Computer-Networks
Question 1 |
Hamming Code | |
CRC | |
VRC | |
None of the above |
→ The design of the CRC polynomial depends on the maximum total length of the block to be protected (data + CRC bits), the desired error protection features, and the type of resources for implementing the CRC, as well as the desired performance.
→ A common misconception is that the "best" CRC polynomials are derived from either irreducible polynomials or irreducible polynomials times the factor 1 + x, which adds to the code the ability to detect all errors affecting an odd number of bits.
→ If we use the generator polynomial g(x)=p(x)(1+x), where p(x) is a primitive polynomial of degree r−1, then the maximal total block length is 2r−1 −1 , and the code is able to detect single, double, triple and an odd number of errors.
Question 2 |
20 | |
40 | |
160 | |
320 |
Round trip delay = 2 * Tp = 80 ms (given)
Optimal window size is = (Tt + 2*Tp) / Tt = 82 / 2 = 41
Option is not given, closest option is 40.
Question 3 |
512 bytes | |
768 bytes | |
1152 bytes | |
1024 bytes |
→ In circuit-switching, the total time required to send data = Setup connection time + Transfer time + Teardown the connection time
→ In datagram-service, the total time required to send data = N*Setup connection time + Transfer time
Question 4 |
WMEKREETSILTWETCOOCYONRU! | |
EETSICOOCYWMEKRONRU!LTWET | |
LTWETONRU!WMEKRCOOCYEETSI | |
ONRU!COOCYLTWETEETSIWMEKR |

Question 5 |
Is desirable property of the cryptographic algorithm | |
Is undesirable property of the cryptographic algorithm | |
Has no effect on the encryption algorithm | |
None of the above |
→ In the case of high-quality block cyphers, such a small change in either the key or the plaintext should cause a drastic change in the ciphertext.
Question 6 |
You can control where traffic goes in the three networks | |
You can do stateful packet filtering | |
You can do load balancing | |
Improve network performance |
1. The first firewall (also called the "front-end" or "perimeter" firewall) must be configured to allow traffic destined to the DMZ only.
2. The second firewall (also called "back-end" or "internal" firewall) only allows traffic from the DMZ to the internal network.
This setup is considered more secure since two devices would need to be compromised. There is even more protection if the two firewalls are provided by two different vendors because it makes it less likely that both devices suffer from the same security vulnerabilities.
Question 7 |
RSA Algorithm | |
Diffie-Hellman Algorithm | |
Electronic Code Book Algorithm | |
None of the above |

Question 8 |
5 kHz | |
10 kHz | |
47 kHz | |
57 kHz |
Highest frequency = Bandwidth+ Lowest Frequency= 5kHz + 52kHz = 57kHz.
Question 9 |
functions as a repeater | |
connects to a digital PBX | |
connects to a token-ring network | |
functions as a gateway |
A repeater is an electronic device that receives a signal and re-transmits it. Repeaters are used to extend transmissions so that the signal can cover longer distances or be received on the other side of an obstruction.
An Ethernet Hub or a Repeater Hub also acts as a repeater and regenerates the signal to avoid its loss.
Question 10 |
Amplitude modulation | |
Carrier modulation | |
Manchester encoding | |
NRZ encoding |
Question 11 |
inserting a 0 in user stream to differentiate it with a flag | |
inserting a 0 in flag stream to avoid ambiguity | |
appending a nipple to the flag sequence | |
appending a nipple to the use data stream |
→ In SDLC the transmitted bit sequence "01111110" containing six adjacent 1 bits is the Flag byte.
→ Bit stuffing ensures that this pattern can never occur in normal data, so it can be used as a marker for the beginning and end of frame without any possibility of being confused with normal data.
→ Bit stuffing is the insertion of non information bits into data. stuffed bits should not be confused with overhead bits. Overhead bits are non-data bits that are necessary for transmission (usually as part of headers, checksums etc).
Question 12 |
1 | |
5 | |
6 | |
11 |
If there are 5 routers, then there should be 5 routing tables as well.
Question 13 |
Push and Pull Model | |
Interface to a storage device | |
Multimedia messaging | |
Hashing |
It is derived from the architecture of the synchronization between HTTP client (browser) and the HTTP server (web server) on the internet. It does the following functions:
SyncML synchronization
WAP push service
MMS service
Question 14 |
Error during transmission | |
File transfer protocol | |
Degrade in TCP performance | |
Interface problem |
Question 15 |
Mutual authentication of client & server | |
Secret communication | |
Data Integrity protection | |
Error detection and correction |
It is utilized to encrypt Web traffic using Hypertext Transfer Protocol (HTTP) and to authenticate Web servers, and to encrypt communications between Web browsers and Web servers etc.
So, other than error detection and correction, all options are correct.
Question 16 |
X.25 | |
X.301 | |
X.409 | |
X.509 |
Question 17 |
his public key | |
his private key | |
receiver’s public key | |
receiver’s private key |
Question 18 |
Reserved for loopback | |
Reserved for broadcast | |
Used for multicast packets | |
Reserved for future addressing |
Question 19 |
Ethernet | |
Bluetooth | |
Broadband Wireless | |
Wireless LANs |
Question 20 |
Port | |
IP | |
Physical | |
Subnet mask |
Physical address i.e. the MAC address is used to recognize a unique host over a network on in a LAN. Port number is required to identify a specific process or application to which a message is to be forwarded when it arrives at a server.
Question 21 |
Create LSAs | |
Flood an internet with information | |
Calculate the routing tables | |
Create a link state database |
Question 22 |
Bus | |
Ring | |
Star | |
None of the above |
FDDI provides a 100 Mbit/s optical standard for data transmission in local area network that can extend in range up to 200 kilometers (120 mi). Although FDDI logical topology is a ring based token network, it did not use the IEEE 802.5 token ring protocol as its basis; instead, its protocol was derived from the IEEE 802.4 token bus timed token protocol.
In addition to covering large geographical areas, FDDI local area networks can support thousands of users. FDDI offers both a Dual-Attached Station (DAS), counter-rotating token ring topology and a Single-Attached Station (SAS), token bus passing ring topology.
Question 23 |
Ubiquitous Teflon port | |
Uniformly terminating port | |
Unshielded twisted pair | |
Unshielded T-connector port |
Question 24 |
255.255.255.0 | |
255.0.0.0 | |
255.255.192.0 | |
255.255.0.0 |
→ Class B IP Addresses range from 128.0.x.x to 191.255.x.x. The default subnet mask for Class B is 255.255.x.x.
→ Class C IP addresses range from 192.0.0.x to 223.255.255.x. The default subnet mask for Class C is 255.255.255.x.
Question 25 |
n(n−1)/ 2, n−1 | |
n, n−1 | |
n−1, n | |
n−1, n(n −1)/2 |
→ In a star topology total number of cable links are (n-1)
Question 26 |
TELNET | |
FTP | |
SNMP | |
SMTP |
FTP→ File transfer protocol
SNMP→ Simple network management protocol
SMTP→ Simple mail transfer protocol Simple Mail Transfer Protocol (SMTP): is an Internet standard for email transmission.
→ Mail servers and other mail transfer agents use SMTP to send and receive mail messages on TCP port 25.
Question 27 |
CDMA | |
CSMA/CA | |
ALOHA | |
None of the above |
CA stands for Collision Avoidance. (In the wireless environment, since a station may not be able to hear all other stations, CD, Collision Detection is not feasible)
Question 28 |
Short frame | |
Small frame | |
Mini frame | |
Runt frame |
Question 29 |

