# Computer Networks | Subject Wise

## Computer-Networks | Subject-wise

 Question 1
One SAN switch has 24 ports. All 24 supports 8 Gbps Fiber Channel technology. What is the aggregate bandwidth of that SAN switch?
 A 96 Gbps B 192 Mbps C 512 Gbps D 192 Gbps
Question 1 Explanation:
Given
SAN switch have Number of ports = 24 ports
Channel capacity of each port = 8Gbps
Aggregate Bandwidth = Number of ports * Channel capacity of each port.
Aggregate Bandwidth = 24 * 8 = 192 Gbps
 Question 2
Data is transmitted continuously at 2.048 Mbps rate for 10 hours and received 512 bits errors. What is the bit error rate?
 A 6.9 e-9 B 6.9 e-6 C 69 e-9 D 4 e-9
Question 2 Explanation:
Bandwidth = 2.048 Mbps
Error bits = 512
Bit error rate is the number of bits sent per data in 10 hours , we sent 512 bit error
10 hours  = 512 bit error
36000 secs = 512 bit error
1 sec      = 512/36000
1 sec      = 0.0142 bit error.

Error rate = Total number of error bits / Bandwidth

Error rate =   (0.0142)/(2.048 x 106)
Error rate = 6.944 x 10-9
 Question 3
Lightweight Directory Access protocol is used for
 A Routing the packets B Authentication C Obtaining IP address D Domain name resolving
Question 3 Explanation:
• The Lightweight Directory Access Protocol (LDAP) is an open, vendor-neutral, industry standard application protocol for accessing and maintaining distributed directory information services over an Internet Protocol (IP) network
• A common use of LDAP is to provide a central place to store usernames and passwords. This allows many different applications and services to connect to the LDAP server to validate users
• When an LDAP session is created, that is, when an LDAP client connects to the server, the authentication state of the session is set to anonymous. The BIND(authenticate) operation establishes the authentication state for a session.
• LDAP is often used by other services for authentication and/or authorization (what actions a given already-authenticated user can do on what service)
 Question 4
What is the maximum number of characters (7 bits + parity ) that can be transmitted in a second on a 19.2 kbps line? This asynchronous transmission requires 1 start bit and 1 stop bit.
 A 192 B 240 C 1920 D 1966
Question 4 Explanation:
In synchronous mode of transfer we don’t require start and stop bits
• Total number of bits per character while transmitting is (7+1)=8 bits
• Bandwidth = 19.2 kbps
• Number of character transmitted in 1 second = (19.2 *1000)/8 = 2400
In Asynchronous transfer, we need start and stop bits in order to synchronize these bits are added to the number of bits in each character.
• Total number of bits per character while transmitting is (7 data bit + 1 parity bit + 1 start bit + 1 stop bit)=10 bits
• Bandwidth = 19.2 kbps
• Number of character transmitted in 1 second = (19.2 *1000)/10 = 1920
 Question 5
IEEE 1394 is related to
 A RS-232 B USB C Firewire D PCI
Question 5 Explanation:
• IEEE 1394 is an interface standard for a serial bus for high-speed communications and isochronous real-time data transfer.
• It was developed in the late 1980s and early 1990s by Apple, which called it FireWire,
• FireWire is also available in Cat 5 and optical fiber versions.
• The 1394 interface is comparable to USB.
• USB was developed subsequently and gained much greater market share. USB requires a master controller whereas IEEE 1394 is cooperatively managed by the connected devices.
Refer : https://en.wikipedia.org/wiki/IEEE_1394
 Question 6
In the Ethernet, which field is actually added at the physical layer and is not part of the frame
 A Preamble B CRC C Address D Location
Question 6 Explanation:
• The first field(Preamble) of 802.3 frame contains 7 bytes of alternating 0's and 1's that alerts the receiving system to the coming frame and enables it to synchronize its input timing
• Preamble belongs to the physical layer and are added at the physical layer only and is not part of the frame.
 Question 7
Ethernet layer-2 switch is a network element type which gives
 A Different collision domain and same broadcast domain B Different collision domain and different broadcast domain C Same collision domain and same broadcast domain D Same collision domain and different broadcast domain
Question 7 Explanation:
Switch act as a collision domain separator not not as a broadcast domain separator.
• Switch is a collision domain separator because it has lookup table associated with it
• Each lookup table entry contains a port associated with the switch and the MAC address of the concerned device(computer) associated with this port.
• Hence collisions domain is separate for each switch.
• Switch is nothing but an active hub.
• Switch is not a broadcast domain separator . But it can be made so using the concept of virtual switching.
• So in that case it will be a broadcast domain separator
 Question 8
What will be the efficiency of a Stop and Wait protocol, if the transmission time for a frame is 20ns and the propagation time is 30ns?
 A 20% B 25% C 40% D 66%
Question 8 Explanation:
Tt(data) : Transmission delay for Data packet
Tp(data) : propagation delay for Data packet

Tt = 20 ns.
Tp = 30 ns

Efficiency of stop and wait protocol =  [ 1 / (1 + 2a) ]

Where  a = Tp / Tt
Tp = propagation delay
Tt = transmission delay

Efficiency = 1 / (1 + 2(30/20))
Efficiency = 1/4 * 100
Efficiency = 25 %
 Question 9
IPv6 does not support which of the following addressing modes?
Question 9 Explanation:
IPv6 support the following Addressing Modes:
• Unicast addressing : In unicast mode of addressing, an IPv6 interface (host) is uniquely identified in a network segment. A packet is delivered to one interface
• Multicast addressing : IPv6 supports multicast features as the single packet data can be sent to multiple destination hosts at a time.
• Anycast addressing :  A packet is delivered to the nearest of multiple interfaces. Multiple interfaces (hosts) are assigned same Anycast IP address.
 Question 10
What is IP class and number of sub-networks if the subnet mask is 255.224.0.0?
 A Class A, 3 B Class A, 8 C Class B, 3 D Class B, 32
Question 10 Explanation:
Equivalent Binary representation will be
• 255 = 11111111
• 224 = 11100000
• 0      = 00000000
• 0      = 00000000
• First 8 1's Represent Class A
• Next 3 1's used for subnet.
Number of subnets possible with 3 bits = 23=8
 Question 11
Which algorithm is used to shape the bursty traffic into a fixed rate traffic by averaging the data rate?
 A Solid bucket algorithm B Spanning tree algorithm C Hocken helm algorithm D Leaky bucket algorithm
Question 11 Explanation:
• Leaky bucket algorithm is used to control transmission rate in a network.
• Leaky bucket algorithm is used to check data transmissions, in the form of packets to define limits on bandwidth.
• By setting the bucket size and output flow rate, the code can be refined to slow, medium and fast.
• It is implemented as a single-server queue with constant service time.
• If the bucket (buffer) overflows then packets are discarded.
• In this algorithm the input rate can vary but the output rate remains constant.
• Algorithm saves busty traffic into fixed rate traffic by averaging the data rate.
 Question 12
A packet filtering firewall can
 A Deny certain users from accessing a service B Block worms and viruses from entering the network C Disallow some files from being accessed through FTP D Block some hosts from accessing the network
Question 12 Explanation:
• Packet filtering firewalls operate at the network layer (Layer 3) of the OSI model.
• Packet filtering firewalls make processing decisions based on network addresses, ports, or protocols
• Packet filtering firewalls examine the packet headers that contain IP addresses and packet options and block or allow traffic through the firewall based on that information
 Question 13
The protocol data unit for the transport layer in the internet stack is
 A Segment B Message C Datagram D Frame
Question 13 Explanation:
The Protocol Data Unit is the unit of communication at a particular layer.
• Physical Layer PDU is the bit or, more generally, symbol
• Data Link Layer PDU is the frame.
• Network Layer PDU is the datagram.
• Transport Layer The transport layer PDU is  segment for TCP
• Application Layer PDU is the data or message.
• link layer PDU is the frame.
A layer lower in the Internet protocol suite, at the Internet layer, the PDU is called a packet, irrespective of its payload type.
 Question 14
How many check bits are required for 16 bit data word to detect 2 bit errors and single bit correction using hamming code?
 A 5 B 6 C 7 D 8
Question 14 Explanation:
For detecting d bit error, we need d+1 bits
Here d=2 So We require = 3 bits
For correcting d bit error we need 2d+1 bits
Here, d=1 bits So we require = 2 * 1 + 1 = 3 bits
∴ Total bits = 3 + 3 = 6
 Question 15
The process of modifying IP address information in IP packet headers while in transit across a traffic routing device is called
Question 15 Explanation:
Port Address Translation (PAT) is an extension to network address translation (NAT) that permits multiple devices on a local area network (LAN) to be mapped to a single public IP address. The goal of PAT is to conserve IP addresses.

Network address translation (NAT)  designed for IP address conservation. Network address translation (NAT) is the process of modifying IP address information in IP packet headers while in transit across a traffic routing device. NAT operates on a router, usually connecting two networks together, and translates the private (not globally unique) addresses in the internal network into legal addresses, before packets are forwarded to another network

Address mapping (also know as pin mapping or geocoding) is the process of assigning map coordinate locations to addresses in a database. The output of address mapping is a point layer attributed with all of the data from the input database.

Port mapping or port forwarding is an application of network address translation (NAT) which redirects a communication request from one address and port number combination to another while the packets are traversing a network gateway, such as a router or firewall.
 Question 16
What is routing algorithm used by OSPF routing protocol?
 A Distance vector B Flooding C Path vector D Link state
Question 16 Explanation:
• Open Shortest Path First (OSPF) is a routing protocol for Internet Protocol (IP) networks.
• It uses a link state routing (LSR) algorithm to calculate the shortest route to a destination through a network based on an algorithm
• Routing Information Protocol (RIP) and Open Shortest Path First (OSPF) are Interior Gateway Protocol, i.e., they both are used within an autonomous system.
• RIP is an old protocol (not used anymore) based on distance vector routing.
• OSPF is based on Link State Routing.
 Question 17
In a system an RSA algorithm with p=5 and q=11, is implemented for data security. What is the value of the decryption key if the value of the encryption key is 27?
 A 3 B 7 C 27 D 40
Question 17 Explanation:
Basic RSA Algorithm:
1. Choose two primes, p and q.
2. Compute n=p*q and z=(p-1)*(q-1).
3. Choose a number relatively prime to z and call it d.
4. Find e such that e*d=1 mod z.

Given two distinct prime numbers p = 5 and q = 11 and  encryption key, e = 27
Compute n = p * q = 5 * 11 = 55
Calculate the totient z = (p-1) * (q-1) = 4 * 10 = 40

Let the value of decryption key be ‘d’ such that :

e * d mod z = 1
27 * d mod 40 = 1
d = 3

Value of decryption key be  d = 3
 Question 18
An IP packet has arrived in which the fragment offset value is 100, the value of HLEN is 5 and the value of total length field is 200. What is the number of last byte in packet.
 A 194 B 394 C 979 D 1179
Question 18 Explanation:
HLEN = 5
Header length = 5 * 4 = 20 bytes
Total length field = 200
So payload size = 200 – 20 = 180
Starting number of first byte of fragment=100*8=800
Therefore, number of last byte in packet = 800 + 180 – 1 = 979
 Question 19
Consider a 50 kbps satellite channel with a 500 milliseconds round trip propagation delay. If the sender wants to transmit 1000 bit frames, how much time will it take for the receiver to receive the frame?
 A 250 milliseconds B 20 milliseconds C 520 milliseconds D 270 milliseconds
Question 19 Explanation:
Given data
Bandwidth =50 kbps
Frame size = 1000 bits

Round trip time = 2 * Propagation time  = Propagation time =500/2= 250 ms
( A ----> satellite -----> B ) + ( B----> satellite --->A )  = 500 ms
A ---> satellite ----> B   equal to B ---->satellite ---->A  = 250ms

Transmission time = Message size / bandwidth  = 1000 bits / 50 kbps = 20 ms

Time to receive the frame by the receiver  = 250 + 20 = 270 ms
 Question 20
An IP packet has arrived with the first 8 bits as 0100 0010. Which of the following is correct?
 A The number of hops this packet can travel is 2. B The total number of bytes in header is 16 bytes C The upper layer protocol is ICMP D The receiver rejects the packet
Question 20 Explanation:
IP Packet header length = 20B to 60B
It is given in the question that Arrived Packet has 8 bits = 0100 0010
First 4 Bits  0100 Represents Version  = IPV4
Next 4 Bits Represents Header Length ( /4) which is range between 20 to 60 bytes.

0010 represents header length  which  is equal to 2 * 4 = 8B
But header length between  20B to 60B

So Receiver will reject the packet because 8B < 20B it should between 20B ≤ x ≤ 60B
 Question 21
Suppose you are browsing the world wide web using a web browser and trying to access the web servers. What is the underlying protocol and port number that are being used?
 A UDP, 80 B TCP, 80 C TCP, 25 D UDP, 25
Question 21 Explanation:
• For a browser to connect and access the web servers HTTP is used and HTTP works on a well defined port number 80.
• TCP is Underlying protocol of HTTP
• TCP is protocol at Transport Layer
• Transport layer which is responsible for creating the end to end Connection between hosts for which it mainly uses TCP and UDP.
• TCP is a secure and connection orientated protocol which uses a handshake protocol to establish a robust connection between two end hosts.
• TCP ensures reliable delivery of messages and is used in various applications.
• UDP  is a stateless and unreliable protocol which ensures best-effort delivery. It is suitable for the applications which have little concern with flow or error control and requires to send the bulk of data like video conferencing.
 Question 22
A mechanism or technology used in Ethernet by which two connected devices choose common transmission parameters such as speed, duplex mode and flow control is called
 A Autosense B Synchronization C Pinging D Auto negotiation
Question 22 Explanation:
• Auto negotiation is a signaling mechanism and procedure used by Ethernet over twisted pair by which two connected devices choose common transmission parameters, such as speed, duplex mode, and flow control.
• Auto negotiation is defined in clause 28 of IEEE 802.3 and was originally a component of Fast Ethernet.
 Question 23
Which of the following is not a valid multicast MAC address?
 A 01:00:5E:00:00:00 B 01:00:5E:00:00:FF C 01:00:5E:00:FF:FF D 01:00:5E:FF:FF:FF
Question 23 Explanation:
• Unicast: from one source to one destination. It means  unicast is One-to-One
• Broadcast: from one source to all possible destinations. It means Broadcast is One-to-All
• Multicast: from one source to multiple destinations stating an interest in receiving the traffic. It means Multicast is One-to-Many. A multicast addressed frame is either flooded out all ports (if no multicast optimization is configured) or sent out only the ports interested in receiving the traffic.
Multicast MAC Address range of 01-00-5E-00-00-00 to 01-00-5E-7F-FF-FF
The broadcast address has the value of FFFF.FFFF.FFFF (all binary ones). The switch will flood broadcast frames out all ports except the port that it was received on.

Multicast frames have a value of 1 in the least-significant bit of the first octet of the destination address. This helps a network switch to distinguish between unicast and multicast addresses. One example of an Ethernet multicast address would be 01:00:0C:CC:CC:CC, which is an address used by CDP (Cisco Discovery Protocol).
 Question 24
An organization is granted the block 130.34.12.64/26. It needs to have 4 subnets. Which of the following is not an address of this organization?
 A 130.34.12.124 B 130.34.12.89 C 130.34.12.70 D 130.34.12.132
Question 24 Explanation:
130.34.12.64/26
232-26 = 26 = So 64 hosts are possible.

