Disk structure and access time in DBMS Theory - Practice Problems & Coding Challenges

Challenge - 5 Problems

🧠 Conceptual

intermediate

Understanding Disk Access Components

Which of the following components primarily determines the seek time when accessing data on a disk?

A The time taken to transfer data from the disk to memory

B The time taken for the disk arm to move to the correct track

C The time taken for the disk to rotate to the correct sector

D The time taken by the CPU to process the data

💡 Hint

Seek time is related to the movement of the disk's mechanical parts.

Seek time is the time required for the disk arm to move the read/write head to the track where the data is stored. It depends on the mechanical movement of the arm.

📋 Factual

intermediate

Components of Disk Access Time

Which of the following correctly lists the three main components of disk access time?

A Seek time, rotational latency, data transfer time

B CPU time, seek time, data transfer time

C Rotational latency, CPU time, memory access time

D Data transfer time, CPU time, network latency

💡 Hint

Think about the mechanical and data movement delays involved in reading from a disk.

Disk access time is made up of seek time (moving the arm), rotational latency (waiting for the sector to rotate under the head), and data transfer time (moving data to memory).

🔍 Analysis

advanced

Calculating Average Rotational Latency

A disk rotates at 7200 revolutions per minute (RPM). What is the average rotational latency in milliseconds?

A 16.67 ms

B 8.33 ms

C 2.08 ms

D 4.17 ms

💡 Hint

Average rotational latency is half the time for one full rotation.

One rotation time = 60 seconds / 7200 = 0.00833 seconds = 8.33 ms. Average latency is half of that = 4.17 ms.

❓ Comparison

advanced

Comparing Access Times of SSD and HDD

Which statement best explains why SSDs generally have faster access times than HDDs?

A SSDs spin faster than HDDs, reducing rotational latency

B SSDs have slower data transfer rates but faster CPU processing

C SSDs have no moving parts, eliminating seek time and rotational latency

D SSDs use larger platters to store data, increasing transfer speed

💡 Hint

Think about the mechanical differences between SSDs and HDDs.

SSDs use flash memory with no moving parts, so they do not have seek time or rotational latency, making access faster than HDDs which rely on mechanical movement.

❓ Reasoning

expert

Effect of Disk Scheduling on Access Time

Consider a disk scheduling algorithm that minimizes seek time by reordering requests. Which of the following is a likely effect of this algorithm on overall disk access time?

A It reduces average seek time but may increase wait time for some requests

B It increases rotational latency by randomizing request order

C It eliminates data transfer time by preloading data

D It increases CPU processing time without affecting disk access

💡 Hint

Think about trade-offs when reordering disk requests to reduce arm movement.

Disk scheduling algorithms like SSTF reduce average seek time by reordering requests, but some requests may wait longer, increasing their individual wait times.

Practice

1. Which of the following components is NOT part of the disk access time?

easy

A. Seek time

B. Rotational latency

C. Transfer time

D. Cache size

Solution

1. Step 1: Understand disk access time components

   Disk access time includes seek time (moving the head), rotational latency (waiting for the sector), and transfer time (reading/writing data).

2. Step 2: Identify the unrelated component

   Cache size is related to memory and buffering, not directly part of disk access time.

3. Final Answer:

   Cache size -> Option D

4. Quick Check:

   Disk access time excludes cache size [OK]

Hint: Remember access time parts: seek, rotate, transfer only [OK]

Common Mistakes:

• Confusing cache size with access time
• Including CPU time as access time
• Mixing memory and disk terms

2. Which of the following correctly describes the structure of a disk?

easy

A. A disk is divided into tracks and sectors

B. A disk is divided into blocks and pages

C. A disk is divided into clusters and bytes

D. A disk is divided into frames and segments

Solution

1. Step 1: Recall disk physical structure

   Disks are physically divided into circular tracks and each track is divided into sectors.

2. Step 2: Match terms with disk structure

   Blocks, pages, clusters, bytes, frames, and segments are terms used in memory or file systems, not the physical disk layout.

3. Final Answer:

   A disk is divided into tracks and sectors -> Option A

4. Quick Check:

   Disk = tracks + sectors [OK]

Hint: Tracks are circles; sectors are slices of those circles [OK]

Common Mistakes:

• Confusing file system units with disk physical units
• Mixing memory terms with disk terms

3. If a disk has a seek time of 4 ms, rotational latency of 3 ms, and transfer time of 2 ms, what is the total access time?

medium

A. 6 ms

B. 9 ms

C. 12 ms

D. 7 ms

Solution

1. Step 1: Add all components of access time

   Total access time = seek time + rotational latency + transfer time = 4 ms + 3 ms + 2 ms.

2. Step 2: Calculate the sum

   4 + 3 + 2 = 9 ms total access time.

3. Final Answer:

   9 ms -> Option B

4. Quick Check:

   4 + 3 + 2 = 9 ms [OK]

Hint: Sum seek, rotate, transfer times for total access [OK]

Common Mistakes:

• Forgetting to add all three times
• Mixing units
• Adding only two components

4. A student calculates disk access time as seek time + transfer time only, ignoring rotational latency. What is the likely effect?

medium

A. The access time will be underestimated

B. The access time will be overestimated

C. The access time will be correct

D. The access time will be zero

Solution

1. Step 1: Understand the components of access time

   Access time includes seek time, rotational latency, and transfer time. Ignoring rotational latency misses part of the delay.

2. Step 2: Effect of ignoring rotational latency

   Ignoring rotational latency means the total time is less than actual, so the access time is underestimated.

3. Final Answer:

   The access time will be underestimated -> Option A

4. Quick Check:

   Missing rotational latency lowers total time [OK]

Hint: Always include rotational latency in access time [OK]

Common Mistakes:

• Ignoring rotational latency
• Assuming transfer time covers all delays
• Confusing seek time with rotational latency

5. A disk has 5000 tracks and the average seek time is proportional to the square root of the number of tracks. If the average seek time for 2500 tracks is 4 ms, what is the average seek time for 5000 tracks?

hard

A. 4 ms

B. 8 ms

C. 5.66 ms

D. 2.83 ms

Solution

1. Step 1: Understand the proportionality

   Average seek time ∝ √(number of tracks). Given seek time for 2500 tracks is 4 ms.

2. Step 2: Calculate seek time for 5000 tracks

   Ratio of seek times = √5000 / √2500 = √2 ≈ 1.414. So, new seek time = 4 ms x 1.414 ≈ 5.66 ms.

3. Final Answer:

   5.66 ms -> Option C

4. Quick Check:

   Seek time scales with sqrt(tracks) [OK]

Hint: Use square root ratio to scale seek time [OK]

Common Mistakes:

• Using linear ratio instead of square root
• Mixing up track counts
• Forgetting to multiply by original time