Category: Computer Networks Objective Questions

Computer Networks Multiple Choice Questions on “Analyzing Subnet Masks”.

1. Which of the following is the broadcast address for a Class B network ID using the default subnetmask?
a) 172.16.10.255
b) 255.255.255.255
c) 172.16.255.255
d) 172.255.255.255

Answer: c
Clarification: 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.

2. You have an IP address of 172.16.13.5 with a 255.255.255.128 subnet mask. What is your class of address, subnet address, and broadcast address?
a) Class A, Subnet 172.16.13.0, Broadcast address 172.16.13.127
b) Class B, Subnet 172.16.13.0, Broadcast address 172.16.13.127
c) Class B, Subnet 172.16.13.0, Broadcast address 172.16.13.255
d) Class B, Subnet 172.16.0.0, Broadcast address 172.16.255.255

Answer: b
Clarification: 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.

3. If you wanted to have 12 subnets with a Class C network ID, which subnet mask would you use?
a) 255.255.255.252
b) 255.255.255.255
c) 255.255.255.240
d) 255.255.255.248

Answer: c
Clarification: 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.

4. The combination of _________ and __________ is often termed the local address of the local portion of the IP address.
a) Network number and host number
b) Network number and subnet number
c) Subnet number and host number
d) Host number

Answer: c
Clarification: It is termed as the local address because the address won’t be applicable outside the subnet. Sub networking is implemented for remote sensing in transparent way from that host which is contained in the sub network which called a local operation.

5. _________ implies that all subnets obtained from the same subnet mask.
a) Static subnetting
b) Dynamic subnetting
c) Variable length subnetting
d) Dynamic length subnetting

Answer: a
Clarification: Static subnetting is used when the requirement is of same number of hosts in each subnet for the institution. The same subnet mask can be used to find the subnet id of each subnet. It is usually used to divide large networks into smaller parts.

6. State whether true or false.
i) A connection oriented protocol can only use unicast addresses.
ii) The any cast service is included in IPV6.
a) True, True
b) True, False
c) False, True
d) False, False

Answer: a
Clarification: In a connection oriented protocol, the host can only establish connection with another host on one unique channel, that’s why it can only use unicast addresses. In IPv6, there is an anycast address in IPv6 which allows sending messages to a group of devices but not all devices in a network.

7. __________ is a high performance fiber optic token ring LAN running at 100 Mbps over distances upto 1000 stations connected.
a) FDDI
b) FDDT
c) FDDR
d) FOTR

Answer: a
Clarification: FDDI stands for Fiber Distributed Data Interface. It is a set of standards for fiber optic token ring LANs running at 100 Mbps over distances up to 200 km in diameter and 1000 stations connected.

8. Which of the following are Gigabit Ethernets?
a) 1000 BASE-SX
b) 1000 BASE-LX
c) 1000 BASE-CX
d) All of the mentioned

Answer: d
Clarification: 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.

9. _________ is a collective term for a number of Ethernet Standards that carry traffic at the nominal rate of 1000 Mbit/s against the original Ethernet speed of 10 Mbit/s.
a) Ethernet
b) Fast Ethernet
c) Gigabit Ethernet
d) Gigabyte Ethernet

Answer: b
Clarification: Fast Ethernet is a set of Ethernet Standards which were introduced in 1995, that carry traffic at the nominal rate of 1000 Mbit/s. 100BASE-TX is the most commonly used Fast Ethernet standard.

10. _________ is another kind of fiber optic network with an active star for switching.
a) S/NET
b) SW/NET
c) NET/SW
d) FS/NET

Answer: a
Clarification: 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.

Computer Networks Question Bank on “Network Management”.

1. Complex networks today are made up of hundreds and sometimes thousands of _________
a) Documents
b) Components
c) Servers
d) Entities

Answer: b
Clarification: Complex networks today are made up of hundreds and sometimes thousands of components. For effective functioning of these thousands of components, good network management is essential.

2. Performance management is closely related to _________
a) Proactive Fault Management
b) Fault management
c) Reactive Fault Management
d) Preventive Fault Management

Answer: b
Clarification: Fault management is really closely related to performance management. It is important to ensure that the network handles faults as effectively as it handles it’s normal functioning to achieve better performance management.

3. Configuration management can be divided into two subsystems: reconfiguration and __________
a) Documentation
b) Information
c) Servers
d) Entity

Answer: a
Clarification: The documentation subsystem of configuration management handles the log making and reporting functions of the configuration management. It also reports the errors in the network caused by the configuration’s failure.