Dividing it into four subnets will give 16 per subnet
• 130.34.12.64 to 130.34.12.79
• 130.34.12.80 to 130.34.12.95
• 130.34.12.96 to 132.34.12.111
• 130.34.12.112 to 130.34.12.127
 Question 25
A supernet has a first address of 205.16.32.0 and a supernet mask of 255.255.248.0. A router receives 4 packets with the following destination addresses. Which packet belongs to this supernet?
 A 205.16.42.56 B 205.17.32.76 C 205.16.31.10 D 205.16.39.44
Question 25 Explanation:
• Perform AND operation between the supernet Mask and the given IP addresses
• Result of the above AND operation matches with the first address of the supernet then given IP address is in the same supernet.
So above result is matching with first address of the supernet So given IP address is in the same supernet.
 Question 26
A slotted ALOHA network transmits 200-bit frames using a shared channel with a 200 Kbps bandwidth. Find the throughput of the system, if the system (all stations put together) produces 250 frames per second :
 A 49 B 368 C 149 D 151
Question 26 Explanation:
For Pure aloha Throughput  (S) = Ge-2G.
For Slotted aloha Throughput  (S) = Ge-G.
 Question 27
The period of a signal is 100 ms. Its frequency is
 A 1003 Hertz B 10−2 KHz C 10−3 KHz D 105 Hertz
Question 27 Explanation:
Given Time Period  = 100 ms
we know that Frequency is the inverse of Time Period.
Frequency = ( 1 / Time Period )
Frequency = 1 / 100ms
Frequency = 1/10-1s   // 100*10-3s = 10-1s
Frequency = 10Hz  // 1 Hz = 10-3 kHz
Frequency =10-2KHz  // 10 * 10-3 kHz  = 10-2KHz
 Question 28
The dotted-decimal notation of the following IPV4 address in binary notation is
10000001 00001011 00001011 11101111
 A 111.56.45.239 B 129.11.10.238 C 129.11.11.239 D 111.56.11.239
Question 28 Explanation:
10000001 00001011 00001011 11101111
(10000001)2 = 27+20=(129)10
(00001011)2  = 20+23+21=(11)10
(00001011)2  = 20+23+21=(11)10
(11101111)2   = 20+21+22+23+25+26+27=(239)10
Given IPV4 address which  is equivalent to 129.11.11.239
 Question 29
Which of the following statements are true ?
(a) Advanced Mobile Phone System (AMPS) is a second generation cellular phone system.
(b) IS - 95 is a second generation cellular phone system based on CDMA and DSSS.
(c) The Third generation cellular phone system will provide universal personnel communication.
 A (a) and (b) only B (b) and (c) only C (a), (b) and (c) D (a) and (c) only
Question 29 Explanation:
• option(A) : False
The first generation (1G) of cellular mobile communication systems, called Advanced Mobile Phone System (AMPS) in North America, used analog FM in 30 kHz channels and accommodated multiple users through frequency division multiple access in the 800–900 MHz band allocated by the 1976 World Allocation Radio Conference
• option(B) : True IS – 95 is a second generation cellular phone system based on CDMA and DSSS.
• option(C) : True The Third generation cellular phone system will provide universal personnel communication.
 Question 30
Match the following symmetric block ciphers with corresponding block and key sizes :
```  List-I                             List-II
(a) DES          (i) block size 64 and key size ranges between 32 and 448
(b) IDEA        (ii) block size 64 and key size 64
(c) BLOWFISH   (iii) block size 128 and key sizes 128, 192, 256
(d) AES         (iv) block size 64 and key size 128```
 A (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii) B (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii) C (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i) D (a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)
Question 30 Explanation:
DES: Data Encryption Standard)
DES(Data Encryption Standard) is an implementation of a Feistel Cipher. It uses 16 round Feistel structure. The block size is 64-bit. Though, key length is 64-bit, DES has an effective key length of 56 bits, since 8 of the 64 bits of the key are not used by the encryption algorithm (function as check bits only).

IDEA: International Data Encryption Algorithm (IDEA)
DEA operates on 64-bit blocks using a 128-bit key

BLOW FISH:  Blowfish Cryptosystem
BLOW FISH is the replacement for DES or IDEA,
64-bit block cipher with variable length key. The key length is variable ,
it can be in the range of 32~448 bits
Default 128 bits key length

AES is based on is a symmetric key algorithm it means the same key is used for both encrypting and decrypting the data.
Advanced Encryption Standard (AES) uses block size 128 and key sizes 128, 192, 256
• Blowfish, AES, RC4, DES, RC5, and RC6 are examples of symmetric encryption. The most widely used symmetric algorithm is AES-128, AES-192, and AES-256.
• Asymmetric encryption Example are EIGamal , RSA, DSA, Elliptic curve techniques, PKCS, Diffie-Hellman, ECC,
 Question 31
Which of the following statements are true ?

(a) Three broad categories of Networks are
(i) Circuit Switched Networks
(ii) Packet Switched Networks
(iii) Message Switched Networks
(b) Circuit Switched Network resources need not be reserved during the set up phase.
(c) In packet switching there is no resource allocation for packets.

Code :
 A (a) and (b) only B (b) and (c) only C (a) and (c) only D (a), (b) and (c)
Question 31 Explanation:
option(A) : True
Three broad categories of Networks are
(i) Circuit Switched Networks
(ii) Packet Switched Networks
(iii) Message Switched Networks

option(B) : False
In Circuit Switched the Network resources need to be reserved during the setup phase; the resources remain dedicated for the entire duration of data transfer phase until the teardown phase.

option(C) : True
In Packet switching, there is no resource allocation for a packet. This means that there is no reserved bandwidth on the links, and there is no scheduled processing time for each packet. Resources are allocated on demand.
 Question 32
Decrypt the message “WTAAD” using the Caesar Cipher with key = 15.
 A LIPPS B HELLO C OLLEH D DAATW
Question 32 Explanation:
The shift cipher is sometimes referred to as the Caesar cipher.

In Caeser cipher algorithm encryption and decryption takes place. During encryption right shift takes place depending upon the key  and in  Decryption left shift takes place depending on key.

The message "WTAAD" is already encrypted. we need to decrypt it by simply shifting the keys by 15 places in reverse order.

Decryption
Use the shift cipher with key = 15 to decrypt the message “WTAAD.”
• We decrypt one character at a time.
• Each character is shifted 15 characters  up.
• Letter W is decrypted to H.
• Letter T is decrypted to E.
• The first A is decrypted to L.
• The second A is  decrypted to L.
• Finally D is decrypted to O.
• The plaintext is HELLO
Encryption
Use the shift cipher with key = 15 to encrypt the message “HELLO.”
• We encrypt one character at a time.
• Each character is shifted 15 characters characters down.
• Letter H is encrypted encrypted to W.
• Letter E is encrypted to T.
• The first L is encrypted to A.
• The second L is also encrypted encrypted to A.
• O is encrypted encrypted to D.
• The cipher text is WTAAD.
 Question 33
Encrypt the Message “HELLO MY DEARZ” using Transposition Cipher with
 A HLLEO YM AEDRZ B EHOLL ZYM RAED C ELHL MDOY AZER D ELHL DOMY ZAER
Question 33 Explanation:
In question encryption of given message using Transposition Cipher with Given message is “HELLO MY DEARZ”
Key - Plain Text -> 2 4 1 3
Key - Cipher Text -> 1 2 3 4

According to key size divide number of character into blocks. Here, key size is 4. So, character block size is 4.

Step 1 : Given message is “HELLO MY DEARZ”  Now arrange them in group of 4
I.e. HELL OMYD EARZ
Step-2: Sort column as per given Plain key as shown in below matrix.
i.e. replace second character in above format to the first place and
fourth character to the first place ,
first character to the third place
third character to the fourth place.
Encrypted message will be :  ELHL MDOY AZER.
Read the matrix horizontally one by one row
 Question 34
Time To Live(TTL) field in the IP datagram is used___
 A To optimize throughput B To prevent packet looping C To reduce delays D To prioritize packets
Question 34 Explanation:
• TTL field is used to limit the lifetime of a IP datagram and to prevent indefinite looping of IP datagrams.
• TTL may be implemented as a counter or timestamp attached to or embedded in the data.
• Once the prescribed event count or timespan has elapsed, data is discarded.
 Question 35
 A 255.255.192.192 B 255.255.255.198 C 255.255.255.240 D 255.255.257.240 E None
Question 35 Explanation:
Direct Broadcast Address is represented as valid Network ID + all hosts bits will be 1's ( series of 1's )

201.15.16.31 = 201.15.00010000.00011111

last 5 bits are the host bits

Last octet of given DBA is 0001 1111.

So, in Subnet mask address all should be 1’s except last 5 digits, i.e., 27 bit for NID which is 255.255.255.224

So none of the options is correct
 Question 36
If the period of a signal is 100 ms. Then its frequency in Hertz is___
 A 10 B 100 C 1000 D 10000
Question 36 Explanation:
Given Time Period  = 100 ms
we know that Frequency is the inverse of Time Period.
Frequency = ( 1 / Time Period )
Frequency = 1 / 100ms
Frequency = 1/10-1s   // 100*10-3s = 10-1s
Frequency = 10Hz  // 1 Hz = 10-3 kHz
Frequency =10-2KHz  // 10 * 10-3 kHz  = 10-2KHz
 Question 37
Identify the true statements from the given statements.
(1) HTTP may use different TCP connection for different objects of a webpage if non persistent connections are used
(2) FTP uses two TCP connections, one for data and another control
(3) TELNET and FTP can only use TWO connection at a time.
 A (1) B (1) and (2) C (2) and (3) D (1),(2) and (3)
Question 37 Explanation:
• HTTP may use different TCP connection for different objects of a webpage if non-persistent connections are used.
• FTP uses data and control connections used with two separate TCP connections
• TELNET and FTP can only use ONE connection at a time
 Question 38
_____ transport layer protocol is used to support electronic mail
 A SNMP B IP C SMTP D TCP
Question 38 Explanation:
Email uses various protocols like SMTP, IMAP and POP. The most prominent one used in application layer is SMTP
• TCP and UDP are two transport layer protocols. SMTP uses TCP as transport layer protocol as TCP is reliable.
• Since TCP is a reliable protocol, it’s more efficient to use TCP protocol for e-mail transfer. TCP also provides more security than other transport layer protocols.

Why TCP not UDP?
SMTP makes more sense to use TCP over UDP. SMTP is a mail transport protocol, and in mail every single packet is important. If you lose several packets in the middle of the message the recipient might not even receive the message and if they do they might be missing key information. This makes TCP more appropriate because it ensures that every packet is delivered.
 Question 39
_____ IP address can be used in WAN
 A 256.0.0.1 B 172.16.0.10 C 15.1.5.6 D 127.256.0.1
Question 39 Explanation:
The Internet Assigned Numbers Authority (IANA) reserves the following IP address blocks for use as private IP addresses:
• 10.0.0.0/8          IP addresses:  10.0.0.0 – 10.255.255.255
• 172.16.0.0/12     IP addresses:  172.16.0.0 – 172.31.255.255
• 192.168.0.0/16  IP addresses: 192.168.0.0 – 192.168.255.255

option(A) and option(D) are not valid IP's  because it is beyond 255. IP address ranges from [0 - 255]

only option(C) is correct
 Question 40
Given message M=1010001101. The CRC for this given message using the divisor polynomial x5 + x4 + x2 +1 is _____
 A 01011 B 10101 C 01110 D 10110
Question 40 Explanation:
Message M = 1010001101
Divisor polynomial = x5 + x4 + x2 +1
Divisor polynomial = 1.x5 +1.x4+0.x3+1.x2+0.x1+1.x0
Divisor polynomial bit= 110101
Bits to be appended to message M= (divisor polynomial bits – 1) = 5
So, Append 5 zeros to message bits M
Modified Message M = 101000110100000

Now perform EX-OR between message M and  divisor polynomial bits.

So now add remainder to the original Message M
∴ M'= 101000110101110 and CRC = 01110.
 Question 41
Three or more devices share a link in ____ connection
 A Unipoint B Polarpoint C Point to point D Multipoint
Question 41 Explanation:
A multipoint communication is established when three or many network nodes are connected to each other. Frame relay, Ethernet, token ring and ATM are some examples of multipoint connections.
 Question 42
Baud rate measures the number ____ transmitted per second.
 A Symbols B Bits C Byte D None of these
Question 42 Explanation:
• Bits per second is straightforward. It is exactly what it sounds like. If I have 1000 bits and am sending them at 1000 bps, it will take exactly one second to transmit them.
• Baud is symbols per second. If these symbols — the indivisible elements of your data encoding — are not bits, the baud rate will be lower than the bit rate by the factor of bits per symbol.
Refer : https://stackoverflow.com/questions/20534417/what-is-the-difference-between-baud-rate-and-bit-rate
 Question 43
 A SMTP B ICMP C ARP D RARP
Question 43 Explanation:
In Network layer the TCP/IP model supports internetworking protocol in short known as IP. The IP uses four protocols internally: ARP, RARP, ICMP & IGMP.

ARP
• ARP stands for Address Resolution Protocol
• ARP protocol is used to find the physical address of a device whose internet address (IP address) is known.
RARP
• RARP stands for Reverse Address Resolution Protocol
• RARP protocol helps to find the internet address of a device whose physical address is known.
ICMP
• ICMP stands for Internet Control Message Protocol
• IP in network layer sends data in form of small packets known as datagrams.
• ICMP protocol sends the datagrams problems back to sender.
• It is used for query and error reporting messages.
IGMP
• IGMP stands for Internet Group Message Protocol
• IGMP protocol is used for simultaneous transmission of a message to a group of recipients.
 Question 44
Bluetooth is an example of
 A Local area network B Virtual private network C Personal area network D None of the mentioned above
Question 44 Explanation:
• PAN or  Personal Area Network  is a computer network that enables communication between computer devices near a person.
• PANs can be wired, such as USB or FireWire, or they can be wireless, such as infrared, ZigBee, Bluetooth and ultrawideband, or UWB. The range of a PAN typically is a few meters
 Question 45
A set of rules that governs data communication:
 A Rule B Medium C Link D Protocol
Question 45 Explanation:
Protocol – A protocol is a set of rules that govern data communication following are the key elements of protocol
1. Timing refers to two characteristics.
2. When data should be sent and
3. How fast they can be sent.
Medium
Transmission medium is the physical path by which a message travels from sender to receiver

A link (or edge) of a network (or graph) is one of the connections between the nodes (or vertices) of the network.
 Question 46
The ___ is the physical path over which a message traversals
 A Path B Protocol C Route D Medium
Question 46 Explanation:
Protocol – A protocol is a set of rules that govern data communication following are the key elements of protocol
1. Timing refers to two characteristics.
2. When data should be sent and
3. How fast they can be sent.
Medium
Transmission medium is the physical path by which a message travels from sender to receiver

Routing
Routing is a process which is performed by network layer devices in order to deliver the packet by choosing an optimal path from one network to another.

Path
Connection between  nodes
 Question 47
Application layer protocol defines:
 A Types of messages exchanged B Rules for when and how processes send and respond to messages C Message format, syntax and semantics D All of above
Question 47 Explanation:
The Application layer is present at the top of the OSI model. It is the layer through which users interact. It provides services to the users. What message is to be sent or the message format, syntax and semantics. A user has access to application layer for sending and receiving messages.

Point to Note
• Application Layer protocol
• TELNET
• FTP
• TFTP
• NFS
• SMTP
• LPD(Line Printer Daemon)
• X window
• SNMP
• DNS
• DHCP
 Question 48
The third generation mobile phone are digital and based on
 A AMPS B Broadband CDMA C CDMA D D-AMPS
Question 48 Explanation:
• AMPS
The first generation (1G) of cellular mobile communication systems, called Advanced Mobile Phone System (AMPS) in North America, used analog FM in 30 kHz channels and accommodated multiple users through frequency division multiple access in the 800–900 MHz band allocated by the 1976 World Allocation Radio Conference
• CDMA
IS – 95 is a second generation cellular phone system based on CDMA and DSSS. IS-54 and IS-136 are second-generation (2G) mobile phone systems, known as Digital AMPS (D-AMPS)
• B-CDMA
The Third generation cellular phone system will provide universal personnel communication.
Broadband code division multiple access (B-CDMA) is a CDMA-based cellphone technology that uses broadband transmission. It is an improvement over the first-generation CDMA, which uses narrowband transmission. This kind of technology allows for better deployment of transmitter signals.
 Question 49
Consider ISO-OSI network architecture reference model. Session layer of this model offer Dialog control, token management and ____________ as services.
 A Synchronization B Asynchronization C Errors D Flow control
Question 49 Explanation:
Session layer provides dialog control, token management and Synchronization, as services.
1. Dialog Control : This layer allows two systems to start communication with each other in half-duplex or full-duplex.
2. Token Management: This layer prevents two parties from attempting the same critical operation at the same time.
3. Synchronization : This layer allows a process to add checkpoints which are considered as synchronization points into stream of data. Example: If a system is sending a file of 800 pages, adding checkpoints after every 50 pages is recommended. This ensures that 50 page unit is successfully received and acknowledged. This is beneficial at the time of crash as if a crash happens at page number 110; there is no need to retransmit 1 to100 page
Refer : https://www.studytonight.com/computer-networks/osi-model-session-layer
 Question 50

An internet service provider (ISP) has following chunk of CIDR-based IP addresses available with it: 245.248.128.0/20 . The ISP want to give half of this chunk of addresses to organization A and a quarter to Organization B while retaining the remaining with itself. Which of the following is a valid allocation of addresses to A and B?

 A 245.248.132.0/22 and 245.248.132.0/21 B 245.248.136.0/21 and 245.248.128.0/22 C 245.248.136.0/24 and 245.248.132.0/21 D 245.248.128.0/21 and 245.248.128.0/22
Question 50 Explanation:
 Question 51
Repeaters function in
 A Physical layer B Data link layer C Network layer D Both (A) and (B)
Question 51 Explanation:
• Repeaters are network devices operating at physical layer of the OSI model that amplify or regenerate an incoming signal before retransmitting it.
• They are incorporated in networks to expand its coverage area. They are also known as signal boosters.
 Question 52
If the frame buffer has 10-bits per pixel  and 8-bits are allocated for each of the R, G and B components then what would be the size of the color lookup table(LUT)
 A (210 + 211) bytes B (210 + 28) bytes C (210 + 224) bytes D (28 + 29) bytes
Question 52 Explanation:
• Number of entries in lookup table = 210 = 1024
• Size of each entry  in lookup table = 8 bits for R component + 8 bits for G component + 8 bits for B component = 24 bits = 3 Bytes
• Size of lookup table = Number of entries in  * size of each entry
• Size of lookup table =1024 * 3 = 3072 Bytes
• 3072 Bytes = 1024 +2048 (210 + 211) bytes
 Question 53
In networking terminology UTP means
 A Unshielded twisted pair B Ubiquitous Teflon port C Uniformly Terminating port D Unshielded T-connector port
Question 53 Explanation:
UTP is the Unshielded Twisted Pair cable. It consists of two unshielded wires twisted together. These are used extensively in Local-Area-Networks LANs and telephone cables
 Question 54
How many bits are required to encode all twenty six letters, ten symbols, and ten numerals?
 A 5 B 6 C 7 D 46
Question 54 Explanation:
• 26 letters ×2 (Capital letters and small letters) + 20 characters(10 symbols +10 numerals) = 52+20=72 .
• log272 = flor(6.something)= 7
• So minimum we require 7 bits to represent 72 symbols
• If you don’t want to codify both capital letters and small letters but only one family of 26 letters and be done with it, then you have 26+20=46 symbols = log246 =6 bits
• So minimum we require 6 bits to represent 46 symbols
 Question 55
Which of the following statement/s is/are true?
(i) Firewall can screen traffic going into or out of an organization.
(ii) Virtual private networks cam simulate an old leased network to provide certain desirable properties.
Choose the correct answer from the code given below:
Code:
 A (i) only B Neither (i) nor (ii) C Both (i) and (ii) D (ii) only
Question 55 Explanation:
option(A) : True

option(B) : True
A virtual private network (VPN) extends a private network across a public network and enables users to send and receive data across shared or public networks as if their computing devices were directly connected to the private network. Applications running across a VPN may therefore benefit from the functionality, security, and management of the private network. Encryption is a common, although not an inherent, part of a VPN connection.
 Question 56
Which register that shift complete binary number in one bit at a time and shift all the stored bits out at a time?
 A Parallel-in parallel-out B Parallel-in Serial-out C Serial-in parallel-out D Serial-in Serial-out
Question 56 Explanation:
Serial-In Serial-Out Shift Register (SISO)
The shift register, which allows serial input (one bit after the other through a single data line) and produces a serial output is known as Serial-In Serial-Out shift register.

Serial-In Parallel-Out shift Register (SIPO)
The shift register, which allows serial input (one bit after the other through a single data line) and produces a parallel output is known as Serial-In Parallel-Out shift register

Parallel-In Serial-Out Shift Register (PISO)
The shift register, which allows parallel input (data is given separately to each flip flop and in a simultaneous manner) and produces a serial output is known as Parallel-In Serial-Out shift register.

Parallel-In Parallel-Out Shift Register (PIPO)
The shift register, which allows parallel input (data is given separately to each flip flop and in a simultaneous manner) and also produces a parallel output is known as Parallel-In parallel-Out shift register.
 Question 57
Simple network management protocol(SNMP) is a protocol that runs on:
 A Application layer B Transport layer C Network layer D Data link layer
Question 57 Explanation:
SNMP
• SNMP stands for Simple Network Management Protocol
• it is an application layer protocol which uses UDP port number 161/162.
• SNMP is used to monitor the network, detect network faults and
• SNMP is used to  to configure remote devices.
 Question 58
Consider the following two statements:

S1: TCP handles both congestion and flow control
S2: UDP handles congestion but not flow control

Which of the following option is correct with respect to the above statements (S1) and (S2)?
Choose the correct answer from the code given below:
Code:
 A Both S1 and S2 are correct B Neither S1 nor S2 is correct C S1 is not correct but S2 is correct D S1 is correct but S2 is not correct
Question 58 Explanation:
• TCP uses congestion window for congestion control and Advertisement window for flow control
• UDP does not handle congestion or flow control because no field are there in UDP header to control flow or congestion
