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All Current Final Term Papers Spring 2013

From 20 Jul , 2013 to 31 Jul 2013 Spring 2013

 

PLZ Share Ur Papers Files Also where u Study

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plz or b papers upload kar dan solution b plz help me

Plz jo current papers share hain wo koi banda solve kr dy uski bari meharbani ho gi

here are current solve papers
diff b/w FAT 16 and FAT 32  pg299

  1. Now lets move our discussion to FAT32. In theory the major difference between FAT 16 and FAT 32 is of course the FAT size. FAT32 evidently will contain more entries and can hence manage a very large disk whereas FAT16 can manage a space of 2 GB maximum practically.

  Or

In the FAT32 there is a special reserved block called FSInfo sector. The block contains some information required by the operating system while cluster allocation/deallocation to files. In FAT12 and 16 this information is calculated when ever required. This calculation at the time of allocation is not feasible in FAT32 as the size of FAT32 is very large and such calculations will consume lots of time, so to save time this information is stored in the FSInfo block and is updated at the time of allocation/deallocation.

 

 

can FAT 12 access 1GB drive?

No

 

  1. 1.     diff b/w s/w interrupt & H/W intrupt?

Software interrupts are invoked by means of some software instruction or statement and hardware interrupt is invoked by means of some hardware controller generally. The only difference between them is the method by which they are invoked

 

  1. 2.     Define defragmentation?

Fragmentation means that clusters of a same file are not contiguously placed, rather they are far apart, increasing seek time hence access time.
• So its desirable that files clusters may be placed contiguously, this can be done by compaction or defragmentation.

 Defragmentation Software reserves space for each file in contiguous block by moving the data in clusters and
re a d j us t i ng .
• As a result of defragmentation the FAT entries will change and data will move from one cluster to other localized cluster to reduce seek time.
• Defragmentation has high computation cost and thus cannot be performed frequently.

 

  1. 3.     State of viruses, 3 name only?

• Dormant State:

A Virus in dormant state has embedded itself within and is observing system activities.

• Activation State:

A Virus when activated would typically perform some unwanted tasks causing data loss. This state may triggered as result of some event.

• Infection State:

A Virus is triggered into this state typically as a result of some disk operation. In this state, the Virus will infect some media or file in order to propagate itself.

  1. 4.      Find the sector# of 1CH

 No of sector/block is 8

         first block is 20

 

1. How large file contain can be managed using FAT?

Larger File Contents
• Larger files would be comprised of numerous clusters.
• The first Cluster # can be read from FCB for rest of the Cluster, a chain is maintained within the FAT.

 

2. What do you mean by mirroring in FAT32?

On all FAT drives, there may be multiple copies of the FAT. If an error occurs reading the primary copy, the file system will attempt to read from the backup copies. On FAT16 and FAT12 drives, the first FAT is always the primary copy and a modification will automatically be written to all copies. However, on FAT32 drives, FAT mirroring can be disabled, and a FAT other than the first one can be the primary (or "active") copy of the FAT.

 

 

3. Enlist all the activities that are to be performed when interrupt 9 occurs?

The service 15H/4FH is called the keyboard hook service. This service does not perform any useful output, it is there to be intercepted by applications which need to alter the keyboard layout. It called by interrupt 9H after it has acquired the scan code of input character from the keyboard port while the scan code is in AL register. When this service returns interrupt 9H translates the scan code into ASCII code and places it in buffer. This service normally does nothing and returns as it is but a programmer can intercept it in order to change the scan code in AL and hence altering the input or keyboard layout.

 

Answer is reads scan code then converts to ASCII and Place it in keyboard buffer and return these are the activities performed when interupt 9 occurs.

4. Difference between COM File and DOS Exe Files?

The main difference in COM File and DOS EXE File is that the COM File starts its execution from the first instruction, whereas the entry point of execution in EXE File can be anywhere in the Program.
• The entry point in case of EXE File is tempered by the Virus which is stored in a 27-byte header in EXE File.

