##### Document information

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Assignment 2

**Question 1: **What is the drawback of VRC**
**

**Vertical Redundancy Check** is a way of error checking by attaching a parity bit to each byte
of data to be transmitted, which is then tested to determine if the transmission is correct.

The resulting parity bit is constructed by XORing the word. The result is a "1" if there is an odd number of 1s, and a "0" if there is an even number of 1s in the word.

**Draw Back:
**

This method is unreliable, because if there are even number of errors in the data, the error

will go undetected.

**Question 2: **Calculate the LRC of the following data: ** [HEX] 864C9BF3
**

In telecommunication, a **longitudinal redundancy check** (LRC) or **horizontal redundancy
check** is a form of redundancy check that is applied independently to each of a parallel group
of bit streams. The data must be divided into transmission blocks, to which the additional
check data is added.

**Data:** 1000 0110 0100 1100 1001 1011 1111 0011

Arranging data:

1 0 0 0
0 1 1 0
0 1 0 0
1 1 0 0
1 0 0 1
1 0 1 1
1 1 1 1
0 0 1 1
**1 0 0 0 (**Parity bit calculated for every column)

**Transmitted Data:** 1000 0110 0100 1100 1001 1011 1111 0011 **1000**

**VRC:
**

**Solution:
**

**LRC:
**

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Assignment 2

**Question 3: ** A: For P=110101 and M=1011010001, Find CRC?

B: In CRC error Detection scheme, choose P(x)=x 4

+x+1. Encode the bits

10111011011?**
**

1 1 0 1 1 0 1

110101 / 1 0 1 1 0 1 0 0 0 1 0 0 0 0 0 1 1 0 1 0 1 : : : : : : : : :

0 1 1 0 0 0 0 : : : : : : : : 1 1 0 1 0 1 : : : : : : : : 0 0 0 1 0 1 0 0 1 : : : : : 1 1 0 1 0 1 : : : : : 0 1 1 1 0 0 0 : : : :

1 1 0 1 0 1 : : : : 0 0 1 1 0 1 0 0 : : 1 1 0 1 0 1 : : 0 0 0 0 0 1 0 0 <= FCS

The FCS is: X2

Transmitted bits: 1 0 1 1 0 1 0 0 0 1 0 0 1 0 0

1 1 0 1 0 1 1

10011 / 1 0 1 1 1 0 1 1 0 1 1 0 0 0 0 1 0 0 1 1 : : : : : : : : : :

0 0 1 0 0 0 1 : : : : : : : : 1 0 0 1 1 : : : : : : : : 0 0 0 1 0 1 0 1 : : : : : 1 0 0 1 1 : : : : : 0 0 1 1 0 1 0 : : :

1 0 0 1 1 : : : 0 1 0 0 1 0 : : 1 0 0 1 1 : : 0 0 0 0 1 0 0 <= FCS

Predetermined Divisor: 10011

FCS = X3

Transmitted bits: 1 0 1 1 1 0 1 1 0 1 1 **0 1 0 0 **

**Part A:
**

**Part B:
**

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Assignment 2

**Question 4: C**alculate the hamming pair wise distance among the following
codewords:
**A:** 00000, 10101, 01010

**B:** 000000, 010101, 101010, 110110**
**

Let V1 = 0 0 0 0 0

V2 = 1 0 1 0 1

V3 = 0 1 0 1 0

D(V1,V2) = 3 D(V1,V3) = 2 D(V2,V3) = 5

Let V1 = 0 0 0 0 0 0

V2 = 0 1 0 1 0 1

V3 = 1 0 1 0 1 0

V4 = 1 1 0 1 1 0

D(V1,V2) = 3 D(V1,V3) = 3 D(V1,V4) = 4 D(V2,V3) = 6 D(V2,V4) = 3

D(V3,V4) = 3

**Question 5: ** Suppose a file of 8,000 bytes is to be sent over a line of 5600 bps.
**A:** Calculate the overhead in bits and time in using asynchronous communication. Assume one
start bit and a stop bit element of length one bit, and 8bits to send the byte itself for each
character. The 8 –bit character consists of all data bits with no parity bit.

