Why disk scheduling reduces seek time in Operating Systems - Challenge Your Understanding

Challenge - 5 Problems

🧠 Conceptual

intermediate

How does disk scheduling reduce seek time?

Disk scheduling algorithms aim to reduce the seek time of the disk head. Which of the following best explains how they achieve this?

A By rearranging the order of pending disk requests to minimize the movement of the disk head.

B By increasing the speed of the disk head motor to move faster between requests.

C By reducing the number of disk requests generated by the system.

D By storing all data in memory to avoid disk access altogether.

💡 Hint

Think about how changing the order of requests affects the distance the disk head travels.

Disk scheduling algorithms reorder the sequence of disk access requests to reduce the total distance the disk head must move. This minimizes seek time, which is the time taken to position the head over the correct track.

📋 Factual

intermediate

Which disk scheduling algorithm is designed to reduce seek time by servicing requests in one direction before reversing?

Among the common disk scheduling algorithms, which one services requests in one direction until the end, then reverses direction, helping reduce seek time?

A Shortest Seek Time First (SSTF)

B SCAN (Elevator Algorithm)

C First-Come, First-Served (FCFS)

D Random Scheduling

💡 Hint

Think of an elevator moving up and down, servicing requests along the way.

The SCAN algorithm moves the disk head in one direction servicing all requests until it reaches the end, then reverses direction. This reduces seek time by minimizing back-and-forth movement.

🔍 Analysis

advanced

Analyzing seek time reduction with SSTF algorithm

Given the disk requests at tracks 10, 22, 20, 2, 40, and the disk head currently at track 15, which request will the Shortest Seek Time First (SSTF) algorithm service next to reduce seek time?

A Track 10

B Track 2

C Track 22

D Track 20

💡 Hint

Calculate the distance from the current head position to each request and pick the smallest.

SSTF selects the request closest to the current head position. From track 15, distances are: 10 (5), 22 (7), 20 (5), 2 (13), 40 (25). Both 10 and 20 are 5 tracks away, but SSTF typically picks the closest request found first, here track 10.

❓ Comparison

advanced

Comparing FCFS and SCAN in terms of seek time

Which statement correctly compares the seek time behavior of First-Come, First-Served (FCFS) and SCAN disk scheduling algorithms?

A Both FCFS and SCAN have the same seek time because they process all requests eventually.

B FCFS always results in lower seek time than SCAN because it processes requests in arrival order.

C SCAN increases seek time because it moves the disk head back and forth unnecessarily.

D SCAN generally reduces seek time compared to FCFS by servicing requests in a sorted order along the disk tracks.

💡 Hint

Consider how the order of servicing requests affects the total distance the disk head travels.

SCAN reduces seek time by moving the disk head in one direction and servicing requests in order, minimizing back-and-forth movement. FCFS can cause more random movement, increasing seek time.

❓ Reasoning

expert

Why does reducing seek time improve overall system performance?

Reducing seek time is a goal of disk scheduling. Why does this reduction improve overall system performance?

A Because it reduces the power consumption of the disk drive significantly.

B Because it increases the disk storage capacity by optimizing data placement.

C Because it decreases the time the CPU waits for data, allowing faster processing of tasks.

D Because it eliminates the need for caching data in memory.

💡 Hint

Think about how disk access speed affects the CPU's ability to continue working.

Reducing seek time means the disk can deliver data faster. This reduces the time the CPU waits for data, improving the speed of running programs and overall system responsiveness.

Practice

1. Why does disk scheduling reduce the seek time in an operating system?

easy

A. It arranges disk requests to minimize the movement of the disk head.

B. It increases the speed of the disk motor.

C. It compresses the data on the disk to reduce size.

D. It duplicates data to multiple disks for faster access.

Solution

1. Step 1: Understand what seek time means

   Seek time is the time taken for the disk head to move to the track where data is stored.

2. Step 2: How scheduling affects seek time

   Disk scheduling arranges requests so the head moves less distance, reducing seek time.