A-2, B-3, C-4, D-1 | |
A-2, B-4, C-3, D-1 | |
A-3, B-4, C-1, D-2 | |
A-3, B-1, C-4, D-2 |
→ Open Shortest Path First (OSPF) is a routing protocol for Internet Protocol (IP) networks. It uses a link state routing (LSR) algorithm and falls into the group of interior gateway protocols (IGPs), operating within a single autonomous system (AS).
→ Border Gateway Protocol (BGP) is a standardized exterior gateway protocol designed to exchange routing and reachability information among autonomous systems (AS) on the Internet.
→ Routing Information Protocol (RIP) is one of the oldest distance vector routing protocols which employ the hop count as a routing metric. RIP implements the split horizon, route poisoning and hold down mechanisms to prevent incorrect routing information from being propagated.
Question 30 |
64 bits | |
128 bits | |
512 bits | |
1024 bits |
→ It can still be used as a checksum to verify data integrity, but only against unintentional corruption. It remains suitable for other non-cryptographic purposes, for example for determining the partition for a particular key in a partitioned database.
Question 31 |
IPsec | |
NetSec | |
PacketSec | |
SSL |
Question 32 |
Browser security | |
FTP security | |
Email security | |
None of the above |
→ It uses Digital Signatures to provide authentication and integrity checking basically to check if the message is actually sent by the person or entity claimed to be the sender.
Question 33 |
wired protected access | |
wi-fi protected access | |
wired process access | |
wi-fi process access |
→ WPA provides data encryption and user authentication to the computing devices connected to a wi-fi network.
→ WPA also includes a Message Integrity Check, which is designed to prevent an attacker from altering and resending data packets. This replaces the cyclic redundancy check (CRC) that was used by the WEP standard.
Question 34 |
Serial Advanced Technology Attachment | |
Serial Advanced Technology Architecture | |
Serial Advanced Technology Adapter | |
Serial Advanced Technology Array |
→ As its name implies, SATA is based on serial signaling technology, unlike Integrated Drive Electronics (IDE) hard drives that use parallel signaling.
Question 35 |
extends the network portion to 16 bits | |
extends the network portion to 26 bits | |
extends the network portion to 36 bits | |
has no effect on the network portion of an IP address |
(192)10 = (11000000)2
Since, 192 is written as 11000000, it has 2 sub-nets and remaining all hosts.
So, for first three octets, 24 bits are fixed and for last octet 2 bits are fixed, i.e. 24 + 2 = 26 bits
Question 36 |
In the LAN header | |
In the application field | |
In the information field of the LAN frame | |
After the TCP header |
Question 37 |
original sender’s physical | |
previous station’s physical | |
next destination’s physical | |
original sender’s service port |
Question 38 |
Radio | |
Optical fibers | |
Coaxial cable | |
Twisted pair |
Question 39 |
can be used to control the flow of information | |
always occurs when the field value is 0 | |
always occurs when the field value is 1 | |
occurs horizontally |
The receiving device should acknowledge each packet it received and after receiving the ACK from the receiving device, the sending device slides the window to right side
Question 40 |
36 MHz | |
360 kHz | |
3.96 MHz | |
396 kHz |
Bandwidth = fh – fi = 4000 - 40 KHz = 3960 KHz = 3.96 MHz
Question 41 |
802.2 | |
802.3 | |
802.5 | |
802.6 |
→ Network devices must acquire the token to transmit data, and may only transmit a single frame before releasing the token to the next station on the ring. When a station has data to transmit, it acquires the token at the earliest opportunity, marks it as busy, and attaches the data and control information to the token to create a data frame, which is then transmitted to the next station on the ring.
→ The frame will be relayed around the ring until it reaches the destination station, which reads the data, marks the frame as having been read, and sends it on around the ring. When the sender receives the acknowledged data frame, it generates a new token, marks it as being available for use, and sends it to the next station.
→ IEEE 802.2 is the logical link control (LLC) as the upper portion of the data link layer of the OSI Model.
→ IEEE 802.6 is a standard governed by the ANSI for Metropolitan Area Networks (MAN).
→ IEEE 802.3 is a working group and a collection of Institute of Electrical and Electronics Engineers (IEEE) standards produced by the working group defining the physical layer and data link layer media access control (MAC) of wired Ethernet.
Question 42 |
109 | |
216 | |
218 | |
219 |
Given 2400 baud value,
Sending maximum of 2400 bits per second with serial port .
Total Data To send = 1 bit(start) + 8 bits(char size) + 2 bits(Stop) = 11 bits.
Number of 8-bit characters that can be transmitted per second = 2400/11 = 218.18
The effective number of characters transmitted = 218 So, the correct option is (C).
Question 43 |
Class A network | |
Class B network | |
Class C network | |
Class D network |
In classful addressing, The range of first octet should be between one of the following:
CLASS A- 0 to 127
CLASS B – 128 to 191
CLASS C – 192 to 223
CLASS D- 224 to 239
CLASS E- 240 to 255
Given IP address = 198.78.41.0
So, it is a class C address.
Question 44 |
allows anyone to decode the transmissions | |
allows only the correct sender to decode the data | |
allows only the correct receiver to decode the data | |
does not encode the data before transmitting it |
A receiver can decode the message using his own private key and the public key of the sender.
Question 45 |
Unshielded T-connector port | |
Unshielded twisted pair | |
Unshielded terminating pair | |
Unshielded transmission process |
There are five types of UTP cables -- identified with the prefix CAT, as in category -- each supporting a different amount of bandwidth.g a different amount of bandwidth.
Question 46 |
Finding the IP address from the DNS | |
Finding the IP address of the default gateway | |
Finding the IP address that corresponds to a MAC address | |
Finding the MAC address that corresponds to an IP address |
Question 47 |
192.166.200.50 | |
00056A:01A01A5CCA7FF60 | |
568, Airport Road | |
01:A5:BB:A7:FF:60 |
MAC addresses are recognizable as six groups of two hexadecimal digits, separated by hyphens, colons, or no separator.
Option (A) is example for IP address
Option (B) and (C ) are not examples for any address representation in computer
01:A5:BB:A7:FF:60 is the MAC address which consists of six groups of two hexadecimal digits, separated by hyphens, colons, or no separator.
Question 48 |
Demonstrating the proper layout for a network | |
Simulating a network | |
To create a virtual private network | |
Segmenting a network inside a switch or device |
2. Higher-end switches allow the functionality and implementation of VLANs.
3. VLANs address issues such as scalability, security, and network management. Network architects set up VLANs to provide network segmentation
Question 49 |
encryption algorithm | |
decryption algorithm | |
key exchange algorithm | |
message digest function |
Question 50 |
Asymmetric key algorithm | |
Symmetric key algorithm | |
Public key algorithm | |
Key exchange |
AES is Symmetric key algorithm
Question 51 |
IP packet with same header
| |
IP packet with new header
| |
IP packet without header
| |
No changes in IP packet |
IPSec Tunnel mode: In IPSec Tunnel mode, the original IP packet (IP header and the Data payload) is encapsulated within another packet.
In IPSec tunnel mode the original IP Datagram from is encapsulated with an AH (provides no confidentiality by encryption) or ESP (provides encryption) header and an additional IP header. The IP addresses of the newly added outer IP header are that of the VPN Gateways.
The traffic between the two VPN Gateways appears to be from the two gateways (in a new IP datagram), with the original IP datagram is encrypted (in case of ESP) inside IPSec packet.
Question 52 |
172.57.88.62 and 172.56.87.23 | |
10.35.28.2 and 10.35.29.4 | |
191.203.31.87 and 191.234.31.88 | |
128.8.129.43 and 128.8.161.55 |
128.8.129.43 (Bitwise AND) 255.255.31.0 = 128.8.1.0
128.8.161.55 (Bitwise AND) 255.255.31.0 = 128.8.1.0
Question 53 |
inserting a ‘0’ in user stream to differentiate it with a flag | |
inserting a ‘0’ in flag stream to avoid ambiguity | |
appending a nibble to the flag sequence | |
appending a nibble to the use data stream |
→ In SDLC the transmitted bit sequence "01111110" containing six adjacent 1 bits is the Flag byte.
→ Bit stuffing ensures that this pattern can never occur in normal data, so it can be used as a marker for the beginning and end of frame without any possibility of being confused with normal data.
Question 54 |
Dynamically discover and maintain routes | |
Distribute routing updates to other routers | |
Reach agreement with other routers about the network topology | |
All of the above |
→ The administrator configure a routing protocol on your network interfaces.
→ Routing protocol learns about other routers automatically.
→ Router and the other routers exchange routes, and each learns about the networks that the other is connected to.
→ When new networks are added or removed, the routers update each other.
Question 55 |
Preamble | |
Postamble | |
Jam | |
None of the above |
→ Assuming you know the collision detection method used by CSMA/CD, the jam signals generated after collision informs the devices that a collision has occurred and invokes a random backoff algorithm which forces the devices on the Ethernet not to send any data until their timer expires. (everyone has a random timer value)
→ Once a transmitting device detects collision which is by examining the data over the Ethernet and identifying that it's not the data it has send, it does receive the jamming signal otherwise collision will keep happening.
→ As the jam signal is generated after collision hence the devices ideally won’t be sending any data during that time.
Question 56 |
DHCP | |
BOOTP | |
RARP | |
ARP |
→ Most residential network routers receive a globally unique IP address within the ISP network.
→ Within a local network, a DHCP server assigns a local IP address to each device connected to the network.
Question 57 |
500 meters of cable | |
200 meters of cable | |
20 meters of cable | |
50 meters of cable |
Given Tp = 200 m / microsec
In, 1 microsec it covers 200m
Therefore, in 0.1 microsec it is 200 * 0.1 = 20 meters
Question 58 |
62 subnets and 262142 hosts | |
64 subnets and 262142 hosts | |
62 subnets and 1022 hosts | |
64 subnets and 1024 hosts |
→ From HID, we took 6-bits for subnetting.
→ Then total subnets possible = ( 26 ) - 2 = 62
→ Total hosts possible for each subnet = (210) - 2 = 1022
Question 59 |
11001001000 | |
11001001011 | |
11001010 | |
110010010011 |
= 1001

∴ Data transmitted is: 11001001011
Question 60 |
Any size | |
216 bytes – size of TCP header | |
216 | |
1500 |
Transport Layer - 65515 byte
Network layer - 65535 byte
Data link layer - 1500 byte
Question 61 |
l = 2 | |
l = 3 | |
l = 4 | |
l = 5 |
→ Maximum number of frames that can be transmit to maximally pack them is
=(Tt+2Tp)/Tx
= (25+1)/1
=26 which is window size
→ Minimum sequence numbers required = 26
→ Minimum number of bits required for sequence number is 5.
Question 62 |
DNS lookup | |
DNS hijacking | |
DNS spoofing | |
None of the mentioned |
→ This results in traffic being diverted to the attacker's computer (or any other computer).
Question 63 |
Differential Manchester encoding
| |
Non return to zero | |
4B/5B encoding
| |
8B/10B encoding |
→ The FC(Fiber Channel) -0 standard defines what encoding scheme is to be used (8b/10b or 64b/66b) in a Fibre Channel system – higher speed variants typically use 64b/66b to optimize bandwidth efficiency.Thus, 8b/10b encoding is used for 4GFC and 8GFC variants;
→ The Fibre Channel 8b/10b coding scheme is also used in other telecommunications systems. Data is expanded using an algorithm that creates one of two possible 10-bit output values for each input 8-bit value.
Question 64 |
480 Mbps
| |
400 Mbps | |
200 Mbps | |
12 Mbps |
Question 65 |
SMTP | |
ARP | |
RIP | |
BOOTP |
→ Reverse Address Resolution Protocol(RARP) is the viceversa case of ARP as it is a protocol by which a physical machine in a local area network can request to learn its IP address from a gateway server.
Question 66 |
P-box | |
S-box | |
Caesar cipher | |
Vigenere cipher |
Question 67 |
IEEE 802.16 | |
IEEE 802.36 | |
IEEE 812.16 | |
IEEE 806.16 |
Question 68 |
physical layer | |
session layer | |
application layer | |
presentation layer |
Question 69 |
96 Gbps | |
192 Mbps | |
512 Gbps | |
192 Gbps |
Each port supports 8 Gbps then all 24 ports have capacity of 24*8 Gbps bandwidth
The aggregate bandwidth is 192 Gbps
Question 70 |
172.16.0.255 | |
172.16.255.255 | |
255.255.255.255 | |
172.255.255.255 |
Convert the above two addresses into binary format and perform the bitwise AND operation

Question 71 |
6.9 e-9 | |
6.9 e-6 | |
69 e-9 | |
4 e-9 |
The amount of time data transferred is = 10 hours = 36000 secs
Total Data transmitted = 2.048 * 106 * 36000 = 2.048 * 36 * 109 bits
The number of Error bits received = 512
Error rate = total number of error bits/ total data transferred per second
= 512 / 73.728 * 109
= 6.944 * 10-9
Question 72 |
Routing the packets | |
Authentication | |
obtaining IP address | |
domain name resolving |
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
Question 73 |
192 | |
240 | |
1920 | |
1966 |
In the given question asynchronous transmission is mentioned and number of bits required is 10 bits. (7 data bits + 1 parity bit + 1 start bit + 1 stop bit )
Bandwidth = 19.2 kbps
Maximum number of characters transmitted in 1 second = (19.2 *1000)/10 = 1920
Question 74 |
RS-232 | |
USB | |
Firewire | |
PCI |
→ It was developed in the late 1980s and early 1990s by Apple, which called it FireWire.
→ The copper cable it uses in its most common implementation can be up to 4.5 metres (15 ft) long. Power is also carried over this cable, allowing devices with moderate power requirements to operate without a separate power supply.
→ FireWire is also available in Cat 5 and optical fiber versions.
→ The 1394 interface is comparable to USB, though USB requires a master controller and has greater market share
Question 75 |
preamble | |
CRC | |
address | |
location |
CRC- A cyclic redundancy check (CRC) is an error-detecting code commonly used in digital networks and storage devices to detect accidental changes to raw data.
Question 76 |
different collision domain and same broadcast domain | |
different collision domain and different broadcast domain | |
same collision domain and same broadcast domain | |
same collision domain and different broadcast domain |
→ Switches for Ethernet are the most common form of network switch.
→ An Ethernet switch operates at the data link layer (layer 2) of the OSI model to create a separate collision domain for each switch port. Each device connected to a switch port can transfer data to any of the other ports at any time and the transmissions will not interfere.
→ Because broadcasts are still being forwarded to all connected devices by the switch, the newly formed network segment continues to be a broadcast domain.
Question 77 |
11010110111011 | |
11010110111101 | |
11010110111110 | |
11010110111001 |
→Blocks of data entering these systems get a short check value attached, based on the remainder of a polynomial division of their contents.
→CRCs can be used for error correction
→Given checksum is x4+ x + 1 so four zeros will append at the end.
Step-1:
Given Data unit = 1101011011, Divisor = 10011,
The input is 11010110110000
Step-2:

Question 78 |
20% | |
25% | |
40% | |
66% |
→The time taken by a sender to send a single packet is equal to transmission time (data) + 2 * propagation delay , out of which only transmission time is useful .
→Then, efficiency =transmission time / transmission time + 2 * propagation delay.
→Given data is transmission time is 20ns
→Propagation delay(time) is 30ns.
→Efficiency =20+(20+2*30)=20/80=0.25 which is equivalent to 25%
Question 79 |
unicast addressing | |
multicast addressing | |
broadcast addressing | |
anycast addressing |
IPv6 does not implement traditional IP broadcast, i.e. the transmission of a packet to all hosts on the attached link using a special broadcast address, and therefore does not define broadcast addresses.
In IPv6, the broadcast addressing t is achieved by sending a packet to the link-local all nodes multicast group at address ff02::1, which is analogous to IPv4 multicasting to address 224.0.0.1.
IPv6 supports the Unicast addressing ,Multicast addressing and Anycast addressing addressing modes:
Question 80 |
Class A, 3 | |
Class A, 8 | |
Class B, 3 | |
Class B, 32 |
The binary number equivalent to 224 is 1110 0000
We can write the subnet mask 255.224.0.0 in binary form as follows
11111111.11100000.00000000.00000000
In the binary representation, the first eight bits represent Class A network address and next three bits( ones) used for represent the number of subnets.
Then the total number of subnets are 23=8.
Question 81 |
solid bucket algorithm | |
spanning tree algorithm | |
hocken helm algorithm | |
leaky bucket algorithm |
The leaky bucket is used to implement traffic policing and traffic shaping in Ethernet and cellular data networks.
The algorithm can also be used to control metered-bandwidth Internet connections to prevent going over the allotted bandwidth for a month, thereby avoiding extra charges.
Question 82 |
deny certain users from accessing a service | |
block worms and viruses from entering the network | |
disallow some files from being accessed through FTP | |
block some hosts from accessing the network |
Packet filtering is a network security mechanism that works by controlling what data can flow to and from a network
A packet filtering firewall can block some hosts from accessing the network.
Question 83 |
segment | |
message | |
datagram | |
frame |
A PDU is composed of protocol specific control information and user data
Protocol data units for the Internet protocol suite are:
The transport layer PDU is the TCP segment for
→TCP, and the datagram for UDP.
→The Internet layer PDU is the packet.
→The link layer PDU is the frame.
Question 84 |
5 | |
6 | |
7 | |
8 |

