QUESTIONS/ANSWERS ON COMPUTER NETWORK

[Karthikeyan B]

1.Does Webmail use POP3, IMAP or neither? If one of these, why was that one chosen?

If neither, which one is it closer to in spirit?

i didnt get a proper answer for this.some say both,some say IMAP.

please clarify

Karthik

[Karthikeyan B]

2.Consider a framing protocol that uses bit-stuffing. Write the bit sequence that will be

transmitted over the link for the given data frame which contains the following

sequence

110101111101011111101011111110.

Now suppose that the following bit sequence arrived over a link:

1101011111010111110010111110110

Show the resulting frame after any stuffed bits have been removed. Also indicate any

errors that have been introduced into the frame?

Can anybody provide some links on how to do bit stuffing?

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[Karthikeyan B]

3.

Q.1 (a).Match the following to one or more layers of TCP/IP model:

i. Flow control

ii. Segmentation and Reassembly

iii. Error correction and retransmission

iv. Logical addressing

Q.1 (b). Give the error detection mechanisms used at various layers in TCP/IP model.

Give reasons why there are different mechanisms applied at different layers.

Q.1 (c). If an application layer packet size is 2 Bytes long, how big is the corresponding

Ethernet packet? Assume that the application uses TCP.

Q.1 (d). Do we need reliable transport service at the transport layer if the data link

transport on all the links in the route is reliable? Justify.

i hope this question will be asked for sure tomorrow.

guys share the answer.

[Ankita Nanwani]

Bit Stuffing is :

 

Sender Side : Insert a '0' after 5 consecutive 1s

Reciver side : Remove a 0 which occurs after 5 consecutive 1's

 

A ) Data To be Transmitted is  :  1101011111     010 11111    1010 11111   110.

Sender will send :              11010 11111 0 010 11111 0 1010 11111 0 110.

Reciever will remove the 0's inserted after 5 consecutive 1s.Thus the reciever will get original data stream as  :

 

                                           1101011111     010 11111    1010 11111   110.

B) Now suppose that the following bit sequence arrived over a link.1101011111010111110010111110110 Show the resulting frame after any stuffed bits have been removed.

Reciever got  : 1101011111010111110010111110110

 

Remove 0's after 5 consecutive 1's

 

So, Reciiever at the end will have  : 1

Initially rceived     :             1101011111 0 1011111  0 01011111 0 110

After removing 0's    :          1101011111    1011111    01011111  110

 

C) Also indicate any

errors that have been introduced into the frame? :  Not sure of this.

[Ankita Nanwani]

WebMail neither uses IMAP nor POP3. But it is fairly similar in spirit to IMAP because

both of them(Webmail and IMAP) allow a remote client to examine and manage a remote mailbox.

In contrast, POP3 just sends the mailbox to the client for processing there.

On Fri, Apr 2, 2010 at 12:01 AM, Karthikeyan B <karthi...@gmail.com> wrote:

--

[Karthikeyan B]

Thanks Ankita..

Additional info on webmail/pop3/imap

What is WebMail?

WebMail is an email client that allows you to check your mail using a web browser. This means you can check your email using any computer that is connected to the Internet. WebMail has several features including address books, personalized display settings, and support for mail folders. Because WebMail includes all of these features, you can choose to use WebMail instead of your normal email software.

Can I use WebMail and my normal email software?

WebMail will work fine on its own or in conjunction with most other email clients, such as Microsoft Outlook Express. However, it is important to understand that most email clients are configured to use the POP protocol, where as WebMail used the IMAP protocol. If you try to check your mail with two clients at the same time (i.e. WebMail and Microsoft Outlook Express) you run the risk of corrupting your mail file. You can avoid this problem by changing you email client software to use the IMAP protocol. FRII Customer Support strongly recommends this option because it will prevent any possible problems between the two different protocols. 

If you have multiple clients (say Outlook at home and WebMail on your laptop) make sure that only one is ever active at a given time.

What happens if I forget to change my email software settings to IMAP?

If your normal software is using the POP protocol instead of the IMAP protocol you will be unable to view messages in one system that have previously been read in the other. Additionally, under specific conditions, all previously read mail may become inaccessible. Another drawback to not using IMAP for all email clients is that any folders created within the IMAP-based client will not be accessible to the software configured for POP. FRII recommends that you use the IMAP protocol whenever checking email from multiple email clients.

[Karthikeyan B]

C) Also indicate any

errors that have been introduced into the frame? :  Not sure of this.

I think there is no errors in the frame since after removing the stuffed bits,we get the original message.

if we didnt get the original message, then there are errors.

