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

Due Date: 31/01/2017

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.


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

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

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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..

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...


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.

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

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



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