4. In Network Management System, the term that is responsible for controlling access to network based on predefined policy is called ___________
a) Fault Management
b) Secured Management
c) Active Management
d) Security Management

Answer: d
Clarification: In Network Management System, the term that is responsible for controlling access to the network based on predefined policy is called security management. The security management ensures authentication, confidentiality and integrity in the network.

5. Control of users’ access to network resources through charges is the main responsibility of ______________
a) Reactive Fault Management
b) Reconfigured Fault Management
c) Accounting Management
d) Security Management

Answer: c
Clarification: Control of users’ access to network resources through charges is the main responsibility of accounting management. The accounting management creates a log of the users activity on the network too and goes hand-in-hand with the configurations management.

6. The physical connection between an end point and a switch or between two switches is __________
a) Transmission path
b) Virtual path
c) Virtual circuit
d) Transmission circuit

Answer: a
Clarification: The physical connection between an end point and a switch or between two switches is transmission path. The transmission path is the physical roadway that the packet needs to propagate in order to travel through the network.

7. Which of the following networks supports pipelining effect?
a) Circuit-switched networks
b) Message-switched networks
c) Packet-switched networks
d) Stream-switched networks

Answer: c
Clarification: Packet switched network is most preferred for pipelining process. Pipelining exponentially reduces the time taken to transmit a large number of packets in the network.

8. In Network Management System, maps track each piece of hardware and its connection to the _________
a) IP Server
b) Domain
c) Network
d) Data

Answer: c
Clarification: Network is the main entity connecting different components in a place. A network map is made to track each component and its connection to the network to ensure better network management.

9. MIB is a collection of groups of objects that can be managed by __________
a) SMTP
b) UDP
c) SNMP
d) TCP/IP

Answer: c
Clarification: MIB stands for Management Information Base. Simple network management controls the group of objects in management information base. It is usually used with SNMP (Simple Network Management Protocol).

10. A network management system can be divided into ___________
a) three categories
b) five broad categories
c) seven broad categories
d) ten broad categories

Answer: b
Clarification: The five broad categories of network management are
• Fault Management
• Configuration Management
• Accounting (Administration)
• Performance Management
• Security Management.

Computer Networks Multiple Choice Questions on “Access Networks”.

1. Which of this is not a constituent of residential telephone line?
a) A high-speed downstream channel
b) A medium-speed downstream channel
c) A low-speed downstream channel
d) An ultra-high speed downstream channel

Answer: c
Clarification: A low-speed downstream channel is not a constituent of a residential telephone line. But it might be just a two-way telephone channel. Internet can be provided through a high-speed downstream channel in a residential telephone line.

2. DSL telcos provide which of the following services?
a) Wired phone access
b) ISP
c) Wired phone access and ISP
d) Network routing and ISP

Answer: c
Clarification: DSL stands for Digital Subscriber Line and ISP stands for Internet Service Provider. In a Digital Subscriber Line system, the same company which provides phone connection is also an ISP. The internet is provided through the pre-installed telephone lines.

3. The function of DSLAM is to __________
a) Convert analog signals into digital signals
b) Convert digital signals into analog signals
c) Amplify digital signals
d) De-amplify digital signals

Answer: a
Clarification: DSLAM stands for Digital Subscriber Line Access Multiplexer and it’s used by Telcos to convert the analog signals to digital signals for the purpose of providing internet. The DSLAM located in a telco’s Central Office does this function.

4. Which of the following terms is not associated with DSL?
a) DSLAM
b) CO
c) Splitter
d) CMTS

Answer: d
Clarification: CMTS stands for Cable modem termination system. It is used in cable internet access. In cable internet access, internet is not provided through telephone lines and the companies that provide such connections don’t necessarily provide telephone access.

5. HFC contains _______
a) Fibre cable
b) Coaxial cable
c) A combination of Fibre cable and Coaxial cable
d) Twisted Pair Cable

Answer: c
Clarification: Hybrid fiber-coaxial (HFC) is a telecommunications industry term for a broadband network that combines optical fiber and coaxial cable. It has been popularly used since the early 1990s. It is stronger than the optical fiber cables and faster than the co-axial cables.

6. Which of the following statements is not applicable for cable internet access?
a) It is a shared broadcast medium
b) It includes HFCs
c) Cable modem connects home PC to Ethernet port
d) Analog signal is converted to digital signal in DSLAM

Answer: d
Clarification: CMTS stands for Cable modem termination system. In cable access analog signal is converted to digital signal by CMTS. In cable internet access, internet is not provided through telephone lines. DSLAM is used by Telecom companies.