Question 85 |
Port address translation (PAT) | |
Network address translation (NAT) | |
Address mapping | |
Port mapping |
It is a method of remapping one IP address space into another by modifying network address information in the IP header of packets while they are in transit across a traffic routing device.The technique was originally used as a shortcut to avoid the need to readdress every host when a network was moved. It has become a popular and essential tool in conserving global address space in the face of IPv4 address exhaustion. One Internet-routable IP address of a NAT gateway can be used for an entire private network.
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.
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 86 |
Distance vector | |
Flooding | |
Path vector | |
Link state |
→ It uses a link state routing (LSR) algorithm and falls into the group of interior gateway protocols (IGPs), operating within a single autonomous system (AS).
→ It implements Dijkstra's algorithm, also known as the shortest path first (SPF) algorithm.
Question 87 |
3 | |
7 | |
27 | |
40 |
1. Choose two distinct prime numbers p and q.
2. Compute n = pq.
3. Compute λ(n) = lcm(λ(p), λ(q)) = lcm(p − 1, q − 1), where λ is Carmichael's totient function. Choose an integer e such that 1 < e < λ(n) and gcd(e, λ(n)) = 1; i.e., e and λ(n) are coprime.
4. Determine d as d ≡ e−1 (mod λ(n)); i.e., d is the modular multiplicative inverse of e modulo λ(n). This means: solve for d the equation d⋅e ≡ 1 (mod λ(n)).
Given two prime numbers are p = 5 and q = 11, encryption key, e = 27
n = p * q = 5 * 11 = 55
λ(n)= (p-1) * (q-1) = 4 * 10 = 40
Let the value of decryption key be ‘d’ such that:
e * d mod λ(n) = 1
27 * d mod 40 = 1
d = 3
Question 88 |
194 | |
394 | |
979 | |
1179 |
→ The total length is 200 bytes and the header length is 20 bytes (5×4), which means that there are 180 bytes in this datagram.
→ If the first byte number is 800, the last byte number must 979 (800+180-1).
Question 89 |
250 milliseconds | |
20 milliseconds | |
520 milliseconds | |
270 milliseconds |
The round-trip time or ping time is the time from the start of the transmission from the sending node until a response is received at the same node. It is affected by packet delivery time as well as the data processing delay, which depends on the load on the responding node. If the sent data packet as well as the response packet have the same length, the round trip time can be expressed as:
Round trip time = 2 × Packet delivery time + processing delay
In case of only one physical link, the above expression corresponds to:
Link round trip time = 2 × packet transmission time + 2 × propagation delay + processing delay
Consider there is no effect of propagation delay and processing delay
round trip time ≈ 2 × packet transmission time
packet transmission time=round trip time/2=500/2= 250 ms
Transmission time= Message (bits) / band width (chanel capacity) = 1000 bits/50kbps =20 ms
Total time to receiver to receive the frame =250+20=270ms
Question 90 |
The number of hops this packet can travel is 2. | |
The total number of bytes in header is 16 bytes | |
The upper layer protocol is ICMP | |
The receiver rejects the packet |
First 4 bits represent Version IPV4
And another 4 bits represent header length (/ 4) which should range between 20 to 60 bytes.
here 0010 represents header length , is equal to 2 * 4 = 8
So, receiver will reject the packet.
Question 91 |
Assume the following information:
Original timestamp value = 46
Receive timestamp value = 59
Transmit timestamp value = 60
Timestamp at the arrival of packet = 69
Which of the following statements is correct?
Receive clock should go back by 3 milliseconds | |
Transmit and Receive clocks are synchronized | |
Transmit clock should go back by 3 milliseconds | |
Receive clock should go ahead by 1 milliseconds |
Sending time =Receive time - stating time= 59 − 46 = 13 milliseconds
Receiving time = Packet arrival time- Packet receive time=67 − 60 = 7 milliseconds
Round-trip time =Sending time+Receiving time= 13 + 7 = 20 milliseconds
Time difference = receive timestamp − ( original timestamp field + one-way time duration)
Time difference = 59 − (46 + 10) = 3
Question 92 |
UDP, 80 | |
TCP, 80 | |
TCP, 25 | |
UDP, 25 |
A browser is an HTTP client because it sends requests to an HTTP server (Web server), which then sends responses back to the client. The standard (and default) port for HTTP servers to listen on is 80, though they can use any port.
HTTP is based on the TCP/IP protocols, and is used commonly on the Internet for transmitting web-pages from servers to browsers.
Question 93 |
Autosense | |
Synchronization | |
Pinging | |
Auto negotiation |
2. Autosense refers to a feature found in network adapters that allows them to automatically recognize the current local network's speed and adjust its own setting accordingly. It is often used with Ethernet, fast Ethernet, switches, hubs and network interface cards
3. Process synchronization refers to the idea that multiple processes are to join up or handshake at a certain point, in order to reach an agreement or commit to a certain sequence of action. Data synchronization refers to the idea of keeping multiple copies of a dataset in coherence with one another, or to maintain data integrity. Process synchronization primitives are commonly used to implement data synchronization.
4. Autonegotiation 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 an optional component in the Fast Ethernet standard.
Question 94 |
01:00:5E:00:00:00 | |
01:00:5E:00:00:FF | |
01:00:5E:00:FF:FF | |
01:00:5E:FF:FF:FF |
Question 95 |
130.34.12.124 | |
130.34.12.89 | |
130.34.12.70 | |
130.34.12.132 |
→ So, addresses from 130.34.12.64 to 130.34.12.127 will be included in this organization.
Question 96 |
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?
205.16.42.56 | |
205.17.32.76 | |
205.16.31.10 | |
205.16.39.44 |
Supernet has a first address=205.16.32.0
Supernet Mask = 255.255.248.0
IP address=?
Step-1: Perform AND operation between supernet mask and IP address. Let us take IP address is 205.16.39.44.
Step-2: AND operation

Question 97 |
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 :
49 | |
368 | |
149 | |
151 |
Given data,
Slotted ALOHA transmits = 200 bit frames
Bandwidth = 200 Kbps
System produces = 250 frames per second
Throughput = ?
Step-2:
Pure ALOHA formula S = G * e-2G
Slotted ALOHA formula S = G * e-G
Frame transmission time = 200/200 kbps = 1ms
Here,
G = ¼ and S = G * e-G
= 0.195 (or) 19.5%
System produces = 250 * 0.195
= 48.75
Question 98 |
The period of a signal is 100 ms. Its frequency is
1003 Hertz | |
10−2 KHz
| |
10−3 KHz | |
105 Hertz |
100 ms = 100 × 10-3s = 10-1s
f = 1/T = 1/10-1Hz = 10 Hz = 10 × 10-3kHz = 10-2kHz
Question 99 |
The dotted-decimal notation of the following IPV4 address in binary notation is
10000001 00001011 00001011 11101111111.56.45.239
| |
129.11.10.238 | |
129.11.11.239
| |
111.56.11.239 |
Since the IP address is of 32 bits and is divided in 4 parts of size 8-bits and then perform simple binary to decimal conversion.

Question 100 |
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) and (b) only | |
(b) and (c) only | |
(a), (b) and (c) | |
(a) and (c) only |
AMPS is a first-generation cellular technology that uses separate frequencies, or "channels", for each conversation (see frequency-division multiple access (FDMA)). It therefore required considerable bandwidth for a large number of users. In general terms, AMPS was very similar to the older "0G" Improved Mobile Telephone Service, but used considerably more computing power in order to select frequencies, hand off conversations to PSTN lines, and handle billing and call setup.
(b) TRUE:
IS - 95 is a second generation cellular phone system based on CDMA and DSSS.
(c) TRUE:
The Third generation cellular phone system will provide universal personnel communication.
Question 101 |
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)-(iv), (b)-(ii), (c)-(i), (d)-(iii)
| |
(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii) | |
(a)-(ii), (b)-(iv), (c)-(iii), (d)-(i) | |
(a)-(iv), (b)-(ii), (c)-(iii), (d)-(i) |
→ International Data Encryption Algorithm (IDEA), originally called Improved Proposed Encryption Standard (IPES), is a symmetric-key block cipher. It uses block size 64 and key size 128.
→ Blowfish is a symmetric-key block cipher used for a large number of cipher suites and encryption products. Blowfish provides a good encryption rate in software and no effective cryptanalysis of it has been found to date. It uses block size 64 and key size ranges between 32 and 448.
→ Advanced Encryption Standard(AES), which uses symmetric key method for encryption of data. It uses block size 128 and key sizes 128, 192, 256.
*****It’s worthy to remember below table.

Question 102 |
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.
(a) and (b) only
| |
(b) and (c) only | |
(a) and (c) only
| |
(a), (b) and (c) |
→ In circuit switching resources are reserved from Setup phase to data transmission phase.
→ When data transmission completes only then the reserved resources are released.
→ Since in packet switching each packet follows different path to reach the destination so no resource reservation is done in case of packet switching.
Question 103 |
Decrypt the message “WTAAD” using the Caesar Cipher with key = 15.
LIPPS
| |
HELLO
| |
OLLEH | |
DAATW |

We decrypt one character at a time. Each character is shifted 15 characters up Letter W is decrypted to H 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 And finally D is The second A is decrypted to L. And, finally, D is decrypted to O. The plain text is HELLO.
Question 104 |
Encrypt the Message “HELLO MY DEARZ” using Transposition Cipher with
HLLEO YM AEDRZ | |
EHOLL ZYM RAED | |
ELHL MDOY AZER | |
ELHL DOMY ZAER |
Here, key size = 4. So, character block size is 4.
Step-2: Remove the spaces in the message and write into sequential order.