**B:** Calculate the overhead in bits and time using synchronous communication. Assume that the

data are sent in frames. Each frame consists of 1000 characters = 8000 bits and overhead of 48

control bits per frame.**
**

**Part A:
**

**Part B:
**

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Assignment 2

Number of Bytes = 8000

Number of bits = 8 x 8000 = 64000

Line capacity = 5600 bps

Start bit length = 1 bit

Stop bit Length = 1 bit

Number of bits for 1 character = 8 bits

Bits apart from data = 2 bits per character (i.e. 2 bits for every byte)

Total bits apart from data = 2 x 8000 = **16000 bits**

Total bits transmitted = 64000 + 16000 = 80000

Therefore, the overhead ratio = (Number of extra bits) / (Total Number of bits Transmitted)

= 16000 / 80000

= 0.2

Percentage overhead = 0.2 x 100 % = 20 % (in bits)**
**

Time overhead = 16000/5600 = **2.85 sec**

Number of data bits per frame = 8000

Total frames in 8000 bytes = 64000/8000 = 8 frames

Control bits per frame = 48

Total number of overhead bits = 48 x 8 = **384 bits**

Therefore, the overhead ratio = (Number of control bits) / (Total Number of bits Transmitted)

= 384 / 64384

= 0.00596 = 0.006

Percentage overhead = 0.006 x 100 % = 0.6 % (in bits)

Time overhead = 384/5600 = **68 ms
**

**Question 6:
**

** A:** Consider the use of 1500-bit frames on a 1-Mbps satellite channel with a 270-ms delay.
What is the maximum link utilization for?

**a:** Stop-and-wait flow control?

**Part A:
**

**Part B:
**

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Assignment 2

**b:** Continuous flow control with a window size of 7?

**c:** Continuous flow control with a window size of 127?

**d:** Continuous flow control with a window size of 255?

**B:** In figure

frames are generated at node A and sent to node C through node B. determine the minimum data rate required between nodes B and C so that the buffers of node B are not flooded, based on the following:

a) The data rate between A and B is 100Kbps.

b) The propagation delay is 5micro seconds per kilometer for both lines.

c) There are full-duplex lines between the nodes.

d) All data frames are 1000 bits long; ACK frames are separate frames of negligible length.

e) Between A and B, a sliding-window protocol with a window size 3 is used.

f) Between B and C, Stop-and-wait is used.

g) There are no errors.

**
**

Bits/frame = 1500 bits / frame

Data rate = R = 1Mbps = 1000000 bits / sec

Propagation Delay = 270 ms = 0.27 sec

Link Utilization for **stop and wait flow control** is given by:

U =

Where a = tprop / tframe

tframe = (Bits/frame) / R = 1.5 x 10 -3 sec / frame

**Part A:
**

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Assignment 2

a = 180 => **U = 0.00277
**

Link Utilization for **continuous flow control** is given by:

U =

Where W is the window size.

For a **window size of 7:
**

=> **U = 0.0193**

For a **window size of 127:
**

=> **U = 0.35**

For a **window size of 255:
**

=> **U = 0.706**

**From A to B:
**

R = 100 kbps, Window size = W = 3, Propagation time / km = 5 usec / km

Frame size = 1000 bits

tprop = 4000 x 5 usec = 20 ms

tframe = 1000 / 100000 = 10 ms

Since the window size between A and B is 3, therefore, A can transmit three frames to B and then

must wait for the acknowledgment of the first frame before transmitting additional frames.

Time taken by A to transmit 3 frame = T

T = (2 x tprop) + tframe = 50 ms

**From B to C:
**

tprop = 1000 x 5 usec = 5 ms

tframe = 1000 / R

B can transmit one frame at a time to C. The time taken to transmit 1 frame and get back the

acknowledgement is T:

T = T = (2 x tprop) + tframe = tframe + 10 ms

**Part B:
**

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Assignment 2

Therefore, it will take 3tframe + 30 ms to transmit 3 frames transmitted by A,

3tframe + 30 ms = 50 ms tframe = 6.66 ms

Since R = 1000 / tframe => **R = 150 kbps**

**
**

**Question 7: **No mention was made of reject (REJ) frames in the stop-and-wait

ARQ discussion. Why it is not necessary to have REJ0 and REJ1 for stop-and-wait

ARQ?**
**

**
**

**Stop-and-wait ARQ** is the simplest kind of automatic repeat-request (ARQ) method. A stop-

and-wait ARQ sender sends one frame at a time. After sending each frame, the sender doesn't

send any further frames until it receives an ACK (acknowledgement) signal. After receiving a

good frame, the receiver sends an ACK. If the ACK does not reach the sender before a certain

time, known as the timeout, the sender sends the same frame again.

