Question
What will be the output of the following code when
implementing the SCAN I/O scheduling algorithm? def scan(requests, head, direction):   seek_sequence = []   distance = 0   cur_track = 0   if direction == "left":     requests.sort(reverse=True)   else:     requests.sort()   for track in requests:     if cur_track < track:       seek_sequence.append(cur_track)       distance += abs(track - cur_track)       cur_track = track   return seek_sequence, distance requests = [98, 180, 37, 122, 14, 124, 65, 67] head = 50 direction = "right" seek_sequence, total_distance = scan(requests, head, direction) print(seek_sequence) print(total_distance)Solution
In the SCAN scheduling algorithm, the head moves in a specific direction and services requests until it reaches the end of the disk. The code sorts the requests based on the specified direction ("right" in this case). The sequence of services starts at the head position of 50 and goes to the lowest request (14), then to the highest request. The output will include the complete sequence of service requests, which is [50, 14, 37, 65, 67, 98, 122, 124, 180]. Why Other Options Are Wrong: B) [50, 14, 37, 67, 65, 98, 124, 122, 180]: This option is incorrect as it misrepresents the order of service requests based on the head movement direction. C) [50, 14, 37, 67, 65, 122, 124, 180]: This option is incorrect because it omits the request for 98, which should be included in the scan order. D) [50, 37, 65, 67, 98, 180]: This option is incorrect as it omits several requests and misrepresents the sequence of servicing. E) [50, 67, 65, 98, 122, 124, 180]: This option is incorrect because it suggests servicing starts from 67 instead of the head position and does not correctly reflect the movement of the head.
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