 Question 59
Given a mask, M=255.255.255.248. How many subnet bits are required for given mask M?
 A 2 B 3 C 4 D 5
Question 59 Explanation:
255.255.255.248
11111111.11111111.11111111.11111000
NID = 29 bits
Host ID =32-29=3 bits
 Question 60
Transparent mode firewall is a:
 A Layer 2 firewall B Layer 3 firewall C Layer 4 firewall D Layer 7 firewall
Question 60 Explanation:
• In transparent mode, the router is, essentially, invisible to the traffic. Traffic goes in one port and out the other "untouched" in the same way that it passes through a switch. It's more accurate to say that a transparent firewall is like a switch with an ability to inspect frames.
• In transparent mode, the firewall is an L2 device and not an L3 or routed hop. Since the transparent mode firewall is not a routed hop, it can be easily introduced into an existing network without IP readdressing
 Question 61
A routing protocol that was used for trail information that gets updated dynamically is:
 A Distance Vector B Path Vector C Link Vector D Multipoint
Question 61 Explanation:
• Distance vector  is a  dynamic routing algorithm in which each router computes distance between itself and each possible destination i.e. its immediate neighbors.
• The router share its knowledge about the whole network to its neighbors and accordingly updates table based on its neighbors.
• The sharing of information with the neighbors takes place at regular intervals.
• It makes use of Bellman Ford Algorithm for making routing tables.
• Problems – Count to infinity problem which can be solved by splitting horizon.
– Persistent looping problem i.e. loop will be there forever.
 Question 62
A packet switching network
 A Is free B Can reduce the cost of using an information utility C Allows communications channel to be shared among more than one user D Boh (B) and (C)
Question 62 Explanation:
• Packet switching is a connectionless network switching technique.
• Message is divided and grouped into a number of units called packets that are individually routed from the source to the destination. There is no need to establish a dedicated circuit for communication.
• Each packet in a packet switching technique has two parts: a header and a payload.
• The header contains the addressing information of the packet and is used by the intermediate routers to direct it towards its destination. The payload carries the actual data.
 Question 63
The four byte IP Address consists of
Question 63 Explanation:
• An IP address is a 32 bit number that uniquely identifies a network interface on a machine. An IP address is typically written in decimal digits, formatted as 4 (8 bit fields that are separated by periods)
• Each 8-bit field represents a byte of the IPv4 address.
• The bytes of the IPv4 address are further classified into two parts: the network address and the host address
• Class A : 1       -  126 (N.H.H.H)
• Class B : 128  -  191 (N.N.H.H)
• Class C : 192  -  223 (N.N.N.H)
• Class D : 224  - 239
• Class E : 240   - 255
 Question 64
In context of TCP/IP computer network models, which of the following is FALSE?
 A Besides span of geographical area, the other major difference between LAN and WAN is that the later uses switching element B A repeater is used just to forward bits from one network to another one C IP layer is connected oriented layer in TCP/IP D A gateway is used to connect incompatible networks
Question 64 Explanation:
• Internet layer is a packet switching network based on connectionless communication
• Host send packets into the network and then the packets travel independently to their destination since the sender is not directly connected to the receiver
• it means IP is connectionless. Data packet can travel from sender to receiver without the recipient having to send an acknowledgement to the sender
 Question 65
Which of the following OS possible in a token passing bus network?
 A In-service expansion B Unlimited number of stations C Both (A) and (B) D Unlimited distance
Question 65 Explanation:
Token bus is a network implementing the token ring protocol over a virtual ring on a coaxial cable. A token is passed around the network nodes and only the node possessing the token may transmit. If a node doesn't have anything to send, the token is passed on to the next node on the virtual ring. Each node must know the address of its neighbour in the ring, so a special protocol is needed to notify the other nodes of connections to, and disconnections from, the ring In a token-ring network, the token and data is passed to the next physical node along the line, but in a token bus network, it does not matter where the nodes are physically located since token-passing is done via a numeric sequence of node addresses.

Source : wiki
 Question 66
Which of the following statement/s is/are true ?

(i) Windows XP supports both peer-peer and client-server networks.
(ii) Windows XP implements Transport protocols as drivers that can be loaded and uploaded from the system dynamically.

Choose the correct answer from the code given below :

Code :
 A Both (i) and (ii) B Neither (i) nor (ii) C (ii) only D (i) only
Question 66 Explanation:
Following screenshot taken from : Silberschatz Galvin and Gagne
 Question 67
The host is connected to a department network which is a part of a university network. The university network, in turn, is part of the internet. The largest network, in which the Ethernet address of the host is unique, is
 A The university network B The internet C The subnet to which the host belongs D The department network
Question 67 Explanation:
• Ethernet address is nothing but MAC Address which is present on NIC and is supposed to be unique. Thus it is unique over the Internet.
• The Main difference is IP address can change with the change in network but your device MAC address is unchanged in different networks.
 Question 68
The decimal floating point number -40.1 represented using IEEE-754 32-bit representation and written in hexadecimal form is
 A 0xC2206000 B 0xC2006666 C 0xC2006000 D 0xC2206666
Question 68 Explanation:

Bias = 2N−1 −1
N−Number of bits to represent exponent in binary

Given -40.1
(40)10 = (101000)2

For converting Fractional part into binary form just multiply with 2 until it get the same number
• 0.1 * 2 =0.2  - 0
• 0.2 * 2 =0.4  - 0
• 0.4 * 2 =0.8  - 0
• 0.8 * 2 =1.6   - 1
• 0.6 * 2 =1.2   - 1
• 0.2 * 2 =0.4  - 1 // stop coz 0.2*2 we already did it is repeating
(0.1)10 =0.00011

(40.1)10 = 101000.00011 =  1.0100000011 * 25
• Biased exponent = actual + bias = 5+bias
• where bias=28−1−1=127
• Biased exponent = 5+127=132 = (1000 0100)2
• sign = 1
• Mantisa = 010100000011 ....00
Therefore, number represented as 1  1000 0100  010100000011 ....00

on converting to hexadecimal we get (0xC2206666)16
 Question 69
In PERT/CPM, the merge event represents___________ of two or more events.
 A Splitting B Completion C Beginning D Joining
Question 69 Explanation:
 Question 70
The first network:
 A ARPANET B NFSNET C CNNET D ASAPNET
Question 70 Explanation:
• The Advanced Research Projects Agency Network (ARPANET) was the first wide-area packet-switching network with distributed control and one of the first networks to implement the TCP/IP protocol suite.
• Both technologies became the technical foundation of the Internet.
• The ARPANET was established by the Advanced Research Projects Agency (ARPA) of the United States Department of Defense.
 Question 71
Communication between a computer and a keyboard involves __ transmission
 A Simplex B Half Duplex C Automatic D Full Duplex
Question 71 Explanation:
• Communication between a computer and a keyboard involves simplex transmission.
• In simplex transmission, data flows in single direction which in this case refers to the data flowing from the keyboard to the computer.
• Another Example would be of the mouse where the data flows from the mouse to the computer only.
•  No data flows from computer to keyboard.
• Simplex Transmission : Data can be transmitted in only one directions
• HALF DUPLEX : Data can be transmitted in both directions on a single carrier but not at the same time
• FULL DUPLEX : Data can be transmitted in both directions on a single carrier  at the same time
 Question 72
Bluetooth is an example of:
 A Personal area network B Virtual private network C Local area network D None of the above
Question 72 Explanation:
• PAN or  Personal Area Network  is a computer network that enables communication between computer devices near a person.
• PANs can be wired, such as USB or FireWire, or they can be wireless, such as infrared, ZigBee, Bluetooth and ultrawideband, or UWB. The range of a PAN typically is a few meters
 Question 73
IPv6 is developed by
 A IETF B ANSI C ISO D IEEE
Question 73 Explanation:
IPv6 was developed by the Internet Engineering Task Force (IETF) to deal with the long-anticipated problem of IPv4 address exhaustion. IPv6 is intended to replace IPv4. In December 1998, IPv6 became a Draft Standard for the IETF, who subsequently ratified it as an Internet Standard on 14 July 2017
 Question 74
Which of the following is NOT a symmetric key algorithm?
 A Ellipse Curve Cryptography B Advanced Encryption standard C Data Encryption Standard D Blowfish
Question 74 Explanation:
Symmetric key Encryption
Blowfish, AES, RC4, DES, RC5, and RC6
The most widely used symmetric algorithm is AES-128, AES-192, and AES-256.

Asymmetric Key Encryption
EIGamal, RSA, DSA, Elliptic curve techniques, PKCS, Diffie-Hellman, ECC(Ellipse Curve Cryptography)

DES: Data Encryption Standard)
DES(Data Encryption Standard) is an implementation of a Feistel Cipher. It uses 16 round Feistel structure. The block size is 64-bit. Though, key length is 64-bit, DES has an effective key length of 56 bits, since 8 of the 64 bits of the key are not used by the encryption algorithm (function as check bits only).

IDEA: International Data Encryption Algorithm (IDEA)
DEA operates on 64-bit blocks using a 128-bit key

BLOW FISH:  Blowfish Cryptosystem
BLOW FISH is the replacement for DES or IDEA,
64-bit block cipher with variable length key. The key length is variable ,
it can be in the range of 32~448 bits
Default 128 bits key length

Advanced Encryption Standard (AES) uses block size 128 and key sizes 128, 192, 256
 Question 75
The DNS maps the IP address to
Question 75 Explanation:
Domain Name System (DNS) maps human readable domain names (in URLs or in email address) to IP addresses and vice versa
For example, DNS translates and maps the domain academyera.com to the IP address 151.162.110.21
 Question 76
How many bits internet address is assigned to each host on a TCP/IP internet which is used in all communication with the host?
 A 16 bits B 32 bits C 48 bits D 64 bits
Question 76 Explanation:
• Internet Protocol version 4 (IPv4) defines an IP address as a 32 bit number
• Internet Protocol version 6 (IPv6) using 128-bits for the IP address, was standardized in 1998
 Question 77
An ACK number of 1000 in TCP always means that
 A 999 bytes have been successfully received B 1000 bytes have been successfully received C 1001 bytes have been successfully received D None of the above
Question 77 Explanation:
ACK number in TCP tells which byte it is expecting to receive next.
• If Initial sequence number was from 0, then ACK number is 1000 means 1000 bytes(0-999) are successfully delivered.
• If Initial sequence number was from 1, then ACK number is 1000 means 999 bytes are successfully delivered.
Initial sequence number is not given in the question so option(d) is correct
Option (C) is not possible because Initial sequence number cannot be negative(-1)
 Question 78
In a class B subnet, we know the IP address of one host and the mask as given below:
 A 125.134.96.0 B 125.134.112.0 C 125.134.112.66 D 125.134.0.0
Question 78 Explanation:
Convert the given decimal number into binary equivalent number and perform AND operation

IP address: 01111101 10000110 01110000 01000010

Subnet Mask: 11111111 11111111 11100000 00000000

Now Perform AND operations between IP and Mask we get

___________________________________________________
NID :                    01111101.10000110.01100000.00000000
___________________________________________________

 Question 79
In a class B subnet, we know the IP address of one host and the mask as given below:
 A 125.134.96.0 B 125.134.112.0 C 125.134.112.66 D 125.134.0.0
Question 79 Explanation:
Convert the given decimal number into binary equivalent number and perform AND operation

IP address: 01111101 10000110 01110000 01000010

Subnet Mask: 11111111 11111111 11100000 00000000

Now Perform AND operations between IP and Mask we get

___________________________________________________
NID :                    01111101.10000110.01100000.00000000
___________________________________________________

 Question 80
In a class B subnet, we know the IP address of one host and the mask as given below:
 A 125.134.96.0 B 125.134.112.0 C 125.134.112.66 D 125.134.0.0
Question 80 Explanation:
Convert the given decimal number into binary equivalent number and perform AND operation

IP address: 01111101 10000110 01110000 01000010

Subnet Mask: 11111111 11111111 11100000 00000000

Now Perform AND operations between IP and Mask we get

___________________________________________________
NID :                    01111101.10000110.01100000.00000000
___________________________________________________

 Question 81
A certain population of ALOHA users manages to generate 70 request/sec. If the time is slotted in units of 50 msec, then channel load would be
 A 4.25 B 3.5 C 350 D 450
Question 81 Explanation:
Given data
No. of requests generated = 70 request/sec
Time slot unit = 50 msec
Number of time slots in 1 second = 1000/50 = 20 seconds
Channel Load = No. of requests in 1 sec / No. of slots in 1 sec
Channel Load = 70/20 = 3.5
 Question 82
Which statement is false?
 A PING is a TCP/IP application that sends datagrams once every second in the hope of an echo response from the machine being PINGED B If the machine is connected and running a TCP/IP protocol stack, it should respond to the PING datagram with a datagram of its own C If PING encounters an error condition, an ICMP message is not returned D PING display the time of the return response in milliseconds or one of several error message
Question 82 Explanation:
 Question 83
A router uses the following routing table:   A packet bearing a destination address 144.16.68.117 arrives at the router. On which interface will it be forwarded?
 A Eth0 B Eth1 C Eth2 D Eth3
Question 83 Explanation:
Given decimal value convert into binary
144. 16. 68. 117
144. 16. 68. 01110101

255.255.255.224
255.255.255. 11100000

Perform AND operation between 144. 16. 68. 117 and 255.255.255.224
144. 16. 68.      01110101
255. 255. 255.  11100000
------------------------------
144. 16. 68.      01100000(96)
-----------------------------
144.16.68.96 is not matching with Destination

Now Checking with 255. 255. 255. 0
Perform AND operation between 144. 16. 68. 117 and 255.255.255.0
144. 16. 68.      01110101
255. 255. 255. 00000000
------------------------------
144. 16. 68.     00000000
-----------------------------
144.16.68.0 is  matching with Destination
 Question 84
Which layers of the OSI reference model are host-to-host layers?
 A Transport, session, presentation, application B Session, presentation, application C Datalink, transport, presentation, application D Physical, datalink, network, transport
Question 84 Explanation:
 Question 85
In time division switches if each memory access takes 100 ns and one frame period is 125 us, then the maximum number of lines that can be supported is
 A 625 lines B 1250 lines C 2300 lines D 318 lines
Question 85 Explanation:
Given
Memory access time =100ns
One Frame Period = 125 us(1 us =1000ns)
In time division switches 2nT = 1 frame period,
where T = memory access time.
Max number of Lines = 125*1000/100*2=625
 Question 86
Suppose a network using CSMA/CD has a bandwidth of 10 Mbps. If the maximum propagation time is 25 microsec, that what will be the minimum frame size?
 A 500 bits B 50 bits C 500 bytes D 4*1011 bits
Question 86 Explanation:
Given
Bandwidth B = 10 Mbps
Propagation time Tp= 25 microsec

For frame size to be minimum, its transmission time should be equal to twice of one way propagation delay.
TFR = 2 x Tp
Let Minimum frame size is L bits.
TFR = L / B
L=TFR * B
where B is the bandwidth

Coming  into the problem

TFR = 2 x 25 microsec
TFR =  50 microsec
So In worst case a station needs to transmit for a period of 50 microseconds to detect the collision
Minimum size of the frame is 10 Mbps * 50 micro seconds = 500 bits or 63 Bytes
So this is actually the minimum size of the frame for standard Ethernet
 Question 87
If 100 users are making 10 request/sec to a slotted ALOHA channel and each slot is of 50 m sec, then what will be the channel load?
 A 10 B 20 C 2 D 50
Question 87 Explanation:
IF 100 user are making 10 request per sec then total number of requests are 1000
Total Number of Requests per second = 1000

Each SLOT is of 50 m sec then the number of slots in 1 second = 1/(50 * 10-3) = 20 slots/sec
Total Number of slots in 1 second = 20
Channel Load = Number of request in 1 sec / Number of slots in 1 sec
 Question 88
If the number of networks and number of hosts in class B are 2m, (2n -2) respectively. Then the relation between m,n is:
 A 3m=2n B 7m=8n C 8m=7n D 2m=3n
Question 88 Explanation:
Class B has 16384 (214) Network addresses

∴2m = 2n-2 = 214 = 216-2
m=14, n=16
Now we have to check all the options
Substitute  8m and 7n
8*14 = 16 = 7*16
112=112
 Question 89
The address of a class B host o be split into subnets with a 3 bit subnet number. What is the maximum number of subnets and maximum number of hosts in each subnet?
 A 8 subnets and 262141 hosts B 6 subnets and 262141 hosts C 6 subnets and 1022 hosts D 8 subnets and 1024 hosts E None
Question 89 Explanation:
In class B (N.N.H.H) NID part is First 2 octet and remaining 2 octets are host id part
So, Here First 3 bits of the 3rd octet are used for subnet and remaining 10 bits for hosts
∴ Maximum number of subnets = 23 = 8
∴ Total Number of host in each subnet = 16-3=13 bits
∴ Total Number of host in each subnet =2(13)-2 =8190
2 is subtracted because, 1st IP of this subnet will be subnet ID and last IP of this subnet will be the Direct Broadcast Address
 Question 90
Chose the option which matches each element of LIST-1 with exactly one element of LIST-2:
 A (i)-(d), (ii)-(c), (iii)-(b), (iv)-(a) B (i)-(b), (ii)-(a), (iii)-(d), (iv)-(c) C (i)-(a), (ii)-(d), (iii)-(c), (iv)-(b) D (i)-(d), (ii)-(b), (iii)-(c), (iv)-(a)
Question 90 Explanation:
Repeater
A repeater operates at the physical layer. Its job is to regenerate the signal over the same network before the signal becomes too weak or corrupted so as to extend the length to which the signal can be transmitted over the same network.

Hub
A hub is basically a multiport repeater. A hub connects multiple wires coming from different branches
Types of Hub : Active Hub , Passive Hub

Gateway
A gateway, as the name suggests, is a passage to connect two networks together that may work upon different networking models. They basically work as the messenger agents that take data from one system, interpret it, and transfer it to another system. Gateways are also called protocol converters and can operate at any network layer.

Routers
A router is a device like a switch that routes data packets based on their IP addresses. Router is mainly a Network Layer device. Routers normally connect LANs and WANs together and have a dynamically updating routing table based on which they make decisions on routing the data packets. Router divide broadcast domains of hosts connected through it.

Bridge
A bridge operates at data link layer. A bridge is a repeater, with add on the functionality of filtering content by reading the MAC addresses of source and destination. It is also used for interconnecting two LANs working on the same protocol. It has a single input and single output port, thus making it a 2 port device.
Types of Bridges : Transparent Bridges, Source Routing Bridges
 Question 91
The maximum data rate of a channel of 3000 Hz bandwidth and SNR of 30 db is
 A 60000 B 15000 C 30000 D 3000
Question 91 Explanation:
Shannon Capacity (Noisy Channel)
In presence of Gaussian band-limited white noise, Shannon-Hartley theorem gives the
maximum data rate capacity
C = B log2 (1 + SNR)
where SNR Signal to noise ration
This theorem gives an upper bound of the data rate which can be reliably transmitted over a thermal-noise limited channel.

Given
Bandwidth = 3000 Hz
SNR = 30 db
Maximum data rate capacity C = B log2 (1 + SNR)
Maximum data rate capacity C= 3000 log2 (1 + 30)
Maximum data rate capacity C= 3000 log2 *(31) = 3000 * 4.9 =14,850 ≈ 15,000
 Question 92
Bit stuffing refers to
 A Inserting a '0' in user data stream to differentiate it with a flag B Inserting a '0' in flag stream to avoid ambiguity C Appending a nibble to the flag sequence D Appending a nibble to the user data stream
Question 92 Explanation:
• Bit stuffing refers to inserting a 0 in user stream to differentiate it with a flag.
• Bit stuffing is required when there is a flag of bits to represent one of the incidents like start of frame, end of frame.
• If same flag of bits appear in the data stream, a zero can be inserted by the sender.
• When the receiver sees this  0 bit from the data stream, it automatically destuffs the 0 bit before sending the data to the network layer.
 Question 93
If the channel is band limited to 6 kHz and signal to noise ratio is 16, what would be the capacity of channel?
 A 16.15kbps B 23.24 kbps C 40.12 kbps D 24.74 kbps
Question 93 Explanation:
Shannon Capacity (Noisy Channel)
In presence of Gaussian band-limited white noise, Shannon-Hartley theorem gives the
maximum data rate capacity
C = B log2 (1 + SNR)
where SNR Signal to noise ration
This theorem gives an upper bound of the data rate which can be reliably transmitted over a thermal-noise limited channel.

Given
Bandwidth = 6 kHz
SNR = 16 db
Maximum data rate capacity C = B log2 (1 + SNR)
Maximum data rate capacity C= 6 log2 (1 + 16)
Maximum data rate capacity C= 6 log2 *(17) = 6 * 4.08 =24.74
 Question 94
Which of the following protocols is used to map IP address to MAC address?
 A ARP B IP C DHCP D RARP
Question 94 Explanation:
ARP
• ARP stands for Address Resolution Protocol
RARP
• RARP stands for Reverse Address Resolution Protocol
• RARP protocol helps to find the internet address(IP Address) of a device whose MAC Address is known.
 Question 95
Correct expression for UDP user datagram length is
 A Length of UDP=length of IP - length of IP header B Length of UDP=length of UDP - length of UDP header C Length of UDP=length of IP + length of IP header D Length of UDP=length of UDP + length of UDP header
Question 95 Explanation:
• A user datagram is encapsulated in an IP datagram.
• There are 2 fields in the IP datagram  that defines total length and length of the header.
• Subtract the length of a UDP datagram that is encapsulated in an IP datagram, then we get the length of UDP user datagram.
 Question 96
Close-loop control mechanism try to:
 A Remove congestion after it occurs B Remove congestion after sometime C Prevent congestion before it occurs D Prevent congestion before sending packets
Question 96 Explanation:
Congestion control refer to mechanism used to control congestion or prevent congestion.

Congestion control mechanism can be broadly classified into 2 categories :
1. Open Loop
2. Closed Loop Congestion Control

Open Loop Congestion Control
Open Loop Congestion Control applied to prevent congestion before it happens.
The congestion control is handled either by the source or the destination.

