sid1221 yes u r right .thnq
I really dont know why its given there like that.. Also it is not a IIT link, it is some Illunosis Institute of Technology..
More proof from Solutions to Galvin 7th edition..
See the f-bonus part.. try to calculate yourself.. youll see the long jump is considered.. PERIOD
Even the method given in this wikipedia link is wrong
"The head is initially at cylinder number 63, moving towards larger cylinder numbers on its servicing pass. "
hope that helps
Answer is 346 as already calculated in answers here.Those having some doubt regarding long jump ,check this image, Now in question Total Head Movements are asked,When Head reaches any End, There is no mechanism for head to jump directly to some arbitrary track.It has to Move.So it has to move along the tracks to reach Track Request on other side.Therefore head will move and we must count it.
since purpose of disk scheduling algorithms is to reduce such Head movements by finding an Optimal algorithm.If you ignore the move which is actually happening in disk,that doesn't serve the purpose of analyzing the algorithms.
Answer is 346.
In CSCAN, we have a longer jump, from one end of disk to the other.
We can use this method
Ans is 165 that huge jump in C-LOOK and C-SCAN doesn't count. That's why it's circular-LOOK and Circular-SCAN.