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Errata (Last Updated: 1/5/2015)
(Notation: “pg” indicated page, “ln” indicates line, negative line number means count
from the bottom up)
pg 7, fig 1-4
replace: fl
with:
f1
(i.e., the number “1” rather than lower case letter “l”)
pg 8, ln 8
replace: presence
with:
the presence
pg 47, ln –3
replace: terminated
with:
terminate
pg 61, ln 9
replace: pi:
with:
p2:
pg 62, ln 9
replace: five functions
with:
five function
pg 65, ln –10
replace: project
with:
process
pg 72, fig 3-1(a)
replace: x_is_positive.wait
with:
X_is_positive.wait
(3 times)
pg 72, fig 3-1(a)
replace: p1
with:
p
(2 times)
pg 72, fig 3-1(b)
replace: p2
with:
q
pg 72, fig 3-1(b)
replace: p1
with:
p
pg 75, ln 18
replace: because of a
with:
because a
pg 83, ln 6
replace: standard input
with:
standard output
pg 83, ln 7
replace: standard output
with:
standard input
pg 92, ln –9
replace: (i+1)%5
with:
i%5+1
pg 99, ln 19, 20, –16
replace: (i+1)%5
with:
i%5+1
pg 104, ln –5
replace: (subscript) a
with:
(subscript) p1
(six times)
replace: (subscript) b
with:
(subscript) p2
(two times)
replace: (subscript) c
with:
(subscript) p3
(six times)
pg 104, ln –4
replace: (e) Assume another message was sent from process c to process a
with:
(b) Assume another message was sent from process p3 to process p1
pg 104, ln –3
replace: 10 according to c’s clock and received at time 2 according to a’s
with:
8 according to p3’s clock and received at time 2 according to p1’s
pg 119, Figure 4-7
add an arrow labeled “p” from p6 to p3
pg 158, ln –5
replace: Thus p2 will
with: Thus p1 will
pg 161, Figure 5-5
replace: SJN
with: SJF
pg 168, ln –11
replace: p2
with:
p3
pg 174, ln –15 and ln –9
replace: four
with:
three
pg 188, fig 6-5(c)
replace: p4
with: p5
(last line of figure)
pg 208, ln –15
replace: references are kept is
with: references are kept in
pg 210, ln 16:
replace: eternal
with:
external
pg 219, ln –9:
replace: H[1]
with:
H[0]
pg 232, ln –7
replace: ever instruction
with:
every instruction
pg 243, ln 9
replace: of 1 KB each
with:
of 4 KB each
pg 246, ln –2:
replace: advantage
with
advantages
pg 251, ln 9
replace: most recently loaded page, P[k]. On a page fault, page P[k+1 % m] is replaced
with:
least recently loaded page, P[k]. On a page fault, page P[k] is replaced
pg 255, ln 2 and ln 14
replace: second-chance
with:
third-chance
pg 262, ln 3
replace: set gradual
with:
set is gradual
pg 309, ln 2
replace: into this function
with: into this table
pg 311, ln 10
replace: can read
with:
can be read
pg 368, Figure 11-3(b)
replace: 34, 35, 36, 37, 38, … 52, 53, 54, 55, 56
with: 36, 37, 38, 39, 40, … 54, 55, 56, 57, 58
pg 371, ln –8
replace: 18.96 MB
with:
18.286 MB
pg 371, ln –18
replace: 0.66 GB
with:
0.64 GB
pg 371, ln –19
replace: 681,984
with:
681,984,000
pg 389 ln 16
replace: read pointer
with:
rear pointer
pg 430, ln 17
replace: where P is
with:
where DEA is the data encryption algorithm, P is
Part 5: Programming Projects
pg 484, ln –19
replace: ’allocated;’
with: ’allocated’;
pg 484, ln –18
replace: r;)
with:
r);
pg 484, ln –16
replace: ’blocked;’
with: ’blocked’;
pg 484, ln –13
replace: self; }
with:
self);
pg 485, ln 1
replace: (Waiting_List
with:
(r->Waiting_list
pg 485, ln 4/5
insert the following new line between lines 4 and 5: insert(q->Other_Resources, r;)
pg 488, ln –10 through –14
replace: (5)
insert(RL, q)
(6)
find highest priority process p;
(6)
p->Status.Type = ‘running’;
(7)
Scheduler();
(8) }
with: (5)
insert(RL, q);
(6)
Scheduler();
(7) }
pg 488, ln –7 through –4
replace: It then changes the status from … it starts the process p.
with:
It then calls the scheduler, which changes the status of the currently highest-priority process p
from ‘ready’ to ‘running’.
pg 489, ln 12
replace: (Ready List, self);
with: (RL, self);
pg 494, ln 18
replace: release(p)
with:
mm_release(p)
GLOSSARY
pg 512, ln –4
replace: unauthorized
with:
authorized
Solutions
Ch 2, problem 11:
ln –2 of program:
replace: P(s4); P(s3); P(s4) p8;
with:
P(s4); P(s4); P(s4) p8;
Ch 3, problem 18(b):
Figure:
move point u marking time step 9 of p1 to time 11 of p1
Ch 4, problem 19(c):
ln –2 of solution:
replace: (p3,50)
with:
(p3,35)
Ch 6, problem 1(a)
ln 3:
replace: B
with: p2
Ch 6, problem 1(b)
ln 4:
replace: B
with: p2
replace: A
with: p1
replace: C
with: p3
Ch 6, problem 17
ln 3:
replace: R5, R5, R2, R4
with: R5, R3, R2, R4
Ch 7, problem 1
right half of figure
replace: 0
with: 1550
replace: 170
with:
1670
replace: 50
with:
1550
ln 4:
replace: Figure 7-3
with: Figure 7-4
Ch8, problem 8a
table
replace: 20 | P1 of S4
with:
20 | P1 of S12
Ch 10, problem 8:
block numbers in the table refer to the original block numbers, i.e., before the deletion or insertion. For
example, when block 4 is deleted, we continue referring to the 3 blocks following block 4 as 5, 6, 7, even
though these blocks get shifted to the left and thus become the logical blocks 4, 5, 6 in the modified file.
delete bl. 4
insert bl. 2A
(a)
(b)
(c)
(e)
(a)
(b)
(c)
(e)
5, 6, 7
0, 1, 2, 3, 4
--3, 5, 6, 7
0, 1, 2
---
5, 6, 7
3
--2A, 3, 5, 6, 7
2, 2A
2A
2A
Ch 10, problem 13(a)
The new scheme needs to access 0.2 x (d-1) blocks for each file (not 0.2 x d) since one of the d blocks is in
the i-node. Thus the number of blocks accessed is reduced to 0.2(d-1)/(0.8+0.2d) = (d-1)/(4+d)
(b) For d=2, the reduction is (2-1)/(4+2) = 1/6 or 16.7%. For d=4, it is (4-1)/(4+4) = 3/8 or 37.5%