首页
网站开发
桌面应用
管理软件
微信开发
App开发
嵌入式软件
工具软件
数据采集与分析
其他
首页
>
> 详细
代做comp3511、代写Python/Java编程
项目预算:
开发周期:
发布时间:
要求地区:
When accessing the logical address at
, the results after
address translation is:
A) No such segment
B) Return address 4400
C) Invalid permission
D) Address out of range
9) In a system with 32-bit address, given the logical address 0x0000F1AE (in
hexadecimal) with a page size of 256 bytes, what is the page offset?
A) 0xAE
B) 0xF1
C) 0xA
D) 0xF100
10) A computer based on dynamic storage memory allocation has a main memory with
the capacity of 55MB (initially empty). A sequence of operations including main
memory allocation and release is as follows: 1. allocate 15MB; 2. allocate 30MB; 3.
release 15MB; 4. allocate 8MB; 5. allocate 6MB. Using the best-fit algorithm, what is the
size of the largest leftover hole in the main memory after the above five operations?
A) 7MB
B) 9MB
C) 10MB
D) 15MB
2. [20 points] Synchronization
You are asked to implement the second reader-writer solution, in which once a writer
is ready, it needs to perform update as soon as possible. There are two classes of
processes accessing shared data, readers and writers. Readers never modify data, thus
multiple readers can access the shared data simultaneously. Writers modify shared
data, so at most one writer can access data (no other writers or readers). This solution
gives priority to writers in the following manner: when a reader tries to access shared
data, if there is a writer accessing the data or if there are any writer(s) waiting to access
shared data, the reader must wait. In another word, readers must wait for all writer(s),
if any, to update shared data, or a reader can access shared data, only when there is no
writer either accessing or waiting.
The following variables will be used:
semaphore mutex =1; /* lock for accessing shared variables */
semaphore readerlock=0; /* readers waiting queue */
semaphore writerlock=0; /* writers waiting queue */
int R_count = 0; /* number of readers accessing data */
int W_count = 0; /* number of writer accessing data */
int WR_count = 0; /* number of readers waiting */
int WW_count = 0; /* number of writers waiting*/
TA Yixin YANG ( yyangfe@cse.ust.hk ) is responsible for questions before the deadline
TA Peng YE ( pyeac@cse.ust.hk ) is responsible for grading and appeal handling after the deadline
Please fill in the blanks to design Writer s and Reader s code.
Writer() {
// Writer tries to enter
wait(mutex);
while (BLANK1) { // Is it safe to write?
WW_count++;
BLANK2
WW_count--;
}
W_count++; // Writer inside
signal(mutex);
// Perform actual read/write access
// Writer finishes update
BLANK3
W_count--; // No longer active
if (BLANK4){ // Give priority to writers
signal(writerlock); // Wake up one writer
} else if (WR_count > 0) { // Otherwise, wake reader
BLANK5
}
signal(mutex);
}
Reader() {
// Reader tries to enter
wait(mutex);
while (BLANK6) { // writer inside or waiting?
BLANK7
wait(readerlock); // reader waits on readerlock
BLANK8
}
Rcount++; // Reader inside!
signal(mutex);
// Perform actual read-only access
// Reader finishes accessing
wait(mutex);
R_count--; // No longer active
if (BLANK9) // No other active readers
BLANK10
signal(mutex);
}
3. [30 points] Deadlocks
Consider the following snapshot of a system:
TA Yixin YANG ( yyangfe@cse.ust.hk ) is responsible for questions before the deadline
TA Peng YE ( pyeac@cse.ust.hk ) is responsible for grading and appeal handling after the deadline
Allocation Max Available
A B C D A B C D A B C D
P0 2 0 0 1 4 2 3 4 3 3 2 1
P1 3 1 2 1 5 2 3 2
P2 2 1 0 3 2 3 1 6
P3 1 3 1 2 1 4 2 4
P4 1 4 3 2 3 6 6 5
Please answer the following questions using the banker s algorithm:
1) (5 points) What is the content of the matrix Need denoting the number of resources
needed by each process?
2) (10 points) Is the system in a safe state? Why?
3) (5 points) If a request from process P1 arrives for (1, 1, 0, 0), can the request be
4) (10 points) If a request from process P4 arrives for (0, 0, 2, 0), can the request be
granted immediately? Why?
TA Yixin YANG ( yyangfe@cse.ust.hk ) is responsible for questions before the deadline
TA Peng YE ( pyeac@cse.ust.hk ) is responsible for grading and appeal handling after the deadline
4. [30 points] Memory Management
1) (15 points) Consider the segment table shown in Table A. Translate each of the
virtual addresses in Table B into physical addresses. Indicate errors (out of range, no
such segment) if an address cannot be translated.