Open loop congestion control Policies are
• Retransmission Policy
• Window Policy
• Acknowledgment Policy
Closed Loop Congestion Control
Closed loop congestion control technique is used to treat or alleviate congestion after it happens.

Some of techniques are :
• Backpressure Technique
• Choke Packet Technique
• Implicit Signaling
• Explicit Signaling
(i) Forward Signaling
(ii) Backward Signaling
 Question 97
Which multiple access technique is used by IEEE 802.11 standard for wireless LAN?
 A CDMA B CSMA/CA C ALOHA D None of the Options
Question 97 Explanation:
• IEEE 802.11 standard for  wireless LANs use a media access control protocol called (CSMA/CA) stands for Carrier-Sense Multiple Access/Collision Avoidance
• While the name is similar to Ethernet's Carrier Sense Multiple Access with Collision Detection (CSMA/CD), the operating concept is totally different.
 Question 98
PGP encrypts data using a block cipher called:
 A International data encryption algorithm B Private data encryption algorithm C Internet data encryption algorithm D None of the options
Question 98 Explanation:
Pretty Good Privacy (PGP) is an encryption program that provides cryptographic privacy and authentication for data communication.

PGP is used for signing, encrypting, and decrypting texts, e-mails, files, directories to increase the security of e-mail communications. Phil Zimmermann developed PGP in 1991

Symmetric key algorithm used in PGP version 2 was IDEA

In cryptography, the International Data Encryption Algorithm (IDEA), originally called Improved Proposed Encryption Standard (IPES), is a symmetric-key block cipher designed by James Massey of ETH Zurich and Xuejia Lai and was first described in 1991.

The algorithm was intended as a replacement for the Data Encryption Standard (DES). IDEA is a minor revision of an earlier cipher Proposed Encryption Standard (PES).

IDEA was used in Pretty Good Privacy (PGP) v2.0 and was incorporated after the original cipher used in v1.0, BassOmatic, was found to be insecure. IDEA is an optional algorithm in the OpenPGP standard.
 Question 99
A subnet mask in class C can have ___ 1's with the remaining bits 0's
 A 10 B 24 C 12 D 7
Question 99 Explanation:
First 3 (N.N.N,H) octet of class C is NID bits and last octet is host bits
In Binary 11111111.11111111.11111111.00000000
Total Number of 1's = 24
Total Number of 0's = 8
 Question 100
What is the value of acknowledgement field in segment?
 A Number of previous bytes to receive B Total number of bytes to receive C Number of next bytes to receive D Sequence of zero's and one's
Question 100 Explanation:
Acknowledgement field in a segment defines the sequence number of the byte which is to be received next
it means sequence number of byte that the sender should transmit next.
 Question 101
Which of the following is a class B host address?
 A 230.0.0.0 B 130.4.5.6 C 230.7.6.5 D 30.4.5.6
Question 101 Explanation:
 Question 102
There is a need to create a network that has 5 subnets, each with at least 16 hosts. which one is used as classful subnet mask?
 A 255.255.255.192 B 255.255.255.248 C 255.255.255.240 D 255.255.255.224
Question 102 Explanation:
we need 5 subnets, each with at least 16 hosts.
∴ Subnet Mask = 255.255.255.240 provides 16 subnets with 14 hosts which is less than 15,
∴ Subnet Mask = 255.255.255.224 provides 8 subnets, each with 30 hosts
∴ Subnet Mask = 255.255.255.192 provides 4 subnets, each with 60 hosts
Comparing all the possible masks, 255.255.255.224 is better
 Question 103
Which NetWare protocol provides link state routing?
 A NLSP B RIP C SAP D NCP
Question 103 Explanation:
NLSP
-NLSP stands for NetWare Link Services Protocol
-NLSP was developed by Novell to replace RIP routing protocols.

SAP

NCP
-NCP stands for  NetWare Core Protocol
-NCP provides client-to-server connections and applications.

RIP
-RIP stands for Routing Information Protocol
-RIP is a dynamic routing protocol which uses hop count as a routing metric to find the best path between the source and the destination network
-RIP is a distance vector routing protocol.
 Question 104
Which layer connects the network support layers and user support layers?
 A Transport layer B Network layer C Data link layer D Session layer
Question 104 Explanation:
Network support layers :
- The network support layers are Physical layer, Data link layer and Network layer.
- These deals with electrical specifications, physical connection, transport timing and reliability.

User support layers :
- The user support layers are: Session layer, Presentation layer, Application layer.
- These allow interoperability among unrelated software system.

Transport layer links these layers by segmenting and rearranging the data. It uses protocols like TCP and UDP.
 Question 105
Question 105 Explanation:
-  Classful addressing is a concept that divides the available address space of IPv4 into five classes namely A, B, C, D and  E.

-  Nowadays, this concept has become obsolete and has been replaced with classless addressing because large ratio of the available addresses in a class(Class A, Class B, Class C, Class D)  in classful addressing is wasted.

-  IP addresses, before 1993 use the classful addressing where classes have a fixed number of blocks and each block has a fixed number of hosts.

-  In classless addressing, one can reserve the number of IP addresses required by modifying the CIDR value and make sure that not many addresses are wasted.
 Question 106
An ethernet destination address 07-01-12-03-04-05 is:
Question 106 Explanation:
Given Ethernet destination Address = 07-01-12-03-04-05
Now to find the type of address we need to check first octant (07)
Convert given number into Binary
07 - 0000 0111

-If LSB bit 0 then Unicast
-If LSB bit 1 then Multicast
-If All Bits are 1 then Broadcast

So ( 07 - 0000 0111 ) LSB bit 1 so it is Multicast
 Question 107
 A It is easy to generate B It cannot be shared C It is different for every access D It can be easily decrypted
Question 107 Explanation:
-OTP stands for One Time Password
-OTP is more secure than a static password, especially a user-created passwords which can be weak and/or reused across multiple accounts.
-One-time password  is an automatically generated per Access and it cannot be brute forced
 Question 108
In classless addressing, there are no classes but addresses are still granted in:
 A Codes B Blocks C IPs D Sizes
Question 108 Explanation:
• In classless addressing  there are no classes, but the addresses are still granted in blocks.

• In classless addressing, when an entity, small or large, needs to be connected to Internet, it is granted a block or range of addresses.

• The size of the block (the number of addresses) varies based on the nature and size of the entry. For example, a household may be given only two addresses; a large organization may be given thousands of addresses. An ISP, may be given thousands or hundreds of thousands based on the number of customer it may serve.

• To simplify the handling of addresses, the Internet authorities impose three restrictions on classless address blocks:

1) The addresses in a block must be contiguous, one after the other.
2) The number of addresses in a block must be a power of 2 (1, 2, 4,8, …. ).
3) The first address must be evenly divisible by the number of address.

For example, a block of addresses (both in binary and dotted decimal notation) granted to a small business that needs 16 addresses.

Below figure contains certain restrictions are applied to this block.
The addresses in a block are contiguous
Number of addresses is a power of 2 (24 = 16)
First address is divisible by 16.
The first address, converted to decimal number, is 3,440,387,360
which divided by 16 results in 215,024,210.
 Question 109
The Demerits of the fragmentation are:
 A Complex routers B Open to DOS attack C No overlapping of fragments D (A) and (B) both
Question 109 Explanation:
fragmentation is the process of breaking the IP packet into multiple packets of smaller size

- The main disadvantage of fragmentation is exit gateway, all the packets should exit via the same gateway.
- It requires overhead to repeatedly fragment and reassemble a large packet.
- When a large packets arrives at a gateway, it breaks the packet into fragments.
- Each fragment is then addressed to the same exit gateway. Because of fragmentation there may be chance of a DoS attack.
- Fragmentation is done at routers which make them complex to implement when routers take too much time to fragment packets this may lead to DOS attack to other packets
 Question 110
In which of the topology, each device has a dedicated point-to-point link to a central controller?
 A Mesh B Bus C Ring D Star
Question 110 Explanation:
 Question 111
The electromagnetic waves ranging in frequencies between 1 GHz and 300 GHz are called____
 A Radio waves B Microwaves C Infrared waves D Light waves
Question 111 Explanation:
Wireless transmission can be categorized into 3 broad groups:
2.Microwaves
3.Infrared

-Electromagnetic waves ranging in frequencies between 3 KHz and 1 GHz are normally called radio waves.
-Electromagnetic waves ranging in frequencies between 1 and 300 GHz are normally called microwaves.
-Infrared signals have frequencies between 300 GHz to 400 THz. They are used for short range communication.
 Question 112
Which of the following protocol is used for transforming electronic mail messages from one machine to another?
 A FTP B SMTP C SNMP D STTP
Question 112 Explanation:
• SMTP stands for Simple Mail Transfer Protocol.
• SMTP is part of the application layer of the TCP/IP protocol.
• SMTP provides a mail exchange between users on the same or different computers
• SMTP Using a process called "store and forward" SMTP moves your email on your machine to other machine  across internet based on e-mail addresses
• SMTP works closely with something called the Mail Transfer Agent (MTA) to send your communication to the right computer and email inbox.
 Question 113
The device bridge is used at ____ layer of OSI reference model.
 A Datalink B Network C Transport D Application
Question 113 Explanation:
Bridges
• Connect two parts of the same network.
• Read only the destination address of each Ethernet packet or Token Ring frame for maximum speed and efficiency.
• Bridges operate at the Data-Link Layer of the OSI Model. They can distinguish between local and remote data, so data traveling from one workstation to another in the same segment doesn't have to cross the bridge.
• Bridges operate on MAC-Layer addresses.
 Question 114
The period of a signal is 100ms. What is its frequency in kilohertz?
 A 10-1 kHz B 10 -2 kHz C 10-3 kHz D 10-4 kHz
Question 114 Explanation:
 Question 115
Machine that places the request to access the data, is generally called as___
 A Server machine B Client machine C Request machine D Intelligent machine
Question 115 Explanation:
Client Machine that places the request to server machine to access the data
Server machine process the request and sends the response to the client machine
 Question 116
The function setcookie() is used to____
Question 116 Explanation:
 Question 117
When a mobile telephone physically moves from one cell to another cell, the base station transfers ownership to the cell getting strongest signal. This process is known as____
 A Handoff B Mobile routing C Mobile switching D Cell Switching
Question 117 Explanation:
 Question 118
Bit stuffing refers to
 A Inserting a '0' in user data stream to differentiate it with a flag B Inserting a '0' in flag stream to avoid ambiguity C Appending a nibble to the flag sequence D Appending a nibble to the user data stream
Question 118 Explanation:
 Question 119
Which of the following would not be specified in a communication protocol?
 A Header contents B Trailer contents C Error Checking D Data content of message
Question 119 Explanation:
• A communication protocol is a system of rules that allow two or more entities of a communications system to transmit information via any kind of variation of a physical quantity.
• The protocol defines the rules, syntax, semantics and synchronization of communication and possible error recovery methods.
• Communicating systems use well-defined formats for exchanging various messages.
• Each message has an exact meaning intended to elicit a response from a range of possible responses pre-determined for that particular situation.
Source : Wifi
 Question 120
Internet Control message protocol(ICMP)
 A Allows gateways to send error control messages to other gateways or hosts B Provides communication between the internet protocol software on one machine and the internet protocol software on another C Only reports error conditions to the original source, the source must relate errors to individual application programs and take action to correct the problem. D All of these
Question 120 Explanation:
 Question 121
Which layers of the OSI model are host to host layers?
 A Transport,session,Presentation, Application B Network,Transport,Session,Presentation C Datalink, Network, Transport, Session D Physical,Data Link,Network,Transport
Question 121 Explanation:
 Question 122
A____is a communication pathway connecting two or more devices. Another of its key characteristics is that it is a shared transmission medium. A signal transmitted by any one device is available for reception by all other devices attached to it.
 A Train B Bus C Tram D Aeroplane
Question 122 Explanation:
 Question 123
Bridge works in which layer of the OSI model?
 A Application layer B Transport layer C Network layer D Data link layer
Question 123 Explanation:
 Question 124
What is the meaning of bandwidth in a network?
 A Transmission capacity of a communication channel B Connected components in a network C Class of IP used in Network D Interconnected by communication channels
Question 124 Explanation:
• Bandwidth in a network is the capacity of a wired or wireless network communications channel to transmit the maximum amount of data from one point to another over a computer network or internet connection in a given amount of time
• In other words Network Bandwidth is the maximum amount of data transmitted over an communications channel in a given amount of time
 Question 125
Which one of the following transmission systems provides the highest data rate to an individual device?
 A Computer bus B Telephone bus C Voice and mode D Lease lines
Question 125 Explanation:
BUS provides the highest data rate to an individual device
BUS is a communication pathway connecting two or more devices
 Question 126
The X.25 standard specifies a
 A Technique for start-stop data B Technique for dial access C DTE/DCE interface D Data bit rate
Question 126 Explanation:
• X.25 is an ITU-T standard protocol suite for packet-switched data communication in wide area networks (WAN).
•  X.25 specification defines only the interface between a subscriber (DTE) and an X.25 network (DCE)
• The X.25 model was based on the traditional telephony concept of establishing reliable circuits through a shared network, but using software to create "virtual calls" through the network. These calls interconnect "data terminal equipment" (DTE) providing endpoints to users, which looked like point-to-point connections. Each endpoint can establish many separate virtual calls to different endpoints.
• One DTE-DCE interface to an X.25 network has a maximum of 4095 logical channels on which it is allowed to establish virtual calls and permanent virtual circuits
 Question 127
Frames from one LAN can be transmitted to another LAN via a device called
 A Router B Bridge C Repeater D Modem
Question 127 Explanation:
 Question 128
With an IP address of 100, you have 80 subnets. What subnet mask should you use to maximize the number of available hosts?
 A 192 B 224 C 248 D 252 E None
Question 128 Explanation:
Question is not valid they have to mention which class type there using without that we cant determine the subnet even 100 is not a valid IP address
 Question 129
The___houses the switches in token ring
 A Transceiver B Nine pin connector C MAU D NIC
Question 129 Explanation:
• Short for Media Access Unit, a MAU is also known as an Ethernet transceiver or MSAU (Multi Station Access Unit) is an adapter, connector, or stand alone device that enables a network device to be connected to a token ring network.
• A MAU is one form of fault tolerance that helps prevent issues if a network device or computer goes down and are commonly available as either active or passive.
• An active MAU is not powered and does not in any way strengthen the signal from a device.
• A passive MAU is powered and repeats and strengthens a signal.
 Question 130
In OSI network architecture, routing is performed by the
 A Network layer B Data link layer C Transport layer D Session layer
Question 130 Explanation:
In OSI network architecture, the routing is performed by network layer which is also called layer 3 with the help of intermediate routers
 Question 131
Which of the following protocol is used for transferring electronic mail messages from one machine to another?
 A HTTP B FTP C SMTP D SNMP
Question 131 Explanation:
 Question 132
Ten signals, each requiring 3000 hz, are multiplexed onto a single channel using FDM. How much minimum bandwidth is required for the multiplexed channel? Assume that the guard bands are 300 Hz wide.
 A 33,700 B 30,000 C 32,700 D 33,000
Question 132 Explanation:
Given
Guard bands are 300 Hz
10 signals each require 3000  HZ
∴ 10 signals requires 10 x 3000 = 30000 Hz
9 (10-1) Guard bands (or gaps) needs =300 x 9 = 2700 Hz
So Minimum Bandwidth is required for the multiplexed channel =30000 + 2700 = 32700Hz
 Question 133
If the period of a signal is 1000ms, then what is its frequency in kilohertz?
 A 10​ -1​ Khz B 1 KHz C 10​ -3​ Khz D 10​ -2​ Khz
Question 133 Explanation:
Given Time Period  is  1000 ms
we already know that Frequency is the inverse of Time Period. so
Frequency = ( 1 / Time Period )
Frequency = ( 1  / 1000 ms )
Frequency =  (1 / 1 s)   // 1000 ms = 1 sec
Frequency = 1Hz  // 1 Hz = 10-3 kHz
Frequency =10-3 KHz
 Question 134
Mechanism that is used to convert domain name into IP address is known___
 A HTTP B URL C FTP D DNS
Question 134 Explanation:
 Question 135
The unit receiving the data item response with another control signal to acknowledgement receipt of the data. This type of argument between two independent units is known as
 A Storage control B Multitasking C Handshaking D Piggybacking
Question 135 Explanation:
 Question 136
In client-server computing vertical scaling means:
 A Adding or removing client workstations with only a slight performance impact B Migrating servers to a new group of client workstations C Migrating client workstations to a larger and faster server machine or multi servers D Combining two or more client workstation groups
Question 136 Explanation:
• In Client/Server architecture, clients, or programs that represent users who need services, and servers, or programs that provide services, are separate logical objects that communicate over a network to perform tasks together.
• A client makes a request for a service and receives a reply to that request; a server receives and processes a request, and sends back the required response.
• Horizontal scaling means adding or removing client workstations with only a slight performance impact.
• Vertical scaling means migrating to a larger and faster server machine or adding server machines.
 Question 137
The 10Base5 cabling scheme of ethernet uses:
 A Twisted pairs B Fiber optics C Thin coax D Thick coax
Question 137 Explanation:
• 10BASE5 (also known as thick Ethernet or thicknet) was the first commercially available variant of Ethernet.
• 10BASE5 uses a thick and stiff coaxial cable up to 500 meters (1,600 ft) in length.
• Up to 100 stations can be connected to the cable using vampire taps and share a single collision domain with 10 Mbit/s of bandwidth shared among them
Source 1: Wiki
Source 2: computer-networking-notes
 Question 138
One of the ad-hoc solutions to count to infinity problem in network routing is:
 A The split horizon hack B Flow based routing C Flooding D Shortest path routing
Question 138 Explanation:
• The problem with Distance Vector Routing (DVR) protocols is Routing Loops
• This routing loop in network causes Count to Infinity Problem.
• Routing loops usually occur when any interface goes down or 2 routers send updates at the same time.
• There are two possible solutions for Count to Infinity problem are
1. Route Poisoning
2. Split horizon
 Question 139
Given a bit rate of b bits/sec, the time required to send 16 bits is:
 A 16*b sec B 16/b sec C 16b sec D B16 sec
Question 139 Explanation:
Given Bandwidth = b bits/sec
it means in 1 sec data can transfer upto b bits
So, There asking Time for Transferring 16 bits
We know that
1 sec = b bits
b bits = 1 sec
1 bit=1/b sec
16 bit=16/b sec
∴ 16/b sec
 Question 140
The built in HTTP request method to request to read a web page is:
 A HEAD B PUT C GET D POST
Question 140 Explanation:
GET
This method requests a representation of the specified resource. Requests using GET should only retrieve data.

This method asks for a response identical to that of a GET request, but without the response body.

POST
Used to submit an entity to the specified resource, often causing a change in state or side effects on the server.

PUT
used to replaces all current representations of the target resource with the request payload.

DELETE
deletes the specified resource.

CONNECT
used to Establishes a tunnel to the server identified by the target resource.

OPTIONS
used to Describe the communication options for the target resource.

TRACE
performs a message loop-back test along the path to the target resource.

PATCH
used to apply partial modifications to a resource.
 Question 141
The traditional cryptographic cipher that reorders the letters but do not disguise them is:
 A Substitute cipher B One-time pads C Secret key algorithms D Transposition cipher
Question 141 Explanation:
Transposition cipher
→In transposition ciphers Reorder the letters but do not disguise them
→The letters of the original message (plaintext) are arranged in a different order to get the ciphertext.
PlaintextRearrange charactersCiphertext
Message M =“HELLOWORLD” distribute the letter up and down between two rows from left to right; then output row-wise.
→ HLOOL  ELWRD  →  Ciphertext C="HLOOLELWRD"
 Question 142
The 10Base5 cabling is also known as___
 A Fast ethernet B Thick ethernet C Thin ethernet D Gigabit ethernet
Question 142 Explanation:
 Question 143
Which of the following devices understands the formats and contents of the data and translate message from one format to another?
 A Gateway B Hub C Switch D Router
Question 143 Explanation:
 Question 144
Which of the following transmission media works on the principle of total internal reflection?
 A Optical fiber cable B Shielded twisted pair cable C Unshielded twisted pair cable D Coaxial cable
Question 144 Explanation:
 Question 145
Which of the following topologies highest reliability?
 A Mesh topology B Bus topology C Star topology D Ring topology
Question 145 Explanation:
Mesh topology is divided into 2 different types
1. Full Mesh
2.Partial Mesh.

Full Mesh
• A Full mesh topology provides a connection from each node to every other node on the network.
• This provides a fully redundant network and is the most reliable of all networks.
• If any link or node in the network fails, then there will be another path that will allow network traffic to continue.
• The major drawback to this type of network is the expense and complexity required to configure this topology.
• This type of topology is only used in small networks with only a few nodes.
• The Number of connections in this network is n(n-1)/2. where n=number of  nodes
Partial Mesh
