Answer: I
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Answer: IV
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Answer: IV
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Answer: III, IV
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Answer: IV
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Answer: II
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Answer: I
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Answer: I
Explanation:
Class B : Out of 32 bits, Class B have
Network id = 16 bits and
Host id = 16 bits
Calculation
Given subnet bits = 6 bits
Maximum number of the subnets = 2⁶ – 2 = 62
Note that 2 is substracted because
* all 0’s is used to represent Network and
* all 1’s is used to represent Broadcast Address of Network.
Maximum number of hosts in each subnet
Host id bits = 32 – (Network id bit + subnet id bits) = 32 – (16+ 6) = 10
Maximum number of hosts in each subnet = 2¹⁰– 2 = 1022
Note that 2 is substracted because
* all 0’s is used to represent Subnet and
* all 1’s is used to represent Broadcast Address of Subnet.
So, Option(I) is correct.

Answer: IV
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Answer: III

Explanation:
RSA key Generation Algorithm:
1) Select 2 primes p & q such that p!= q.
2) Calculate n = p*q.
3) Calculate ϕ(n) = (p-1)*(q-1)
4) Select e such that 1< e< ϕ (n) & gcd (e, ϕ(n)) == 1
5) Calculate d , e*d mod ϕ(n) ==1
6) Public key = (e , n)
Private key = d
7) Return public key(e,n) & private key(d,n)

Calculate Ciphertext & Plaintext using key e and d:
Ciphertext = (Plaintext)^e mod n
Plaintext = (Ciphertext)^d mod n

Given,
p=3, q=11
d=7
Plaintext =19
Ask,
Ciphertext=?

Calculation,
1) Two prime numbers p =3 , q=11
2) Calculate n = p*q = 3*11 = 33
3) Calculate ϕ(n) = (p-1)*(q-1) = (3-1)*(11-1) = 2*10=20
4) Calculate d ,
=>e*d mod ϕ(n) ==1
=>e*7 mod 20 ==1
=> 3*7 mod 20 ==1 {if we take value of e is 3 , then reminder will be 1}
5) public key (e) = 3

Now, Calculate Ciphertext using key :
Ciphertext = (Plaintext)^e mod n
= 19³ mod 33
=28

So, Option(III) is correct.

Answer: I
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Answer: IV
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Answer: II
Explanation:

Answer: III
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Answer: IV
Explanation:

POP has two modes:
i) Delete Mode – A mail is deleted from the mailbox on the mail server after successful retrieval. (Delete after reading)
ii) Keep Mode – A mail remains in the mailbox on the mail server after successful retrieval. (Store even after reading)

Stetement(A)- True.
Stetement(B)- True.
Stetement(C)- False.
Stetement(D)- False.
Stetement(E)- True.

So, option(iv) is correct.

Answer: I
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Answer: IV

Explanation:
Concept:
In cryptography,
• To encrypt a message, we need an encryption algorithm & To decrypt a message, we need a decryption algorithm.
• Encryption algorithm convert plaintext into ciphertext & Decryption algorithm convert ciphertext into plaintext.
• For Encryption algorithm, we need an encryption key, and the plaintext.
• For Decryption algorithm, we need a decryption key, and the ciphertext.

Ciphertext & user details are not needed by encryption algorithm.

So, Option(IV) is correct.

Answer: II
Explanation:
Given,
Bandwidth = 10 Mbps = 10 x 106bps
Frame per minutes = 12,000 Frame per min
Frame size = 10,000 bits

Ask,
Throughput = ?

Throughput::
In data Transmission, network throughput tells how much data was transferred from a source at any given time & typically measured in bps, as in Mbps or Gbps.

Caculation:
Throughput = Frames per minute * each Frame size
= 12,000 per minute* 10,000 bits
= 12,000 * 10,000 bits per minute
= (12,000 * 10,000 / 60) bits per second
= 2,000,000 bps
= 2 Mbps

So, Option(ii) is correct.