3. Final Answer:

   It arranges disk requests to minimize the movement of the disk head. -> Option A

4. Quick Check:

   Less head movement = less seek time [OK]

Hint: Disk scheduling reduces head movement to cut seek time [OK]

Common Mistakes:

• Confusing seek time with data transfer speed
• Thinking disk speed changes physically
• Assuming data compression affects seek time

2. Which of the following is a correct reason why disk scheduling algorithms reduce seek time?

easy

A. They process requests in the order they arrive without change.

B. They increase the disk's rotational speed.

C. They reorder requests to reduce the distance the disk head moves.

D. They store all data in memory to avoid disk access.

Solution

1. Step 1: Identify how disk scheduling works

   Disk scheduling algorithms reorder requests to optimize head movement.

2. Step 2: Why reordering helps

   By reordering, the disk head travels less distance, reducing seek time.

3. Final Answer:

   They reorder requests to reduce the distance the disk head moves. -> Option C

4. Quick Check:

   Reordering requests = less seek time [OK]

Hint: Reordering requests reduces head travel distance [OK]

Common Mistakes:

• Thinking requests are always processed as received
• Confusing seek time with rotational speed
• Assuming all data is cached in memory

3. Consider a disk with requests at tracks 10, 22, 20, 2, and 40. The disk head is currently at track 20. Which scheduling approach will likely reduce seek time the most?

medium

A. Processing requests in the order: 10, 22, 20, 2, 40

B. Processing requests in the order: 20, 10, 2, 22, 40

C. Processing requests in the order: 40, 22, 20, 10, 2

D. Processing requests in the order: 20, 22, 40, 10, 2

Solution

1. Step 1: Identify current head position and requests

   The head is at 20; requests are 10, 22, 20, 2, 40.

2. Step 2: Calculate seek distances for each option

   Processing requests in the order: 20, 10, 2, 22, 40 moves from 20 to 10 (10 tracks), then 10 to 2 (8 tracks), then 2 to 22 (20 tracks), then 22 to 40 (18 tracks). Total = 56 tracks, which is less than other options.

3. Final Answer:

   Processing requests in the order: 20, 10, 2, 22, 40 -> Option B

4. Quick Check:

   Minimized head movement = Processing requests in the order: 20, 10, 2, 22, 40 [OK]

Hint: Choose order minimizing total head travel [OK]

Common Mistakes:

• Ignoring current head position
• Not calculating total seek distance
• Assuming original request order is best

4. A disk scheduling algorithm is not reducing seek time as expected. Which of the following could be a likely cause?

medium

A. The algorithm processes requests strictly in arrival order without reordering.

B. The disk head speed is too fast.

C. The disk has too much free space.

D. The disk uses a cache to store frequently accessed data.

Solution

1. Step 1: Identify what reduces seek time

   Seek time reduces when requests are reordered to minimize head movement.

2. Step 2: Analyze the cause of no improvement

   If requests are processed strictly in arrival order, no optimization happens, so seek time stays high.

3. Final Answer:

   The algorithm processes requests strictly in arrival order without reordering. -> Option A

4. Quick Check:

   No reordering = no seek time reduction [OK]

Hint: No reordering means no seek time improvement [OK]

Common Mistakes:

• Blaming disk speed or free space incorrectly
• Confusing caching with seek time reduction
• Ignoring request processing order

5. A disk scheduling algorithm uses the Shortest Seek Time First (SSTF) method. If the disk head is at track 50 and requests come in at tracks 45, 10, 55, 60, and 48, what is the order of servicing requests to minimize seek time?

hard

A. 50 -> 48 -> 45 -> 55 -> 60 -> 10

B. 50 -> 10 -> 45 -> 48 -> 55 -> 60

C. 50 -> 55 -> 60 -> 48 -> 45 -> 10

D. 50 -> 48 -> 45 -> 10 -> 55 -> 60

Solution

1. Step 1: Identify closest request to current head

   From 50, closest is 48 (distance 2), then from 48 closest 45 (3), then from 45 closest 10 (35), then from 10 closest 55 (45), then 60.

2. Step 2: Follow SSTF order

   Service 48 first, then 45, then 10, then 55, then 60.

3. Final Answer:

   50 -> 48 -> 45 -> 10 -> 55 -> 60 -> Option D

4. Quick Check:

   SSTF picks nearest request each time [OK]

Hint: Always pick nearest request next in SSTF [OK]

Common Mistakes:

• Choosing requests in arrival order
• Ignoring distance when selecting next request
• Mixing up SSTF with other algorithms