• A partial mesh topology provides alternate routes from each node to some of the other nodes on the network.
• This type of topology provides some redundancy and is commonly used in backbone environments, networks where services are vital, and in wide area networks, WANs.
• The most notable partial mesh network is the Internet.
 Question 146
What is the length of an IP address in (Pre-1PU6)?
 A 8 B 1 C 2 D 4
Question 146 Explanation:
Printing Mistake it should be Pre-IPV6
IPV4 is 32 Bits    =  4 Bytes
IPV6 is 128 Bits  = 16 Bytes
 Question 147
An analog signal carries 4 bits in each signal unit. if 1000 signal units are sent per second, then baud rate and bit rate of the signal are___ and ___
 A 1000 bauds/sec, 4000 bps B 1000 bauds/sec, 500 bps C 4000 bauds/sec, 1000 bps D 2000 bauds/sec, 1000 bps
Question 147 Explanation:
Bit rate is the number of bits per second.
Baud rate is the number of signal elements per second.
In the analog transmission of digital data, the baud rate is less than or equal to the bit rate

In this case, r = 4, S = 1000, and N is unknown.
We can find the value of N from
S = N x (1/r)  or  N=S*r=1000*4=4000 bps

Baud rate = 1000 bauds per second (baud/s)
 Question 148
Which of the following devices takes data sent from one network device and forwards it to the destination node based on MAC address ?
 A Hub B Modem C Switch D Gateway
Question 148 Explanation:
• Switch takes data from one network and forward it to other network based on MAC address.
• Switches are networking devices operating at layer 2 or a data link layer of the OSI model.
• They connect devices in a network and use packet switching to send, receive or forward data packets or data frames over the network.
• When a data frame arrives at any port of a network switch, it examines the destination address, performs necessary checks and sends the frame to the corresponding device(s).
• It supports unicast, multicast as well as broadcast communications.
 Question 149
__________ do not take their decisions on measurements or estimates of the current traffic and topology.
 A Static algorithms B Adaptive algorithms C Non - adaptive algorithms D Recursive algorithms
Question 149 Explanation:
Non-Adaptive Routing Algorithm do not base their routing decisions on measurements and estimates of the current traffic and topology
Refer : https://www.cse.iitk.ac.in/users/dheeraj/cs425/lec11.html
 Question 150
The number of bits used for addressing in Gigabit Ethernet is __________.
 A 32 bit B 48 bit C 64 bit D 128 bit
Question 150 Explanation:
For Addressing in gigabit Ethernet 6 Bytes (48 bits)  will be used
 Question 151
Which of the following layer of OSI Reference model is also called end-to-end layer?
 A Network layer B Data layer C Session layer D Transport layer
Question 151 Explanation:
→ Transport layer is responsible for End to End communication.
→ Network layer is responsible for Host to Host communication.
→ Data Link layer is responsible for Process to Process communication.
→ Presentation layer is responsible for Encryption and Decryption
 Question 152
The IP address __________ is used by hosts when they are being booted
 A 0.0.0.0 B 1.0.0.0 C 1.1.1.1 D 255.255.255.255
Question 152 Explanation:
0.0.0.0 has several uses, one of the use is: It is used when systems being booted.

"0.0.0.0" is a valid address syntax. So it should parse as valid wherever an IP address in traditional dotted-decimal notation is expected. Once parsed, and converted to workable numeric form, then its value determines what happens next.

The all-zero value does have a special meaning. So it is "valid", but has a meaning that may not be appropriate (and thus treated as not valid) for particular circumstances. It is basically the "no particular address" placeholder. For things like address binding of network connections, the result can be to assign an appropriate interface address to the connection. If you are using it to configure an interface, it can remove an address from the interface, instead. It depends on the context of use to determine what "no particular address" really does.

In the context of a route entry, it usually means the default route. That happens as a result more of the address mask, which selects the bits to compare. A mask of "0.0.0.0" selects no bits, so the compare will always succeed. So when such a route is configured, there is always somewhere for packets to go (if configured with a valid destination).

In some cases, merely "0" will also work and have the same effect. But this is not guaranteed. The "0.0.0.0" form is the standard way to say "no particular address" (in IPv6 that is "::0" or just "::").

 Question 153
Which of the given wireless technologies used in IoT, consumes the least amount of power ?
 A Zigbee B Bluetooth C Wi-Fi D GSM/CDMA
Question 153 Explanation:
→Bluetooth devices consume less power than WiFi in most of the cases.
→Bluetooth devices was specifically designed to have low energy consumption.
→Bluetooth devices is an ultra low energy wireless technology for IOT devices which consumes the very smaller amount of power
 Question 154
Which speed up could be achieved according to Amdahl’s Law for infinite number of processes if 5% of a program is sequential and the remaining part is ideally parallel ?
 A Infinite B 5 C 20 D 50
Question 154 Explanation:
Amdahl’s law speed up for infinite number of process
S = 1 / (1-P)
where P = Parallel part of the program

Given
Sequential part = 5%.
So Parallel part  (P) = 1 - Sequential Part
Parallel part  (P) = 1 - 5%
Parallel part  (P) = 1 - 0.05
Parallel part  (P) = 0.95 (95%)

Speed Up
S = 1 / (1-P)
S = 1 / (1-0.95)
S = 1 / 0.05
S = 20
 Question 155
Time taken by a packet to travel from client to server and then back to client is called
 A STT B RTT C PTT D Total time
Question 155 Explanation:
• RTT stands for Round-Trip Time.
• Round-Trip delay (RTD) or Round-trip time (RTT) is the length of time it takes for a signal to be sent plus the length of time it takes for an acknowledgement of that signal to be received. This time delay includes the propagation times for the paths between the two communication endpoints.
Propagation Delay is  Amount of time taken by a packet to make a physical journey from one router to another router.
Propagation Delay = (Distance between routers) / (Velocity of propagation)
RoundTripTime (RTT) = 2* Propagation Delay
TimeOut (TO) = 2* RTT
Time To Live (TTL) = 2* TimeOut. (Maximum TTL is 180 seconds)
 Question 156
A network with bandwidth of 10Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network?
 A 2 Mbps B 4 Mbps C 8 Mbps D 12 Mbps
Question 156 Explanation:
In data transmission, network throughput is the amount of data moved successfully from one place to another in a given time period, and typically measured in bits per second (bps), as in Megabits per second (Mbps) or Gigabits per second (Gbps).

Given That each frame per minute. So, convert into seconds
i.e. 12,000 frames per minute. Each Frame is carrying 10,000 bits
Throughput = (12,000 * 10,000) / (60 seconds)
Throughput= 2 * 1000000 bits / seconds.
Throughput= 2 Mbps.
 Question 157
The most common connector used with fiber-optic cable is
 A RJ45 B RJ54 C BNC D MT-RJ
Question 157 Explanation:
Most common connectors are
 Question 158
In which of the following topologies, each device has a dedicated point-to-point connection with only two devices on either side of it?
 A Mesh topology B Star topology C Bus topology D Ring topology
Question 158 Explanation:
• In a Ring Topology each device has a dedicated point to point connection with only the two devices on either side of it.
• In a Ring Topology A signal is passed along the ring in one direction, from device to device, until it reaches its destination.
• Each device in the ring incorporates a repeater. When a device receives a signal intended for another device, its repeater regenerates the bits and passes them along.
 Question 159
Which of the following IEEE standards is NOT a standard for WAN in mobile communication?
 A IEEE 802.11a B IEEE 802.11b C IEEE 802.11d D IEEE 802.11g
Question 159 Explanation:
• IEEE 802.11d is an IEEE 802.11 amendment that adds geographical regulations to the original standard.
• IEEE 802.11d facilitates the development of wireless local area network (WLAN) devices that comply with the wireless communications regulations of their respective countries.
• IEEE 802.11d is also known as IEEE 802.11d-2001.
 Question 160
Which of the following address is used to deliver a message to the correct application program running on a host?
 A Logical B Physical C IP D Port
Question 160 Explanation:
→IP address lets you know where the network is located.

→ MAC address also known as a media access control address is a unique 48 bit number

→Port address identifies a process or service you want to carry on.
→With in a network on a particular Host A port number is a way to identify a specific process
 Question 161
Ten signals, each requiring 3000Hz, are multiplexed on to a single channel using FDM. What is the minimum bandwidth required for the multiplexed channel?Assume that the guard bands are 300 Hz wide.
 A 30,000 B 30,300 C 32,700 D 33,000
Question 161 Explanation:
 Question 162
Which of the following TCP ports are used by File Transfer Protocol(FTP)?
 A 20 and 21 B 20 and 23 C 21 and 25 D 23 and 25
Question 162 Explanation:
 Question 163
The maximum size of the data that the application layer can pass on to the TCP layer below is __________.
 A 2​ 16​ bytes B 2​ 16​ bytes + TCP header length C 2​ 16​ bytes - TCP header length D 2​ 15​ byte. E None
Question 163 Explanation:
→There is no limitation on the amount of data Application layer can pass on to the TCP Layer. Application layer can send any size of data. The lower layers divides the data if needed.

→Transport layer divides the data into several segment and can forward maximum 65,515 byte data to network layer. Network layer divides the data into packets.

→ Network layer can only send a packet of maximum size 65,535 byte to Data Link Layer

→Data link Layer maximum length of the payload field is 1500 bytes.
 Question 164
A packet whose destination is outside the local TCP/IP network segment is sent to _____.
 A File server B DNS server C DHCP server D Default gateway
Question 164 Explanation:
• A packet whose destination is outside the local TCP/IP network segment is sent to the default gateway.
• A default gateway serves as an access point or IP router that a networked computer uses to send information to a computer in another network or the Internet
 Question 165
Distance vector routing algorithm is a dynamic routing algorithm. The routing tables in distance vector routing algorithm are updated _____.
 A Automatically B By server C By exchanging information with neighbour nodes D With backup database
Question 165 Explanation:
• A distance-vector routing protocol in data networks determines the best route for data packets based on distance.
• Distance-vector routing protocols measure the distance by the number of routers a packet has to pass, one router counts as one hop.
•  By their  own local knowledge each router prepares its own  routing table.
• Each router knows about all the routers present in the network
• Distance to its neighboring routers
• Each router exchanges its distance vector with its neighboring routers.
• Each router prepares a new routing table using the distance vectors it has obtained from its neighbors.
• This step is repeated for (n-2) times if there are n routers in the network. now routing tables converge / become stable.
• Distance vector routing protocols use the Bellman–Ford algorithm and Ford–Fulkerson algorithm to calculate the best route.
• Another way of calculating the best route across a network is based on link cost, and is implemented through link-state routing protocols.
 Question 166
In link state routing algorithm after construction of link state packets, new routes are computed using:
 A DES algorithm B Dijkstra’s algorithm C RSA algorithm D Packets
Question 166 Explanation:
• Link-state routing protocols is used in  distance-vector routing protocols.
• link-state routing protocols include Open Shortest Path First (OSPF) and Intermediate System to Intermediate System (IS-IS).
• The link-state protocol is performed by every switching node in the network
• The basic concept of link-state routing is that every node constructs a map of the connectivity to the network, in the form of a graph, showing which nodes are connected to which other nodes.
• Each node then independently calculates the next best logical path from it to every possible destination in the network.
• Each collection of best paths will then form each node's routing table.
• This contrasts with distance-vector routing protocols, which work by having each node share its routing table with its neighbours
• In a link-state protocol the only information passed between nodes is connectivity related.
• Link-state algorithms are sometimes characterized informally as each router, "telling the world about its neighbors."
• Each node independently runs an algorithm over the map to determine the shortest path from itself to every other node in the network; generally some variant of Dijkstra's algorithm is used.
• This is based around a link cost across each path which includes available bandwidth among other things
Source : wiki
 Question 167
In 3G network, W-CDMA is also known as UMTS. The minimum spectrum allocation required for W-CDMA is _______.
 A 2 MHz B 20 KHz C 5 KHz D 5 MHz
Question 167 Explanation:
• The Universal Mobile Telecommunications System (UMTS) is a third generation mobile cellular system for networks based on the GSM standard.
• Developed and maintained by the 3rd Generation Partnership Project
• UMTS uses wideband code division multiple access (W-CDMA) radio access technology to offer greater spectral efficiency and bandwidth to mobile network operators.
• W-CDMA/UMTS requires a minimum spectrum allocation of 5 MHz.
• Using this bandwidth, it has the capacity to carry over 100 simultaneous voice calls.
• It is able to carry data at speeds up to 2 Mbps in its original format.
• 20 MHz is the bandwidth defined for LTE.
• Cdmaone technology uses a bandwidth of 1.25 MHz.
• GSM’s bandwidth is 200 KHz.
 Question 168
Which of the following statements is not true with respect to microwaves?
 A Electromagnetic waves with frequencies from 300 GHz to 400 Thz. B Propagation is line-of-sight. C Very high-frequency waves cannot penetrate walls. D Use of certain portions of the band requires permission from authorities.
Question 168 Explanation:
 Question 169
In a fast Ethernet cabling, 100 Base-TX uses ____ cable and maximum segment size is_____.
 A Twisted pair, 100 metres B Twisted pair, 200 metres C Fibre optics, 1000 metres D Fibre optics, 2000 metres
Question 169 Explanation:

100 Base TX, Fast Ethernet, transmits data at 100 Mbps.
100 Base TX uses catagory5 UTP which is twisted pair cable.
100 Base TX  Segment size is 100 meter
 Question 170
A network with bandwidth of 10 Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network ?
 A 1 Mbps B 2 Mbps C 10 Mbps D 12 Mbps
Question 170 Explanation:
 Question 171
Which of the following protocols is used by email server to maintain a central repository that can be accessed from any machine ?
 A POP3 B IMAP C SMTP D DMSP
Question 171 Explanation:
• POP3 is post office protocol Version 3.POP is a protocol which listens on port 110 and is responsible for accessing the mail service on a client machine. POP3 works in two modes such as Delete Mode and Keep Mode.
• The basic idea behind IMAP is for the email server to maintain a central repository that can be accessed from any machine.
• Thus, unlike POP3, IMAP does not copy email to the users personal machine because the user may have several. In other words, the mailbox can be viewed as a relational database system rather than a linear sequence of messages
• SMTP stands for simple mail transfer protocol.
• DMSP stands for distributed mail service protocol.
 Question 172
An attacker sits between the sender and receiver and captures the information and retransmits to the receiver after some time without altering the information. This attack is called as _____.
 A Denial of service attack B Masquerade attack C Simple attack D Complex attack
Question 172 Explanation:
• An attacker sits between the sender and receiver and captures the information and re-transmits to the receiver after some time without altering the information. This attack is called as Denial of service attack.
• A Denial-of-Service (DoS) attack is an attack meant to shut down a machine or network, making it inaccessible to its intended users. DoS attacks accomplish this by flooding the target with traffic, or sending it information that triggers a crash.
• A Masquerade attack is an attack that uses a fake identity, such as a network identity, to gain unauthorized access to personal computer information through legitimate access identification. If an authorization process is not fully protected, it can become extremely vulnerable to a masquerade attack.
 Question 173
A multiplexer combines four 100-Kbps channels using a time slot of 2 bits. What is the bit rate?
 A 100 Kbps B 200 Kbps C 400 Kbps D 1000 Kbps
Question 173 Explanation:
The link carries 50000 frames per second as each frame contains 2 bits per channel
∴ 100 kbps/2=50kbps
Frame duration  = 1/50,000 sec= 0.00002 sec=20 microseconds
Bit rate = Frame rate * Number of bits per frame
Bit rate=50000 * 8 = 400 kbps
 Question 174
In a fully-connected mesh network with 10 computers, total ______ number of cables are required and ______ number of ports are required for each device.
 A 40,9 B 45,10 C 45,9 D 50,10
Question 174 Explanation:
Total Number of cables in fully-Connected Mesh network for N computer = n *(n-1) / 2
Number of ports are required = n-1
∴Total Number of cables = 10(10 – 1) / 2 = 5 * 9 = 45 cables.
∴ Number of ports are required for each device=10-1=9
 Question 175
In TCP/IP Reference model, the job of _______ layer is to permit hosts to inject packets into any network and travel them independently to the destination.
 A Physical B Transport C Application D Host-to-network E None
Question 175 Explanation:
Internet Layer

→ The job of the internet layer is to permit hosts to inject packets into any network and have them travel independently to the destination.

→ It provides only connectionless services between hosts.

→ It defines an official packet format & protocol called Internet Protocol.

→ The job of the internet layer is to deliver IP packets where they are supposed to go.

→ Packet routing is a major issue another is Avoiding Congestion

→ TCP/IP internet layer is very similar in functionality to the OSI network layer.

→ TCP/IP supports the internetworking protocol (IP). IP in turn uses 4 supporting protocols: ARP, RARP, ICMP, IGMP.

→ Internet layer ensure that every packet originated at the host will reach the destination.
 Question 176
If there are N people in the world and are using secret key encryption/decryption for privacy purpose, then number of secret keys required will be:
 A N B (N - 1) C N(N - 1) / 2 D N(N + 1) / 2
Question 176 Explanation:
→Each pair of users must have a secret key. If the number of people wants to use this method in the world is N, then there are N(N-1)/2 secret keys.

→The distribution of keys among different parties can be very difficult. This problem can be resolved by combining the Secret Key Encryption/Decryption with the Public Key Encryption/Decryption algorithm.
 Question 177
Optical fiber uses reflection to guide light through a channel, in which angle of incidence is ________ the critical angle.
 A Equal to B Less than C Greater than D Less than or equal to
Question 177 Explanation:
• Optical fiber works on the principle of Total internal reflection
• Total internal reflection is a phenomenon which occurs when a propagating wave strikes a medium boundary at an angle larger than a particular critical angle with respect to the normal to the surface.
• Optical fiber uses reflection to guide light through a channel, in which angle of incidence is must be greater than critical angle
• if angle is equal or less then critical angle it will bend along or inside the surface respectively
Refer : https://en.wikipedia.org/wiki/Total_internal_reflection
 Question 178
An attacker sits between customer and Banker, and captures the information from the customer and retransmits to the banker by altering the information. This attack is called as ______.
 A Masquerade Attack B Replay Attack C Passive Attack D Denial of Service Attack
Question 178 Explanation:
Replay attack: An attacker spies the communication  between sender and receiver and retransmits the information later. Perfect example is question itself. An attacker sits between customer and banker  and captures the information from the customer and retransmits  to the banker by altering the information .

Masquerade attack  : A Masquerade attack is an attack that uses a fake identity, such as a network identity, to gain unauthorized access to personal computer information through legitimate access identification. If an authorization process is not fully protected, it can become extremely vulnerable to a masquerade attack.

Passive Attack : Attacker monitors the target system for its vulnerabilities. (ex:open ports).  The purpose is solely to gain information about the target and no data is changed on the target.
Example : Stealing Neighbour's Wifi if it is not password protected.

Denial of service attack
• An Attacker sits between the sender and receiver and captures the information and re-transmits to the receiver after some time without altering the information. This attack is called as Denial of service attack.
• A Denial-of-Service (DoS) attack is an attack meant to shut down a machine or network, making it inaccessible to its intended users. DoS attacks accomplish this by flooding the target with traffic, or sending it information that triggers a crash.
 Question 179
Which of the following fields in IPv4 datagram is not related to fragmentation?
 A Type of service B Fragment offset C Flags D Identification
Question 179 Explanation:
• Type of service identifies the type of packets. It is not related to fragmentation but is used to request specific treatment such as high throughput, high reliability or low latency for the IP packet depending upon the type of service it belongs to.
• The Fragment Offset field specifies where the fragment fits in the original datagram. The offset of the first fragment will always be 0. The size of the field (13 bits) is 3-bits shorter than the size of the total length field (16 bits).
• Identification - uniquely identifies the datagram. usually incremented by 1 each time a datagram is sent.
• Flags and Fragmentation Offset - used for fragmentation
 Question 180
Which of the following is not a congestion policy at network layer?
 A Flow Control Policy B Packet Discard Policy C Packet Lifetime Management Policy D Routing Algorithm
Question 180 Explanation:
Below table contains different policies at various layers can affect the congestion
 Question 181
Which of the following protocols is an application layer protocol that establishes, manages and terminates multimedia sessions ?
 A Session Maintenance Protocol B Real time Streaming Protocol C Real time Transport Control Protocol D Session Initiation Protocol
Question 181 Explanation:
Session Maintenance Protocol coordinates the initiation and termination of communication sessions.

Real – time Streaming Protocol designed for use in entertainment and communications systems to control streaming media servers.

Real – time Transport Control Protocol works with Real-Time Protocol (RTP) to monitor data delivery on large multicast networks.