7. Among the optical-distribution architectures that are essentially switched ethernet is __________
a) AON
b) PON
c) NON
d) MON

Answer:a
Clarification: AON stands for Active optical networks which are essentially switched Ethernets. Each user has his/her own dedicated optical fiber line connecting to the ISP in an AON.

8. StarBand provides __________
a) FTTH internet access
b) Cable access
c) Telephone access
d) Satellite access

Answer: d
Clarification: StarBand was a two-way satellite broadband Internet service available in the U.S. from 2000–2015. It was discontinued from September 30 2015 due to increasing competition from other ISPs.

9. Home Access is provided by __________
a) DSL
b) FTTP
c) Cable
d) All of the mentioned

Answer: d
Clarification: Home Internet Access is provided by DSL, FTTP, and Cable. FTTP provides the fastest speeds followed by the cable connections and then the DSLs. FTTP is popularly used in modern connections.

10. ONT is connected to splitter using _________
a) High speed fibre cable
b) HFC
c) Optical cable
d) Twisted pair cable

Answer: c
Clarification: ONT stands for Optical Network Terminal. The ONT connects to the Termination Point (TP) with an optical fibre cable. It translates light signals from the fibre optic line to electric signals that the router can read.

11. Which of the following factors affect transmission rate in DSL?
a) The gauge of the twisted-pair line
b) Degree of electrical interfernece
c) Shadow fading
d) The gauge of the twisted-pair line and degree of electrical interference

Answer: d
Clarification: Because DSL is made of twisted wire copper pair, the gauge of twisted pair line i.e. the protection and electrical interference would affect the transmission rate in DSL. Unlike DSL, FTTP is not really affected by these factors.

Computer Networks Multiple Choice Questions on “HTTP”.

1. The number of objects in a Web page which consists of 4 jpeg images and HTML text is ________
a) 4
b) 1
c) 5
d) 7

Answer: c
Clarification: 4 jpeg images + 1 base HTML file.

2. The default connection type used by HTTP is _________
a) Persistent
b) Non-persistent
c) Can be either persistent or non-persistent depending on connection request
d) None of the mentioned

Answer: a
Clarification: By default the http connection is issued with persistent connection. In persistent connection server leaves connection open after sending response. As little as one RTT (Time for a small packet to travel from client to server and back) is required for all referenced objects.

3. The time taken by a packet to travel from client to server and then back to the client is called __________
a) STT
b) RTT
c) PTT
d) JTT

Answer: b
Clarification: RTT stands for round-trip time.

4. The HTTP request message is sent in _________ part of three-way handshake.
a) First
b) Second
c) Third
d) Fourth

Answer: c
Clarification: In first step client sends a segment to establish a connection with the server. In the second the step the client waits for the acknowledgement to be received from the server. After receiving the acknowledgement, the client sends actual data in the third step.

5. In the process of fetching a web page from a server the HTTP request/response takes __________ RTTs.
a) 2
b) 1
c) 4
d) 3

Answer: b
Clarification: By default the http connection will be persistent connection. Hence it will take only 1 RTT to fetch a webpage from a server.

6. The first line of HTTP request message is called _____________
a) Request line
b) Header line
c) Status line
d) Entity line

Answer: a
Clarification: The line followed by request line are called header lines and status line is the initial part of response message.

7. The values GET, POST, HEAD etc are specified in ____________ of HTTP message
a) Request line
b) Header line
c) Status line
d) Entity body

Answer: a
Clarification: It is specified in the method field of request line in the HTTP request message.

8. The __________ method when used in the method field, leaves entity body empty.
a) POST
b) SEND
c) GET
d) PUT

Answer: c
Clarification: There are two methods which help to request a response from a server. Those are GET and POST. In GET method, the client requests data from server. In POST method the client submits data to be processed to the server.

9. The HTTP response message leaves out the requested object when ____________ method is used
a) GET
b) POST
c) HEAD
d) PUT

Answer: c
Clarification: HEAD method is much faster than GET method. In HEAD method much smaller amount of data is transferred. The HEAD method asks only for information about a document and not for the document itself.

10. Find the oddly matched HTTP status codes
a) 200 OK
b) 400 Bad Request
c) 301 Moved permanently
d) 304 Not Found

Answer: d
Clarification: 404 Not Found.

11. Which of the following is not correct?
a) Web cache doesnt has its own disk space
b) Web cache can act both like server and client
c) Web cache might reduce the response time
d) Web cache contains copies of recently requested objects

Answer: a
Clarification: Web cache or also known as HTTP cache is a temporary storage where HTML pages and images are stored temporarily so that server lag could be reduced.

