Why disk scheduling reduces seek time in Operating Systems - See It in Action

Why Disk Scheduling Reduces Seek Time

📖 Scenario: Imagine you work in a library where you need to fetch books from different shelves. Moving from one shelf to another takes time. You want to find a way to pick books in an order that saves your walking time.

🎯 Goal: Build a simple example that shows how ordering requests (disk scheduling) helps reduce the total movement (seek time) when accessing data on a disk.

📋 What You'll Learn

Create a list of disk requests representing positions on the disk.

Set a starting position of the disk head.

Apply a simple scheduling method to reorder requests to minimize movement.

Calculate the total seek time based on the reordered requests.

💡 Why This Matters

🌍 Real World

Disk scheduling is used in computers to decide the order of reading or writing data on a hard drive, which helps the system work faster and more efficiently.

💼 Career

Understanding disk scheduling is important for roles in system administration, software development, and IT support, where optimizing hardware performance is valuable.

1

Create the list of disk requests

Create a list called requests with these exact disk positions: 95, 180, 34, 119, 11, 123, 62, 64.

Operating Systems

# Create the list of disk requests
# Your code here

Hint

Use square brackets to create a list and separate numbers with commas.

2

Set the starting position of the disk head

Create a variable called head_position and set it to 50 to represent the current position of the disk head.

Operating Systems

requests = [95, 180, 34, 119, 11, 123, 62, 64]
# Set the starting position of the disk head
# Your code here

Hint

Use a simple assignment to create the variable.

3

Reorder requests to reduce seek time

Create a new list called scheduled_requests that contains the requests sorted by their distance from head_position. Use the sorted() function with a key that calculates the absolute difference from head_position.

Operating Systems

requests = [95, 180, 34, 119, 11, 123, 62, 64]
head_position = 50
# Reorder requests to minimize seek time
# Your code here

Hint

Use sorted() with a lambda function to sort by distance.

4

Calculate total seek time after scheduling

Create a variable called total_seek_time and set it to 0. Then use a for loop with variables i from 0 to len(scheduled_requests) - 1 to add the absolute difference between consecutive requests (starting from head_position) to total_seek_time. Use previous_position to track the last position, starting with head_position.

Operating Systems

requests = [95, 180, 34, 119, 11, 123, 62, 64]
head_position = 50
scheduled_requests = sorted(requests, key=lambda x: abs(x - head_position))
# Calculate total seek time
# Your code here

Hint

Use a loop to add distances between each request and the previous position.

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