Session Initiation Protocol
• SIP is a signaling protocol in which its function includes initiating, maintaining and terminating real time sessions. SIP is used for signaling and controlling multimedia sessions.
• SIP  is an application-layer control protocol that can establish, modify, and terminate multimedia sessions such as Internet telephony calls.
 Question 182
Match the following port numbers with their uses:
 A (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii) B (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii) C (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i) D (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
Question 182 Explanation:
 Question 183
Which of the following is not associated with the session layer ?
 A Dialog control B Token management C Semantics of the information transmitted D Synchronization
Question 183 Explanation:
• Dialog control, Session checkpoint and recovery, Token management and Synchronization is associated with session layer.
• Semantics of the information transmitted is  associated with Presentation layer.
 Question 184
What is the size of the ‘total length’ field in IPv4 datagram?
 A 4 bit B 8 bit C 16 bit D 32 bit
Question 184 Explanation:
 Question 185
Which of the following is/are restriction(s) in classless addressing?
 A The number of addresses needs to be a power of 2. B The mask needs to be included in the address to define the block. C The starting address must be divisible by the number of addresses in the block. D All of the above
Question 185 Explanation:
• Addresses in a block must be contiguous, one after the other
• Number of addresses in a block must be a power of 2
• The Mask needs to be included in the address to define the block
• The Starting address must be divisible by the number of addresses in the block.
 Question 186
Error control is needed at the transport layer because of potential error occuring___.
 A From transmission line noise B In router C From out of sequence delivery D From packet losses
Question 186 Explanation:
The error control in the transport layer usually refers to the guaranteed delivery mechanism with TCP, which attempts to safe guard against frames/packets getting lost entirely.
 Question 187
Making sure that all the data packets of a message are delivered to the destination is _________ control.
 A Error B Loss C Sequence D Duplication
Question 187 Explanation:
Error control in data link layer is the process of detecting and correcting data frames that have been corrupted or lost during transmission.

In case of frames are corrupted or lost the receiver does not receive the correct data-frame and sender is ignorant about the loss. Data link layer follows a technique to detect transit errors and take necessary actions, which is retransmission of frames whenever error is detected or frame is lost. The process is called Automatic Repeat Request (ARQ).
 Question 188
Which transport class should be used with a perfect network layer ?
 A TP0 and TP2 B TP1 and TP3 C TP0, TP1, TP3 D TP0, TP1, TP2, TP3, TP4
Question 188 Explanation:
The OSI model defines 5 transport layer classes: TP0, TP1, TP2, TP3, TP4.
• TP0 and TP2 are used with perfect network layers.
• TP1 and TP3 are used with residual-error network layers.
• TP4 is used with unreliable network layers.
TPO  : Simple class
TP     : Basic error recovery class
TP2   : Multiplexing class
TP3   : Error recovery and Multiplexing class
TP4   : Error detection and recovery class
 Question 189
Which transport class should be used with residual-error network layer ?
 A TP0, TP2 B TP1, TP3 C TP1, TP3, TP4 D TP0, TP1, TP2, TP3, TP4
Question 189 Explanation:
 Question 190
Virtual circuit is associated with a __________ service.
 A Connectionless B Error-free C Segmentation D Connection-oriented
Question 190 Explanation:
• Virtual Circuits are connection-oriented service  means there is a reservation of resources like buffers, CPU , ,  ,  , in advance
• Datagram networks are connection-less services means there is no need for resources reservation in advance
 Question 191
The data unit in the TCP/IP application layer is called a __________ .
 A Message B Segment C Datagram D Frame
Question 191 Explanation:
PDU for
Application layer = message
Transport Layer = segment
Network Layer = datagram/packet
Physical Layer = Bits
 Question 192
Which of following is an example of a client - server model :
 A DNS B FTP C TELNET D All the above
Question 192 Explanation:
In Client-Server Model Servers provide the services to the clients to perform a user based tasks.
Example of Client-Server Model are DNS,FTP,TELNET
 Question 193
Encryption and decryption are the functions of the __________ layer of OSI model :
 A Transport B Session C Router D Presentation
Question 193 Explanation:
Presentation layer is responsible for
Encryption and Decryption,
compression and decompression,
translation
Authentication

 Question 194
Which of the following are Data link layer standard ?
(1) Ethernet
(2) HSSI
(3) Frame Relay
(4) 10base T
(5) Token ring
 A 1, 2 B 1, 3, 5 C 1, 3, 4, 5 D 1, 2, 3, 4, 5
Question 194 Explanation:
 Question 195
Which type of Bridge would be used to connect an Ethernet Segment with a token ring Segment ?
 A Transparent Bridge B Source-Route Bridge C Translation Bridge D None of these
Question 195 Explanation:
• Transparent bridge used to interconnect two LANs running the same type of protocol
• Translation Bridges are also used to interconnect 2 LANs that are operating 2 different networking protocols. For example : LAN A could be an Ethernet LAN and LAN B could be a token ring.
• Translation bridge allows data from one LAN to be transferred to another
Transparent Bridge
Interconnects two LANs running the same type of protocol
Translation Bridge
Used to interconnect two LANs that are operating two different networking protocols
 Question 196
Which protocol is used to encapsulate a data pocket created of a higher OSI model layer ?
 A HDLC B SDLC C LAPB D LAPD
Question 196 Explanation:
LAP-D is data link layer protocol across an ISDN channel to encapsulate data into frames
LAP-D : Link Access Procedure D channel is the second layer protocol on the ISDN(Integrated Services Digital Network) protocol stack. Its development was heavily based on High-Level Data Link Control. Data transmissions take place on B channels. LAPD is the ITU Q.921 protocol
• HDLC :  High-Level Data Link Control
• SDLC : Synchronous Data Link Control developed by IBM in 1970 as a replacement for its binary synchronous (BSC) protocol.
• LAP-B : Link Access Protocol – Balanced
• LAP-D : Link Access Procedure D channel
 Question 197
What is the correct subnet mask to use for a class-B address to support 30 Networks and also have the most hosts possible ?
 A 255​ .​ 255​ .​ 255​ .​ 0 B 255​ .​ 255​ .​ 192​ .​ 0 C 255​ .​ 255​ .​ 240​ .​ 0 D 255​ .​ 255​ .​ 248​ .​ 0
Question 197 Explanation:
Class B(N.N.H.H)
NID part is 16 bits
HID Part is 16 bits

In the given questions they are asking about 30 networks so in order to represent 30 networks minimum we require (log2 30 = 5 bits )

25= 32 networks can be supported at most
Now NID Part became 21 bits and HID Part became 11 bits
Now Subnet Masking 255.255.11111000 = 245.255.248.0
 Question 198
Another name of IEEE 802.11a is :
 A WECA B Fast Ethernet C Wi-Fi 5 D 802​ .​ 11g
Question 198 Explanation:
IEEE 802.11a was one of the first Wi-Fi standards to be launched
IEEE 802.11a provided the capability for raw data speeds of up to 54 Mbps at 5 GHz.

 Question 199
The network 198:78:41:0 is a :
 A Class A Network B Class B Network C Class C Network D Class D Network
Question 199 Explanation:
• CLASS  A  ranges from  :  0 to 127
• CLASS  B  ranges from  :  128 to 191
• CLASS  C  ranges from  :  192 to 223
• CLASS  D  ranges from  :  224 to 239
• CLASS  E  ranges from  :  240 to 255
So it is from Class  C network
 Question 200
 A Extends the network portion to 16 bits B Extends the network portion to 26 bits C Extends the network portion to 36 bits D Has no effect on the network portion of an IP address
Question 200 Explanation:
In binary representation = 11111111.11111111.11111111.1100000

Number of bits for the NID Part(Network portion) = 8 + 8 + 8 + 2 = 26 bits
Number of bits for the HID Part(Host's portion) = 32 – 26 = 6 bits
Number of addresses available = 26 = 64
 Question 201
What is the difference between the Ethernet frame preamble field and the IEEE 802.3 preamble and start of frame Delimiter fields ?
 A 1 byte B 1 bit C 4 bits D 16 bits
Question 201 Explanation:

Note :
SFD/SOF stands for Start of Frame Delimiter
FCS stands for Frame check sequence
Figure 1 is for Ethernet
Figure 2 is for Ethernet 802.3

Figure 1 = Ethernet frame preamble field = 7 byte
Figure 2 =IEEE 802.3 Preamble and SFD = 7 + 1 = 8 byte
Difference is 8 – 7 = 1 byte.
 Question 202
What is the function of a translating bridge ?
 A Connect similar remote LANs B Connect similar local LANs C Connect different types of LANs D Translate the network addresses into a layer 2 address
Question 202 Explanation:
 Question 203
The program used to determine the round - trip delay between a workstation and a destination address is :
 A Tracert B Traceroute C Ping D Pop
Question 203 Explanation:
Ping
Ping (also written as PING or ping) is a utility that you use to determine whether or not a specific IP address is accessible.

Traceroute
Traceroute is a utility that traces a packet from your computer to an Internet host, but it will show you how many hops the packet requires to reach the host and how long each hop takes. If you're visiting a Web site and pages are appearing slowly, you can use traceroute to figure out where the longest delays are occurring.

Tracert
Traceroute, also called tracert, is a utility that uses ICMP packets to record the route through the internet from one computer to another. It calculates the time taken for each hop as the packet is routed to the destination.

POP is an application layer protocol in the OSI model that provides end users the ability to extracts and retrieves email from a remote mail server for access by the host machine..
 Question 204
Which of the following algorithms is not a broadcast routing algorithm ?
 A Flooding B Multi Destination routing C Reverse path forwarding D All of the above
Question 204 Explanation:
– Sending distinct packets
– Flooding
– Multi-destination routing
– Using spanning tree
– Reverse path forwarding

In some applications, hosts need to send messages to many or all other hosts.
• Sending a packet to all destinations simultaneously is called broadcasting

Flooding is used by a switch at layer-2 to send unknown unicast frames to all other interfaces. If a frame is not destined for a host which receives it, the host will ignore it and not be interrupted. This, too, is limited to a broadcast domain.

Multi-destination Routing
• Each packet contains either a list of destinations or a bit map indicating the desired destinations. When a packet arrives at a router, the router checks all the destinations to determine the set of output lines that will be needed.
• The router generates a new copy of the packet for each output line to be used and includes in each packet only those destinations that are to use the line.

 Question 205
An analog signal has a bit rate of 6000 bps and a baud rate of 2000 baud. How many data elements are carried by each signal element ?
 A 0.336 bits/baud B 3 bits/baud C 120,00,000 bits/baud D None of the above
Question 205 Explanation:
Given An analog signal has a
Bit rate = 6000 bps
Baud rate = 2000 baud
Bits per baud = (bit rate / baud rate)
Bits per baud = 6000 / 2000
Bits per baud = 3 bits/ baud is transmitted
 Question 206
How many distinct stages are there in DES algorithm, which is parameterized by a 56-bit key ?
 A 16 B 17 C 18 D 19
Question 206 Explanation:
DES stands for Data Encryption Standard Algorithm which is parameterized by a 56 bit key has 19 distinct stages including 16 rounds and plain text (here, password) is encrypted in block of 64 bits.
 Question 207
Match the following :
 LIST-I LIST-II a. Call control protocol i. Interface between Base Transceiver Station (BTS) and Base Station Controller (BSC) b. A-bis ii. Spread spectrum c. BSMAP iii. Connection management d. CDMA iv. Works between Mobile Switching Centre (MSC) and Base Station Subsystem (BSS)
 A A-iii, b-iv, c-i, d-ii B A-iii, b-i, c-iv, d-ii C A-i, b-ii, c-iii, d-iv D A-iv, b-iii, c-ii, d-i
Question 207 Explanation:
Call control protocol do Connection management.
Call control
process used in telecommunications networks to monitor and maintain connections once they have been established

A-bis is an Interface between Base Transceiver Station (BTS) and Base Station Controller (BSC)

BSMAP stands for Base Station Management Application Part.
BSMAP supports all Radio Resource Management and Facility Management procedures between the MSC and the BSS or to a cell(s) within the BSS

CDMA stands for Code-Division Multiple Access
CDMA is a digital cellular technology that uses Spread Spectrum Technique
 Question 208
​The third generation mobile phone are digital and based on
 A AMPS B Broadband CDMA C CDMA D D-AMPS
Question 208 Explanation:
 Question 209
Consider ISO-OSI network architecture reference model. Session layer of this model offer Dialog control, token management and ____________ as services.
 A Synchronization B Asynchronization C Errors D Flow control
Question 209 Explanation:
 Question 210
​An internet service provider (ISP) has following chunk of CIDR-based IP addresses available with it: 245.248.128.0/20 . The ISP want to give half of this chunk of addresses to organization A and a quarter to Organization B while retaining the remaining with itself. Which of the following is a valid allocation of addresses to A and B?
 A 245.248.132.0/22 and 245.248.132.0/21 B 245.248.136.0/21 and 245.248.128.0/22 C 245.248.136.0/24 and 245.248.132.0/21 D 245.248.128.0/21 and 245.248.128.0/22
Question 210 Explanation:
 Question 211
 A Physical B Network C Datalink D Application
Question 211 Explanation:
A bridge operates at the data link layer and uses MAC address giving it access to the physical address of all stations connected to it in the same network
Note :
 Question 212
The minimum frame length for 10 Mbps Ethernet is _______ bytes and maximum is _______ bytes.
 A 64 & 128 B 128 & 1518 C 1518 & 3036 D 64 & 1518
Question 212 Explanation:
The original Ethernet IEEE 802.3 standard defined the Minimum frame size 64 bytes and the maximum frame size as 1518 bytes.
 Question 213
​Which of the following statement/s is/are true?
(i) Firewall can screen traffic going into or out of an organization.
(ii) Virtual private networks cam simulate an old leased network to provide certain desirable properties.
Choose the correct answer from the code given below:
 A (i) only B Neither (i) nor(ii) C Both (i) and (ii) D (ii) only
Question 213 Explanation:
 Question 214
​Identify the correct sequence in which the following packets are transmitted on the network by a host when a browser requests a web page from a remote server, assuming that the host has been restarted.
 A HTTP GET request, DNS query, TCP SYN B DNS query, TCP SYN, HTTP GET request C TCP SYN, DNS query, HTTP GET request D DNS query, HTTP Get request, TCP SYN
Question 214 Explanation:
When a web browser requests any web page from a remote server using URL let say academyera.com then DNS converts domain name into IP address (113.20.0.123) then a TCP connection will be established using Handshaking procedure by sending out TCP SYN and finally HTTP Get Request will be sent to access the webpage
 Question 215
Consider the following two statements:
S1: TCP handles both congestion and flow control
S2: UDP handles congestion but not flow control
Which of the following option is correct with respect to the above statements (S1) and (S2)?
 A Both S1 and S2 are correct B Neither S1 nor S2 is correct C S1 is not correct but S2 is correct D S1 is correct but S2 is not correct
Question 215 Explanation:
 Question 216
Match the following secret key algorithm (List 1) with the corresponding key lengths (List 2) and choose the correct answer from the code given below,
 A (a)-(ii),(b)-(iii), (c)- (iv), (d)-(i) B (a)-(iv),(b)-(iii), (c)- (ii), (d)-(i) C (a)-(iii),(b)-(iv), (c)- (ii), (d)-(i) D (a)-(iii),(b)-(iv), (c)- (i), (d)-(ii)
Question 216 Explanation:
 Question 217
The bit rate of a signal is 3000 bps. If each signal unit carries 6 bits, the baud rate of the signal is _______.
 A 500 baud/sec B 1000 baud/sec C 3000 baud/sec D 18000 baud/sec.
Question 217 Explanation:
•Bit rate is the number of bits per second.
•Baud rate is the number of signal units per second.
•Baud rate is less than or equal to the bit rate.
• Bit rate is important in computer efficiency
• Baud rate is important in data transmission.
• Baud rate determines the bandwidth required to send signal
• Baud rate = Bit rate / Number of bits per signal unit

Given
Bit rate of a signal is 3000 bps
Number of bits per signal unit = 6 bits
Baud rate = Bit rate / Number of bits per signal unit
Baud rate = 3000/6 =500 bauds/sec
 Question 218
Match the following :
List – I    List – II
a. Physical layer   i. Allow resources to network access
b. Datalink layer   ii. Move packets from one destination to other
c. Network layer iii. Process to process message delivery
d. Transport layer iv. Transmission of bit stream
e. Application Layer v. Formation of frames
 A A-iv, b-v, c-ii, d-iii, e-i B A-v, b-iv, c-i, d-ii, e-iii C A-i, b-iii, c-ii, d-v, e-iv D A-i, b-ii, c-iv, d-iii, e-v
Question 218 Explanation:
a. Physical layer       -       iv. Transmission of bit stream
b. Datalink layer       -       v. Formation of frames
c. Network layer        -       ii. Move packets from one destination to other
d. Transport layer     -      iii. Process to process message delivery
e. Application Layer   -    i. Allow resources to network access
 Question 219
The four byte IP Address consists of
Question 219 Explanation:
 Question 220
SET, an open encryption and security specification model that is designed forprotecting credit card transactions on the internet, stands for
 A Secure Electronic Transaction B Secular Enterprise for Transaction C Security Electronic Transmission D Secured Electronic Termination
Question 220 Explanation:
• SET stands for Secure Electronic Transaction
• SET is an open encryption and security specification designed to protect credit card transactions on the Internet
• SET supported by major corporations such as VISA Inc. and MasterCard. SET has been designed to operate in a wired and wireless infrastructure
• SET protocol is an evolution of the existing credit-card based payment system and provides enhanced security for information transfer as well as authentication of transaction participant identities by registration and certification.
• SET permits customers to make credit-card payment to any merchant offering web-based services, customers also have the option of paying for other types of services using the on-line banking facilities.
 Question 221
Which of the following statement/s is/are true ?
(i) windows XP supports both peer-peer and client-server networks.
(ii) Windows XP implements Transport protocols as drivers that can be loaded and uploaded from the system dynamically.
 A Both (i) and (ii) B Neither (i) nor (ii) C (ii) only D (i) only
Question 221 Explanation:
 Question 222
The host is connected to a department network which is a part of a university network. The university network, in turn, is part of the internet. The largest network, in which the Ethernet address of the host is unique, is
 A The university network B The internet C The subnet to which the host belongs D The department network
Question 222 Explanation:
 Question 223
​Suppose that everyone in a group of N people wants to communicate secretly with (N-1) other people using symmetric key cryptographic system. The communication between any two persons should not be decodable by the others in the group. The number of keys required in the system as a whole to satisfy the confidentiality requirement is
 A 2N B N(N-1) C N(N-1)/2 D (N-1)​ 2
Question 223 Explanation:
It implies Every person in a group needs to communicate with remaining (N-1) other people using different keys
i.e , Total number of keys required = 1+2+3+4 ..….N-2+N-1 = N(N-1)/2
This is similar to number of edges needed in a complete graph with N vertices is N(N-1)/2.
 Question 224
In ______ CSMA protocol, after the station finds the line idle, it sends or refrains from sending based on the outcome of a random number generator.
 A Non-persistent B 0-persistent C 1-persistent D P-persistent
Question 224 Explanation:
step 1 : In p-Persistent CSMA Protocol when station found channel is ideal then it send or refrain from sending based on probability p, and delay for one time unit with probability (1 - p)

step 2 : In p-Persistent CSMA Protocol when station found channel is busy then it continue to listen until medium becomes idle, then go to Step 1

step 3 : If transmission is delayed by one time unit, continue with Step 1
 Question 225
Which of the following substitution technique have the relationship between a character in the plain text and a character in the ciphertext as one-to-many ?
 A Monoalphabetic B Polyalphabetic C Transpositional D None of the above
Question 225 Explanation:
• Polyalphabetic is  one-to-many
• Monoalphabetic is  one to one
• Transpositional is  one to one
Monoalphabetic cipher is one where each symbol in plain text is mapped to a fixed symbol in cipher text.
Relationship between a character in the plain text and the characters in the cipher text is one-to-one
Each alphabetic character of plain text is mapped onto a unique alphabetic character of a cipher text.

Polyalphabetic cipher is any cipher based on substitution, using multiple substitution alphabets.
Relationship between a character in the plain text and the characters in the cipher text is one-to-many.
Each alphabetic character of plain text can be mapped onto ‘m’ alphabetic characters of a cipher text.

In Transposition Technique identity of the characters remains unchanged but their positions are changed to create the ciphertext.
 Question 226
What is the maximum length of CAT-5 UTP cable in Fast Ethernet network ?
 A 100 meters B 200 meters C 1000 meters D 2000 meters
Question 226 Explanation:
 Question 227
The count-to-infinity problem is associated with
 A Flooding algorithm B Hierarchical routing algorithm C Distance vector routing algorithm D Link state routing algorithm
Question 227 Explanation:
 Question 228
A slotted ALOHA network transmits 200-bit frames using a shared channel with a 200 Kbps bandwidth. Find the throughput of the system, if the system (all stations put together) produces 250 frames per second :
 A 49 B 368 C 149 D 151
Question 228 Explanation:
 Question 229
The dotted-decimal notation of the following IPV4 address in binary notation is . 10000001 00001011 00001011 11101111
 A 111.56.45.239 B 129.11.10.238 C 129.11.11.239 D 111.56.11.239
Question 229 Explanation:
 Question 230
Match the following symmetric block ciphers with corresponding block and key sizes :
```  List-I                             List-II
(a) DES          (i) block size 64 and key size ranges between 32 and 448
(b) IDEA        (ii) block size 64 and key size 64
(c) BLOWFISH   (iii) block size 128 and key sizes 128, 192, 256
(d) AES         (iv) block size 64 and key size 128```
 A (a)-(iv), (b)-(ii), (c)-(i), (d)-(iii) B (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii) C (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i) D (a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)
Question 230 Explanation:
 Question 231
Which of the following statements are true ?