12. The conditional GET mechanism
a) Imposes conditions on the objects to be requested
b) Limits the number of response from a server
c) Helps to keep a cache upto date
d) None of the mentioned

Answer: c
Clarification: The HTTP protocol requests the server of the website its trying to access so that it can store its files, images etc. in cache memory. This request of asking the server for a document considering a specific parameter is called conditional GET Request.

13. Which of the following is present in both an HTTP request line and a status line?
a) HTTP version number
b) URL
c) Method
d) None of the mentioned

Answer: a
Clarification: Status line is the the start line of an HTTP response. It contains the information such as the protocol version, a status text, status code.

Computer Networks Multiple Choice Questions on “UDP”.

1. Which of the following is false with respect to UDP?
a) Connection-oriented
b) Unreliable
c) Transport layer protocol
d) Low overhead

Answer: a
Clarification: UDP is an unreliable, connectionless transport layer protocol that provides message-based data transmission. TCP is an example of connection-oriented protocols.

2. Return value of the UDP port “Chargen” is _______
a) String of characters
b) String of integers
c) Array of characters with integers
d) Array of zero’s and one’s

Answer: a
Clarification: Using Chargen with UDP on port 19, the server sends a UDP datagram containing a random number of characters every time it receives a datagram from the connecting host. The number of characters is between 0 and 512.

3. Beyond IP, UDP provides additional services such as _______
a) Routing and switching
b) Sending and receiving of packets
c) Multiplexing and demultiplexing
d) Demultiplexing and error checking

Answer: d
Clarification: De-multiplexing is the delivering of received segments to the correct application layer processes at the recipients end using UDP. Error checking is done through checksum in UDP.

4. What is the main advantage of UDP?
a) More overload
b) Reliable
c) Low overhead
d) Fast

Answer: c
Clarification: As UDP does not provide assurance of delivery of packet, reliability and other services, the overhead taken to provide these services is reduced in UDP’s operation. Thus, UDP provides low overhead, and higher speed.

5. Port number used by Network Time Protocol (NTP) with UDP is ________
a) 161
b) 123
c) 162
d) 124

Answer: b
Clarification: The Network Time Protocol is a clock synchronization network protocol implemented by using UDP port number 123 to send and receive time stamps.

6. What is the header size of a UDP packet?
a) 8 bytes
b) 8 bits
c) 16 bytes
d) 124 bytes

Answer: a
Clarification: The fixed size of the UDP packet header is 8 bytes. It contains four two-byte fields: Source port address, Destination port address, Length of packet, and checksum.

7. The port number is “ephemeral port number”, if the source host is _______
a) NTP
b) Echo
c) Server
d) Client

Answer: d
Clarification: Port numbers from 1025 to 5000 are used as ephemeral port numbers in Windows Operating System. Ephemeral port numbers are short-lived port numbers which can be used for clients in a UDP system where there are temporary clients all the time.

8. “Total length” field in UDP packet header is the length of _________
a) Only UDP header
b) Only data
c) Only checksum
d) UDP header plus data

Answer: d
Clarification: Total length is the 16 bit field which contains the length of UDP header and the data. The maximum value of the Total length field and the maximum size of a UDP datagram is 65,535 bytes (8 byte header + 65,527 bytes of data).

9. Which is the correct expression for the length of UDP datagram?
a) UDP length = IP length – IP header’s length
b) UDP length = UDP length – UDP header’s length
c) UDP length = IP length + IP header’s length
d) UDP length = UDP length + UDP header’s length

Answer: a
Clarification: A user datagram is encapsulated in an IP datagram. There is a field in the IP header that defines the total length of the IP packet. There is another field in the IP header that defines the length of the header. So if we subtract the length of the IP header that is encapsulated in the IP packet, we get the length of UDP datagram.

10. The ______ field is used to detect errors over the entire user datagram.
a) udp header
b) checksum
c) source port
d) destination port

Answer: b
Clarification: Checksum field is used to detect errors over the entire user datagram. Though it is not as efficient as CRC which is used in TCP, it gets the job done for the UDP datagram as UDP doesn’t have to ensure the delivery of the packet.

Computer Networks MCQs on “Designing Subnet Masks”.

1. The network address of 172.16.0.0/19 provides how many subnets and hosts?
a) 7 subnets, 30 hosts each
b) 8 subnets, 8,190 hosts each
c) 8 subnets, 2,046 hosts each
d) 7 subnets, 2,046 hosts each

Answer: b
Clarification: A CIDR address of /19 is 255.255.224.0. This is a Class B address, so that is only 3 subnet bits, but it provides 13 host bits, or 8 subnets, each with 8,190 hosts.