Table A
Segment number Starting address Segment length
0 260 90
1 1466 160
2 2656 130
3 146 50
4 2064 370
Table B
Segment number Offset
0 420
1 144
2 198
3 296
4 50
5 32
2) (15 points) Consider a virtual memory system providing 128 pages for each user
program; the size of each page is 8KB. The size of main memory is 256KB. Consider
one user program occupied 4 pages, and the page table of this program is shown as
below:
Logical page number Physical block number
TA Yixin YANG ( yyangfe@cse.ust.hk ) is responsible for questions before the deadline
TA Peng YE ( pyeac@cse.ust.hk ) is responsible for grading and appeal handling after the deadline
Assume there are three requests on logical address 040AFH, 140AFH, 240AFH. First
please describe the format of the logical address and the physical address. Then please
illustrate how the virtual memory system will deal with these requests.
软件开发、广告设计客服
QQ:99515681
邮箱:99515681@qq.com
工作时间:8:00-23:00
微信:codinghelp
热点项目
更多
代写dts207tc、sql编程语言代做
2024-12-25
cs209a代做、java程序设计代写
2024-12-25
cs305程序代做、代写python程序...
2024-12-25
代写csc1001、代做python设计程...
2024-12-24
代写practice test preparatio...
2024-12-24
代写bre2031 – environmental...
2024-12-24
代写ece5550: applied kalman ...
2024-12-24
代做conmgnt 7049 – measurem...
2024-12-24
代写ece3700j introduction to...
2024-12-24
代做adad9311 designing the e...
2024-12-24
代做comp5618 - applied cyber...
2024-12-24
代做ece5550: applied kalman ...
2024-12-24
代做cp1402 assignment - netw...
2024-12-24
热点标签
mktg2509
csci 2600
38170
lng302
csse3010
phas3226
77938
arch1162
engn4536/engn6536
acx5903
comp151101
phl245
cse12
comp9312
stat3016/6016
phas0038
comp2140
6qqmb312
xjco3011
rest0005
ematm0051
5qqmn219
lubs5062m
eee8155
cege0100
eap033
artd1109
mat246
etc3430
ecmm462
mis102
inft6800
ddes9903
comp6521
comp9517
comp3331/9331
comp4337
comp6008
comp9414
bu.231.790.81
man00150m
csb352h
math1041
eengm4100
isys1002
08
6057cem
mktg3504
mthm036
mtrx1701
mth3241
eeee3086
cmp-7038b
cmp-7000a
ints4010
econ2151
infs5710
fins5516
fin3309
fins5510
gsoe9340
math2007
math2036
soee5010
mark3088
infs3605
elec9714
comp2271
ma214
comp2211
infs3604
600426
sit254
acct3091
bbt405
msin0116
com107/com113
mark5826
sit120
comp9021
eco2101
eeen40700
cs253
ece3114
ecmm447
chns3000
math377
itd102
comp9444
comp(2041|9044)
econ0060
econ7230
mgt001371
ecs-323
cs6250
mgdi60012
mdia2012
comm221001
comm5000
ma1008
engl642
econ241
com333
math367
mis201
nbs-7041x
meek16104
econ2003
comm1190
mbas902
comp-1027
dpst1091
comp7315
eppd1033
m06
ee3025
msci231
bb113/bbs1063
fc709
comp3425
comp9417
econ42915
cb9101
math1102e
chme0017
fc307
mkt60104
5522usst
litr1-uc6201.200
ee1102
cosc2803
math39512
omp9727
int2067/int5051
bsb151
mgt253
fc021
babs2202
mis2002s
phya21
18-213
cege0012
mdia1002
math38032
mech5125
07
cisc102
mgx3110
cs240
11175
fin3020s
eco3420
ictten622
comp9727
cpt111
de114102d
mgm320h5s
bafi1019
math21112
efim20036
mn-3503
fins5568
110.807
bcpm000028
info6030
bma0092
bcpm0054
math20212
ce335
cs365
cenv6141
ftec5580
math2010
ec3450
comm1170
ecmt1010
csci-ua.0480-003
econ12-200
ib3960
ectb60h3f
cs247—assignment
tk3163
ics3u
ib3j80
comp20008
comp9334
eppd1063
acct2343
cct109
isys1055/3412
math350-real
math2014
eec180
stat141b
econ2101
msinm014/msing014/msing014b
fit2004
comp643
bu1002
cm2030
联系我们
- QQ: 9951568
© 2021
www.rj363.com
软件定制开发网!