Gate 2015
Duration: 4 min
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AI Summary
An AI-generated summary of this video lecture.
The video presents a solution to a GATE 2015 problem concerning disk scheduling algorithms, specifically the optimized C-SCAN algorithm. The instructor begins by reading the problem statement, which involves a disk with tracks 0-255, a current head position at track 45 moving in the positive direction, and a queue of requests: 40, 67, 11, 240, 87. The goal is to determine the service order and the total seek distance. He draws a number line to visualize the tracks and marks the current head position and the request locations. He explains that C-SCAN moves in one direction (positive) servicing requests until the end of the disk, then jumps to the beginning and continues in the same direction. He identifies the requests in the positive direction (67, 87, 240) and those in the negative direction (11, 40) which will be serviced after the jump to track 0.
Chapters
0:00 – 2:00 00:00-02:00
The instructor introduces the problem statement, highlighting the key parameters: head at track 45, moving positive, requests {40, 67, 11, 240, 87}. He draws a horizontal line representing the disk tracks from 0 to 255. He marks the current head position at 45. He then plots the request tracks on the line: 11, 40, 67, 87, and 240. He explains the C-SCAN logic: the head moves towards the positive direction, servicing requests 67, 87, and 240. He notes that the head will continue to the end of the disk (track 255) before jumping to the start (track 0). He then identifies the remaining requests (11, 40) that will be serviced after the jump. He begins writing the calculation for the total seek distance on the board, starting with the distance from the current head to the end of the disk. He writes `(255 - 45)`.
2:00 – 4:03 02:00-04:03
The instructor continues the calculation for the total seek distance. He writes `(255 - 45)`, representing the distance from the current head to the end of the disk. He then adds `255` to this value, representing the distance of the jump from the end (255) to the start (0). Finally, he adds `40`, representing the distance from the start (0) to the last serviced request (40). He sums these values: 210 + 255 + 40 = 505. He circles the final answer 505 and selects option (C) as the correct choice. He briefly confirms the service order as 67, 87, 240, 11, 40, although the primary focus of this segment is the numerical calculation of the seek distance. He explicitly writes the sum `210 + 255 + 40` and the result `505` on the board, circling it to indicate the final answer.
The lecture effectively demonstrates how to apply the C-SCAN disk scheduling algorithm to a specific problem. By visualizing the track layout and the head movement, the instructor clarifies the service order and the calculation of seek distance. The key takeaway is the specific method for calculating C-SCAN seek distance, which includes the full traversal to the end, the jump to the start, and the traversal to the last request, resulting in a total distance of 505 units. This approach highlights the importance of understanding the specific variations of disk scheduling algorithms for exam purposes.