Step-3: Get cipher text according to ascending order is ELHL MDOY AZER.
Question 105 |
To optimize throughput | |
To prevent packet looping | |
To reduce delays | |
To prioritize packets |
→ If the limit is not defined then the packets can go into an indefinite loop.
→ The packet is discarded when the time to live field reaches 0 to prevent looping.
Question 106 |
255.255.192.192 | |
255.255.255.198 | |
255.255.255.240 | |
255.255.257.240 |

Question 107 |
10 | |
100 | |
1000 | |
10000 |
Frequency = 1/Period.
Frequency = 1/100 x 10-3 Hz
= 10 Hz
Question 108 |
(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.
(1) | |
(1) and (2) | |
(2) and (3) | |
(1),(2) and (3) |

Question 109 |
SNMP | |
IP | |
SMTP | |
TCP |
Primary protocols for E-Mail management
Post Office Protocol (POP)
Simple Mail Transfer Protocol (SMTP)
Internet Message Access Protocol (IMAP)
The above protocols are Application Layer Protocols
→ Once a client connects to the E-Mail Server, there may be 0(zero) or more SMTP transactions. If the client has no mail to send, then there are no SMTP transactions. Every e-mail message sent is an SMTP transfer.
→ SMTP is only used to send (push) messages to the server. POP and IMAP are used to receive messages as well as manage the mailbox contents(which includes tasks such as deleting, moving messages etc.).
Question 110 |
256.0.0.1 | |
172.16.0.10 | |
15.1.5.6 | |
127.256.0.1 |
→ 172.16.x.x is private address which is usen with in local area network.
→ So,option C is correct.
→ Option D, eliminated because the IP address atmost consists of 255.
Question 111 |
01011 | |
10101 | |
01110 | |
10110 |
M = 1010001101
Divisor polynomial: 1.x5 +1.x4+0.x3+1.x2+0.x2+1.x0
Divisor polynomial bit= 110101
append 5 zeroes = M = 101000110100000

∴ CRC = 01110
Question 112 |
Symbols | |
Bits
| |
Byte | |
None of these |
Baud rate is nothing but number of signals units transmitted per unit time.
Baud, or baud rate, is used to describe the maximum oscillation rate of an electronic signal. For example, if a signal changes (or could change) 1200 times in one second, it would be measured at 1200 baud.
Baud unit symbol "Bd is synonymous to symbols per second or pulses per second. It is the unit of symbol rate, also known as baud or modulation rate; the number of distinct symbol changes (signalling events) made to the transmission medium per second in a digitally modulated signal or a line code.
Question 113 |
Unipoint | |
Polarpoint | |
Point to point | |
Multipoint |

Question 114 |
SMTP | |
ICMP | |
ARP | |
RARP |
● The Internet Control Message Protocol (ICMP) is a supporting protocol in the Internet protocol suite. It is used by network devices, including routers, to send error messages and operational information indicating
●Address Resolution Protocol (ARP) is a protocol for mapping an Internet Protocol address (IP address) to a physical machine address that is recognized in the local network.
●RARP (Reverse Address Resolution Protocol) is a protocol by which a physical machine in a local area network can request to learn its IP address from a gateway server's Address Resolution Protocol (ARP) table or cache.
Question 115 |
Local area network | |
Virtual private network | |
Personal area network | |
none of the mentioned above |
→ A wireless personal area network (WPAN) is a low-powered PAN carried over a short-distance wireless network technology such as IrDA, Wireless USB, Bluetooth and ZigBee.
Question 116 |
rule | |
medium | |
Link | |
Protocol |
Question 117 |
Path | |
Protocol | |
Route | |
Medium |
● Routing is the process of moving a packet of data from one network to another network based on the destination IP address.
● The physical channel used for transmission in the network is called medium.
● Message travel from sender to receiver via a medium using a protocol
Question 118 |
types of messages exchanged | |
rules for when and how processes send and respond to messages | |
message format, syntax and semantics | |
all of above |
● The application layer abstraction is used in both of the standard models of computer networking: the Internet Protocol Suite (TCP/IP) and the OSI model.
● Although both models use the same term for their respective highest level layer, the detailed definitions and purposes are different.
Question 119 |
Physical layer | |
Data link layer | |
Network layer | |
Both (A) and (B) |
● Repeaters are used to extend transmissions so that the signal can cover longer distances or be received on the other side of an obstruction.
● Physical layer in the OSI model plays the role of interacting with actual hardware and signaling mechanism
Question 120 |
Unshielded twisted pair | |
Ubiquitous Teflon port | |
Uniformly Terminating port | |
Unshielded T-connector port |
● Due to its low cost, UTP cabling is used extensively for local-area networks (LANs) and telephone connections
Question 121 |
5 | |
6 | |
7 | |
46 |
● To find how many bits you would need, you would need to find the power of 2 that is equal to or the one immediately larger than your number. The number of bits would be the exponent of the power.
● In our case, since 128=2 7 >72>64=2 6 , the power immediately larger than 72 would be 128=27, so you would need 7 bits, which can codify 128 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. This time 2 6 =64>46>32=2 5 , so the power immediately larger than 46 would be 64=2 6 ; so 6 bits needed .
Question 122 |
The third generation mobile phone are digital and based on
AMPS | |
Broadband CDMA
| |
CDMA | |
D-AMPS |
→ AMPS(Advanced Mobile Phone Service): It uses frequency division multiple access(FDMA) to provide access to the channel by multiple stations at the same time. In FDMA whole available bandwidth is divided into small frequency bands and each station is allocated to one frequency band which it will use to send and receive data.
→ D-AMPS(Digital- Advanced Mobile Phone Service) : It uses time division multiple access(TDMA). In this kind of medium access all stations uses the whole available bandwidth for a fixed time slots.
→ CDMA(Code Division Multiple Access) : In CDMA all stations are allowed to use the channel simultaneously and can use the the whole bandwidth. In this technique the sender and receiver uses a code to distinguish there data from others.
→ Broadband CDMA : Broadband code division multiple access (B-CDMA) is a CDMA-based cell phone 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 123 |
Consider ISO-OSI network architecture reference model. Session layer of this model offer Dialog control, token management and ____________ as services.
Synchronization | |
Asynchronization
| |
Errors | |
Flow control
|
→ Dialog control: The session layer allows two systems to enter into a dialog. It allows the communication between two processes to take place in either half-duplex (one way at a time) or full-duplex (two ways at a time) mode.
→ Synchronization: The session layer allows a process to add checkpoints, or synchronization points, to a stream of data. For example, if a system is sending a file of 2000 pages, it is advisable to insert checkpoints after every 100 pages to ensure that each 100-page unit is received and acknowledged independently. In this case, if a crash happens during the transmission of page 523, the only pages that need to be reset after system recovery are pages 501 to 523. Pages previous to 501 need not be resent.
Question 124 |
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?
245.248.132.0/22 and 245.248.132.0/21
| |
245.248.136.0/21 and 245.248.128.0/22 | |
245.248.136.0/24 and 245.248.132.0/21
| |
245.248.128.0/21 and 245.248.128.0/22 |

Question 125 |
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)
(210 + 211) bytes | |
(210 + 28) bytes
| |
(210 + 224) bytes | |
(28 + 29) bytes
|
8-bits are allocated for each of the R, G and B components, means each entry in color lookup table is of 24-bits (8-bits for each of the R, G and B component).
24-bits = 3 Bytes
So the size of lookup table is = Number of entries * size of each entry
So the size of lookup table is = (210 * 3) Bytes
the size of lookup table is = 3072 Bytes = (210 + 211) Bytes
Question 126 |
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:(i) only | |
Neither (i) nor (ii) | |
Both (i) and (ii)
| |
(ii) only |
Statement 2 is correct. Virtual private networks cam simulate an old leased network to provide certain desirable properties.
Question 127 |
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:Both S1 and S2 are correct
| |
Neither S1 nor S2 is correct
| |
S1 is not correct but S2 is correct
| |
S1 is correct but S2 is not correct
|
TCP uses advertisement window for flow control and uses Slow start, Congestion Avoidance, Congestion Detection mechanism for congestion control.
UDP is an unreliable data transmission protocol. It does not perform congestion control or flow control.
Question 128 |
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,
LIST-1 List-2 a) Blow Fish (i) 128-256 bits b) DES (ii) 128 bits c) IDEA (iii) 1-448 bits d) RC5 (iv) 56 bitsCodes:
(a)-(ii), (b)-(iii), (c)- (iv), (d)-(i)
| |
(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
| |
(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i) | |
(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
|
• DES uses a 64-bit key, but eight of those bits are used for parity checks, effectively limiting the key to 56-bits.
• IDEA (International Data Encryption Algorithm) is an encryption algorithm which uses a block cipher with a 128-bit key, and is generally considered to be very secure. It is considered among the best publicly known algorithms.
• RC5 has a variable block size (32, 64 or 128 bits), key size (0 to 2040 bits) and number of rounds (0 to 255). The original suggested choice of parameters were a block size of 64 bits, a 128-bit key and 12 rounds.
Question 129 |
The four byte IP Address consists of
Neither network nor Host Address | |
Network Address | |
Both Network and Host Address | |
Host Address |

Question 130 |
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 :Both (i) and (ii) | |
Neither (i) nor (ii)
| |
(ii) only
| |
(i) only
|
→ Windows XP Professional known as peer-to-peer networking. This feature allows these micro-businesses to easily deploy a functional network without a server-class machine.
Question 131 |
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
The university network
| |
The internet | |
The subnet to which the host belongs
| |
The department network
|
But within the network any two devices should have different MAC addresses.
Question 132 |
The decimal floating point number -40.1 represented using IEEE-754 32-bit representation and written in hexadecimal form is
0xC2206000
| |
0xC2006666
| |
0xC2006000 | |
0xC2206666
|
and corresponding binary value is 1100 0010 0010 0000 0110 0110 0110 0110
Now, we will write Hexadecimal value by grouping four bits from least significant bits is 0xc2206666.
Question 133 |
In PERT/CPM, the merge event represents___________ of two or more events.
splitting
| |
completion
| |
beginning
| |
joining
|
→ The critical path method (CPM), or critical path analysis (CPA), is an algorithm for scheduling a set of project activities. It is commonly used in conjunction with the program evaluation and review technique (PERT).
→ A critical path is determined by identifying the longest stretch of dependent activities and measuring the time required to complete them from start to finish.
→ Event:
An event represents a point in time signifying the completion of some activities and the beginning of new ones. This is usually represented by a circle in a network which is also called a node or connector. The events are classified into three categories
1. Merge event – When more than one activity comes and joins an event such an event is known as merge event.
2. Burst event – When more than one activity leaves an event such an event is known as burst event.
3. Merge and Burst event – An activity may be merge and burst event at the same time as with respect to some activities it can be a merge event and with respect to some other activities it may be a burst event.

Question 134 |
Parallel-in parallel-out | |
Parallel-in Serial-out | |
Serial-in parallel-out | |
Serial-in Serial-out |
SIPO:
A serial-in, parallel-out shift register is similar to the serial-in, serial-out shift register in that it shifts data into internal storage elements and shifts data out at the serial-out, data-out, pin. It is different in that it makes all the internal stages available as outputs. Therefore, a serial-in, parallel-out shift register converts data from serial format to parallel format.
More Information
Serial-in to Serial-out (SISO) - the data is shifted serially “IN” and “OUT” of the register, one bit at a time in either a left or right direction under clock control.
Parallel-in to Serial-out (PISO) - the parallel data is loaded into the register simultaneously and is shifted out of the register serially one bit at a time under clock control.
Parallel-in to Parallel-out (PIPO) - the parallel data is loaded simultaneously into the register, and transferred together to their respective outputs by the same clock pulse.
Question 135 |
Application layer | |
Transport layer | |
Network layer | |
Data link layer |
→ SNMP is a component of the Internet Protocol Suite as defined by the Internet Engineering Task Force (IETF). It consists of a set of standards for network management, including an application layer protocol, a database schema, and a set of data objects.
Question 136 |
2 | |
3 | |
4 | |
5 |
11111111.11111111.11111111.11111000
255.255.255.248/29 mask.
→ It is 32 bit address and number of bits in the Network ID is 29
→ Number of bits in host ID=32-29= 3
Question 137 |
Layer 2 firewall | |
Layer 3 firewall | |
Layer 4 firewall | |
Layer 7 firewall |
→ A transparent firewall, on the other hand, is a Layer 2 firewall that acts like a "bump in the wire," or a "stealth firewall," and is not seen as a router hop to connected devices.
Question 138 |
Distance Vector | |
Path Vector | |
Link Vector | |
Multipoint |
→ A distance vector routing protocol sends its updates to neighboring routers and depends on them to pass the update information along to their neighbors.
→ For this reason, distance vector routing is said to use hop-by-hop updates.
Question 139 |
is free | |
can reduce the cost of using an information utility | |
allows communications channel to be shared among more than one user | |
boh (B) and (C) |
Question 140 |
Besides span of geographical area, the other major difference between LAN and WAN is that the later uses switching element | |
A repeater is used just to forward bits from one network to another one | |
IP layer is connected oriented layer in TCP/IP | |
A gateway is used to connect incompatible networks |
→ IP is connectionless, in that a data packet can travel from a sender to a recipient without the recipient having to send an acknowledgement.
Question 141 |
in-service expansion | |
unlimited number of stations | |
both (A) and (B) | |
unlimited distance |
Question 142 |
AMPS | |
Broadband CDMA | |
CDMA | |
D-AMPS |
→ AMPS(Advanced Mobile Phone Service): It uses frequency division multiple access(FDMA) to provide access to the channel by multiple stations at the same time. In FDMA whole available bandwidth is divided into small frequency bands and each station is allocated to one frequency band which it will use to send and receive data.
→ D-AMPS(Digital- Advanced Mobile Phone Service) : It uses time division multiple access(TDMA). In this kind of medium access all stations uses the whole available bandwidth for a fixed time slots.
→ CDMA(Code Division Multiple Access) : In CDMA all stations are allowed to use the channel simultaneously and can use the the whole bandwidth. In this technique the sender and receiver uses a code to distinguish there data from others.
→ Broadband CDMA : Broadband code division multiple access (B-CDMA) is a CDMA-based cell phone 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 143 |
Synchronization | |
Asynchronization | |
Errors | |
Flow control |
→ Dialog control: The session layer allows two systems to enter into a dialog. It allows the communication between two processes to take place in either half-duplex (one way at a time) or full-duplex (two ways at a time) mode.
→ Synchronization : The session layer allows a process to add checkpoints, or synchronization points, to a stream of data. For example, if a system is sending a file of 2000 pages, it is advisable to insert checkpoints after every 100 pages to ensure that each 100-page unit is received and acknowledged independently. In this case, if a crash happens during the transmission of page 523, the only pages that need to be reset after system recovery are pages 501 to 523. Pages previous to 501 need not be resent.
Question 144 |
245.248.132.0/22 and 245.248.132.0/21 | |
245.248.136.0/21 and 245.248.128.0/22 | |
245.248.136.0/24 and 245.248.132.0/21 | |
245.248.128.0/21 and 245.248.128.0/22 |

