UGC NET Computer Science Data Communication and Computer Networks Previous Year Questions (PYQs)
Practise 170 Data Communication and Computer Networks questions asked in UGC NET Computer Science from 2018–2026. Questions cover Transmission media, Error detection and correction, OSI reference model and the TCP/IP protocol suite, Data link layer, Medium access control and Network layer. Every question is shown with its options and the correct answer, free to read.
170 PYQs2018–2026Answers includedFree
Practise Data Communication and Computer Networks as timed sets
Focused 10-question sets with instant scoring, explanations, and weak-area analysis.
Arrange the following in the correct order in the context of the OSI model, from bottom to top.
A. Network layer
B. Physical layer
C. Presentation layer
D. Data link layer
E. Application layer
AD, A, C, B, E
BA, C, E, B, D
CB, D, A, C, E
DC, A, B, E, D
Answer:(C) B, D, A, C, E
Explanation
The OSI stack rises from Physical to Data Link to Network to Presentation to Application. Replacing these with the supplied letters gives B, D, A, C, E.
Which topology requires a central controller or a hub?
ABus
BStar
CMesh
DRing
Answer:(B) Star
Explanation
A star topology connects every device to a central hub or switch. Bus and ring topologies use a shared backbone or loop, while a mesh uses multiple direct links.
I. Service ready II. Transaction failed III. Help message IV. Service not available
AA-III, B-IV, C-I, D-II
BA-IV, B-III, C-I, D-II
CA-IV, B-I, C-III, D-II
DA-II, B-I, C-IV, D-III
Answer:(A) A-III, B-IV, C-I, D-II
Explanation
SMTP code 214 is a help message, 421 means service not available, 220 means service ready, and 554 indicates a transaction failed. These pairings produce A-III, B-IV, C-I, D-II.
Identify the incorrect statements about protocols used in the TCP/IP protocol suite.
A. IP is a reliable connectionless protocol.
B. ARP finds a node's physical address from its logical address.
C. ICMP sends notification of datagram problems to the recipient of the message.
D. IGMP facilitates simultaneous transmission of a message to a group of recipients.
AB and D only
BD only
CA and C only
DA, C and D only
Answer:(C) A and C only
Explanation
IP is connectionless but does not provide reliable delivery, so A is incorrect. ICMP reports errors to the sender/source, not the recipient, so C is incorrect. ARP and IGMP are correctly described.
Choose the correct bottom-to-top order in the OSI model.
A. Transforming datagrams into frames for transmission
B. Transmission of bits over a communication channel
C. Handling the user interface
D. Control and monitoring of the subnet
E. Transmission using a connection-oriented or connectionless protocol
AA, B, C, D, E
BB, C, D, E, A
CB, A, D, E, C
DA, B, D, E, C
Answer:(C) B, A, D, E, C
Explanation
Physical-layer work is B, data-link framing is A, network-layer subnet control is D, transport-layer service is E, and user-facing handling is C. This gives B, A, D, E, C.
Which item is a problem in piggybacking acknowledgement frames?
A. There is no need to send separate acknowledgement frames.
B. It requires more bandwidth.
C. The data-link layer must decide how long to wait for a packet to which the acknowledgement can be attached.
D. The number of acknowledgement frames decreases.
AA and D only
BB only
CB and C only
DC only
Answer:(D) C only
Explanation
Piggybacking reduces separate acknowledgement frames and saves bandwidth, so A and D are benefits and B is false. Its drawback is deciding how long to delay an acknowledgement while waiting for outgoing data; that is C only.
What is the maximum Hamming distance among all pairs of the following binary codes?
Code
Binary value
C₁
00000
C₂
01011
C₃
10101
A2
B3
C4
D5
Answer:(C) 4
Explanation
C₁ and C₂ differ in three positions, C₁ and C₃ differ in three positions, while C₂ = 01011 and C₃ = 10101 differ in four positions. Thus the maximum pairwise Hamming distance is 4.
Choose the right sequence of stages for data transmission using CSMA/CD.