(a) Three broad categories of Networks are
(i) Circuit Switched Networks
(ii) Packet Switched Networks
(iii) Message Switched Networks
(b) Circuit Switched Network resources need not be reserved during the set up phase.
(c) In packet switching there is no resource allocation for packets.

Code :
 A (a) and (b) only B (b) and (c) only C (a) and (c) only D (a), (b) and C
Question 231 Explanation:
 Question 232
​In Challenge-Response authentication the claimant
 A Proves that she knows the secret without revealing it B Proves that she doesn’t know the secret C Reveals the secret D Gives a challenge
Question 232 Explanation:
Challenge-Response Authentication is a computer security protocol in which one entity sending a challenge to another entity. The second entity must respond with the appropriate answer to be authenticated.

An simple Example of Challenge-Response Authentication is password authentication. The challenge is from a server asking the client for a password to authenticate the client's identity so that the client can be served. i.e. client knows the secret password without revealing it or will provide the secret password when it is asked.
 Question 233
Decrypt the message “WTAAD” using the Caesar Cipher with key=15.
 A LIPPS B HELLO C OLLEH D DAATW
Question 233 Explanation:
 Question 234
To guarantee correction of upto t errors, the minimum Hamming distance dmin in a block code must be .
 A T+1 B T−2 C 2t−1 D 2t+1
Question 234 Explanation:
• To guarantee correction of up to t-bit errors in all cases, the minimum Hamming distance in a block code must be dmin = 2t + 1 bit is needed.
• For Error detecting of t-bit errors in all cases, the minimum Hamming distance is t + 1 bit is needed.
 Question 235
Encrypt the Message “HELLO MY DEARZ” using Transposition Cipher with
 A HLLEO YM AEDRZ B EHOLL ZYM RAED C ELHL MDOY AZER D ELHL DOMY ZAER
Question 235 Explanation:
 Question 236
The Mobile Application Protocol (MAP) typically runs on top of which protocol ?
 A SNMP (Simple Network Management Protocol) B SMTP (Simple Mail Transfer Protocol) C SS7 (Signalling System 7) D HTTP (Hypertext Transfer Protocol)
Question 236 Explanation:
MAP stands for Mobile Application Protocol
MAP is an SS7 protocol that provides an application layer for the various nodes in GSM and UMTS mobile core networks and GPRS core networks to communicate with each other in order to provide services to users.