Question 145 |
(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:
(i) only | |
Neither (i) nor(ii) | |
Both (i) and (ii) | |
(ii) only |
Statement 2 is correct.Virtual private networks cam simulate an old leased network to provide certain desirable properties.
Question 146 |
HTTP GET request, DNS query, TCP SYN | |
DNS query, TCP SYN, HTTP GET request | |
TCP SYN, DNS query, HTTP GET request | |
DNS query, HTTP Get request, TCP SYN |
Step 1: DNS query is sent to Domain name server to convert the Domain name into it’s IP address.
Step 2: TCP SYN request packet is sent by sender to establish a connection for data transmission.
Step 3: HTTP GET request is used to request data from a specified IP address.
Question 147 |
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)?
Both S1 and S2 are correct | |
Neither S1 nor S2 is correct | |
S1 is not correct but S2 is correct | |
S1 is correct but S2 is not correct |
TCP uses advertisement window for flow control and uses Slow start, Congestion Avoidance, Congestion Detection mechanism for congestion control.
UDP is an unreliable data transmission protocol. It does not perform congestion control or flow control.
Question 148 |

(a)-(ii),(b)-(iii), (c)- (iv), (d)-(i) | |
(a)-(iv),(b)-(iii), (c)- (ii), (d)-(i) | |
(a)-(iii),(b)-(iv), (c)- (ii), (d)-(i) | |
(a)-(iii),(b)-(iv), (c)- (i), (d)-(ii) |
● DES uses a 64-bit key, but eight of those bits are used for parity checks, effectively limiting the key to 56-bits. ● IDEA (International Data Encryption Algorithm) is an encryption algorithm which uses a block cipher with a 128-bit key, and is generally considered to be very secure. It is considered among the best publicly known algorithms
● RC5 has a variable block size (32, 64 or 128 bits), key size (0 to 2040 bits) and number of rounds (0 to 255). The original suggested choice of parameters were a block size of 64 bits, a 128-bit key and 12 rounds.
Question 149 |
Neither network nor Host Address | |
Network Address | |
Both Network and Host Address | |
Host Address |

Question 150 |
(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.
Both (i) and (ii) | |
Neither (i) nor (ii) | |
(ii) only | |
(i) only |
→ Windows XP Professional known as peer-to-peer networking. This feature allows these micro-businesses to easily deploy a functional network without a server-class machine
Question 151 |
The university network | |
The internet | |
The subnet to which the host belongs | |
The department network |
But within the network any two devices should have different MAC addresses.
Question 152 |
2N | |
N(N-1) | |
N(N-1)/2 | |
(N-1) 2 |
→ We have N people in group so number of keys needed are N(N-1)
→ But two people in a group can use same keys then no need to use 2(N-1) keys they can communicate using (N-1) keys.
So total number of keys needed [N(N-1)]/2
Question 153 |
ARPANET | |
NFSNET | |
CNNET | |
ASAPNET |
Question 154 |
Simplex | |
Half Duplex | |
Automatic | |
Full Duplex |
● Half-duplex data transmission means that data can be transmitted in both directions on a signal carrier, but not at the same time
● Full-duplex data transmission means that data can be transmitted in both directions on a signal carrier at the same time
Question 155 |
Personal area network | |
Virtual private network | |
Local area network | |
None of the above |
Question 156 |
IETF | |
ANSI | |
ISO | |
IEEE |
Question 157 |
A binary address as strings | |
An alphanumeric address | |
A hierarchy of domain names | |
A hexadecimal address |
→ DNS is a host name to IP address translation service.
→ DNS is a distributed database implemented in a hierarchy of name servers.
→ It is an application layer protocol for message exchange between clients and servers.
Question 158 |
16 bits | |
32 bits | |
48 bits | |
64 bits |
→ Planning to shift IPv6 address, it has 128 bits to communicate with the host
Question 159 |
100111001101 | |
100111001011 | |
100111001 | |
100111001110 |
→ Given data unit is 100111001 and divisor is 1011
→ The polynomial is written in binary as the coefficients; a 3rd-degree polynomial has 4 coefficients (1x3 + 0x2 + 1x + 1). In this case, the coefficients are 1, 0, 1 and 1. The result of the calculation is 3 bits long.
→ First padded with zeros corresponding to the bit length n of the CRC. Here we used 3-bit CRC
→ After padding zeros to the data unit, Data unit = 100111001000
→ Divide data unit by divisor

→ Append the remainder(011) to the data unit where zero’s are padded Then data unit becomes 100111001011 at receiver end
Question 160 |
999 bytes have been successfully received | |
1000 bytes have been successfully received | |
1001 bytes have been successfully received | |
None of the above |
2. The acknowledgement number would be the total number of bytes received by the destination, FROM the sender, plus one (i.e. the one being the next byte it expects to get). Initial Sequence number is needed in order to identify how many number of bytes received by receiver.
So the answer is option-D because of not mentioning of sequence number.
Question 161 |
IP address: 125.134.112.66
Mask: 255.255.224.0
What is the first address(Network address)?
125.134.96.0 | |
125.134.112.0 | |
125.134.112.66 | |
125.134.0.0 |
IP address: 125.134.112.66
Subnet Mask: 255.255.224.0
IP address in binary Form = 01111101 10000110 01110000 01000010
Subnet Mask in binary Form = 11111111 11111111 11100000 00000000
Every IP address has Network id and Host id,
We will get Network id by doing bitwise -AND (&) operation with IP address and Subnet Mask

Question 162 |
4.25 | |
3.5 | |
350 | |
450 |
The amount of time slotted in unit is 50 msec
The number of time slots for generating user requests in 1 second are 1000/50=20
The Channel load can be calculated dividing number of requests to the time slots Channel Load = 70/20=3.5
Question 163 |
PING is a TCP/IP application that sends datagrams once every second in the hope of an echo response from the machine being PINGED | |
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 | |
If PING encounters an error condition, an ICMP message is not returned | |
PING display the time of the return response in milliseconds or one of several error message |
→ The program reports errors, packet loss, and a statistical summary of the results, typically including the minimum, maximum, the mean round-trip times, and standard deviation of the mean.
Question 164 |

A packet bearing a destination address 144.16.68.117 arrives at the router. On which interface will it be forwarded?
eth0 | |
eth1 | |
eth2 | |
eth3 |

Question 165 |
Transport, session, presentation, application | |
Session, presentation, application | |
Datalink, transport, presentation, application | |
Physical, datalink, network, transport |

Question 166 |
Which of the following is NOT a symmetric key algorithm?
Ellipse Curve Cryptography
| |
Advanced Encryption standard
| |
Data Encryption Standard
| |
Blowfish |

Question 167 |
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?
500 bits | |
50 bits | |
500 bytes
| |
4*1011 bits
|
→ This means, in the worst case, a station needs to transmit for a period of 50 μs to detect the collision.
→ The minimum size of the frame is 10 Mbps × 50 μs = 500 bits or 62.5 bytes.
→ This is actually the minimum size of the frame for Standard Ethernet.
Question 168 |
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?
10 | |
20 | |
2 | |
50 |
→Total user = 100 making 10 request/sec, Then total number of requests are 1000
Each slot time is 50 ms, so the number of slots in one second = 1/(50 X 10-3) = 20 slots/sec
Requests per second = 1000
Time slots per second = 20
Channel load = Number of Requests / Number of Slots = 1000 / 20 = 50
Question 169 |
Chose the option which matches each element of LIST-1 with exactly one element of LIST-2:
(i)-(d), (ii)-(c), (iii)-(b), (iv)-(a)
| |
(i)-(b), (ii)-(a), (iii)-(d), (iv)-(c)
| |
(i)-(a), (ii)-(d), (iii)-(c), (iv)-(b)
| |
(i)-(d), (ii)-(b), (iii)-(c), (iv)-(a)
|
2. Gateway→ All seven layers: A gateway, as the name suggests, is a passage to connect two networks together that may work upon different networking models. They basically works 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. Gateways are generally more complex than switch or router.
3. Router→ Network layer: 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→ Data link layer: A bridge is a repeater, with add on 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.
Question 170 |
Which of the following protocols is used to map IP address to MAC address?
ARP | |
IP | |
DHCP | |
RARP
|
→ MAC address to IP address using Reverse Address resolution protocol(RARP).
Question 171 |
625 lines | |
1250 lines | |
2300 lines | |
318 lines |
Number of lines=125*1000/100*2=625 (1us =1000ns)
Question 172 |
8 subnets and 262141 hosts | |
6 subnets and 262141 hosts | |
6 subnets and 1022 hosts | |
8 subnets and 1024 hosts | |
None of the Above |
→ Class-B addressing is given

Question 173 |
60000 | |
15000 | |
30000 | |
3000 |
=3000*log2(1+30)
=3000*4.95
=14850. So approximately, 15000
Question 174 |
Inserting a '0' in user data stream to differentiate it with a flag | |
Inserting a '0' in flag stream to avoid ambiguity | |
Appending a nibble to the flag sequence | |
Appending a nibble to the user data stream |
●The receiver knows how to detect and remove or disregard the stuffed bits.
Question 175 |
16.15kbps | |
23.24 kbps | |
40.12 kbps | |
24.74 kbps |
In presence of Gaussian band-limited white noise, Shannon Hartley theorem gives the
maximum data rate capacity C = B log 2 (1+ S/N)
=6log 2 (1+16)
=6*log 2 (17)
=6*4.08=24.48
Question 176 |
3m=2n | |
7m=8n | |
8m=7n | |
2m=3n |
Total number of hosts per network is 216-2
Substitute 8m and 7n
⇒ 14*8=16*7
112=112
Question 177 |
length of UDP=length of IP - length of IP header | |
length of UDP=length of UDP - length of UDP header | |
length of UDP=length of IP + length of IP header | |
length of UDP=length of UDP + length of UDP header |
→ The size of the data by computing "total length - header length"
→ A user datagram is encapsulated in an IP datagram. There is a field in the IP datagram the defines the total length. There is another field in the IP datagram that defines the length of the header. So if we subtract the length of a UDP datagram that is encapsulated in an IP datagram, we get the length of UDP user datagram.
Question 178 |
Remove congestion after it occurs | |
Remove congestion after sometime | |
Prevent congestion before it occurs | |
Prevent congestion before sending packets |
open-loop congestion control (prevention)
closed-loop congestion control (removal).

