Operating System

Chapter 7
Exercise #3 – Given a disk pack with 10 platters yielding 18 recordable surfaces…
A = Rotational speed = 10ms
B = Transfer rate = 0.1ms/track
C = Density per track = 19,000 bytes
D = Number of records to be store = 200,000 records
E = Size of each record = 160 bytes
F = Block size = 10 logical records
G = Number of tracks per surface = 500

a) Number of blocks per track:
(C) / (E*F) 19,000 / (160*10)
19,000 / 1600
11.875
11 BLOCKS
b) Waste per track:
C – (E*F*11)
19,000 – (10*160*11)
19,000 – 17,600
1,400 BYTES
c) Number of tracks required to store the entire file:
(D*E) / C
(200,000*160)/19,000
32,.00,000/19,000
1,684.2105263157894
1,684 TRACKS
d) Total waste to store the entire file:
1,400 * 1,684
2,357,600 BYTES
e) Time to write all of the blocks:
(A+((B*11)*F))*(11*1,684)
(10+((.1*11)*10))*(11*1,684)
(10+(1.1*10))*18,524
(10+11)*18,524
21*18,524
389,004 MS
f) Time to write all of the records if they’re not blocked:
(A+(C*B))*D
(10+(19,000*.1))*200,000
(10+1,900)*200,000
1,910*200,000
380,000,000 MS
g) Optimal blocking factor to minimize waste:
IDK.
h) What would be the answer to (e) if the time it takes to move to the next track were 5ms?
389,004 + (5*1,684)
*** NOTE: ME THINKS MY FORMULA IS WRONG DEPENDING ON HOW ROTATIONAL SPEED IS DEFINED. I NOTICED I HAVEN’T USED THE GIVEN INFO OF 500 TRACKS PER SURFACE SO MY HUNCH IS THAT IT SHOULD BE CONFINED WITH VARIABLE IN COMPUTING TIME…

Exercise #4 – Given that it takes 1 ms to travel from one track to the next, and that the arm is originally positioned at Track 15 moving toward the low-numbered tracks, compute how long it will take to satisfy the following requests:
4, 40, 11, 35, 7, and 14
Using the LOOK policy
All request are present in the wait queue
Ignore rotational time and transfer time; just consider seek time

1. How does the result compare with Fig. 7.16?

START
FINISH
TRAVELED
5
10
15
20
25
30
35
40
15
14
1

X

14
11
3

X

11
7
4

X

7
4
3
X

4
35
31

X

35
40
5

X
Total # of Tracks Traveled:
47
Here’s my stating the obvious: less both distance traveled as well as average travel between jerks
Average # of Tracks Traveled:
7.833333333333333333333

2. Why do you think there is a difference between the two? The reverse direction made the difference: the path the arm had taken was identical to Fig. 15 which was operating under the SSTF (Shortest Seek Time Policy) policy.

Exercise #8 – Compute how long it will take to process then records…
A = Time to transfer a record 1ms
B = Time to process a record 2ms
C = Time to access the next record 1ms (10ms divided by ten records)
D = Number of records
The only thing that I noticed is that it takes the disk 10ms to make a complete rotation and since a track also holds 10records, I took the straight forward approach and assumed that by the time the F has finished a particular task, the disk is already in the right position to start the next series of task with A. Hence, ((A+B+C)*D)-C) with the - C to account for the last step which doesn’t need to access anything since, well, it is the last.
((A+B+C)*D) - C
((1+2+1)*10) – 1
4*10-1
39ms