12.2 explain why SSTF scheduling has tendency to favor the center cylinders of a disk over the innermost and outermost cylinders.

You are watching: None of the disk-scheduling disciplines, except fcfs, is truly fair (starvation may occur).

The SSTF algorithm is biased toward the middle cylinders in much the same way the SCAN algorithm is. Since, in the SSTF algorithm, the closest cylinder is always chosen, climate all middle cylinder references will serviced on the way to the finish of the disk.
I current the book's answer come this question, also though i think the is an ext muddled... Just FYI:
The center of the decaying is the location having actually the smallest median distance come all other tracks. For this reason the disc head tends to relocate away indigenous the edge of the disk.
Here is another method to think of it. The existing location that the head divides the cylinders into two groups. If the head is not in the center of the disk and a brand-new request arrives, the new request is more likely to it is in in the group that includes the center of the disk; thus, the head is an ext likely to relocate in that direction.
12.15 nobody of the disk-scheduling disciplines, except FCFS, is truly fair
(starvation may occur).
a. Define why this delinquent is true.
b. Explain a way to change algorithms such as SCAN come ensure fairness.
c. Define why fairness is an essential goal in a time-sharing system.
d. Offer three or an ext examples of circumstances in which it is essential that the operating device be
unfair in offer I/O requests.
(a) new requests for the track end which the head at this time resides can theoretically arrive as easily as
these requests room being serviced.
(b) every requests larger than part predetermined age might be required to the height of the queue, and also an
associated bit for each could be set to indicate that no brand-new request can be relocated ahead of this requests.
For SSTF, the rest of the queue would need to be reorganized through respect come the last of this old requests.
(c) To prevent unusually long response times.
(d) Paging and swapping should take priority end user requests. It may be desirable for various other kernelinitiated
I/O, such as the writing of ¯le system metadata, to take precedence end user I/O. If the kernel
supports real-time procedure priorities, the I/O inquiry of those processes have to be favored.
12.16 expect that a disc drive has actually 5000 cylinders, numbered 0 come 4999.
The journey is currently serving a inquiry at cylinder 143, and also the previous request was in ~ cylinder 125. The
queue of pending requests, in FIFO order, is
86, 1470, 913, 1774, 948, 1509, 1022, 1750, 130
Starting native the present head position, what is the complete distance (in cylinders) that the disk eight moves
to satisfy all the pending requests, because that each of the adhering to disk scheduling algorithms? a. FCFS b. SSTF
c. SCAN d. Look at e. C-SCAN.
a. The FCFS schedule is 143, 86, 1470, 913, 1774, 948, 1509, 1022, 1750, 130. The total seek street is
b. The SSTF schedule is 143, 130, 86, 913, 948, 1022, 1470, 1509, 1750, 1774. The complete seek distance is
c. The SCAN schedule is 143, 913, 948, 1022, 1470, 1509, 1750, 1774, 4999, 130, 86. The complete seek distance
is 9769.
d. The look at schedule is 143, 913, 948, 1022, 1470, 1509, 1750, 1774, 130, 86. The total seek street is
e. The C-SCAN schedule is 143, 913, 948, 1022, 1470, 1509, 1750, 1774, 4999, 86, 130. The full seek
distance is 9813.
f. (Bonus) The C-LOOK schedule is 143, 913, 948, 1022, 1470, 1509, 1750, 1774, 86, 130. The complete seek
distance is 3363.
12.17 muy larga...
12.18 se toma en cuenta el 12.17...
Suppose the the disc in practice 12.17 rotates in ~ 7200 RPM.
a. What is the median rotational latency that this disk drive?
b. What look for distance deserve to be covered while that you discovered
for component a?
a. 7200 rpm offers 120 rotations every second. Thus, a complete rotation
take away 8.33 ms, and the typical rotational latency (a fifty percent rotation)
take away 4.167 ms.
b. Fixing t =0.7561 + 0.2439 pL because that t =4.167 provides L = 195.58,
for this reason we can seek over 195 tracks (about 4% that the disk) throughout
an average rotational latency.
12.19 The speeding up seek defined in practice 12.17 is common of hard-disk
drives. By contrast, floppy disks (and plenty of hard disks manufactured
before the mid-1980s) generally seek at a addressed rate. Expect that the
disk in exercise 12.3 has a constant-rate seek rather than a constantacceleration
seek, for this reason the look for time is the the type t = x + yL, whereby t
is the time in milliseconds and L is the look for distance. Mean that the
time to look for to an adjacent cylinder is 1 millisecond, together before, and also is
0.5 millisecond for each extr cylinder.
a. Write an equation for this seek time as a role of the seek
b. Utilizing the seek-time role from component a, calculate the total seek
time for each that the schedules in exercise 12.2. Is her answer
the very same as it to be for practice 12.3(c)?
c. What is the portion speedup the the faster schedule over FCFS
in this case?
a. T = 0.95 + 0.05L
b. FCFS 362.60; SSTF 95.80; SCAN 497.95; watch 174.50; C-SCAN 500.15;
(and C-LOOK 176.70). SSTF is still the winner, and LOOK is the

See more: Who Sang The Song Hooked On A Feeling ': The Story Behind B

c. (362.60 − 95.80)/362.60 = 0.74 The portion speedup the SSTF
over FCFS is 74%, through respect to the seek time. If we encompass the
overhead of rotational latency and also data transfer, the percentage
speedup will be less.
Copy lines Copy permalink