[Karthikeyan B]

sorry, since the transmitted and received frames differ in some bits,errors are occurred.

sent->1101011111     010 11111    1010 11111   110
received->1101011111     010111110    0101 11110  110

[Ankita Nanwani]

Ye..I was drafting the mail for the same :)

[Karthikeyan B]

4,Consider the following data stream: 11000100100111010100010010011100. Using

this data stream determine the corresponding 8 bit Checksum.

Anybody solved this problem.is it same as CRC?

How to solve if 16bit and 32 bit checksum is given?

[Ankita Nanwani]

Q.1 (a).Match the following to one or more layers of TCP/IP model:

i. Flow control  ---> Transport Layer

ii. Segmentation and Reassembly  -----> Transport layer

iii. Error correction and retransmission  ----> Transport layer

iv. Logical addressing  ----> Internet Layer

 

Q.1 (b). Give the error detection mechanisms used at various layers in TCP/IP model.

Give reasons why there are different mechanisms applied at different layers.

 

Transport layer  :  TCP protocol : uses Sequence numbers and acknowledgement numbers and also weak checksum

                              UDP : best effort "unreliable" protocol (uses weak checksum)

Internet layer : Like UDP, IP is best effort "unreliable" protocol

 

Network Link layer :  Can use CRC for to compensate for weak checksum on the transport layer.

 

 

 

Q.1 (c). If an application layer packet size is 2 Bytes long, how big is the corresponding

Ethernet packet? Assume that the application uses TCP.

 

At Application layer =  data is 2 bytes

At Transport layer  = TCP is used so, 20 bytes header will be added : so 20 + 2 = 22

At Internet layer = 20 bytes IP header will be added .so 22 +  20 = 42

 

At Network link layer, ethernet is used.valid ethernet packet length should be of minimum 64 bytes.

 

So, 6 byte Destination address +  6 bytes source address +  2 byte for Type + 42 bytes data(from above layers) + x bytes of padding + 4 bytes checksum = 64

 

So, padding field of 4 bytes will be added.

 

And also, 8 byte of preamble will be added.

 

So, ethernet packet will be 72 bytes long.

 

Please correct me if I am wrong.

 

 

 

Q.1 (d). Do we need reliable transport service at the transport layer if the data link

transport on all the links in the route is reliable? Justify.

 

Some data link protocols like ARQ(Automatic Repeat Request ) implements a reliable transmission scheme.TCP also provides end-to-end transmission scheme.This overlap, which ensures that transmission is reliable,will lead to overhead at both layers and hence, lead to degraded performance.

 

 

Please feel free to correct me at places, where you feel I have not answered correctly. :)

[Ankita Nanwani]

I think, here to calculate the 8 bit checksum, the given bit stream would be divided into segments of 8 bits

 

11000100

10011101

01000100

10011100

 

Add all the  8 bit segments

 

So, the answer would be

 

11000100

10011101

01000100

10011100

-------------

111111001

Take 1's complement  of the sum :  0000 0110 which would be the checksum

 

If 16 bit checksum is asked, then I guess, we should divide the data stream in 16 bit segments and repeat the above calculation.

[Karthikeyan B]

Thanks a lot!!!!

Regarding checksum and CRC.both are different algorithms right.

Then i have a doubt on CRC.

consider  a transmitted message is 10001000 of polynomial x3+x2+1..

then we have the divider 1101 and the message 10001000 ...

in some places i see, they are adding some number 0's or 1's at the end of message and start divding..

on what basis we need to decide that number ??

[Karthikeyan B]

SOLUTION FOR SUBNET PROBLEM..

(15 MARKS)

1 Assume that you have been assigned the 200.35.1.0/24 network

block. Define an extended network prefix that allows the creation of

20 hosts on each subnet.

A minimum of 5 bits are required to define 20 hosts so the extended

network prefix is /27 (27 = 32-5). Mask is 255.255.255.224.

2 What is the maximum number of hosts that can be assigned to each

subnet?

The maximum number of hosts on each subnet is 2^5 -2, or 30.

3 What is the maximum number of subnets that can be defined?

The maximum number of subnets is 23, or 8.

4 Specify the subnets of 200.35.1.0/24 in binary format and dotted-decimal

notation.

Subnet #0: 11001000.00100011.00000001. 000 00000 = 200.35.1.0/27

Subnet #1: 11001000.00100011.00000001. 001 00000 = 200.35.1.32/27

Subnet #2: 11001000.00100011.00000001. 010 00000 = 200.35.1.64/27

Subnet #3: 11001000.00100011.00000001. 011 00000 = 200.35.1.96/27

Subnet #4: 11001000.00100011.00000001. 100 00000 = 200.35.1.128/27

Subnet #5: 11001000.00100011.00000001. 101 00000 = 200.35.1.160/27

Subnet #6: 11001000.00100011.00000001. 110 00000 = 200.35.1.192/27

Subnet #7: 11001000.00100011.00000001. 111 00000 = 200.35.1.224/27

5 List the range of host addresses that can be assigned to Subnet #6

(200.35.1.192/27).