2. Your router has the following IP address on Ethernet0: 172.16.2.1/23. Which of the following can be valid host IDs on the LAN interface attached to the router?

i. 172.16.1.100
ii. 172.16.1.198
iii. 172.16.2.255
iv. 172.16.3.0

a) i only
b) ii and iii only
c) iii and iv only
d) ii only

Answer: c
Clarification: The router’s IP address on the E0 interface is 172.16.2.1/23, which is 255.255.254.0. This makes the third octet a block size of 2. The router’s interface is in the 172.16.2.0 subnet, and the broadcast address is 172.16.3.255 because the next subnet is 172.16.4.0. The valid host range is 172.16.2.1 to 172.16.3.254. The router is using the first valid host address in the range.

3. Which two statements describe the IP address 10.16.3.65/23?

i. The subnet address is 10.16.3.0 255.255.254.0.
ii. The lowest host address in the subnet is 10.16.2.1 255.255.254.0.
iii. The last valid host address in the subnet is 10.16.2.254 255.255.254.0.
iv. The broadcast address of the subnet is 10.16.3.255 255.255.254.0.

a) i and iii
b) ii and iv
c) i, ii and iv
d) ii, iii and iv

Answer: b
Clarification: The mask 255.255.254.0 (/23) used with a Class A address means that there are 15 subnet bits and 9 host bits. The block size in the third octet is 2 (256 – 254). So this makes the subnets in the interesting octet 0, 2, 4, 6, etc., all the way to 254. The host 10.16.3.65 is in the 10.16.2.0 subnet. The next subnet is 10.16.4.0, so the broadcast address for the 10.16.2.0 subnet is 10.16.3.255. The valid host addresses are 10.16.2.1 to 10.16.3.254.

4. What is the maximum number of IP addresses that can be assigned to hosts on a local subnet that uses the 255.255.255.224 subnet mask?
a) 14
b) 15
c) 16
d) 30

Answer: d
Clarification: 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.

5. You need to subnet a network into 5 subnets, each with at least 16 hosts. Which classful subnet mask would you use?
a) 255.255.255.192
b) 255.255.255.224
c) 255.255.255.240
d) 255.255.255.248

Answer: b
Clarification: You need 5 subnets, each with at least 16 hosts. The mask 255.255.255.240 provides 16 subnets with 14 hosts which is less than 15, so this will not work. The mask 255.255.255.224 provides 8 subnets, each with 30 hosts so this may work. The mask 255.255.255.192 provides 4 subnets, each with 60 hosts so this may work. Comparing both the possible masks, 255.255.255.224 provides the best answer.

6. You have a network that needs 29 subnets while maximizing the number of host addresses available on each subnet. How many bits must you borrow from the host field to provide the correct subnet mask?
a) 2
b) 3
c) 4
d) 5

Answer: d
Clarification: A 240 mask is 4 subnet bits and provides 16 subnets, each with 14 hosts. We need more subnets, so let’s add subnet bits. One more subnet bit would be a 248 mask. This provides 5 subnet bits (32 subnets) with 3 host bits (6 hosts per subnet). This is the best answer.

7. If an Ethernet port on a router were assigned an IP address of 172.16.112.1/25, what would be the valid subnet address of this host?
a) 172.16.112.0
b) 172.16.0.0
c) 172.16.96.0
d) 172.16.255.0

Answer: a
Clarification: A /25 mask is 255.255.255.128. Used with a Class B network, the third and fourth octets are used for subnetting with a total of 9 subnet bits, 8 bits in the third octet and 1 bit in the fourth octet. Since there is only 1 bit in the fourth octet, the bit is either off or on-which is a value of 0 or 128. The host in the question is in the 0 subnet, which has a broadcast address of 127 since 128 is the next subnet

8. You have an interface on a router with the IP address of 192.168.192.10/29. Including the router interface, how many hosts can have IP addresses on the LAN attached to the router interface?
a) 6
b) 8
c) 30
d) 32

Answer: a
Clarification: A /29 (255.255.255.248), regardless of the class of address, has only 3 host bits. Six hosts are the maximum number of hosts on this LAN, including the router interface. Out of the 8 addresses possible with the host bits, the first and the last address are for the subnet id and broadcast address respectively.

9. What is the subnet id of a host with an IP address 172.16.66.0/21?
a) 172.16.36.0
b) 172.16.48.0
c) 172.16.64.0
d) 172.16.0.0

Answer: c
Clarification: A /21 is 255.255.248.0, which means we have a block size of 8 in the third octet, so we just count by 8 until we reach 66. The subnet in this question is 64.0. The next subnet is 72.0, so the broadcast address of the 64 subnet is 71.255.