A. If the channel is clear, the sender sends data.
B. The sender senses the channel for activity.
C. The sender waits for a random back-off time before retransmitting.
D. Collision detection occurs if two devices transmit at the same time.
AA, B, D, C
BB, A, D, C
CB, A, C, D
DC, B, A, D
Answer:(B) B, A, D, C
Explanation
CSMA/CD first senses the medium, then transmits if it is idle. A collision may be detected during transmission, after which a random back-off precedes retransmission. This is B, A, D, C.
SMTP is an application-layer protocol, UDP is transport-layer, Ethernet is data-link-layer, and IP is network-layer. Thus the match is A-III, B-II, C-IV, D-I.
A single bit travels at 3 × 10⁸ m/s through a 12,000 km fibre-optic cable. What is its one-way propagation delay?
A20 ms
B5 ms
C50 ms
D40 ms
Answer:(D) 40 ms
Explanation
Convert 12,000 km to 1.2 × 10⁷ m. Divide distance by the stated speed: (1.2 × 10⁷)/(3 × 10⁸) = 0.04 seconds. Multiplying by 1,000 converts this to 40 ms, so option D is correct; 20 ms and 5 ms result from incorrect unit conversion or arithmetic.
A. It is called Uniform Data Protocol.
B. It is called User Datagram Protocol.
C. It belongs to the Transport layer.
D. It belongs to the Network layer.
AA and C only
BB and C only
CA and D only
DB and D only
Answer:(B) B and C only
Explanation
UDP expands to User Datagram Protocol and operates at the transport layer. IP is a network-layer protocol; “Uniform Data Protocol” is not UDP’s expansion.
What is the maximum bit rate of a noiseless channel with bandwidth 2 kHz transmitting a signal with 8 levels?
A12000 bps
B16000 bps
C4000 bps
D32000 bps
Answer:(A) 12000 bps
Explanation
By the Nyquist formula, maximum bit rate = 2B log₂L = 2 × 2000 × log₂8 = 4000 × 3 = 12,000 bps. Using 8 directly rather than log₂8 produces the larger distractors.
A. Data-link layer
B. Network layer
C. Transport layer
D. Presentation layer
List-I
List-II
A. Data-link B. Network C. Transport D. Presentation
I. Encryption II. Connection control III. Routing IV. Framing
AA-IV, B-III, C-I, D-II
BA-III, B-IV, C-II, D-I
CA-IV, B-II, C-III, D-I
DA-IV, B-III, C-II, D-I
Answer:(D) A-IV, B-III, C-II, D-I
Explanation
The data-link layer performs framing (IV), the network layer performs routing (III), the transport layer provides connection control (II), and the presentation layer handles encryption/representation (I).
Arrange the following steps for Pulse Code Modulation (PCM).
A. Sampling
B. Low-Pass filter
C. Encoding
D. Quantization
AA, B, C, D
BB, C, D, A
CB, A, D, C
DD, C, B, A
Answer:(C) B, A, D, C
Explanation
PCM first band-limits the analog signal with a low-pass filter, then samples it at discrete time intervals, then quantizes each sample to the nearest available level, and finally encodes each quantized value as a binary code.
B-A-D-C — Low-Pass filter, Sampling, Quantization, Encoding — is the correct sequence.
Which of the following statements are true for link state routing?
A. Link state Routing is suffer from count-to-infinity problem.
B. Link state Routing uses flooding in the network to distribute link state packets.
C. Each router creates its own link state packet.
D. Every router has exactly the same link state database.
E. Each router finds its own shortest paths to the other routers by using distance vector routing.
AA & E Only
BB, C & D Only
CA, B, C & D Only
DA, B & E Only
Answer:(B) B, C & D Only
Explanation
Link-state routing and distance-vector routing solve the same problem with very different mechanisms, and several of these statements swap one algorithm's property for the other's.
A. False. Count-to-infinity is a distance-vector problem, caused by routers slowly updating hop counts through each other, and link-state routing avoids it because every router computes paths from a complete, identical topology map.
B. True. Each router floods its link state packet to every other router in the network so that everyone converges on the same topology view.
C. True. Every router is responsible for creating the link state packet that describes its own directly connected links.
D. True. Because flooding delivers every router's link state packet to every other router, all routers end up building an identical link state database once the network converges.
