, and deadlock handling.

Option A

The rotational speed of the hard disk is 6000 RPM, which means the disk completes 6000 rotations in one minute. To find the time for one complete rotation, we divide 60 seconds by 6000, giving 60/6000=0.01 seconds per rotation.
The average rotational latency is the time the system waits for the required sector to come under the read/write head. On average, this waiting time is half of one full rotation.
So, the average latency time is 0.01/2=0.005 seconds, which can also be written as 5*10^-3 seconds.

Option B

In SSTF (Shortest Seek Time First), at every step we choose the request which is closest to the current head position.
Given:

• Requests: 98, 183, 37, 122, 14, 124, 65, 67
• Initial head: 53

At each step, keep track of:

• Current head position.
• Remaining requests.
• Distance to each pending request.
• Choose the one with minimum distance.

Step 1: Head at 53
Pending: {98, 183, 37, 122, 14, 124, 65, 67}
Distances from 53:

• 65 → |65‑53| = 12
• 37 → |37‑53| = 16
• 67 → |67‑53| = 14
• 98 → |98‑53| = 45
• 14 → |14‑53| = 39
• 122 → |122‑53| = 69
• 124 → |124‑53| = 71
• 183 → |183‑53| = 130

Closest: 65 (Distance 12).
Movement: 53 → 65
Remaining: 98, 183, 37, 122, 14, 124, 67
Step 2: Head at 65
Pending: {98, 183, 37, 122, 14, 124, 67}
Distances from 65:

• 67 → 2, 37 → 28, 98 → 33, 122 → 57, 124 → 59, 14 → 51, 183 → 118

Closest: 67 (Distance 2).
Movement: 65 → 67
Remaining: 98, 183, 37, 122, 14, 124
Step 3: Head at 67
Pending: {98, 183, 37, 122, 14, 124}
Distances from 67:

• 37 → 30, 98 → 31, 122 → 55, 124 → 57, 14 → 53, 183 → 116

Closest: 37 (Distance 30).
Movement: 67 → 37
Remaining: 98, 183, 122, 14, 124
Step 4: Head at 37
Pending: {98, 183, 122, 14, 124}
Distances from 37:

• 14 → 23, 98 → 61, 122 → 85, 124 → 87, 183 → 146

Closest: 14 (Distance 23).
Movement: 37 → 14
Remaining: 98, 183, 122, 124
Step 5: Head at 14
Pending: {98, 183, 122, 124}
Distances from 14:

• 98 → 84, 122 → 108, 124 → 110, 183 → 169

Closest: 98 (Distance 84).
Movement: 14 → 98
Remaining: 183, 122, 124
Step 6: Head at 98
Pending: {183, 122, 124}
Distances from 98:

• 122 → 24, 124 → 26, 183 → 85

Closest: 122 (Distance 24).
Movement: 98 → 122
Remaining: 183, 124
Step 7: Head at 122
Pending: {183, 124}
Distances from 122:

• 124 → 2, 183 → 61

Closest: 124 (Distance 2).
Movement: 122 → 124
Remaining: 183
Step 8: Head at 124
Pending: {183}
Distances from 124:

• 183 → 59

Movement: 124 → 183
Total Head Movement: 
12+2+30+23+84+24+2+59 = 236 Cylinders.

Option A

Each sector of the disk stores 512 bytes, which means 1 KB = 2 sectors. Therefore, a file of size 42797 KB requires 42797*2 = 85594 sectors in total.
The disk has 16 recording surfaces and 64 sectors per surface, so each cylinder contains 16*64 = 1024 sectors. This means that once 1024 sectors are filled, the storage moves to the next cylinder.
The file starts at disk location <1200, 9, 40>. The starting sector number within cylinder 1200 is calculated as (9*64)+40 = 616. Hence, the number of free sectors remaining in cylinder 1200 is 1024-616 = 408 sectors.
After filling these 408 sectors, the remaining sectors to be stored are 85594-408 = 85186 sectors. These remaining sectors occupy 83 complete cylinders because 83*1024 = 84992 sectors.
Since some sectors are still left after filling these 83 cylinders, they go into the next cylinder, which is cylinder number 1284.

Option B

In a disk scheduling system, the disk head moves from one cylinder to another to satisfy disk access requests. The time taken depends on the total head movement.
In the Shortest Seek Time First (SSTF) algorithm, the disk head always moves to the request that is closest to the current head position. This minimizes seek time at each step.
Given:

• Total Cylinders = 100
• Initial Head Position = 50
• Request Queue = 34, 4, 7, 19, 10, 83, 2, 6, 20, 15
• Time taken to Move 1 Cylinder = 1 Second

Now process the requests step-by-step using SSTF.

Now Add all Head Movements : 16+14+1+4+5+3+1+2+2+81 = 129
Therefore, the total time taken is 129 seconds.

Option A

In an operating system, multiple processes often request disk I/O operations at the same time. To manage these requests efficiently, the OS uses Disk Scheduling Algorithms, which decide the order in which the disk head (arm) moves to service requests. The main goal is to reduce Seek Time, which is the time taken by the disk arm to move to the track where data is located.

Among the given options, the Shortest-Service-Time-First (SSTF) algorithm selects the disk I/O request that is closest to the current position of the disk arm. This means the arm moves the least possible distance, which minimizes seek time and improves performance.

Option C

• FAT32 generally has lower disk utilization compared to NTFS because it uses larger cluster sizes, which leads to more unused space.
• NTFS supports advanced features like better caching, journaling, and efficient space management, which usually makes its read and write performance better than FAT32, especially on large disks.
• FAT32 does not support individual files up to 32 GB. The maximum file size supported by FAT32 is only 4 GB.
• NTFS stands for New Technology File System.

Option C

In a disk system, data is stored on circular tracks, and a read/write head is used to access this data. When a request is made, the head must first move to the correct track where the data is located. The time taken for this movement is an important performance factor.
Seek Time is defined as the time required to move the read/write head to the desired track.

Option C

RAID (Redundant Array of Independent Disks) is a storage technology that combines multiple physical disks into one logical unit for better performance, reliability, or both. Each RAID level has a specific method for distributing and storing data and parity across disks.