Subnet #6: 11001000.00100011.00000001. 110 00000 = 200.35.1.192/27

Host #1: 11001000.00100011.00000001.110 00001 = 200.35.1.193/27

Host #2: 11001000.00100011.00000001.110 00010 = 200.35.1.194/27

Host #3: 11001000.00100011.00000001.110 00011 = 200.35.1.195/27

:

Host #29: 11001000.00100011.00000001.110 11101 = 200.35.1.221/27

Host #30: 11001000.00100011.00000001.110 11110 = 200.35.1.222/27

6 What is the broadcast address for subnet 200.35.1.192/27?

11001000.00100011.00000001.110 11111 = 200.35.1.223

CORRECT ME IF I AM WRONG

[Karthikeyan B]

Hi Ankita,

i think the addition performed is wrong.just want to make u aware of tat.

Karthik

On Fri, Apr 2, 2010 at 12:08 PM, Ankita Nanwani <ankita....@gmail.com> wrote:

[Karthikeyan B]

Some of the concepts on which we can expect problems:

1.CRC 2.CHECKSUM 3.BIT STUFFING 4.SUBNET MASK 5.DIJKSTRAS ALGORITHM 6.CIDR ENTRIES

7.HAMMING CODE 8.ENCODING TECHNIQUES

if you find some more please add to the list

[ajay karthik]

@ Ankita Nanwani

Q.1 (c). If an application layer packet size is 2 Bytes long, how big is the corresponding

Ethernet packet? Assume that the application uses TCP.

 

At Application layer =  data is 2 bytes

At Transport layer  = TCP is used so, 20 bytes header will be added : so 20 + 2 = 22

At Internet layer = 20 bytes IP header will be added .so 22 +  20 = 42

 

At Network link layer, ethernet is used.valid ethernet packet length should be of minimum 64 bytes.

 

So, 6 byte Destination address +  6 bytes source address +  2 byte for Type + 42 bytes data(from above layers) + x bytes of padding + 4 bytes checksum = 64

 

So, padding field of 4 bytes will be added.

 

And also, 8 byte of preamble will be added.

If it is 10MBPS line the source and destination addresses must be compulsarily 6 byte else it also can be 2 byte.... the solutions for these type of questions use 2 byte... so its better we mention this and give two possible values... given in page 275 tanenbaum 4.3.3

[Arun Vijaya Natarajan]

Guys,

Im just trying to solve all the old question papers..

I dint get an answer for the below qun..

Suppose you are designing a selective repeat sliding windows protocol for a 1Mbps
point-to point link to the moon, which has a one way latency of 1.25 seconds.
Assuming that each frame carries 1 Kbytes of data, what is the minimum number of
bits you need for the sequence number?

Did anyone solve it.. pls share..

Thanks,
Arun

[ANKIT SOMANI]

Which of the following schedules is (conflict) serializable? For each serializable
schedule, determine the equivalent serial schedules. [3 X 4 = 12]
4.1. R1(X);R3(X);W(X);R2(X);W3(X);
4.2. R1(X);R3(X);W3(X);W1(X);R2(X);
4.3. R3(X);R2(X);W3(X);R1(X);W1(X);
4.4. R3(X);R2(X);R1(X);W3(X);W1(X);



Please share..Any body know how to solve this kind of DB problem.. ?


Regards
~Ankit Somani

[Ankita Nanwani]

Hi Arun,

 

 

Here is the answer to your question :

 

http://pages.cs.wisc.edu/~akella/CS640/F07/assign/HW2/homework2.pdf

http://pages.cs.wisc.edu/~akella/CS640/F07/assign/HW2/homework2_soln.pdf

[Arun Vijaya Natarajan]

Thanks Ankita.. :)

Thanks,
Arun

[Ankita Nanwani]

But, is the solution clear to you? coz I did not get that +1 thing in it..If its clear to you..pls explain..Thanks in advance!

Thanks Ankita.. :)

Thanks,
Arun

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[Arun Vijaya Natarajan]

Hi Ankita,

I guess it is the ack bit which takes a free ride with the data instead of a separate frame just for acknowledge. This concept is called piggybacking..

Im not sure yet on this.. Ppl, correct me if Im wrong..

Thanks,
Arun

--
Regards,
Arun V N