Why disk scheduling reduces seek time in Operating Systems - Visual Breakdown

Concept Flow - Why disk scheduling reduces seek time

Requests arrive

↓

Add to queue

↓

Disk scheduler selects next request

↓

Disk head moves to request location

↓

Request serviced

↓

Repeat for next request

Disk scheduling organizes requests to minimize disk head movement, reducing seek time by choosing the next request smartly.

Execution Sample

Operating Systems

Requests = [98, 183, 37, 122, 14, 124, 65, 67]
Head = 53
Schedule = SCAN
# Process requests in order minimizing head movement

This example shows how SCAN scheduling processes disk requests to reduce seek time by moving the head in one direction servicing requests.

Analysis Table

Step	Current Head Position	Next Request Chosen	Seek Distance	Total Seek Time
1	53	65	12	12
2	65	67	2	14
3	67	98	31	45
4	98	122	24	69
5	122	124	2	71
6	124	183	59	130
7	183	37	146	276
8	37	14	23	299

💡 All requests serviced; total seek time is 299 units

State Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	After 6	After 7	Final
Head Position	53	65	67	98	122	124	183	37	14
Total Seek Time	0	12	14	45	69	71	130	276	299

Key Insights - 2 Insights

Why does the scheduler choose request 65 after starting at 53 instead of 37?

Because SCAN moves the head in one direction (upwards), it picks the closest request in that direction (65) to reduce back-and-forth movement, as shown in execution_table step 1.

Why is the seek distance between 183 and 37 so large?

After reaching the highest request (183), the head reverses direction to service remaining requests (37), causing a large seek distance, visible in execution_table step 7.

Visual Quiz - 3 Questions

Look at the execution_table, what is the head position after servicing the 4th request?

A 122

B 98

C 124

D 67

💡 Hint

Check the 'Current Head Position' column at step 4 in execution_table.

At which step does the total seek time first exceed 50 units?

A Step 5

B Step 3

C Step 4

D Step 6

💡 Hint

Look at the 'Total Seek Time' column in execution_table to find when it passes 50.

If the scheduler did not use SCAN and picked requests randomly, how would the total seek time likely change?

A It would decrease

B It would increase

C It would stay the same

D It would be zero

💡 Hint

Refer to the concept that SCAN reduces seek time by ordering requests to minimize head movement.

Concept Snapshot

Disk scheduling reduces seek time by ordering requests to minimize disk head movement.
SCAN moves the head in one direction servicing requests in order.
This avoids unnecessary back-and-forth movement.
Result: faster access and less wear on the disk.
Total seek time is sum of all head movements between requests.

Full Transcript

Disk scheduling helps reduce the time the disk head spends moving between requests. When multiple requests come in, the scheduler arranges them to minimize the distance the head travels. For example, the SCAN algorithm moves the head in one direction, servicing requests as it goes, then reverses direction. This reduces back-and-forth movement. In the example, starting at position 53, the head moves to requests 65, 67, 98, and so on, accumulating seek time. The total seek time after all requests is 299 units. This organized approach is faster than random servicing and reduces disk wear.

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