 

 

5. Write a Formula to transfer the cluster # in LSN for FAT32 file System.

In reflection of the anatomy of FAT32 based system the method used to translate the
cluster # into LSN also varies. The following formula is used for this purpose.
Starting Sector # for a Cluster
Starting Sector = Reserved Sect. + FatSize *
FatCopies + (cluster # - 2) *
size of cluster  page no ....293

 

7. How many maximum possible entries are there in FAT32 and FAT16?

In FAT16 we can have 2^16=65536 entries.

FAT32 evidently will contain more entries and can hence manage a very large disk whereas FAT16 can manage a space of 2 GB maximum practically.

 

8. Cross reference of Cluster can be Detected?

• If a cluster lie in more than one file chain, then its said to be Cross Referenced.
• Cross references can be detected easily
by traversing through the chain of all files
and marking the cluster # during traversal.
• If a cluster is referenced more than once then it indicates a cross reference.

• To solve the problem only one reference should be maintained.

 

9. Write down the Structure of boot Block.

Inside a Boot Block

• Contains Code and Data

jmp codepart

OSName

BIOS

Parameter Block

codepart:

• Boot Block executes at Booting time.             Pg244

 

 

10 How portion table virus fools dos about conventional memory?

• The transient part of Command.Com loads itself such that its last byte is loaded in the last byte of Conventional Memory. If somehow there is some Memory beyond Command.Com’s transient part it will not be accessible by DOS.

• At 40:13H a word contains the amount of KBs in Conventional Memory which is typically 640.
• If the value at 40:13H is somehow reduced to 638 the transient part of Command.Com will load itself such that its last byte is loaded at the last byte of 638KB mark in Conventional RAM.
• In this way last 2KB will be left unused by DOS. This amount of memory is used by the Virus according to its own size.

 

 

11. Anatomy of FAT32?

No fixed space reserved for root directory.
• FCB of root directory is saved in a cluster and the cluster # for root directory is saved in BPB.

 

12. TSR program that sets the caps lock bit in the keyboard status bytes whenever interrupt 8 occurs.

Another Example

#include <dos.h>

 void interrupt (*old)();

void interrupt new();

char far *scr=(char far* ) 0x00400017;

void main() {

old=getvect(0x08);

setvect(0x08,new);

keep(0,1000); }

void interrupt new (){

*scr=64;

(*old)(); }


2.  Can we send the data to keyboard yes or not?
Keyboard is a typically an input device but some data can also be send to the keyboard device. This data is used as some control information by the keyboard. One such information is the typematic rate.


3. Write a Formula to transfer the cluster # in LSN for FAT32 file System.

In reflection of the anatomy of FAT32 based system the method used to translate the
cluster # into LSN also varies. The following formula is used for this purpose.
Starting Sector # for a Cluster
Starting Sector = Reserved Sect. + FatSize *
FatCopies + (cluster # - 2) *
size of cluster  page no ....293


1. calculate the sector # for following cluster 44H. we have following information block per cluster=8 first user block no is =20.................... 2 marks

 

Sector # = Cluster # * Sector Per Cluster

 

2. ntfs store file name in ___225_________ and fat store file name in _______31___ characters? .... 2 marks not sure

 

 

3. what is paging in context of non contiguous memory system? ...... 2 marks

  1. A process is subdivided into segment of variable size and each segment or few segments of the process can be placed anywhere in memory

80386 and higher processors also support Paging, i.e. a Process may be divided into fixed size Pages and then only few pages may be loaded any where in memory for Process Execution.

 

 

4. what type of operations performed by system software ? ........... 2 marks

 

 

5. what is usage of cd. and cd..      3 marks pg281

6. three different types of computer virus?.......... 3 marks

There are various types of viruses but we will discuss those which embed themselves within the program or executable code which are

Executable file viruses
Partition Table or boot sector viruses

 

 

7. what is difference in physical and logical view of file ? .......... 3 marks

File

• Is logically viewed as an organization of Data.
• Physically it can be a collection of clusters or blocks.
• O.S. needs to maintain information about the cluster numbers of which a file may be comprised of.

 

 

8. discuss all the step performed by following c statement 

unsigned int far *lpt= (unsigned int far *)0x00400008;

outport((*lpt+2), inport (*lpt+2)|0x10;

outport (0x21, inport(0x21)|0x80).................... 3 marks

 

 

9. how communication of modem can be improved in term of data transfer?.......... 5 marks 

  • Run regular anti-virus and anti-malware scans on all computers using the DSL modem.
  • Contact your ISP. Ask them to run a speed test. This test can help you determine the source of any

faults in your connection, and whether they originate at your ISP at at your location.