MAP is used to access the Home Location Register, Visitor Location Register, Mobile Switching Center, Equipment Identity Register, Authentication Centre, Short message service center and Serving GPRS Support Node (SGSN).

Source : Wifi
 Question 237
If a packet arrive with an M-bit value is ‘1’ and a fragmentation offset value ‘0’, then it is ______ fragment.
 A First B Middle C Last D All of the above
Question 237 Explanation:
• M bit = 0 it means that there are no more fragments and it is the last fragment
• M bit = 1 it means that this packet is NOT the last packet among all fragments
• offset = 0 Identify the 1st fragment
Coming to the questions
Because of M bit = 1 it is either the first fragment or a middle one.
Because of offset = 0 it is the first fragment.
 Question 238
Coaxial cables are categorized by Radio Government rating are adapted for specialized functions. Category RG-59 with impedance 75Ω used for
 A Cable TV B Ethernet C Thin Ethernet D Thick Ethernet
Question 238 Explanation:
RG-59/U is a specific type of coaxial cable, often used for low-power video and RF signal connections. The cable has a characteristic impedance of 75 ohms, and a capacitance of around 20pF/ft (60pF/m).The 75 ohm impedance matches a dipole antenna in free space.

RG-59 is often used at baseband video frequencies, such as composite video. It may also be used for broadcast frequencies

RG-59 coaxial cable is commonly packed with consumer equipment, such as VCRs or digital cable/satellite receivers.

Source : Wiki
 Question 239
AES is a round cipher based on the Rijndael Algorithm that uses a 128-bit block of data. AES has three different configurations. ______ rounds with a key size of 128 bits, ______ rounds with a key size of 192 bits and ______ rounds with a key size of 256 bits.
 A 5, 7, 15 B 10, 12, 14 C 5, 6, 7 D 20, 12, 14
Question 239 Explanation:
AES allows for three different key lengths: 128, 192, or 256 bits. Encryption consists of 10 rounds of processing for 128-bit keys, 12 rounds for 192-bit keys, and 14 rounds for 256-bit keys.
 Question 240
Data Encryption Techniques are particularly used for ______.
 A Protecting data in Data Communication System. B Reduce Storage Space Requirement. C Enhances Data Integrity. D Decreases Data Integrity.
Question 240 Explanation:
Data Encryption Techniques are particularly used for protecting data in Data Communication System

while sending data from sender to receivers in between hacker can steal your data using various hacking techniques so if you encrypt  the data before sending  to the receiver even if hacker steal the data they won't understand anything because data is already encrypted. So, at receiver side they can decrypt the data by using key
 Question 241
When data and acknowledgement are sent in the same frame, this is called as
 A Piggy packing B Piggy backing C Backpacking D Good packing
Question 241 Explanation:
• Piggy backing is a technique that improves the efficiency of the bidirectional protocols.
• In this technique data frame and ACK are combined.
• The Major advantage of piggybacking is better use of available channel bandwidth.
• When a frame is carrying data from A to B, it can also carry control information about arrived (or lost) frames from B
• When a frame is carrying data from B to A, it can also carry control information about the arrived (or lost) frames from A.
 Question 242
Encryption and Decryption is the responsibility of _______ Layer.
 A Physical B Network C Application D Datalink
Question 242 Explanation:
Presentation layer is responsible for
• Encryption and Decryption
• Compression and Decompression
• Translation
• Authentication
Encryption and decryption is the responsibility of Application Layer( in TCP/IP model Application Layer + Presentation Layer + Session Layer)
 Question 243
An analog signal carries 4 bits in each signal unit. If 1000 signal units are sent per second, then baud rate and bit rate of the signal are and
 A 4000 bauds / sec & 1000 bps B 2000 bauds / sec & 1000 bps C 1000 bauds / sec & 500 bps D 1000 bauds / sec & 4000 bps
Question 243 Explanation:
 Question 244
Using the RSA public key cryptosystem, if p = 13, q = 31 and d = 7, then the value of ‘e’ is
 A 101 B 103 C 105 D 107
Question 244 Explanation:
Basic RSA Algorithm:
1. Choose two primes, p and q.
2. Compute n=p*q and z=(p-1)*(q-1).
3. Choose a number relatively prime to z and call it d.
4. Find e such that e*d=1 mod z.

Given two distinct prime numbers p = 13 and q = 31 and  decryption key, d = 7
Compute n = p * q = 13 * 31 = 403
Calculate the totient z = (p-1) * (q-1) = 12 * 30 = 360

Let the value of encryption key be ‘e’ such that :
e * d mod z = 1
e * d = 1 mod z
e * 7 = 1 mod z  // d=7
e * 7 = 1 mod 360 // z=360 already computed

then e * 7 must be 361, 721, 1081, 1441, etc. Dividing each of these in turn by 7 to see which is divisible by 7, we find that 721/7 = 103, hence e = 103.

Value of encryption key be  e = 103
 Question 245
Match the following :
 List-I List-II a. Wireless Application Environment i. HTTP b. Wireless Transaction Protocol ii. IP c. Wireless Datagram Protocol iii. Scripts d. Wireless iv. UDP
 A A-ii, b-iv, c-i, d-iii B A-iv, b-iii, c-ii, d-i C A-iv, b-iii, c-i, d-ii D A-iii, b-i, c-iv, d-ii
Question 245 Explanation:
• Wireless application Environment provides an architecture for communication between wireless devices and web servers and full applications can be made using scripts
• Wireless transaction is a standard Protocol used in HTTP
• In Wireless datagram protocol UDP can be used
• Wireless is a internet protocol IP
 Question 246
Networks that use different technologies can be connected by using
 A Packets B Switches C Bridges D Routers
Question 246 Explanation:
• Packets are the units of exchanging information in network. A bridge connects two or more networks, or segments of the same network.
• Switches are like bridges, except that they have multiple ports with the same type of connection (bridges generally have only two ports) and have been described as nothing more than fast bridges
• Bridge is a device used to connect two separate Ethernet networks into one extended Ethernet network.
• Routers forward data packets from one place to another. They forward data depending on the network, not the hardware(MAC)address. Router is an network layer device and networks connected by a router can be two LANs (local area networks) or WANs (wide area networks) or a LAN and its ISP's (Internet service provider's) network.
• Routers can connect networks using different media and architectures. They do not care about the type of data they handle, and they thus perform very little filtering of data, except for broadcasts.
 Question 247
Both hosts and routers are TCP/IP protocol software. However, routers do not use protocol from all layers. The layer for which protocol software is not needed by a router is
 A Layer – 5 (Application) B Layer – 1 (Physical) C Layer – 3 (Internet) D Layer – 2 (Network Interface)
Question 247 Explanation:
• Routers mainly work in 3 layers i.e physical layer and data link layer and network layer it does not work in above layers which include application layer which is Layer 5
• Application layer includes all the higher-level protocols so the Layer for which protocol software is not needed by a router is Layer - 5 is  the Application Layer
 Question 248
Which of the following TCP/IP Internet protocol is diskless machine uses to obtain its IP address from a server ?
 A RAP B RIP C ARP D X.25
Question 248 Explanation:
• RAP stands for Route Access Protocol, RAP is a protocol that utilizes port 38 and is used for distributing routing information at all levels of the Internet. RAP is further defined in RFC 1476.
• RIP used in distance vector routing
• X.25  an ITU-T standard protocol suite for packet switched wide area network (WAN) communication.
ARP
• ARP stands for Address Resolution Protocol
• ARP protocol is used to find the physical address of a device whose internet address (IP address) is known.
RARP
• RARP stands for Reverse Address Resolution Protocol
• RARP protocol helps to find the internet address of a device whose physical address is known.
 Question 249
Decryption and encryption of data are the responsibility of which of the following layer ?
 A Physical layer B Data Link layer C Presentation layer D Session layer
Question 249 Explanation:
 Question 250
In which circuit switching, delivery of data is delayed because data must be stored and retrieved from RAM ?
 A Space division B Time division C Virtual D Packet
Question 250 Explanation:
• Time Division Switching uses Time division multiplexing (TDM) to achieve switching
• Methods used in Time Division Switching
• Time Slot interchanger (TSI)
• TDM bus
TSI (Time Slot Interchanger) in time division switches consists of random access memory (RAM) with several memory locations.
The RAM ills up with incoming data from the time slots in the order received. Hence in time division switches deliver of data is delayed because data must be stored and retrieved from RAM.
 Question 251
In which Routing Method do all the routers have a common database ?
 A Distance vector B Link state C Link vector D Dijkstra method
Question 251 Explanation:
• Link state protocols are all based on the idea of a distributed map of the network.
• All of the routers that run a link-state protocol have the same copy of this network map, which is built up by the routing protocol itself and not imposed on the network from an outside source.
• The network map and all of the information about the routers and links (and the routes) are kept in a link-state database on each router.
 Question 252
The station to hub distance in which it is 2000 metres.
 A 100 Base-Tx B 100 Base-Fx C 100 Base-T4 D 100 Base-T1
Question 252 Explanation:
Following are the 4 standards for 10Mbps Traditional Ethernet technologies
1. 10BaseT - 100 meters
2. 10Base2 - 185 meters
3. 10Base5 - 500 meters
4. 10BaseFL - 2000 meters
Fast Ethernet is referred by 100BaseX standard. 100BaseX has 3 specifications. Here 100 refers to the speed 100Mbps. Base refers to Baseband. The specifications are
1. 100BaseT4
2. 100BaseTX
3. 100BaseFX
100BaseFx uses the 2-strand fiber-optic cable and the station to hub distance is 2000 metros
 Question 253
Mobile IP provides two basic functions.
 A Route discovery and registration B Agent discovery and registration C IP binding and registration D None of the above
Question 253 Explanation:
Mobile IP provides 2 basic functions
1. Agent Discovery
2. Registration

 Question 254
Checksum used along with each packet computes the sum of the data, where data is treated as a sequence of
 A Integer B Character C Real numbers D Bits
Question 254 Explanation:
• In checksum error detection scheme, the data is divided into k segments each of m bits.
• In the sender’s end the segments are added using 1’s complement arithmetic to get the sum. The sum is complemented to get the checksum.
• The checksum segment is sent along with the data segments.
• At the receiver’s end, all received segments are added using 1’s complement arithmetic to get the sum. The sum is complemented.
• If the result is zero, the received data is accepted otherwise discarded.
 Question 255
For the transmission of the signal, Bluetooth wireless technology uses
 A Time division multiplexing B Frequency division multiplexing C Time division duplex D Frequency division duplex
Question 255 Explanation:
• In a piconet one unit acts as a master and the others act as slaves (a master can have up to 7 slaves).
• Bluetooth channels use a Frequency-Hop/Time-Division-Duplex (FH/TDD) scheme in which the time is divided into 625–µsec intervals called slots.
• The master-to-slave transmission starts in even numbered slots, while the slave-to-master transmission starts in odd-numbered slots.
• Masters and slaves are allowed to send 1, 3, or 5–slot packets, which are transmitted in consecutive slots.
• Packets can carry synchronous information (voice link) or asynchronous information (data link) Information can only be exchanged between a master and a slave
 Question 256
What is the routing algorithm used by RIP and IGRP ?
 A OSPF B Link-state C Dynamic D Dijkstra vector E None
Question 256 Explanation:
RIP :
RIP stands for Routing Information Protocol
RIP  in which distance vector routing protocol is used for data/packet transmission.
RIP works on Bellman Ford algorithm.
RIP is a distance vector protocol.

IGRP :
IGRP Stands For Interior Gateway Routing protocol.
IGRP in which uses distance vector protocol (interior) to exchange data within a system.
IGRP also works on Bellman ford Algorithm
IGRP is also a distance vector protocol.

Dijkstra algorithm is used by link state and it should be distance vector not Dijkstra vector
 Question 257
Hub is a term used with
 A A Star Networks B A Ring Networks C A Router D A Bridge
Question 257 Explanation:
• In star topology each device in the network is connected to a central device called hub.
• Unlike Mesh topology, star topology doesn’t allow direct communication between devices, a device must have to communicate through hub.
• If one device wants to send data to other device, it has to first send the data to hub and then the hub transmit that data to the designated device.
 Question 258
Which of the following network access standard disassembler is used for connection station to a packet switched network ?
 A X.3 B X.21 C X.25 D X.75
Question 258 Explanation:
 Question 259
A station in a network in a network forward incoming packets by placing them on its shortest output queue. What routing algorithm is being used ?
 A Hot potato routing B Flooding C Static routing D Delta routing
Question 259 Explanation:
• Hot Potato Routing is also called as Deflection Routing. Hot Potato routing is the routing technique enabling packet routing without storing them in buffers.
• The Hot Potato routing technique continuously transfers data until the packet reaches the destination without the packets having to wait or stored in the buffer.
 Question 260
Identify the incorrect statement :
 A The ATM adaptation layer is not service dependent. B Logical connections in ATM are referred to as virtual channel connections. C ATM is streamlined protocol with minimal error and flow control capabilities D ATM is also known as cell delays.
Question 260 Explanation:
Asynchronous Transfer Mode (ATM) technology and services creates the need for an adaptation layer in order to support information transfer protocols,

which are not based on ATM. This adaptation layer defines how to segment higher-layer packets into cells and the reassembly of these packets. Additionally, it defines how to handle various transmission aspects in the ATM layer.

Examples of services that need adaptations are Gigabit Ethernet, IP, Frame Relay, SONET/SDH, UMTS/Wireless, etc.
 Question 261
The number of bits required for an IPv6 address is
 A 16 B 32 C 64 D 128
Question 261 Explanation:
IPv6 address is 128 bits in length
IPv4 address is 32 bits in length

∴number of bits required for an IPv6 address is 128
∴number of bits required for an IPv4 address is 32
 Question 262
Mesh analysis is applicable to only
 A Non-planar network B Planar network C Circuits containing voltage sources D Circuits containing current sources
Question 262 Explanation:
• mesh means a smallest loop which is closed one and formed by using circuit components. The mesh must not have any other loop inside it.
• Mesh analysis is based on Kirchhoff Voltage Law
• Mesh analysis is applicable only for planar networks. A circuit is said to be planar if it can be drawn on a plane surface without crossovers.
 Question 263
Which of the following switching techniques is most suitable for interactive traffic ?
 A Circuit switching B Message switching C Packet switching D All of the above
Question 263 Explanation:
Packet switching
• Messages broken into smaller units (packets)
• Connectionless, packets routed independently (datagram)
• Packet may arrive out of order
• Pipelining of packets across network can reduce delay, increase throughput
• Lower delay that message switching, suitable for interactive traffic
 Question 264
Which of the following addresses is used to deliver a message to the correct application program running on a host ?
 A Port B IP C Logical D Physical
Question 264 Explanation:
 Question 265
In ________ substitution, a character in the plaintext is always changed to the same character in the ciphertext, regardless of its position in the text.
 A Polyalphabetic B Monoalphabetic C Transpositional D Multi alphabetic
Question 265 Explanation:
Monoalphabetic cipher is one where each symbol in plain text is mapped to a fixed symbol in cipher text.
Relationship between a character in the plain text and the characters in the cipher text is one-to-one
Each alphabetic character of plain text is mapped onto a unique alphabetic character of a cipher text.
For Example : if "Z" is encrypted to "M", then every time we see the letter "Z" in the plain-text, we replace it with the letter "M" in the cipher-text.

 Question 266
 A Class A B Class B C Class C D Class D
Question 266 Explanation:
• CLASS  A  ranges from  :  0 to 127
• CLASS  B  ranges from  :  128 to 191
• CLASS  C  ranges from  :  192 to 223
• CLASS  D  ranges from  :  224 to 239
• CLASS  E  ranges from  :  240 to 255
So it is from Class  B network
 Question 267
In hierarchical routing with 4800 routers, what region and cluster sizes should be chosen to minimize the size of the routing table for a three layer hierarchy ?
 A 10 clusters, 24 regions and 20 routers B 12 clusters, 20 regions and 20 routers C 16 clusters, 12 regions and 25 routers D 15 clusters, 16 regions and 20 routers
Question 267 Explanation:
Assume that there are 'a' clusters, 'b' regions in each region and 'c' routers per region.
Then a*b*c=4800. All the options qualify multiplication

The formula to minimize is (a - 1) + (b - 1) + c.

option A : 10 clusters, 24 regions and 20 routers
=(a - 1) + (b - 1) + c
=9+23+20 =52

option B : 12 clusters, 20 regions and 20 routers
=(a - 1) + (b - 1) + c
= 11+19+20=50

option c : 16 clusters, 12 regions and 25 routers
=(a - 1) + (b - 1) + c
=15+11+25=51

option D : 15 clusters, 16 regions and 20 routers
=(a - 1) + (b - 1) + c
= 14+15+20=49
 Question 268
In IPv4 header, the ______ field is needed to allow the destination host to determine which datagram a newly arrived fragments belongs to.
 A Identification B Fragment offset C Time to live D Header checksum
Question 268 Explanation:

Identification
• Uniquely identifies the datagram.
• Usually incremented by 1 each time a datagram is sent.
• All fragments of a datagram contain the same identification value.
• This allows the destination host to determine which fragment belongs to which datagram.
Fragment Offset
• This offset tells the exact position of the fragment in the original IP Packet.
• Offset of the payload of the current fragment in the original datagram
Time to Live
• Upper limit of routers
• usually set to 32 or 64.
• decremented by each router that processes the datagram,
• router discards the datagram when TTL reaches 0.