Disk scheduling algorithm - SCAN and C-SCAN I'm studying several disk scheduling algorithms; however, I'm kinda confused with SCAN. I understand the concept, but I don't understand the example given in the book. I think that the author made a mistake, but I want to be sure.
The example is the following:

Description of SCAN: The disk arm starts at one end of the disk, and moves toward the other
end, servicing requests until it gets to the other end of the disk,
where the head movement is reversed and servicing continues.
Suppose that the cylinder range goes from 0 to 199 and you have the following queue:
98, 183, 37, 122, 14, 124, 65, 67
Head pointer: 53
As seen in the image, the total head movement is of 208 cylinders


I don't understand why 208 cylinders. If I understand correctly, the head will start at 53 and then move to 0 (that's 53 cylinders), then the head will move up and the last service requested is at 183 (that's another 183 cylinders). So the total head movement should be 53 + 183 = 236.
The book also has an image and description of C-SCAN:

C-SCAN moves the head from one end of the disk to the other, servicing requests
along the way. When the head reaches the other end, however, it immediately returns
to the beginning of the disk without servicing any requests on the return trip.
The C-SCAN scheduling algorithm essentially treats the cylinders as a circular
list that wraps around from the ﬁnal cylinder to the ﬁrst one.


Using the same example, with C-SCAN the cylinder would start at 53, then move up to 199 (146 cylinder movements), then go to 0 (199 movements), and then the last request would be at 37 (37 cylinder movements). The total cylinder movements would be 146+199+37=382. My question here is, do we also have to count the cylinders when the head goes back to 0 since it isn't servicing any requests?
Note: This information was extracted from Operating System Concepts with Java - 8th Edition by Silberschatz, Galvin, Gagne.
You can also find this part of the chapter here.
Here are the examples of various disk scheduling algorithms.

Yes you are correct in both the case. The author here has made mistake. 
For SCAN: The answer should be 236.
For C-SCAN: The answer should be 146+37=183.(Note: The head movement when its not servicing any requests is not counted.)
Refer the link, it gives a clear concept of how the different kinds of disk scheduling algorithms are used in practice.