Operating System

Homework-3

Submitted By:

Name: Suraj Kumar Singh

Section: E - 3004

Roll_No: RE3004B77

Group: 2

Course_code: 1604

Subject_code: CAP316

Submitted To:

Miss. jasleen

PART A

1 Consider the techniques of Swapping and overlays. Do they

have anything in common? Where do they differ in

implementation? Justify your answer.

Ans- Overlays:- Overlays is a memory management scheme. We can use

overlays to enable a process to be larger than the amount of memory

allocated to it. The idea of overlays is to keep in memory only those

instructions and data that are needed at any given time. When other

instructions are needed, they are loaded into space occupied previously by

instructions that are no longer needed, or In operating systems, an overlay is

when a process replaces itself with the code of another program

As in dynamic loading, overlays do not require any special

support from the operating system. They can be implemented completely by

user with simple file structures. The use of overlays is currently limited to

microcomputer and other systems that have limited amounts of physical

memory and that lack hardware support for more advanced techniques.

Swapping:- Swapping is a simple memory/process management

technique used by the operating system(os) to increase the utilization of the

processor by moving some blocked process from the main memory to the

secondary memory(hard disk). When you load a file or program, the file is

stored in the random access memory (RAM). Since RAM is finite, some files

cannot fit on it. These files are stored in a special section of the hard drive

called the "swap file". "Swapping" is the act of using this swap file.

A swapping is a mechanism in which a process can be swapped temporarily

out of memory to a backing store and then brought back into memory for

continued execution.

A variant of this swapping policy is used for priority-based scheduling

algorithms. If a higher-priority process arrives and wants service, the

memory manager can swap out the lower-priority process and then load and

execute the higher-priority process. When the higher-priority process

finishes, the lower-priority process can be swapped back in and continued.

This variant of swapping is sometimes called roll out, roll in.

Normally, a process that is swapped out will be swapped back into the same

memory space that it occupied previously. This restriction is dictated by the

method of address binding. If binding is done at assembly or load time, then

the process cannot be easily moved to a different location.

Implementation of overlay:-

Implementation of swapping:-

Read () 20KB

Function1()

50KB

Function2() 70

KB

Display () 40KB

Read()

(20kb)

Display ()

(40kb)

Function1()

(50kb)Function1()

(70kb)

Operating system

User

process/application

Process1

If we assume a data transfer rate of 1 megabyte/sec and access time of 10

milliseconds, then to actually transfer a 100Kbyte process we require:

Transfer Time = 100K / 1,000 = 1/10 seconds

= 100 milliseconds

Access time = 10 milliseconds

Total time = 110 milliseconds

As both the swap out and swap in should take place, the total swap time is

then about 220 milliseconds (above time is doubled). A round robin CPU

scheduling should have a time slot size much larger relative to swap time of

220 milliseconds. Also if process is not utilising memory space and just

waiting for I/O operation or blocked, it should be swapped.

2. A system uses multi-programming with variable no. of

tasks, with a memory of 110K for user processes. Draw the

memory map corresponding to the current memory

allocation table:

Proces2

A new job D arrives needing 15K of memory. Show the

memory allocation table entry for job D

using each the allocation strategy of first fit, best fit and worst

Ans-First fit:-

Job Base address Length

D 0 15

A 20 10

B 46 18

C 90 20

Best fit:-

Job Base address Length

A 20 10

D 31 15

B 46 18

C 90 20

Worst fit:-

Job Base address Length

A 20 10

B 46 18

D 65 15

C 90 20

3. Suppose a fixed partitioning memory system with partitions

of 100K, 500K, 200K, 300K and 600K (in memory order)

exists with all 5 partitions currently unused (or free).

a) Using the best fit algorithm, show the state of

memory after processes of 212K, 417K,

112K and 350K (in request order) arrive.

b) Using the best available algorithm, show the state of

memory after processes of 212K,

417K, 112K and 350K (in request order) arrive.

c) Compute the total internal

fragmentation for part (a), if any. d)

Compute the total external

fragmentation for part (b), if any.

Ans:- (a) Best fit:-

In the above figure, at memory space 500k, 83k memory space is left, at

200k, 88k memory space is left, at 300k, 82k memory space is left, at 600k,

250k memory space is left.

(b) from FIFO:-

Process 350 must be wait until release the memory space from other

process.

From best fit:-

From worst fit:-

Process 350 must be wait until release the memory space from other

process.

In the above figure “Best fit” is the best algorithm for memory allocation.

After allocated memory to the arrival processes the memory state is, at

memory space 500k, 83k memory space is left, at 200k, 88k memory space

is left, at 300k, 82k memory space is left, at 600k, 250k memory space is

left.

(c) Total internal fragmentation in part (a):-

(500K – 417K) + (200K – 112K) + (300K – 212K) +(600-350)

=83k+88k+88k+250k=509k.

(d) There is no any external fragmentation because there is no request in

pending. 

P A R T B

4 Consider a logical address space of 8 pages of 1024 bytes

each, mapped onto a physical memory of 32 frames.

Determine the no. of bits in the logical address and that in

physical address.

Ans-

As was given for a 8 Page, 1024 words & 32 frames (shown below)

8 pages -> 2^3 bits

1024 bytes -> 2^10 bits

32 frames -> 2^5 bits

Therefore:

Logical memory = 3+10=13 bits (Page + Word)

Physical memory = 10 + 5 =15 bits (Word + Frame)

5. Consider the page reference string: 0, 1, 7, 0, 1, 2, 0, 1, 2, 3,

2, 7, 1, 0, 3, 1, 0, 3. Calculate the total no. of page faults

(with no pre-paging at all) using OPT, FIFO, LRU and Clock

algorithms such that 3 frames are allocated to the process

Ans-FIFO:-

0 1 7 0 1 2 0 1 2 3 2 7 1 0 3 1 0 3

0 0 0 2 2 2 3 3 3 1 1 1

1 1 1 0 0 0 2 2 2 0 0

7 7 7 1 1 1 7 7 7 3

total page faults=12.

OPT:-

0 1 7 0 1 2 0 1 2 3 2 7 1 0 3 1 0 3

0 0 0 0 3 3 3

1 1 1 1 1 1

7 2 2 7 0

Total page fault=7

LRU:-

0 1 7 0 1 2 0 1 2 3 2 7 1 0 3 1 0 3

0 0 0 0 3 3 1 1 3 3

1 1 1 1 7 7 7 7 1

7 2 2 2 2 0 0 0

Total page faults=10

6. A virtual memory system exhibits the follow trace of page nos.

as: 1,2,3,2,6,3,4,1,5,6,1,6,4,2.

Simulate the page replacements for the following scenarios:

a) FIFO

with 3

page

frames b)

FIFO with

4 page

frames

Determine the total no. of page faults for each. State whether

Belady’s anomaly occur?

Ans-

a) FIFO with 3 page frames:

1 2 3 2 6 3 4 1 5 6 1 6 4 2

1 1 1 6 6 6 5 5 5 2

2 2 2 4 4 4 6 6 6

3 3 3 1 1 1 4 4

Here Total Number of page faults = 10

a)FIFO With 4 page Frames :

1 2 3 2 6 3 4 1 5 6 1 6 4 2

1 1 1 1 4 4 4 4

2 2 2 2 1 1 1

3 3 3 3 5 5

6 6 6 6 2

Here Total Number of page Faults = 8

According to Belady’s Anomaly, When more frames are allotted to a

process(to accommodate more pages of the process), it generates more

page faults whereas it should be the other way round. As it should be the

other way round. But in the above case there is no Belady’s anomaly occure.