Answer: IV
Explanation:

Confidentiality
• Confidentiality means that only the authorized individuals/systems can view sensitive or classified information.
• The data being sent over the network should not be accessed by unauthorized individuals.
• Confidentiality means that the sender and the receiver expect privacy.

Integrity
• Data integrity means maintaining and assuring the accuracy and completeness of data over its entire lifecycle.
• Integrity means that the data must arrive at the receiver exactly as they were sent.

Authentication
• Authentication is the process of determining whether someone is who it is declared to be.
• Authentication means the receiver is ensured that the message is coming from the intended sender.

So, Option(IV) is correct.

Answer: II

Explanation:
• A piconet is a wireless set of devices make of use Bluetooth technology protocols.
• A piconet is connection of two or more devices occupying the same physical.
• A piconet is a combination of one master device( any device whis share the resource) and seven active slave devices(devices connected to the master device like phone,printer,speaker,scanner,mic etc) which has maximum 255 further slave devices can be inactive.

Other topic related to Piconet is Scatternet.

Scatternet -> sactternet is combination of Piconet.

So the correct answer is Option(II). (One Master and Seven Slave Devices )



For more details about Piconet Please Visit the More Discussion

Answer: I
Explanation:

ICANN (Internet Corporation for Assigned Name and Numbers)is non-profit organization that was formed in 1998. ICANN maintains the central repository for IP Addresses & help coordinate the supply the IP Addresses. It also manages the domain name system & root servers.

Answer: I

Explanation:
Let’s try to understand the basic,
In distance vector & link state routing, router have routing table & router have to store information about other router. As networks grow in size, number of router increases in network. Due to this, router can’t handle network traffic efficiently. So, routing will have to done hierarchically.

In three level hierarchically routing,
Group of routers are called region & Group of regions are called cluster.

Given:
Number of cluster = 8 cluster
Each cluster have 9 regions.
Each region have 10 routers.

Total number of entries in hierarchically table of each router
=10 entries for router + 8 entries for routing to other regions within own cluster + 7 entries for distant clusters
=10 + 8 + 7
=25

or

Total number of entries in hierarchically table of each router
= Total nmber of router in a region + (Number of regions – 1) +(Number of clusters -1)
=10 + 8 + 7
=25

So, Option(I) is correct.

Answer: I
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Answer: III
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Answer: I

Explanation:
Classful IP addresses is divided into classes. The leading bits of classes are:
Class A – 0
Class B – 10
Class C – 110
Class D – 1110
Class E – 1111

On the basis of percentage, address space consume by all classes are:
Class A – 50%
Class B – 25%
Class C – 12.5%
Class D – 6.25%
Class E – 6.25%

Answer: IV
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Answer: III

Explanation:
ARP :
• Address Resolution Protocol [ARP] is a network layer protocol that is used to convert IP address into MAC address.
• ARP is protocol that allows a client to send a broadcast message with its IP address and receive an MAC address in reply.

DNS:
• DNS stands for Domain Name System.
• DNS is a service that translates the domain name into IP addresses.
• For example, the domain name www.samagraCS.com has a IP address 162.0.215.21(Suppose)
• Where, www is a particular host server.
• The com part of the domain name reflects the purpose of the organization or entity (in this example, commercial) and is called the top-level domain name.
• The samagraCS part of the domain name defines the organization and together with the top-level is called the second-level domain name.

RARP:
• Reverse Address Resolution Protocol (RARP) is a network layer protocol used to convert an IP address from a given physical address.
• RARP is the complement of ARP.
• “RARP is a protocol that allows a client to send a broadcast message with its MAC address and receive an IP address in reply”.

ICMP:
• The ICMP stands for Internet Control Message Protocol.
• ICMP is a network level protocol.
• ICMP (Internet Control Message Protocol) is an error-reporting protocol that network devices such as routers use to generate error messages to the source IP address when network problems prevent delivery of IP packets.

So, Option(III) is correct.