Question 179 |
CDMA | |
CSMA/CA
| |
ALOHA | |
None of the Options |
A wireless LAN can be characterized by the following attributes
Architecture
Physical layer
Medium access control layer
Architecture
Based on IEEE 802.11 standards, wireless LANs can be classified as follows: infrastructure and ad hoc. With the ad hoc architecture, wireless stations communicate directly with each other on an ad hoc basis.
Physical Layer
The physical layer is characterized by the following three elements: the frequency band, the multiple access method (which are similar to "multiplexing" in the wired environment, and the data rate.
Medium Access Control (MAC) Layer
This layer provide the same function as the MAC layer in the wired LAN environment. IEEE 802.11, instead of using Ethernet's CSMA/CD, specifies a scheme called CSMA/CA.
CA stands for Collision Avoidance. (In the wireless environment, since a station may not be able to hear all other stations, CD, Collision Detection is not feasible)
Question 180 |
International data encryption algorithm | |
private data encryption algorithm | |
Internet data encryption algorithm | |
none of the options |
→ IDEA is an optional algorithm in the OpenPGP standard. PGP encrypts data by using a block cipher called international data encryption algorithm.
→ IDEA operates on 64-bit blocks using a 128-bit key and consists of a series of 8 identical transformations (a round, see the illustration) and an output transformation (the half-round).
→ The processes for encryption and decryption are similar.
Question 181 |
10 | |
24 | |
12 | |
7 |
→ Convert 255.255.255.0 into binary format.
11111111.11111111.11111111.00000000.
→ Total number of 1’s are 24. And total number of 0’s are 8.
Question 182 |
Number of previous bytes to receive | |
Total number of bytes to receive | |
Number of next bytes to receive | |
Sequence of zero's and one's |
Question 183 |
230.0.0.0 | |
130.4.5.6 | |
230.7.6.5 | |
30.4.5.6 |
→ Class B IP Addresses range from 128.0.x.x to 191.255.x.x. The default subnet mask for Class B is 255.255.x.x.
→ Class C IP addresses range from 192.0.0.x to 223.255.255.x. The default subnet mask for Class C is 255.255.255.x.

Note: Class A addresses 127.0.0.0 to 127.255.255.255 cannot be used and is reserved for loopback and diagnostic functions.
Private IP Addresses:

Question 184 |
255.255.255.192 | |
255.255.255.248 | |
255.255.255.240 | |
255.255.255.224
|
Question 185 |
NLSP | |
RIP | |
SAP | |
NCP |
→ NCP (NetWare Core Protocol)provides client-to-server connections and applications.
→ RIP is a distance vector routing protocol.
Question 186 |
Transport layer | |
Network layer | |
Data link layer | |
Session layer |
→ Session, presentation and application layers are user support layers.
→ Transport layer layer connects the network support layers and user support layers
Question 187 |
Additively | |
Multiplicatively | |
Exponentially | |
None of the options |
→ TCP slow start is an algorithm which balances the speed of a network connection. Slow start gradually increases the amount of data transmitted until it finds the network’s maximum carrying capacity.
→ TCP slow start is one of the first steps in the congestion control process. It balances the amount of data a sender can transmit (known as the congestion window) with the amount of data the receiver can accept (known as the receiver window).
→ The lower of the two values becomes the maximum amount of data that the sender is allowed to transmit before receiving an acknowledgment from the receiver.
Question 188 |
Classless addressing | |
Classful addressing | |
Subnet advertising | |
None of the options |
→ The problem with this classful addressing method is that millions of class A address are wasted, many of the class B address are wasted, whereas, number of addresses available in class C is so small that it cannot cater the needs of organizations.
→ Class D addresses are used for multicast routing, and are therefore available as a single block only. Class E addresses are reserved.
Note: Because of these problems, Classful networking was replaced by Classless Inter-Domain Routing (CIDR) in 1993.
Question 189 |
Unicast address | |
Multicast address | |
Broadcast address | |
All of the options |
The first octant of MAC address written in binary format.
07=0000 0111.
The least significant bit 0 then unicast, 1 means Multicast and entire 8 bits are 1’s then broadcast.
Question 190 |
it is easy to generate | |
it cannot be shared | |
it is different for every access | |
it can be easily decrypted |
Question 191 |
Codes | |
Blocks | |
IPs | |
Sizes |
→ To overcome the problem of address depletion and give more organizations access to internet, classless addressing was designed and implemented.
→ 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.
Question 192 |
Complex routers | |
Open to DOS attack | |
No overlapping of fragments | |
(A) and (B) both |
→ The support for fragmentation of larger packets provides a protocol allowing routers to fragment a packet into smaller packets when the original packet is too large for the supporting data link frames.
→ IP fragmentation exploits (attacks) use the fragmentation protocol within IP as an attack vector.
IP fragment over lapped:
The IP fragment overlapped exploit occurs when two fragments contained within the same IP packet have offsets that indicate that they overlap each other in positioning within the packet. This could mean that either fragment A is being completely overwritten by fragment B, or that fragment A is partially being overwritten by fragment B.
Overlapping fragments may also be used in an attempt to bypass Intrusion Detection Systems
Question 193 |
Mesh | |
Bus | |
Ring | |
Star |
→ The devices are not directly linked to one another.
Question 194 |
Radio waves | |
Microwaves | |
Infrared waves | |
Light waves |
Question 195 |
FTP | |
SMTP | |
SNMP | |
STTP |
SMTP is an application-layer protocol that enables the transmission and delivery of email over the Internet
Question 196 |
Datalink | |
Network | |
Transport | |
Application |
→ Bridges are important in networks because the networks are divided into many parts geographically remote from one another.
→ Bridge works at Data link layer
Question 197 |
10-1 kHz | |
10 -2 kHz | |
10-3 kHz | |
10-4 kHz |
100 ms=100*10-3s
= 10-1 s
f=1/T
=1/10-1Hz
=10 Hz
=10*10-3 kHz
=10-2 kHz
Question 198 |
Server machine | |
Client machine | |
Request machine | |
Intelligent machine |
→ The server is often (but not always) on another computer system, in which case the client accesses the service by way of a network.
Question 199 |
Enable or disable cookie support
| |
Declare cookie variables | |
Store data in cookie variables | |
Clear data from cookie variables |
→ A cookie is often used to identify a user. A cookie is a small file that the server embeds on the user's computer. Each time the same computer requests a page with a browser, it will send the cookie too. With PHP, you can both create and retrieve cookie values.
→ Syntax
→ setcookie(name,value,expire,path,domain,secure,httponly);
Question 200 |
Handoff | |
Mobile routing | |
Mobile switching
| |
Cell Switching
|
→ A well-implemented handoff is important for delivering uninterrupted service to a caller or data session user.
→ In satellite communications it is the process of transferring satellite control responsibility from one earth station to another without loss or interruption of service.
Question 201 |
Inserting a '0' in user data stream to differentiate it with a flag | |
Inserting a '0' in flag stream to avoid ambiguity | |
Appending a nibble to the flag sequence | |
Appending a nibble to the user data stream |
● The receiver knows how to detect and remove or disregard the stuffed bits.

Question 202 |
Header contents | |
Trailer contents | |
Error Checking | |
Data content of message |
→ There are many properties of a transmission that a protocol can define. Common ones include: packet size, transmission speed, error correction types, handshaking and synchronization techniques, address mapping, acknowledgement processes, flow control, packet sequence controls, routing, address formatting
Question 203 |
Allows gateways to send error control messages to other gateways or hosts | |
Provides communication between the internet protocol software on one machine and the internet protocol software on another | |
Only reports error conditions to the original source, the source must relate errors to individual application programs and take action to correct the problem. | |
All of these |
→ ICMP creates and sends messages to the source IP address indicating that a gateway to the Internet that a router, service or host cannot be reached for packet delivery.
Question 204 |
transport,session,Presentation, Application | |
Network,Transport,Session,Presentation | |
Datalink, Network, Transport, Session | |
Physical,Data Link,Network,Transport |
Question 205 |
Train | |
Bus | |
Tram | |
Aeroplane |
Question 206 |
Application layer | |
Transport layer | |
Network layer | |
Data link layer |
●Bridges are important in networks because the networks are divided into many parts geographically remote from one another.
● Bridge works at Data link layer
Question 207 |
Transmission capacity of a communication channel | |
Connected components in a network | |
Class of IP used in Network | |
Interconnected by communication channels |
●For digital devices, the bandwidth is usually expressed in bits per second(bps) or bytes per second.
● For analog devices, the bandwidth is expressed in cycles per second, or Hertz (Hz).
Question 208 |
Computer bus | |
Telephone bus | |
Voice and mode | |
Lease lines |
Question 209 |
Technique for start-stop data | |
Technique for dial access | |
DTE/DCE interface | |
Data bit rate |
→ X.25 does not specify how the network operates internally – many X.25 network implementations used something very similar to X.25 or X.75 internally
Question 210 |
Router | |
Bridge | |
Repeater | |
modem |
→ Bridges are important in networks because the networks are divided into many parts geographically remote from one another. Something is required to join these networks so that they can become part of the whole network.
→ Take for example a divided LAN, if there is no medium to join these separate LAN parts an enterprise may be limited in its growth potential. The bridge is one of the tools to join these LANS.
Question 211 |
192 | |
224 | |
248 | |
252 | |
None of the above |
Question 212 |
Transceiver | |
Nine pin connector | |
MAU | |
NIC |
Question 213 |
Network layer | |
Data link layer | |
Transport layer | |
Session layer |
→ Data is transferred in the form of packets via logical network paths in an ordered format controlled by the network layer
Question 214 |
HTTP | |
FTP | |
SMTP | |
SNMP |
→ SMTP (Simple Mail Transfer Protocol) is a TCP/IP protocol used in sending and receiving e-mail.
→ However, since it is limited in its ability to queue messages at the receiving end, it is usually used with one of two other protocols, POP3 or IMAP, that let the user save messages in a server mailbox and download them periodically from the server.
→ In other words, users typically use a program that uses SMTP for sending e-mail and either POP3 or IMAP for receiving email.
Question 215 |
33,700 | |
30,000 | |
32,700 | |
33,000 |
Step-2: Guard band (or gaps)=(10-9)=9.
300x9 =2700 Kz
Step-3: Minimum bandwidth is required for the multiplexed channel 30000+2700= 32700Hz
Question 216 |
10 -1 Khz | |
1 KHz | |
10 -3 Khz | |
10 -2 Khz |
Step-2: Time=1000 ms which is equals to =1 sec
Step-3: Frequency=1/1 sec
=1000/1000 sec
=1K/1000 sec
=10 -3 Khz
Note: 1 sec = 10 -3 ms
Question 217 |
HTTP | |
URL | |
FTP | |
DNS |
Question 218 |
Storage control | |
Multitasking | |
Handshaking | |
Piggybacking |
Piggybacking In two-way communication, wherever a frame is received, the receiver waits and does not send the control frame back to the sender immediately. The receiver waits until its network layer passes in the next data packet. The delayed acknowledgement is then attached to this outgoing data frame.
Question 219 |
fast ethernet | |
thick ethernet | |
thin ethernet | |
gigabit ethernet |
→ 10BASE5 uses a thick and stiff coaxial cable up to 500 metres (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. The system is difficult to install and maintain.
Question 220 |
Gateway | |
Hub | |
Switch | |
Router |
→ Besides routing packets, gateways also possess information about the host network's internal paths and the learned path of different remote networks.
→ If a network node wants to communicate with a foreign network, it will pass the data packet to the gateway, which then routes it to the destination using the best possible path. holds its current state. There will be no change.
Question 221 |
Optical fiber cable | |
Shielded twisted pair cable | |
Unshielded twisted pair cable | |
coaxial cable |
→ In practical fibers, the cladding is usually coated with a layer of acrylate polymer or polyimide. This coating protects the fiber from damage but does not contribute to its optical waveguide properties.
Question 222 |
Mesh topology | |
Bus topology | |
Star topology | |
Ring topology |
→ A mesh topology can be a full mesh topology or a partially connected mesh topology. In a full mesh topology, every computer in the network has a connection to each of the other computers in that network.
→ The number of connections in this network can be calculated using the following formula (n is the number of computers in the network): n(n-1)/2.
→ This network having highest reliability.
Question 223 |
8 | |
1 | |
2 | |
4 |
→ Pre-IPv6 is nothing but IPv4. So, total size in 32 bits=4 bytes
Question 224 |
1000 bauds/sec, 4000 bps | |
1000 bauds/sec, 500 bps | |
4000 bauds/sec, 1000 bps | |
2000 bauds/sec, 1000 bps |
Step-1: Important formulas to find a solution is
Bit rate=baud rate*no.of bits per signal
Baud rate(or signal rate)=bit rate/no.of bits per signal
Step-2:They are expecting to find bit rate. Given data, 1000 bauds(signal)per second and number of bits per signal=4
Step-3: Bit rate=1000*4=4000 bits per second.
Question 225 |
The 10Base5 cabling scheme of ethernet uses:
Twisted pairs
| |
Fiber optics | |
Thin coax
| |
Thick coax |
10Base5 is also called thick Ethernet, ThickWire, and ThickNet.
→ The number 10: At the front of each identifier, 10 denotes the standard data transfer speed over these media - ten megabits per second (10Mbps).
→ The word Base: Short for Baseband, this part of the identifier signifies a type of network that uses only one carrier frequency for signaling and requires all network stations to share its use.
→ The segment type or segment length: This part of the identifier can be a digit or a letter:
- Digit - shorthand for how long (in meters) a cable segment may be before attenuation sets in.
For example, a 10Base5 segment can be no more than 500 meters long.
- Letter - identifies a specific physical type of cable.
For example, the T at the end of 10BaseT stands for twisted-pair.