**
**

If the receiver sees that the frame is good, it sends an ACK. If the receiver sees that the frame

is damaged, the receiver discards it and does not send an ACK -- pretending that the frame

was completely lost, not merely damaged. Hence this timeout acts as a reject signal for the

transmitter. The transmitter simply sends the same frame.

**
**

**Question 8: **Find link utilization of the following scenarios

**A:** WAN using ATM with two stations a thousand kilometers apart. Frame size is 424 bits and

data rate is 155.52Mbps. Speed of signal is 2*10 8

m/s.

**B:** LAN with two stations 100 meters apart. Frame size is 1000bits and data rate is 100Mbps.

Speed of signal is 2*10 8

m/s.

**Stop and Wait ARQ:
**

**REJ0 and REJ1 Not necessary:
**

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Assignment 2

Given that: Data rate = R = 155.52 Mbps Distance = d = 1000 km Speed of the signal = V = 2 x 108 m/s Frame size = L = 424 bits length of the link in bits: B = R * (d / V) = 720000000 bits

a = (B / L) = 720000000 / 424 = 1.69 x

**
**

U = ( ) = ( ) => **U = 2.95 x 10-7**

Given That: Data rate = R = 100 Mbps Distance = d = 100 m Speed of the signal = V = 2 x 108 m/s Frame size = L = 1000 bits Now, for length of the link in bits:

B = R * (d / V) = 46.29 x bits

a = (B / L) = 720000000 / 1000 = 46.29 x

**
**

U = ( ) = ( ) => **U = 1.08 x 10-5**

**Question 9: **“When will you reach home”

Generate Huffman code?

**
**

Total Characters = 24 Frequencies: a = 1, c = 1, e = 3, h = 3, i = 1, l = 2, m = 1, n = 1, o = 2, r = 1, u = 1, y = 1, w = 2, space = 4 Respective Probabilities: = 1/24, 1/24, 1/8, 1/8, 1/24, 1/12, 1/24, 1/24, 1/12, 1/24, 1/24, 1/24, 1/12

**Part A:
**

**Part B:
**

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Assignment 2

Arranging according to frequencies: Space : 4 e: 3 h: 3 l: 2 0: 2 w: 2 a: 1 c: 1 i: 1 m: 1 n: 1 r: 1 u: 1 y: 1

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Assignment 2

**Question 10: **What are the dangers involved in internet communication? Clarify
with example?
**
**

The Internet has its origins in a research project of the American administration in the late
sixties; the development goal was to establish the most robust and reliable networking
possible between relatively few university computer centres. The security and confidentiality
of communication played a minor role at that time. This still has an effect on the Internet
today with its hundreds of millions of users worldwide.
**1. Spying out information
**Information is generally transported across a large number of network nodes and transmission
links on the Internet, where it can be read relatively easily by third parties.
**2. Corrupting information, pretending a false identity
**During their transport through the Internet data can be modified unnoticed by the recipient.
Sender details in e-mails and computer addresses can also be falsified. You thus cannot be
sure of the identity of your communication partner and the authenticity of the information
received.
**3. Implanting harmful programs
**Together with seemingly harmless information (e-mail, websites) harmful programs (viruses,
worms, Trojan horses, ...) are sometimes sent or offered on the Internet, which in most cases
can implant themselves unnoticed on your computer and cause damage there.
**4. Unauthorized penetration into computers
**Even if you do not actively communicate yourself, your computer is in danger of being misused
by unauthorized third parties (’hackers’), if there is a connection to the Internet. Hackers
often exploit faults in the operating and application software to clandestinely implant harmful
programs through the network, which often enable complete control of the computer. Via
such Trojan horses it is also possible, for example, to log keyboard strokes and send them to
the hacker, who can thus obtain information extremely worth being protected (e.g.
passwords,
PIN).
**5. Disclosure of personal information
**In communicating on the Internet, on the one hand, you always leave information at different
places, which allows conclusions to be drawn concerning your person and your communication
behaviour (name, Internet address, type and amount of retrieved information etc.). On the
other hand, you are often actively requested to enter sensitive information. It should be noted

**Internet Communication:
**

**Dangers Involved:
**

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Assignment 2

here that adequate data protection provisions penalizing a misuse of personal data are not effective in all countries.

**Question 11: **Illustrate how digital signatures work by giving an example?
**
**
A digital code that can be attached to an electronically transmitted message that uniquely
identifies the sender. Like a written signature, the purpose of a digital signature is to
guarantee that the individual sending the message really is who he or she claims to be. Digital
signatures are especially important for electronic commerce and are a key component of most
authentication schemes. To be effective, digital signatures must be unforgeable. There are a
number of different encryption techniques to guarantee this level of security.