  • Reset network equipment like the modem and any routers in use if you notice any slowdown in

speed. If your networking hardware is experiencing any difficulties, this reset may resolve them.

  • Install DSL line filters on any other telephone jacks in your house or office that use the same line as the DSL modem. Plug the line filter in the phone jack, and then plug your devices into the line

filter. These get rid of any interference from these other devices that may be robbing your modem of its top speeds.

 

10. how can a certain volume with moderate space and another volume with large space can be managed by same fat system ?................. 5 marks

how can a certain volume with moderate space and another volume with large space can be managed by the same FAT system. The answer is that the number of entries might be same but the size of cluster may be different. The cluster size can vary from 512 bytes to 32K in powers of 2 depending upon the volume size. The above slide shows how the cluster size and the exact number of required FAT entries can be determined.

 

 

11. how disk partitioning software works? ............. 5 marks

Write the code part of partition table to appropriately load the Boot Block of active partition in primary partition table.
• Places data in the partition table regarding primary and extended partitions.

• As per specification of the user assigns a appropriate size to primary and extended partition by modifying their data part.

 

12. how can we recover the deleted content of file ? explain each step of recovery ....... 5 marks 

0xE5 at the start of file entry is used to mark the file as deleted. • The contents of file still remain on disk.

• The contents can be recovered by placing a valid file name, character in place of E5 and then recovering the chain of file in FAT.

• If somehow the clusters used by deleted file has been overwritten by some other file, it cannot be recovered.

 

 

1:Disadvantages of using undocumented service of Itnterrupt?(2)

This may not be available in all versions of DOS.

 

(2)Purpose of this block statement (2)

char hi,char lo;

hi=hi-0x30;

lo=lo-0x30;

hi=hi4;

hi=hi|li;

 

3: How much space is reserverd for root directory in FAT 32? (2)

No fixed space reserved for root directory.

 

4:What is contiguous memory management system?(2)

Earlier Operating System like DOS has contiguous memory management system i.e. a program cannot be loaded in memory if a

contiguous block of memory is not available to accommodate it

 

6:Translate the cluster#into LSN in FAT 32(3)

 

  1. After calculating the sector number for the cluster the contents of the file can be accessed by reading all the blocks within the cluster. In this way only the starting cluster will be read. If the file contains a number of cluster the subsequent clusters numbers within the file chain can be accessed from the FAT.

For NTFS simply the following formula will be used to translate the sector number into cluster number. Sector # = Cluster # * Sector Per Cluster

 

8:Meaning of this C ststement . (3)

Unsigned int far*lpt=(unsigned in far*)0x00400008;

If ((inport((*lpt)+1)&0x80)==80)

Check the status of the printer if it is idle or not

 

9:Why we need to convert a cluster number into sector number?(5)

All the system calls use the LSN address. If the cluster number is known it should be converted into LSN to access the blocks within the cluster. Moreover all the information about file management uses the cluster number rather than the LSN for simplicity and for the purpose of managing large disk space.

 

10:Using port 61h write the C language statement that should connect interval timer to the speaker and in next ststement it should turn off the speaker.(5)-

page74

 ✿Bîã✿ gud keep it up & thanks for sharing 

Attention Related Final Term papers Spring 2013: All Fellows You don’t need to go at any other site for current Final Term papers Spring  2013, Because All discussed data/sharing of our members in this discussion are going from here to other sites. (Other sites Admins/mods Copy from here & posted at their sites with fake IDs or original name: p). you can judge this at other sites yourself. So don’t waste your precious time with different links.

plz anyone tell the answer of this question soon.. please

discuss all the step performed by following c statement

unsigned int far *lpt= (unsigned int far *)0x00400008;

outport((*lpt+2), inport (*lpt+2)|0x10;

outport (0x21, inport(0x21)|0x80).................... 3 marks

 

meray pas mid term ky video lectures nae hein due to memory problem ..please share if anybody know the ans

mera paper jitna yad hy,utna share kar rhe houn

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The first cluster number of a file can be found in--------- 

►BPB 

►DPB 

►FCB (page 265)

►None

In memory map of first 1 MB of ram , the higher -------- is called system memory. 