Answer: II
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Answer: IV

Explanation:
Given :
Bandwidth (Bw) = 1Gbps = 10⁹ bps
Link Length or Distance (d) = 800 Km
Speed or velocity (v) = 200,000 Km / Sec

Propogation Delay: It is a measure of the time required for asignal to propogate from one end of circuit to other end.

Formula used :


Calculation:


= 0.004 sec
= 4 ms

{1 second = 1000 millisecond}

So, Option(IV) is correct.

Answer: II

Explanation:
Loopback address
• A loopback address is a special IP address, 127.0.0.1, reserved by InterNIC for use in testing network cards.
• A loopback address works is that a data packet will get sent through a network and routed back to the same device where it originated.
• In IPv4, 127.0.0.1 is the most commonly used loopback address, however, this can range be extended to 127.255.255.255.
• The loopback (IP) address is used to send a packet from host to itself & check NIC working or not.
• The loopback (IP) address is a member of class A network. (IP Address of Class A network range from 0.0.0.0 to 127.255.255.255)

Statement P: The loopback(IP) address is a member of class B network (False).
Because, The loopback(IP) address is a member of class A network.

Statement Q: The loopback(IP) address is used to send a packet from host to itself (True)

So, Option(II) is correct.

Answer: I
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Answer: II

Explanation:
We know,
In a Class B network, the first 16 bits are the network part of the address. & remaining 16 bits indicate the host within the network.

Subnet mask:
A subnet mask always consists of a series of contiguous 1 bits followed by a series of contiguous 0 bits.Where,
• 1’s represent Network ID & Subnet ID
• 0’s represent Host ID

Given that,
Number of Subnet = 500 subnet
So, Number of bits required to represent subnet = ceil value of(500) = 9bits

Number of Host per subnet = 100 subnet
So, Number of bits required to represent subnet = ceil value of(100) = 7 bits

Now,
We borrow 9bits of Subnet ID bits from Host ID part(16 bits is HID part in class B) to represent 500 subnet.
Network ID bits = 16 bits
Subnet ID bits = 9 bits
Host ID bits = 16-9 = 7bits

Now its obvious,
Number of 1’s in subnet mask = Network ID bits + Subnet ID bits = 16 + 9 = 25
Number of 0’s in subnet mask = Host ID bits = 7

Hence, Subnet Mask is
1111 1111 . 1111 1111 . 1111 1111 . 1000 0000
In decimal form it is represented as:
255 : 255 : 255 : 128

So, Option(II) is correct.

Answer: IV
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Answer: IV

Explanation:
Given,
The Euler’s phi function

Where, p runs over all primes dividing n that means
p = prime factors of n

Now, given n = 45
Then, p (prime factor of 45) = {3,5}

Calculation,
Put values of p in given Euler’s phi function


So, Option(IV) is correct.

Answer: IV
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Answer: IV
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Answer: I
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Answer: II

Explanation:






According to given Option,
valid allocation of addresses to A = 245.248.136.0/21 and
valid allocation of addresses to B = 245.248.128.0/22

So, Option(II) is correct.

Answer: III
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Answer: II
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Answer: IV

Explanation:
UDP Operation:
• Connectionless Services
• Unreliable Services
• No flow Control
• No Error Control Except Checksum
• No Congestion Control Protocol
• No retransmission
• UDP is a message-oriented protocol.

TCP Operation:
• Connection Oriented Services
• reliable Services
• flow Control
• Error Control
• Congestion Control Protocol
• Retransmission
• TCP is a byte-oriented protocol.

Statements S1: TCP handles both congestion and flow control(True)
Statements S2: UDP handles congestion but not flow control(False)

So, Option(IV) is correct.

Answer: III
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Answer: III

Explanation:
IP Address: An IP address is a 32-bit number. The 32 bits are divided into four octets. an IP address is divided into Two parts named network id
and host id. The first part (Network id)of an IP address identifies the network on which the host resides, while the second part(Host id) identifies
the particular host on the given network.