**
**

**Step 1: Calculate the Message Digest
**

In the first step of the process, a **hash-value** of the message (often called the **message digest**)
is calculated by applying some cryptographic **hashing algorithm** (for example, MD2, MD4,
MD5, SHA1, or other). The calculated hash-value of a message is a sequence of bits, usually
with a fixed length, extracted in some manner from the message.

**Step 2: Calculate the Digital Signature
**

In the second step of digitally signing a message, the information obtained in the first step
hash-value of the message (the message digest) is encrypted with the private key of the
person who signs the message and thus an encrypted hash-value, also called **digital signature**,
is obtained. For this purpose, some mathematical cryptographic **encrypting algorithm** for
calculating digital signatures from given message digest is used. The most often used
algorithms are **RSA** (based on the number theory), **DSA** (based on the theory of the discrete
logarithms), and **ECDSA** (based on the elliptic curves theory).

**Digital Signatures:
**

**Working:
**

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Assignment 2

**
**

**
**

**Step 1: Calculate the Current Hash-Value
**

In the first step, a hash-value of the signed message is calculated. For this calculation, the
same hashing algorithm is used as was used during the signing process. The obtained hash-
value is called the **current hash-value** because it is calculated from the current state of the
message.

**Step 2: Calculate the Original Hash-Value
**

In the second step of the digital signature verification process, the digital signature is
decrypted with the same encryption algorithm that was used during the signing process. The
decryption is done by the public key that corresponds to the private key used during the
signing of the message. As a result, we obtain the **original hash-value** that was calculated
from the original message during the first step of the signing process (the original message
digests).

**Step 3: Compare the Current and the Original Hash-Values
**

In the third step, we compare the current hash-value obtained in the first step with the original hash-value obtained in the second step. If the two values are identical, the verification if successful and proves that the message has been signed with the private key that corresponds to the public key used in the verification process. If the two values differ from onr another, this means that the digital signature is invalid and the verification is unsuccessful.

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Assignment 2

**Question 12: **Search on the web to find out exact procedure of encryption using

**A:** DES (Even Reg Number)

**B:** RSA (Odd Reg Number)

Show all steps by using example. You can use any data to demonstrate?
**
**

**
**

The DES, which is a block cipher, is the most widely known encryption algorithm. A "block
cipher" refers to a cipher that encrypts a block of data all at once, and then goes on to the
next block.
In block encryption algorithms, the plaintext is divided into blocks of fixed length which are
then enciphered using the secret key. The DES is the algorithm in which a **64-bit block** of
plaintext is transformed (encrypted/enciphered) into a **64-bit ciphertext** under the control of
a **56-bit internal** **key**, by means of permutation and substitution.
The key is given in a 64-bit word, of which eight are parity bits, at locations 8, 16, 24,…, 64.
The algorithm consists of 16 "rounds" of operations that mix the data and key together in a
prescribed manner using the fundamental operations of permutation and substitution.

**Part A:
**

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Assignment 2

**
**

**Key Generation Algorithm
**

1. Generate two large random primes, *p* and *q*, of approximately equal size such that their
product n = pq is of the required bit length, e.g. 1024 bits. [See note 1].

2. Compute n = pq and (φ) phi = (p-1)(q-1).
3. Choose an integer *e*, 1 < e < phi, such that gcd(e, phi) = 1. [See note 2].
4. Compute the secret exponent *d*, 1 < d < phi, such that ed ≡ 1 (mod phi). [See note 3].
5. The public key is (n, e) and the private key is (n, d). Keep all the values d, p, q and phi

secret.

n is known as the *modulus*.
e is known as the *public exponent* or *encryption exponent* or just the *exponent*.
d is known as the *secret exponent* or *decryption exponent*.

**Encryption
**

Sender A does the following:-

1. Obtains the recipient B's public key (n, e).
2. Represents the plaintext message as a positive integer *m* [see note 4].
3. Computes the ciphertext c = me mod n.
4. Sends the ciphertext *c* to B.

**Part B:
**

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Assignment 2

**Decryption
**

Recipient B does the following:-

1. Uses his private key (n, d) to compute m = cd mod n.
2. Extracts the plaintext from the message representative *m*.

**
**

**
**

**
**

**
**

**
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