E. False. Once a router has the complete topology database, it computes shortest paths using Dijkstra's algorithm, not distance vector routing, which is a separate algorithm that never builds a full topology map.
Match List-I (Network-layer address protocols) with List-II.
List-I
List-II
A. Address resolution protocol
I. Map physical address to IP address
B. Reverse address resolution protocol
II. Used by host and router to send notification of datagram problem back to the sender
C. Internet control message protocol
III. Help a multicast router identify hosts in a LAN
D. Internet Group message protocol
IV. Map IP address to physical address
AA-I, B-II, C-III, D-IV
BA-II, B-I, C-IV, D-III
CA-IV, B-II, C-III, D-I
DA-IV, B-I, C-II, D-III
Answer:(D) A-IV, B-I, C-II, D-III
Explanation
Each protocol resolves one direction of the address mapping or one specific network-layer notification role.
A-IV — Address Resolution Protocol maps a known IP address to the corresponding physical (MAC) address.
B-I — Reverse Address Resolution Protocol does the opposite, mapping a known physical address to an IP address.
C-II — Internet Control Message Protocol carries error and diagnostic notifications, such as an undeliverable datagram, back to the sending host or router.
D-III — Internet Group Management Protocol lets a multicast router discover which hosts on a LAN belong to a multicast group.
A-IV, B-I, C-II, D-III is therefore the correct match.
A signal travels through a transmission medium and its power is reduced to one-half of its original value, so that P₂ = 1/2 P₁. What is the power loss in attenuation?
A-6 dB
B-3 dB
C-1.5 dB
D-1 dB
Answer:(B) -3 dB
Explanation
Attenuation in decibels for a power ratio uses the formula
dB = 10 log₁₀(P₂ / P₁).
The negative sign shows a loss, so halving the power gives an attenuation of about -3 dB, matching option B.
Option A (-6 dB) comes from mistakenly using the amplitude/voltage formula dB = 20 log₁₀(P₂/P₁) instead of the power formula, which doubles the result to about -6.02 dB. Options C and D do not correspond to any standard relationship for a 50% power loss and are too small in magnitude.
Which of the following statements are true for digital-to-digital encoding schemes?
A. Unipolar encoding uses one level of value.
B. Unipolar encoded signal contains a direct current component (a component with zero frequency).
C. In NRZ-L, the level of the signal is dependent upon the state of the bit.
D. In NRZ-I, the signal is not inverted of a 1 is encountered.
E. RZ encoding scheme removes both DC components and problem of synchronization.
AA, B, C & E Only
BA, C, D & E Only
CB, C, D & E Only
DA, B, D & E Only
Answer:(A) A, B, C & E Only
Explanation
Evaluate each statement against the standard digital-to-digital encoding rules.
A. Unipolar encoding uses one level of value. — correct. Unipolar signaling represents one bit value with a single non-zero voltage level and the other bit value with zero voltage, so only one active signal level is used.
B. Unipolar encoded signal contains a direct current component. — correct. Because the signal never swings to a negative voltage, its average value is nonzero, which shows up as a DC component in the frequency spectrum.
C. In NRZ-L, the level of the signal is dependent upon the state of the bit. — correct. NRZ-L stands for non-return-to-zero level, where one voltage level directly represents bit 1 and the other level represents bit 0.
D. In NRZ-I, the signal is not inverted of a 1 is encountered. — incorrect. The defining rule of NRZ-I (non-return-to-zero invert) is the opposite: the signal level inverts whenever a 1 occurs and stays the same for a 0, so this statement misstates the standard rule.
E. RZ encoding scheme removes both DC components and problem of synchronization. — correct. Because the signal returns to zero at the middle of every bit interval, it produces a transition in every bit period, which supplies a clocking reference and avoids the long DC runs that unipolar and NRZ schemes suffer from.
Assertion (A): 1-persistent CSMA provides low delay under low load conditions.
Reason (R): When the channel is free, stations transmit immediately without any random delay, ensuring quick access and minimal delay.
