We are here with you hands in hands to facilitate your learning & don't appreciate the idea of copying or replicating solutions. Read More>>

Looking For Something at vustudents.ning.com? Click Here to Search

www.bit.ly/vucodes

+ Link For Assignments, GDBs & Online Quizzes Solution

www.bit.ly/papersvu

+ Link For Past Papers, Solved MCQs, Short Notes & More

Assignment No. 04
Semester: Fall 2016
CS604: Operating System Total Marks: 15

Due Date: 31/01/2017

Instructions:
Please read the following instructions carefully before submitting the assignment. It should be clear that your assignment will not get any credit if:

 The assignment is submitted after the due date.
 The submitted assignment does not open or file is corrupt.
 Assignment is copied (partial or full) from any source (websites, forums, students, etc)

Note: You have to upload only .doc file. Assignment in any other format (extension) will not be accepted and will be awarded with zero marks.

Objective:

The objective of this assignment is to provide hands on experience of:
 Understanding of Memory Management
 To know the algorithm of deadlock handling mechanisms

For any query about the assignment, contact at CS604@vu.edu.pk
Assignment

Question No. 1 (10 marks):

Suppose a logical address space of 32 pages of 2048 words. There are 64 frames in the main memory. According to given information calculate the various parameters related to paging.

Calculate the size of Logical Address?
Calculate the size of Physical Address?

Question No. 2 (05 marks):

There is a system with 14 tape drives and three processes.To follow the deadlock avoidance algorithm create a Safe Sequence of the following processes by calculating the available tape drives.

Processes Max Need of tape drives Allocated tape drives Available tape drives
P0 12 5
P1 5 3
P2 9 3

Be Best of Luck st of Luck

+ Click Here To Join also Our facebook study Group.

..How to Join Subject Study Groups & Get Helping Material?..


See Your Saved Posts Timeline

Views: 2960

.

+ http://bit.ly/vucodes (Link for Assignments, GDBs & Online Quizzes Solution)

+ http://bit.ly/papersvu (Link for Past Papers, Solved MCQs, Short Notes & More)

+ Click Here to Search (Looking For something at vustudents.ning.com?)

Attachments:

Replies to This Discussion

Please Discuss here about this assignment.Thanks

Our main purpose here discussion not just Solution

We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions. Read More>>

 

Note:-

For Important Helping Material related to this subject (Solved MCQs, Short Notes, Solved past Papers, E-Books, FAQ,Short Questions Answers & more). You must view all the featured Discussion in this subject group.

For how you can view all the Featured discussions click on the Back to Subject Name Discussions link below the title of this Discussion & then under featured Discussion corner click on the view all link.

Or visit this link 

Click Here For Detail.

&

.•°How to Download past papers from study groups°•.

 

Please Click on the below link to see…

.... How to Find Your Subject Study Group & Join .... 

How to solve question 2 ..i did't understand it from handouts....some guide me plz

2nd me total 14 tapes hn,,, jin me se toatl allocate 11 hui,, bki 3 reh gyi... so initail P0 me 3 ho ga,,, r phr 1st P1 process ho ga, phr available 3 r P1 k allocate wala 3 dono add ho k 6 ho jyen ge,,, phr p2 process ho ga.. P2 ka allocate 3 or available 6 add ho k 9 ho jye ga...thn last p0.. last pr P2 k available 9 me P0 k allocated tapes 5 add ho kr (9+5) toatl 14 tapes ho jyen gi//// solved


Mujhe lecture se itni smjh i.. me n kr liya.. dnt knw theek ya ni..
Attachments:

App nay Q1 ma word glat likhy hain ya 2048 hain app nay ziada likha ha so answers b glt hain

no mano.. d (words) k bits ko find krne k liye "[log2 2k]" means 2048 dbl ho jye ga... 2k (2048 +2048= 4096) ..... [log2 4096]... bk pr b ese ta.. r lec me b /// ab deh lo

1st to blkl corrct.. 2nd chk kr ln...btayen 2nd theek ya ni...

Attachments:

salam kiya yeh solution thk hay ?

nai wasifa ap ka 2nd qstn thik nai lg ra mjy. Ap ny just Allocated tape drives ko samny rkhty hvy qstn solve kiya hy jb k Max Need of tape drives ko b dykhna hy . mera ap sb ko mshwra hy k lec 27 dykhain jst last 10 mints ka pura nai. system safe nai hy mery hisab sy to ap log b lec dykhain plz or apna cmnt krain.

Solution:
p = [log2 32] bits = 5 bits
f = [log2 64] bits = 6 bits
d = [log2 4096] bits = 12 bits
Logical Address Size = |p| + |d| = 5 + 12 = 17 bits
Physical Address Size = |f| + |d| = 6 + 12 = 18 bits

Solution:
Safe Sequence:
<P1, P2, P0>


Processes Max Need of tape drives Allocated tape drives Available tape drives
P0 12 5 3
P1 5 3 6
P2 9 3 9

1st question men Logical Address or Physical Address pocha  na k L.A ka size :/

size he pucha hy qstn ma.
Calculate the size of Logical Address?
Calculate the size of Physical Address?

plz check again

CS604 Assignment #4 Idea Solution

Attachments:

RSS

© 2019   Created by + M.Tariq Malik.   Powered by

Promote Us  |  Report an Issue  |  Privacy Policy  |  Terms of Service