C-Scan Algorithm

Question

Consider the following track requests in the disk queue:

95, 180, 34, 119, 11, 123, 62, 64

The C-scan scheduling algorithm is used and the read/write head is positioned at location 50. If tracks are numbered from 0 to 199 and head moving toward  smaller track number on its servicing pass, then total seek time needed with 2 msec time to move from one track to another while servicing these requests is  ________ msec. [Assume moving from one end to another end will take 10 msec]

Comments

Total Head movement from 50 to 0 = 50 and 199 to 62 137 and switch time from track 0 to track 199 = 10msec so total time is 50*2 +137*2+10 = 384.

More Ref :- http://www.cs.iit.edu/~cs561/cs450/disksched/disksched.html

Answer 1

Total Head movement from 50 to 0 = 50 and 199 to 62 137 and switch time from track 0 to track 199 = 10msec so total time is 50*2 +137*2+10 = 384.

More Ref :- http://www.cs.iit.edu/~cs561/cs450/disksched/disksched.html

Comments

why multiply with 2

Answer 2

in the question they given that, " head moving toward  smaller track number on its servicing pass "

current head is 50 ==> move 34 then 11 then 0 ---> circular ---> at 199 with cost (50+10 = 60) ---> come up to 62 ( 199-62 = 137)

total movements = 60+137 = 197

total seek time = 197 * 2 msec = 394ms