Question 226 |
One of the ad-hoc solutions to count to infinity problem in network routing is:
The split horizon hack
| |
Flow based routing | |
Flooding
| |
Shortest path routing
|
1. Route Poison
2. The split horizon hack
Question 227 |
Given a bit rate of b bits/sec, the time required to send 16 bits is:
16*b sec
| |
16/b sec
| |
16b sec | |
b16 sec
|
The data rate R is a function of the duration of the bit or bit time (TB).
R = 1/TB
Question 228 |
The built in HTTP request method to request to read a web page is:
HEAD | |
PUT
| |
GET
| |
POST |
The request method indicates the method to be performed on the resource identified by the given Request-URI. The method is case-sensitive and should always be mentioned in uppercase. The following table lists all the supported methods in HTTP/1.1.

Question 229 |
The traditional cryptographic cipher that records the letters but do not disguise them is:
Substitute cipher | |
One-time pads
| |
Secret key algorithms
| |
Transposition cipher |
Also this code can be easily broken.
Question 230 |
In client-server computing vertical scaling means:
Adding or removing client workstations with only a slight performance impact
| |
Migrating servers to a new group of client workstations | |
Migrating client workstations to a larger and faster server machine or multi servers | |
Combining two or more client workstation groups
|
→ Vertical Scaling is most commonly used in applications and products of middle-range as well as small and middle-sized companies. One of the most common examples of Virtual Scaling is to buy an expensive hardware and use it as a Virtual Machine hypervisor (VMWare ESX).
→ Vertical Scaling usually means upgrade of server hardware. Some of the reasons to scale vertically includes increasing IOPS (Input / Output Operations), amplifying CPU/RAM capacity, as well as disk capacity.
→ However, even after using virtualization, whenever an improved performance is targeted, the risk for downtimes with it is much higher than using horizontal scaling.
Question 231 |
STT | |
RTT | |
PTT | |
Total time |
→In this context, the source is the computer initiating the signal and the destination is a remote computer or system that receives the signal and retransmits it
Question 232 |
2 Mbps | |
4 Mbps | |
8 Mbps | |
12 Mbps |
Given bandwidth is 10Mbps
Number of average frames 12,000 per minute.
Each Frame is carrying 10,000 bits
So, 12,000 * 10,000 / 60 seconds.
= 2 * 1000000 bits / seconds.
= 2 Mbps.
Question 233 |
RJ45 | |
RJ54 | |
BNC | |
MT-RJ |
→All of these types of connectors can be used with either multimode or single mode fiber.
Question 234 |
Mesh topology | |
Star topology | |
Bus topology
| |
Ring topology |
→Data travels from node to node, with each node along the way handling every packet.
Question 235 |
IEEE 802.11a | |
IEEE 802.11b | |
IEEE 802.11d | |
IEEE 802.11g |
802.11 — applies to wireless LANs and provides 1 or 2 Mbps transmission in the 2.4 GHz band using either frequency hopping spread spectrum (FHSS) or direct sequence spread spectrum (DSSS).
802.11a — an extension to 802.11 that applies to wireless LANs and provides up to 54-Mbps in the 5GHz band. 802.11a uses an orthogonal frequency division multiplexing encoding scheme rather than FHSS or DSSS.
802.11b (also referred to as 802.11 High Rate or Wi-Fi) — an extension to 802.11 that applies to wireless LANS and provides 11 Mbps transmission (with a fallback to 5.5, 2 and 1-Mbps) in the 2.4 GHz band. 802.11b uses only DSSS. 802.11b was a 1999 ratification to the original 802.11 standard, allowing wireless functionality comparable to Ethernet.
802.11e — a wireless draft standard that defines the Quality of Service (QoS) support for LANs, and is an enhancement to the 802.11a and 802.11b wireless LAN (WLAN) specifications.
802.11g — applies to wireless LANs and is used for transmission over short distances at up to 54-Mbps in the 2.4 GHz bands.
802.11n — 802.11n builds upon previous 802.11 standards by adding multiple-input multiple-output (MIMO). The real speed would be 100 Mbit/s (even 250 Mbit/s in PHY level), and so up to 4-5 times faster than 802.11g.
802.11ac — 802.11ac builds upon previous 802.11 standards, particularly the 802.11n standard, to deliver data rates of 433Mbps per spatial stream, or 1.3Gbps in a three-antenna (three stream) design. The 802.11ac specification operates only in the 5 GHz frequency range and features support for wider channels (80MHz and 160MHz) and beamforming capabilities by default to help achieve its higher wireless speeds.
802.11ac Wave 2 — 802.11ac Wave 2 is an update for the original 802.11ac spec that uses MU-MIMO technology and other advancements to help increase theoretical maximum wireless speeds for the spec to 6.93 Gbps.
802.11ad — 802.11ad is a wireless specification under development that will operate in the 60GHz frequency band and offer much higher transfer rates than previous 802.11 specs, with a theoretical maximum transfer rate of up to 7Gbps (Gigabits per second).
802.11ah— Also known as Wi-Fi HaLow, 802.11ah is the first Wi-Fi specification to operate in frequency bands below one gigahertz (900 MHz), and it has a range of nearly twice that of other Wi-Fi technologies. It's also able to penetrate walls and other barriers considerably better than previous Wi-Fi standards.
802.11r - 802.11r, also called Fast Basic Service Set (BSS) Transition, supports VoWi-Fi handoff between access points to enable VoIP roaming on a Wi-Fi network with 802.1X authentication.
802.1X — Not to be confused with 802.11x (which is the term used to describe the family of 802.11 standards) 802.1X is an IEEE standard for port-based Network Access Control that allows network administrators to restricted use of IEEE 802 LAN service access points to secure communication between authenticated and authorized devices.
Question 236 |
Logical | |
Physical | |
IP | |
Port |
Logical Address:
A logical address is the address at which an item (memory cell, storage element, network host) appears to reside from the perspective of an executing application program.
Physical address: Each system having a NIC(Network Interface Card) through which two systems physically connected with each other with cables. The address of the NIC is called Physical address or mac address. This is specified by the manufacturer company of the card. This address is used by data link layer.
Port Address: There are many application running on the computer. Each application run with a port no.(logically) on the computer. This port no. for application is decided by the Kernel of the OS. This port no. is called port address.
Question 237 |
30,000 | |
30,300 | |
32,700 | |
33,000 |
There are total 10 channels and 9 guard bands.Each guard band is 300Hz wide 9 guard band occupies 2700 hz.
10 channels occupies 30000 hz.
So total bandwidth is 32700.
Question 238 |
20 and 21 | |
20 and 23 | |
21 and 25 | |
23 and 25 |

Question 239 |
Hub | |
Modem | |
Switch | |
Gateway |
→ Physical layer uses HUB and repeater
→ Data link layer uses Switch. MAC address comes into data link layer.
Question 240 |
Static algorithms | |
Adaptive algorithms | |
Non - adaptive algorithms | |
Recursive algorithms |
1. Flooding
2. Random Walk
Question 241 |
32 bit | |
48 bit | |
64 bit | |
128 bit |
→ The Ethernet frame is the packet that delivers data between stations.
→ Each frame is made up of bits which are organized into different fields. The four main fields are the destination Ethernet address field, source Ethernet address field, data field and frame check sequence field.
Question 242 |
Network layer | |
Data layer | |
Session layer | |
Transport layer |
→ Network layer-Host to Host connectivity
→ Transport layer-End to End connectivity
→ Presentation layer-Encryption and Decryption
Question 243 |
0.0.0.0 | |
1.0.0.0 | |
1.1.1.1 | |
255.255.255.255 |
If a packet destination address is not known, then if the default route is not known, then if the default route is present in the routing table of the router, then the packet is not discarded but forwarded to the next router. 0.0.0.0
Question 244 |
Zigbee | |
Bluetooth | |
Wi-Fi | |
GSM/CDMA |
Question 245 |
Infinite | |
5 | |
20 | |
50 |
S = 1 / (1-P)
where,
P is parallel part of program,
sequential part of program is 5%
. = 1 - sequential part
= 1 - 0.05 (or 5%)
= 0.95 (or 95%)
Step-2: Apply into the formula is
S = 1 / (1-P)
S = 1 / (1-0.95)
S = 1 / 0.05
S = 20
Question 246 |
2 16 bytes | |
2 16 bytes + TCP header length | |
2 16 bytes - TCP header length | |
2 15 byte. | |
None of the above |
Transport Layer - 65515 byte
Network layer - 65535 byte
Data link layer - 1500 byte
Note: Actually it works on any size, but given options are wrong. Excluded for evaluation.
Question 247 |
File server | |
DNS server | |
DHCP server | |
Default gateway |
→ More narrowly defined, a router merely forwards packets between networks with different network prefixes. The networking software stack of each computer contains a routing table that specifies which interface is used for transmission and which router on the network is responsible for forwarding to a specific set of addresses.
→ If none of these forwarding rules is appropriate for a given destination address, the default gateway is chosen as the router of last resort.
→ The default gateway is specified by the configuration setting often called default route.
Question 248 |
automatically | |
by server | |
by exchanging information with neighbour nodes | |
with backup database |
→ Distance-vector protocols update the routing tables of routers and determine the route on which a packet will be sent by the next hop which is the exit interface of the router and the IP address of the interface of the receiving router.
→ Distance is a measure of the cost to reach a certain node. The least cost route between any two nodes is the route with minimum distance.
Question 249 |
DES algorithm | |
Dijkstra’s algorithm | |
RSA algorithm | |
Packets |
→ 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.
Question 250 |
2 MHz | |
20 KHz | |
5 KHz | |
5 MHz |
→ IMT-DS uses a version of CDMA called wideband CDMA or W-CDMA. W-CDMA uses a 5.MHz bandwidth. It was in europe, and it is compatible with the CDMA used in IS-95.
→ IMT-TC uses a combination of W-CDMA and TDMA. The standard tries to reach the IMT-2000 goals by adding TDMA multiplexing to W-CDMA.
Question 251 |
Electromagnetic waves with frequencies from 300 GHz to 400 Thz. | |
Propagation is line-of-sight. | |
Very high-frequency waves cannot penetrate walls. | |
Use of certain portions of the band requires permission from authorities. |
→ The electromagnetic spectrum covers electromagnetic waves with frequencies ranging from below one hertz to above 10 25 hertz, corresponding to wavelengths from thousands of kilometers down to a fraction of the size of an atomic nucleus.
Question 252 |
twisted pair, 100 metres | |
twisted pair, 200 metres | |
fibre optics, 1000 metres | |
fibre optics, 2000 metres |
→ The letter following the dash ("T" or "F") refers to the physical medium that carries the signal (twisted pair or fiber, respectively), while the last character ("X", "4", etc.) refers to the line code method used.
→ Fast Ethernet is sometimes referred to as 100BASE-X, where "X" is a placeholder for the FX and TX variants.
Question 253 |
1 Mbps | |
2 Mbps | |
10 Mbps | |
12 Mbps |
Bandwidth= 10 Mbps
Frames per minute=12000
Each frame per bits=10000
Throughput=?
Step-1: Here, they given each frame per minute. So, convert into seconds is
(12000*10000) / 60
Step-2: 120000000 / 60
= 2000000
It is nothing but 2Mbps
Question 254 |
POP3 | |
IMAP | |
SMTP | |
DMSP |
→ Internet Message Access Protocol (IMAP) is an Internet standard protocol used by email clients to retrieve email messages from a mail server over a TCP/IP connection.
→ IMAP was designed with the goal of permitting complete management of an email box by multiple email clients, therefore clients generally leave messages on the server until the user explicitly deletes them.
→ An IMAP server typically listens on port number 143. IMAP over SSL (IMAPS) is assigned the port number 993.
→ POP3 is designed to delete mail on the server as soon as the user has downloaded it. However, some implementations allow users or an administrator to specify that mail be saved for some period of time. POP can be thought of as a "store-and-forward" service.
→ SMTP is used for connecting to outbound servers to send email while POP3 and IMAP are used to connect to incoming servers to retrieve messages.
→ DMSP is Distributed Mail System for Personal Computers.
Question 255 |
Denial of service attack | |
Masquerade attack | |
Simple attack | |
Complex attack |
→ A denial-of-service attack (DoS attack) is a cyber-attack in which the perpetrator seeks to make a machine or network resource unavailable to its intended users by temporarily or indefinitely disrupting services of a host connected to the Internet.
Question 256 |
100 Kbps | |
200 Kbps | |
400 Kbps | |
1000 Kbps |
= 20 sec
Frame size= No.of bits per frame
= 8 bits
Bit Rate= Frame Rate * No.of bits per frame
= 50000*8
= 400 kbps