ABoth A and R are correct and R is the correct explanation of A
BBoth A and R are correct but R is NOT the correct explanation of A
CA is correct but R is not correct
DA is not correct but R is correct
Answer:(A) Both A and R are correct and R is the correct explanation of A
Explanation
In 1-persistent CSMA, a station that wants to transmit senses the channel and, as soon as the channel becomes idle, transmits with probability 1, that is, immediately and without waiting any random backoff period.
Under low load, few stations are competing for the channel at any given time, so the chance that two stations sense the channel idle and transmit at the same instant is small. This lets stations gain access and finish transmitting quickly, giving low delay, so Assertion A is correct.
Reason R correctly describes this exact mechanism: transmitting immediately once the channel is sensed free, with no random delay, is precisely why access is quick under low contention. So R is both correct and the correct explanation of A.
Dialog control refers to managing whose turn it is to transmit in a conversation between two communicating systems, deciding whether communication is half-duplex or full-duplex.
The session layer of the OSI model is responsible for establishing, managing, and terminating sessions between applications, and dialog control is one of its core functions.
A. Network Layer — handles routing and logical addressing, not turn-taking between applications.
B. Transport Layer — handles end-to-end delivery and reliability, not who transmits next.
C. Session Layer — correct. It manages dialog control between communicating applications.
D. Presentation Layer — handles data translation, encryption, and compression, not dialog turn-taking.
Arrange the following interframe time interval in increasing order.
A. PIFS (PCF Interframe Spacing)
B. DIFS (DCF Interframe Spacing)
C. SIFS (Short Interframe Spacing)
D. EIFS (Extended Interframe Spacing)
AA, B, C, D
BB, C, A, D
CD, A, B, C
DC, A, B, D
Answer:(D) C, A, B, D
Explanation
IEEE 802.11 defines four interframe spacing intervals that control how long a station waits before transmitting, and their durations increase in a fixed order.
SIFS (Short Interframe Space) is used for the highest-priority frames such as ACKs, so it must be the smallest gap.
PIFS (PCF Interframe Space) is slightly longer and lets the point coordinator gain priority access over ordinary DCF stations.
DIFS (DCF Interframe Space) is longer still and is the standard wait time for contention-based access.
EIFS (Extended Interframe Space) is the longest, used after a station receives a corrupted frame it cannot interpret, giving other stations time to recover.
Mapping the labels to this order gives C (SIFS), A (PIFS), B (DIFS), D (EIFS), which is C, A, B, D.
Match List-I (IP addresses) with List-II (Address classes).
List-I
List-II
A. 190.10.75.250
I. Class E IP address
B. 230.41.10.20
II. Class B IP address
C. 242.9.20.80
III. Class A IP address
D. 127.10.125.230
IV. Class D IP address
AA-I, B-II, C-III, D-IV
BA-II, B-I, C-IV, D-III
CA-II, B-IV, C-I, D-III
DA-III, B-I, C-II, D-IV
Answer:(C) A-II, B-IV, C-I, D-III
Explanation
IP address classes are determined by the value of the first octet.
A. 190.10.75.250 falls in the 128–191 range, which is Class B, so A pairs with II.
B. 230.41.10.20 falls in the 224–239 range, which is Class D (multicast), so B pairs with IV.
C. 242.9.20.80 falls in the 240–255 range, which is Class E, so C pairs with I.
D. 127.10.125.230 falls in the 0–127 range, which is conventionally taught as Class A, so D pairs with III.
Assertion (A): Asymmetric key cryptography is significantly slower than symmetric key cryptography.
Reason (R): Asymmetric key cryptography is based on complex mathematical structure to secure data.
ABoth A and R are correct and R is the correct explanation of A
BBoth A and R are correct but R is NOT the correct explanation of A
CA is correct but R is not correct
DA is not correct but R is correct
Answer:(A) Both A and R are correct and R is the correct explanation of A
Explanation
Asymmetric (public-key) cryptography relies on computationally expensive operations such as large-integer modular exponentiation or elliptic-curve point multiplication, which makes it markedly slower than symmetric algorithms that use simple substitution and permutation on fixed-size blocks. Assertion A is therefore correct.
