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Networking - Quiz(MCQ)
A)
Multiplexing
B)
Error correction
C)
Delivery of packets
D)
Demultiplexing

Correct Answer : Option (D) :   Demultiplexing


Explanation : Demultiplexing is the process of error checking and dropping of the header, delivering messages to appropriate process based on port number. The transport layer does this on the receiver’s end after the packet is received and takes help of the header attached by the sender’s side transport layer during multiplexing.

A)
8-bytes
B)
8-bits
C)
16-bytes
D)
16-bits

Correct Answer : Option (A) :   8-bytes


Explanation : An ICMP message has an 8-byte header and a variable size data section. Out of the 8 bytes, the first 4 bytes are of a fixed format having the type, code and checksum fields and the next 4 bytes depend upon the type of the message.

A)
flow control
B)
router control
C)
switch control
D)
error control

Correct Answer : Option (A) :   flow control


Explanation : Firstly, it informs the source that the datagram has been discarded. Secondly, it warns the source that there is congestion in the network. It’s type 4 error reporting message after which the source is expected to reduce the flow of packets.

A)
java
B)
ping
C)
shell
D)
traceroute

Correct Answer : Option (B) :   ping


Explanation : Ping program is used to find if a host is alive and responding. It is to be entered into a command line with the syntax “ping (IP address)” to be executed. Traceroute is a program used to find the shortest route to the destination IP.

A)
172.16.10.255
B)
172.255.255.255
C)
172.16.255.255
D)
255.255.255.255

Correct Answer : Option (C) :   172.16.255.255


Explanation : In this case, the class B network ID is 172.16.0.0. We know that the default mask of a class B network is 255.255.0.0. If we OR any address in a network with the complement of the default mask (0.0.255.255), we get the broadcast address of the network. In this case, the result of OR would be 172.16.255.255.

A)
Class A, Subnet 172.16.13.0, Broadcast address 172.16.13.127
B)
Class B, Subnet 172.16.13.0, Broadcast address 172.16.13.255
C)
Class B, Subnet 172.16.0.0, Broadcast address 172.16.255.255
D)
Class B, Subnet 172.16.13.0, Broadcast address 172.16.13.127

Correct Answer : Option (D) :   Class B, Subnet 172.16.13.0, Broadcast address 172.16.13.127


Explanation : We know that the prefix 172 lies in class B (128 to 191) of IPv4 addresses. From the subnet mask, we get that the class is divided into 2 subnets: 172.16.13.0 to 172.16.13.127 and 172.16.13.128 to 172.16.13.255. The IP 172.16.13.5 lies in the first subnet. So the starting address 172.16.13.0 is the subnet address and last address 172.16.13.127 is the broadcast address.

A)
255.255.255.240
B)
255.255.255.248
C)
255.255.255.252
D)
255.255.255.255

Correct Answer : Option (A) :   255.255.255.240


Explanation :

If you have eight networks and each requires 10 hosts, you would use the Class C mask of 255.255.255.240. Why? Because 240 in binary is 11110000, which means you have four subnet bits and four host bits. Using our math, we’d get the following:
24-2=14 subnets
24-2=14 hosts.

A)
1000 BASE-CX
B)
1000 BASE-LX
C)
1000 BASE-SX
D)
All of the above

Correct Answer : Option (D) :   All of the above


Explanation : In computer networking, Gigabit Ethernet (GbE or 1 GigE) is a term describing various technologies for transmitting Ethernet frames at a rate of a gigabit per second (1,000,000,000 bits per second), as defined by the IEEE 802.3-2008 standard. It came into use beginning in 1999, gradually supplanting Fast Ethernet in wired local networks, as a result of being considerably faster.

A)
SW/NET
B)
S/NET
C)
NET/SW
D)
FS/NET

Correct Answer : Option (B) :   S/NET


Explanation : A 50-MBd active star fiber optical Local area network (LAN) and its optical combiner and mixing rod splitter are presented. The limited power budget and relatively large tapping losses of light wave technology, which limit the use of fiber optics in tapped bus LAN topologies, are examined and proven tolerable in optical star topologies.

A)
16
B)
24
C)
30
D)
35

Correct Answer : Option (C) :   30


Explanation : A /27 (255.255.255.224) is 3 bits on and 5 bits off. This provides 8 subnets, each with 30 hosts. Does it matter if this mask is used with a Class A, B, or C network address? Not at all. The number of host bits would never change.