SCAN (elevator algorithm) in Operating Systems - Time & Space Complexity

Time Complexity: SCAN (elevator algorithm)

O(n)

Understanding Time Complexity

We want to understand how the SCAN disk scheduling algorithm's running time changes as the number of disk requests grows.

Specifically, how does the number of requests affect the total work done by SCAN?

Scenario Under Consideration

Analyze the time complexity of the following SCAN algorithm snippet.

function SCAN(requests, head, direction) {
  sort requests in ascending order
  while requests remain {
    if direction is up {
      service requests >= head in order
      move head to max request
      direction = down
    } else {
      service requests <= head in reverse order
      move head to min request
      direction = up
    }
  }
}
    

This code moves the disk head back and forth, servicing requests in order along the way.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Traversing the sorted list of requests to service them in order.
• How many times: Each request is visited once during the head's movement in one direction.

How Execution Grows With Input

As the number of requests increases, the algorithm must visit each request once while moving the head back and forth.

Input Size (n)	Approx. Operations
10	About 10 moves to service all requests
100	About 100 moves to service all requests
1000	About 1000 moves to service all requests

Pattern observation: The total work grows roughly in direct proportion to the number of requests.

Final Time Complexity

Time Complexity: O(n)

This means the time to service all requests grows linearly with the number of requests.

Common Mistake

[X] Wrong: "SCAN takes longer than linear time because the head moves back and forth multiple times."

[OK] Correct: Even though the head moves back and forth, each request is only serviced once, so the total work still grows linearly with the number of requests.

Interview Connect

Understanding SCAN's time complexity helps you explain how disk scheduling balances efficiency and fairness, a useful skill in system design discussions.

Self-Check

What if the requests were not sorted before servicing? How would that affect the time complexity?

Practice

1. What is the main idea behind the SCAN (elevator) disk scheduling algorithm?

easy

A. Requests are served in the exact order they arrive without reordering.

B. The disk arm moves in one direction, servicing requests until it reaches the end, then reverses direction.

C. The disk arm jumps directly to the closest request regardless of direction.

D. The disk arm moves randomly to service requests.

Solution

1. Step 1: Understand SCAN movement

   SCAN moves the disk arm in one direction, serving all requests along the way until it reaches the last track in that direction.

2. Step 2: Direction reversal

   After reaching the end, the arm reverses direction and serves requests on the way back, similar to an elevator moving up and down.

3. Final Answer:

   The disk arm moves in one direction, servicing requests until it reaches the end, then reverses direction. -> Option B

4. Quick Check:

   SCAN = moves one way then reverses [OK]

Hint: Think of SCAN like an elevator moving up and down [OK]

Common Mistakes:

• Confusing SCAN with FCFS (first-come, first-served)
• Thinking SCAN jumps directly to closest request
• Assuming SCAN moves randomly

2. Which of the following correctly describes the order in which SCAN services disk requests?

easy

A. It services the closest request next, regardless of direction.

B. It services requests in the order they arrive without reordering.

C. It services requests by moving the arm in one direction, then reverses at the end.

D. It randomly selects requests to service next.

Solution

1. Step 1: Identify SCAN's servicing order

   SCAN moves the disk arm in one direction, servicing all requests in that path.

2. Step 2: Direction reversal at end

   When the arm reaches the last request or the disk end, it reverses direction and services requests on the return path.

3. Final Answer:

   It services requests by moving the arm in one direction, then reverses at the end. -> Option C

4. Quick Check:

   SCAN = one direction then reverse [OK]

Hint: Remember SCAN moves like an elevator, up then down [OK]

Common Mistakes:

• Confusing SCAN with SSTF (closest request first)
• Assuming SCAN ignores direction
• Thinking SCAN services requests randomly

3. Given the disk requests at tracks [10, 22, 20, 2, 40, 6] and the disk arm currently at track 15 moving towards higher tracks, which request will SCAN service first?

medium

A. Track 20

B. Track 10

C. Track 6

D. Track 2

Solution

1. Step 1: Identify current arm position and direction

   The arm is at track 15 and moving towards higher tracks (upwards).

2. Step 2: Find the closest request in the moving direction

   Requests above 15 are 20, 22, and 40. The closest is track 20.

3. Final Answer:

   Track 20 -> Option A

4. Quick Check:

   SCAN services requests ahead in current direction first [OK]

Hint: Pick closest request ahead in current direction [OK]

Common Mistakes:

• Choosing requests behind current arm position
• Ignoring arm's moving direction
• Picking the closest request overall instead of in direction

4. A disk arm is at track 50 moving towards track 0. The requests are at tracks [10, 20, 30, 40, 60]. Which request will SCAN service next? Identify the error in this reasoning: "Since the arm is moving towards 0, it will service track 60 next because it is the closest."

medium

A. The arm will service track 60 next; no error in reasoning.

B. The arm will service track 20 next; the error is ignoring request order.

C. The arm will service track 10 next; the error is assuming direction matters.

D. The arm will service track 40 next; the error is ignoring the arm's direction.

Solution

1. Step 1: Understand arm direction and position

   The arm is at track 50 moving towards 0 (downwards).

2. Step 2: Identify requests in arm's direction

   Requests below 50 are 10, 20, 30, 40; above 50 is 60.

3. Step 3: Find next request in direction

   The closest request in the moving direction is track 40, not 60.

4. Final Answer:

   The arm will service track 40 next; the error is ignoring the arm's direction. -> Option D

5. Quick Check:

   SCAN respects arm direction when choosing requests [OK]

Hint: Always consider arm direction before picking requests [OK]

Common Mistakes:

• Ignoring arm direction and picking closest request overall
• Assuming SCAN jumps to farthest request
• Confusing SCAN with SSTF

5. A disk arm is at track 50 on a disk with tracks numbered 0 to 199. Requests are at tracks [95, 180, 34, 119, 11, 123, 62, 64]. The arm is moving towards higher tracks. Using the SCAN algorithm, what is the order in which the requests will be serviced?

hard

A. [62, 64, 95, 119, 123, 180, 199, 34, 11]

B. [62, 64, 95, 119, 123, 180, 34, 11]

C. [95, 119, 123, 180, 199, 64, 62, 34, 11]

D. [95, 119, 123, 180, 64, 62, 34, 11]

Solution

1. Step 1: Identify requests above current arm position

   Arm at 50 moving up; requests above 50 are 62, 64, 95, 119, 123, 180.

2. Step 2: Service requests upwards in order

   Requests are serviced in ascending order: 62, 64, 95, 119, 123, 180.

3. Step 3: Reach end of disk and reverse

   SCAN continues to the last track (199) before reversing direction.

4. Step 4: Service remaining requests on return

   On return, service requests below 50 in descending order: 34, then 11.

5. Final Answer:

   [62, 64, 95, 119, 123, 180, 199, 34, 11] -> Option A

6. Quick Check:

   SCAN = up servicing, end, then down servicing [OK]

Hint: Service all requests up, go to end, then service down [OK]

Common Mistakes:

• Skipping the last track before reversing
• Not servicing requests in sorted order
• Ignoring requests on return path