Reason R is also correct. Asymmetric schemes such as RSA and elliptic-curve cryptography derive their security from hard mathematical problems, such as integer factorization or the discrete logarithm problem, and these same complex structures are what make the underlying operations computationally heavy.
Because the heavy mathematical machinery described in R is precisely why asymmetric cryptography runs slower, R correctly explains A.
Match List-I (Network protocols) with List-II (OSI/TCP-IP layers).
List-I
List-II
A. SMTP
I. Application layer
B. BGP
II. Transport layer
C. TCP
III. Data Link Layer
D. PPP
IV. Network Layer
AA-I, B-II, C-III, D-IV
BA-II, B-I, C-IV, D-III
CA-IV, B-II, C-III, D-I
DA-I, B-IV, C-II, D-III
Answer:(D) A-I, B-IV, C-II, D-III
Explanation
SMTP operates at the Application layer since it defines how mail clients and servers exchange messages, so A pairs with I.
TCP is a core Transport-layer protocol responsible for reliable end-to-end delivery, so C pairs with II.
PPP is a Data Link-layer protocol used to establish point-to-point serial connections, so D pairs with III.
BGP is the protocol routers use to exchange reachability information between autonomous systems, and it is classified here as the remaining Network-layer routing protocol, so B pairs with IV.
A-I, B-IV, C-II, D-III is the only option that correctly places SMTP at Application, TCP at Transport, and PPP at Data Link, making it the internally consistent match.
A packet sent by a router or node to the source to inform it of congestion is called a:
AExplicit packet
BBackpressure packet
CChoke packet
DRetransmission packet
Answer:(C) Choke packet
Explanation
A choke packet explicitly tells a source to reduce its sending rate because congestion has been detected. Backpressure instead propagates congestion control hop by hop toward the source.
Which is not a characteristic of packet switching?
AData is broken into packets.
BEach packet may take a different route.
CIt requires a dedicated path.
DIt supports dynamic routing.
Answer:(C) It requires a dedicated path.
Explanation
Packet switching shares network links and routes packets independently, often dynamically. Requiring an exclusive end-to-end path is the defining property of circuit switching, not packet switching.
Arrange the steps of accessing a website after a user enters a URL: DNS resolution, HTTP request, URL entry, IP-address acquisition, and webpage display.
AB, A, C, E, D
BC, B, A, E, D
CB, C, D, A, E
DC, A, D, B, E
Answer:(D) C, A, D, B, E
Explanation
The user first enters the URL. The browser resolves the domain name, obtains an IP address from that resolution, sends the HTTP request to the server, and finally renders the received webpage. The order is C, A, D, B, E.
Editorially repaired version of an ambiguous NTA item. Consider these XML statements:
A. XML has no predefined tag set, and users can define their own tags.
B. XML is used primarily for presentation.
C. XML is case-sensitive.
D. XML supports dynamic data representation.
Which statements are true?
AB, C, and D only
BA, B, and D only
CA, C, and D only
DB and D only
Answer:(C) A, C, and D only
Explanation
XML has no fixed, predefined vocabulary, so applications may define their own element names. XML names are case-sensitive, and XML can represent data whose values or structure change over time. XML itself is not a presentation language; presentation is normally handled by CSS, XSLT, or another rendering layer. Thus A, C, and D are true.
Which are controlled-access protocols: reservation, polling, TDMA, token passing, and CSMA/CA?
AA and D only
BA, B, and D only
CC, D, and E only
DA, D, and E only
Answer:(B) A, B, and D only
Explanation
Controlled access coordinates stations so they take turns. Reservation, polling, and token passing are the standard controlled-access methods. TDMA is a channelization technique, while CSMA/CA is contention based.
Which are data-encoding schemes: NRZ, Manchester encoding, amplitude modulation, Hamming code, and bipolar AMI?
AA, C, and D only
BA, B, and D only
CC, D, and E only
DA, B, and E only
Answer:(D) A, B, and E only
Explanation
NRZ, Manchester, and bipolar AMI are line-encoding schemes. Amplitude modulation is a modulation technique and Hamming code is an error-control code, so neither belongs in the list of data-encoding schemes here.