Answer: I
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Answer: II
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Answer: III
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Answer: I

Explanation:
Given,
Frame size = 200 bits
Channel Bandwidth= 200 Kbps
System Produces = 250 frame per second

Ask,
Throughput of the system = ?

Formula,
The Throughput of slotted ALOHA is S = G * e^-G
The maximum Throughput S_max = 0.368 (or 36.8%) when G =1
Where, G is the average number of frames produced by the system during one frame transmission time.
The Frame Transmission time =

Calculation,
The Frame Transmission time = = 10^-3 second = 1 millisecond
System Produces = 250 frame per second = (250 * 10³ / 10³) frame per second = (1/4) frame per millisecond
so, The Load (G) = 1/4 frame

S = G * e^-G
= (1/4) * e^-(1/4)
= 0.195 (or 19.5%) {where, e = 2.71}

Throughput = 250 * 0.195 = 49 Frames
Only 49 out of 259 frames will survive.

So, Option(I) is correct.

Answer: II

Explanation:
Given,
Period(T) = 100ms

Formula used,

Calculation:

= second
= second
= 10 Hz
= Hz
= 10^-2 KHz

So, Option(II) is correct.

Answer: III

Explanation:
Dotted Decimal Representation: An IP address is a 32-bit number written in dotted decimal notation, four 8-bit fields (octets) converted from binary to decimal numbers, separated by dots.

10000001 00001011 00001011 11101111

convert 8-bit binary to decimal & seperate them by dots

129.11.11.239

So, Option(III) is correct.

Answer: II
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Answer: II
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Answer: III

Explanation:
Switching: In large networks, there can be multiple paths from sender to receiver. The switching technique will decide the best route for data transmission.
Three broad categories of Networks are
(i) Circuit Switched Networks
(ii) Packet Switched Networks
(iii) Message Switched Networks

Circuit Switched Networks
• A dedicated path is establish between sender and receiver.
• Resources need to be reserved during the set up phase.

Packet Switched Networks
• Itis a switching technique in which the message is sent in one go, but it is divided into smaller pieces, and they are sent individually.
• There are two approaches to Packet Switching:
1) Datagram Packet switching:
• Resourse is reserved
• Connection oriented

2) Virtual Circuit Switching
• no Resourse is reserved
• Connectionless

Message Switching
• It is a switching technique in which a message is transferred as a complete unit and routed through intermediate nodes at which it is stored and forwarded.
• In Message Switching technique, there is no establishment of a dedicated path between the sender and receiver.
• Resources need not reserved during the set up phase.

Statement(a) : True
Statement(b) : False
Statement(c) : True

So, Option(III) is correct.

Answer: I
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Answer: II

Explanation:
Basic Concept
A substitution cipher replaces one symbol with another. Substitution ciphers can be categorized as
• Mono alphabetic
• Polyalphabetic ciphers

Shift Cipher
• The simplest monoalphabetic cipher is probably the shift cipher.
• The shift cipher is sometimes referred to as the Caesar cipher.
• The encryption algorithm is “shift key characters down,” with key equal to some number. The decryption algorithm is “shift key characters up.”

Calculation
W = 23 -15 = 8 (h)
T = 20 -15 =5 (e)
A = 1 – 15 = -14 + 26 = 12 (l) {when get negative number then add 26 to wrap arround}
A = 1 – 15 = -14 + 26 = 12 (l)
D = 4 – 15 = -11 + 26 = 15 (o)

Decrypted message = hello

So, Option(II) is correct.

Answer: IV
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Answer: III

Explanation:
Given,
PainText = HELLO MY DEARZ

In encryption, we move the character at position 2 to position 1, the character at Position 4 to position 2, and so on.

Ask,
CipherText =?

Calculation


So, Option(III) is correct.

Answer: III
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Answer: III
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Answer: II
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Answer: IV

Explanation:

So, Option(IV) is correct.