Question 257 |
40,9 | |
45,10 | |
45,9 | |
50,10 |
Here, n=10(total number of computers)
= n(n-1) / 2
= 10(9) / 2
= 45
→ To accommodate that many links, every device on the network must have n-1 input/output(I/O) ports
n=10(total number of computers)
= n-1 ports
= 10-1
= 9
Question 258 |
Physical | |
Transport | |
Application | |
Host-to-network | |
None of the above |
Question 259 |
N | |
(N - 1) | |
N(N - 1) / 2 | |
N(N + 1) / 2 |
→ To communicate using symmetric cryptography, both parties have to agree on a secret key. After that, each message is encrypted with that key, transmitted, and decrypted with the same key. Key distribution must be secret. If it is compromised, messages can be decrypted and users can be impersonated. However, if a separate key is used for each pair of users, the total number of keys increases rapidly as the number of users increases. With n users, we would need [n(n-1)]/2 keys.
Question 260 |
equal to | |
less than | |
greater than | |
less than or equal to |
→ Total internal reflection is the phenomenon which occurs when a propagated wave strikes a medium boundary at an angle larger than a particular critical angle with respect to the normal to the surface.
→ If the refractive index is lower on the other side of the boundary and the incident angle is greater than the critical angle, the wave cannot pass through and is entirely reflected.
Question 261 |
Masquerade Attack | |
Replay Attack | |
Passive Attack | |
Denial of Service Attack |
→ Replay involves the passive capture of a data unit and its subsequent retransmission to produce an unauthorized effect.
→ Passive attacks are very difficult to detect, because they do not involve any alteration of the data.Typically,the message traffic is sent and received in an apparently normal fashion,and neither the sender nor receiver is aware that a third party has read the messages or observed the traffic pattern.However,it is feasible to prevent the success of these attacks,usually by means of encryption.Thus,the emphasis in dealing with passive attacks is on prevention rather than detection.
→ The denial of service prevents or inhibits the normal use or management of communications facilities.This attack may have a specific target; for example, an entity may suppress all messages directed to a particular destination (e.g.,the security audit service).Another form of service denial is the disruption of an entire network,either by disabling the network or by overloading it with messages so as to degrade performance
Question 262 |
Type of service | |
Fragment offset | |
Flags | |
Identification |
→ Fragment offset field along with Don't Fragment and More Fragment flags in the IP protocol header are used for fragmentation and reassembly of IP packets.
→ The More Fragment(MF) flag is set for all the fragment packets except the last one.
Question 263 |
Flow Control Policy | |
Packet Discard Policy | |
Packet Lifetime Management Policy | |
Routing Algorithm |
1.Preallocation of Resources
2.Traffic Shaping
3.Discarding Packets (No Preallocation)
4.Isarithmic Congestion Control
5.Virtual Circuits Admission Control
6.Choke Packets
7.Packet Discard
8. Packet Lifetime Management
9. Routing Algorithm
Transport layer congestion policies:
1. Flow Control, etc..,
Question 264 |
Session Maintenance Protocol | |
Real time Streaming Protocol | |
Real time Transport Control Protocol | |
Session Initiation Protocol |
→ SIP is used for signaling and controlling multimedia communication sessions in applications of Internet telephony for voice and video calls, in private IP telephone systems, in instant messaging over Internet Protocol (IP) networks as well as mobile phone calling over LTE (VoLTE).
Question 265 |

(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii) | |
(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii) | |
(a)-(ii), (b)-(iv), (c)-(iii), (d)-(i) | |
(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii) |

Question 266 |
Dialog control | |
Token management | |
Semantics of the information transmitted | |
Synchronization |
1. Session checkpointing and recovery
2. Dialog control
3. Token management
4. Synchronization
Question 267 |
4 bit | |
8 bit | |
16 bit | |
32 bit |

Question 268 |
The number of addresses needs to 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. | |
All of the above |
1. The number of addresses needs to be a power of 2.
2. The mask needs to be included in the address to define the block.
3. The starting address must be divisible by the number of addresses in the block.
Question 269 |
from transmission line noise | |
in router | |
from out of sequence delivery | |
from packet losses |
→ The error control in the transport layer usually refers to the guaranteed delivery mechanism with TCP, which attempts to safeguard against frames/packets getting lost entirely.
Question 270 |
Error | |
Loss | |
Sequence | |
Duplication |
Error Control is performed end to end in this layer to ensure that the complete message arrives at the receiving transport layer without any error. Error Correction is done through retransmission.
Question 271 |
TP0 and TP2 | |
TP1 and TP3 | |
TP0, TP1, TP3 | |
TP0, TP1, TP2, TP3, TP4 |
→ TP0 to TP3 work only with connection-oriented communications, in which a session connection must be established before any data is sent.
→ TP4 also works with both connection-oriented and connectionless communications. Transport Protocol Class 0 (TP0) performs segmentation (fragmentation) and reassembly functions. TP0 discerns the size of the smallest maximum protocol data unit (PDU) supported by any of the underlying networks, and segments the packets accordingly. The packet segments are reassembled at the receiver.
Transport Protocol Class 2 (TP2) performs segmentation and reassembly, as well as multiplexing and demultiplexing of data streams over a single virtual circuit.
Transport Protocol Class 4 (TP4) offers error recovery, performs segmentation and reassembly, and supplies multiplexing and demultiplexing of data streams over a single virtual circuit. TP4 sequences PDUs and retransmits them or reinitiates the connection if an excessive number are unacknowledged. TP4 provides reliable transport service and functions with either connection-oriented or connectionless network service. TP4 is the most commonly used of all the OSI transport protocols.
Question 272 |
TP0, TP2 | |
TP1, TP3 | |
TP1, TP3, TP4 | |
TP0, TP1, TP2, TP3, TP4 |
Transport Protocol Class 3 (TP3) offers error recovery, segmentation and reassembly, and multiplexing and demultiplexing of data streams over a single virtual circuit. TP3 also sequences PDUs and retransmits them or re-initiates the connection if an excessive number are unacknowledged.
Question 273 |
Connectionless | |
Error-free | |
Segmentation | |
Connection-oriented |
Question 274 |
message | |
segment | |
datagram | |
frame |
Question 275 |
DNS | |
FTP | |
TELNET | |
All the above |
→ Clients and servers exchange messages in a request–response messaging pattern. The client sends a request, and the server returns a response. This exchange of messages is an example of inter-process communication.
Example:
1. FTP
2. DNS
3. TELNET
Question 276 |
transport | |
session | |
router | |
presentation |
1. Encryption and decryption
2. Compression and decompression
3. Translation
Question 277 |
(1) Ethernet
(2) HSSI
(3) Frame Relay
(4) 10base T
(5) Token ring
1, 2 | |
1, 3, 5 | |
1, 3, 4, 5 | |
1, 2, 3, 4, 5 |
The High-Speed Serial Interface (HSSI) is a differential ECL serial interface standard developed by Cisco Systems and T3plus Networking primarily for use in WAN router connections. It is capable of speeds up to 52 Mbit/s with cables up to 50 feet (15 m) in length.
Question 278 |
Transparent Bridge | |
Source-Route Bridge | |
Translation Bridge | |
None of these |
Source-Route Bridge: To prevent loops in a system with redundant bridges is to use source routing bridges. A transparent bridge's duties include filtering frames, forwarding, and blocking.
In a system that has source routing bridges, these duties are performed by the source station and, to some extent, the destination station. Theoretically a bridge should be able to connect LANs using different protocols at the data link layer, such as an Ethernet
Segment with a token ring Segment but practically it can not connect two LANs with different protocols. One of the reasons of this is Maximum data size. If an incoming frame's size is too large for the destination LAN, the data must be fragmented into several frames. The data then need to be reassembled at the destination. However, no protocol at the data link layer allows the fragmentation and reassembly of frames. The other issues because of which a bridge is not able to connect LANs using different protocols are data rate, bit order, security, multimedia support.
So, the answer is Option(D)
Question 279 |
HDLC | |
SDLC | |
LAPB | |
LAPD |
LAPD is the second layer protocol on the ISDN protocol stack in the D channel (the ISDN channel in which the control and signalling information is carried).
LAPD standards are specified in ITU-T Q.920 and Q.921. Its development was heavily based on High-Level Data Link Control.
Question 280 |
255 . 255 . 255 . 0 | |
255 . 255 . 192 . 0 | |
255 . 255 . 240 . 0 | |
255 . 255 . 248 . 0 |

→ In class-B addressing NID’s are having 16-bits and host IDs are having 16 bits.
Step-1: Given data, to support for 30 networks is required. So number of bits taken from host ID to uniquely identify each of these network is (log 30/ log 2) i.e. 5-bits.
Step-2: NID’s are required for network is 21 bits and host id is 11-bits.
Step-3: Subnet mask of a network have NID as all 1's and Host ID as all 0's.
Network subnet mask is : 11111111. 11111111. 11111000. 00000000
Decimal representation of above subnet mask is : 255.255.248.0
Question 281 |
WECA | |
Fast Ethernet | |
Wi-Fi 5 | |
802 . 11g |
802.11b typically has a higher range at low speeds (802.11b will reduce speed to 5.5 Mbit/s or even 1 Mbit/s at low signal strengths). 802.11a also suffers from interference.
Question 282 |
Class A Network | |
Class B Network | |
Class C Network | |
Class D Network |

Question 283 |
Extends the network portion to 16 bits | |
Extends the network portion to 26 bits | |
Extends the network portion to 36 bits | |
Has no effect on the network portion of an IP address |
(192) 10 = (11000000) 2
Since, 192 is written as 11000000, it has 2 sub-nets and remaining all hosts.
So, for first three octets, 24 bits are fixed and for last octet 2 bits are fixed, i.e. 24 + 2 = 26 bits
Question 284 |
1 byte | |
1 bit | |
4 bits | |
16 bits |

Question 285 |
Connect similar remote LANs | |
Connect similar local LANs | |
Connect different types of LANs | |
Translate the network addresses into a layer 2 address |
A translating bridge provides a connection capability between two local area networks that employ different protocols at the data link layer. Because networks using different data link layer protocols normally use different media, a translating bridge also provides support for different physical layer connections.
A translating bridge operation, A translating bridge connects local area networks that employ different protocols at the data link layer. In this example, the translating bridge is used to connect an ethernet local area network to a token network.

Question 286 |
Tracert | |
Traceroute | |
Ping | |
Pop |
Question 287 |
Flooding | |
Multi Destination routing | |
Reverse path forwarding | |
All of the above |