 ►64kb 

►384kb      (page 317)

►640kb 

►None

The -------- of boot block constitutes of BPB. 

 ►Code part 

►Data part   (page 242)

►Both 

►None

Data part is the partition table signature whose value should be _______ indicating that the code part contains valid executable code.

►00AA

►0055

►050A

►AA55 (Page 219)

The entry point of execution in EXE File can be

  ►Start of the first instruction

►Start of the last instruction

►Anywhere in the Program   (Page 335)

►Can be in the middle of the program

BPB stands for _________.

 ►BIOS parameter block     (Page 243)

►BIOS processing block 

►Base processing block 

►BIOS partition block 

The keyboard input character scan code is received at ___ port.

 ►60H   (Page 179)

►61H

►62H

►63H

Maximum possible entries in FAT12 are ________.

  ►1024

►2048

►4096   (Page 264)

►65536

To access drive parameter block we use undocumented service ______

►09H/32H

►11H/32H

►17H/32H

►21H/32H (Page 249)

________ is an absolute address relative to the start of physical drive.

 ►LBA      (Page 240)

►LSN

►CHS

►None of the above

 For supporting long file names, _________ fragments can be supported.

►12

►20

►26

►32

FAT based file system can store file name in ________ form.

►ASCII

►UNICODE

►Both ASCII and UNICODE

►None of the given

File system used in CD’s is_______ file system

►Contiguous       

►Chained

►Indexed

►None

  Int 14H____________ can be used to send a byte

 ►Service#0

►Service#1   (Page 121)

►Service#2

►None of the given

 _________ is used to identify the cause of interrupt.

 ►Interrupt ID Register     (Page 116)

►PC Register

►AC Register

►None of All These

 In NTFS, up to ________ characters are used to store files names,

  ►30

►48

►255   (Page 283)

►510

  NMI Stand for 

►Non Maskable Interrupt (Page 46)

►Non Multitude Interrupt

►Non Mask able Instruction

►None of given

BIOS DO NOT support ______. 

 ►LPT1 

►LPT2 

►LPT3 

►LPT4    (Page 91)

 ___________ file system is used in NTFS based systems. 

►Chained

► Contiguous

► Indexed

►None of the given

 PPI stands for

►Parallel Programmable interface 

►Peripheral Programmable interface    (Page 76)

►Port Programmable interface 

►None of the given

Using which interrupt and services, we can access DPB? (2 marks)

How much space is reserved for root directory in FAT32? (2 marks)

What is the advantage of defragmentation? (2 marks)

Write down the names of three data structure used for memory management? (3 marks)

Answer:- (Page 321)

• DOS makes use of various Data Structures for Memory Management:

• MCB ( Memory Control Block )

• EB ( Environment Block )

• PSP ( Program Segment Prefix )

Write down the Structure of boot Block.

How can we check in-service interrupt?

Suppose we read the contents of Drive parameter block and get the following information .

 Number of reserved blocks=2  

Number of blocks in FAT= 7 

Number of blocks in root directory=32 

Find the number of systems blocks .make the appropriate assumptions when needed. 

What are cross reference and how this problem can be solved? (5

Answer: - (Page 316)

• If a cluster lie in more than one file chain, then its said to be Cross Referenced.

• Cross references can pose great problems.

• Cross references can be detected easily by traversing through the chain of all files and marking the cluster #

during traversal.

• If a cluster is referenced more than once then it indicates a cross reference.

• To solve the problem only one reference should be maintained.

Describe the advantages and disadvantages of FAT32?

 

Dua Zahra thanks for sharing ur paper & best of luck for ur result 

Attention Related Final Term papers Spring 2013: All Fellows You don’t need to go at any other site for current Final Term papers Spring  2013, Because All discussed data/sharing of our members in this discussion are going from here to other sites. (Other sites Admins/mods Copy from here & posted at their sites with fake IDs or original name: p). you can judge this at other sites yourself. So don’t waste your precious time with different links.

these are the files that you want 

ok

remember me in your prayer

jazak Allah

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