The source prints 196 bits for Twofish; that typo is corrected to 256 bits below. Which encryption statements are true: symmetric-key cryptography requires secrecy of the key-generating function above the decrypting function; DES uses a 64-bit block and 56-bit key; Twofish uses a 128-bit block with variable keys up to 256 bits; and RC4 is a stream cipher operating on bits or bytes rather than blocks?
AB and C only
BA and D only
CA and C only
DB, C, and D only
Answer:(D) B, C, and D only
Explanation
Symmetric encryption depends on keeping the key secret, not the algorithm or a key-generating function. DES uses a 64-bit block and a 56-bit key; Twofish has a 128-bit block and supports keys up to 256 bits; and RC4 is a stream cipher. Therefore B, C, and D are true. The source's printed 196-bit maximum has been explicitly corrected to 256 bits in this learner-facing version.
Match count-to-infinity, IEEE 802.16, IEEE 802.15, and AODV with their networking associations.
List I
List II
A. Count-to-infinity problem
I. Mobile Adhoc Networks
B. IEEE 802.16
II. Distance-vector routing
C. IEEE 802.15
III. WiMAX
D. AODV
IV. Bluetooth
AA-II, B-III, C-I, D-IV
BA-II, B-III, C-IV, D-I
CA-II, B-IV, C-III, D-I
DA-IV, B-III, C-I, D-II
Answer:(B) A-II, B-III, C-IV, D-I
Explanation
Count-to-infinity is a distance-vector-routing problem. IEEE 802.16 is WiMAX, IEEE 802.15 includes Bluetooth personal-area networking, and AODV is used in mobile ad hoc networks. The mapping is A-II, B-III, C-IV, D-I.
Match digital signature, hash function, AES, and RSA with their security roles.
List I
List II
A. Digital Signature
I. Asymmetric encryption algorithm
B. Hash Function
II. Confirms authenticity and integrity
C. AES
III. Produces a fixed-size digest
D. RSA
IV. Symmetric encryption algorithm
AA-II, B-III, C-I, D-IV
BA-II, B-III, C-IV, D-I
CA-III, B-II, C-I, D-IV
DA-IV, B-II, C-I, D-III
Answer:(B) A-II, B-III, C-IV, D-I
Explanation
A digital signature confirms authenticity and integrity; a hash function produces a fixed-size digest; AES is a symmetric encryption algorithm; and RSA is asymmetric. The matching is A-II, B-III, C-IV, D-I.
The primary function of a firewall in network security is
Ato monitor network traffic and detect anomalies
Bto create virtual private networks for secure communication
Cto filter incoming and outgoing network traffic based on security rules
Dto encrypt data before transmission
Answer:(C) to filter incoming and outgoing network traffic based on security rules
Explanation
A firewall enforces access-control rules by allowing, denying or inspecting traffic at a network boundary. VPN creation, intrusion detection and encryption are separate security functions. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.
Compare corresponding bit positions: the second, fourth and fifth bits differ. Hence three substitutions are needed, so the Hamming distance is 3. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.
A. Service-level agreement B. Virtualisation C. Internet of Things D. Platform as a service
I. Connecting physical devices over the Internet II. Development environment over the Internet III. Contracts specifying performance and availability IV. Multiple operating systems on one physical server
AA-III, B-II, C-I, D-IV
BA-III, B-IV, C-I, D-II
CA-III, B-I, C-IV, D-I
DA-I, B-IV, C-III, D-II
Answer:(B) A-III, B-IV, C-I, D-II
Explanation
A service-level agreement specifies performance and availability, virtualisation permits multiple operating systems on one physical server, IoT connects physical devices, and PaaS provides an online development/deployment platform.
A. In connection-oriented service, the destination address is specified only during connection setup.
B. Packet sequencing is not guaranteed in connection-oriented service.
C. Flooding sends an incoming packet on every outgoing line except the arrival line.
D. The presentation layer performs error detection and recovery.
AA, B and C only
BB, C and D only
CA and C only
DA, C and D only
Answer:(C) A and C only
Explanation
Connection-oriented delivery keeps connection state and preserves ordering, so B is false. Flooding is correctly described by C, and error recovery belongs to lower layers rather than the presentation layer; hence A and C are correct.
A. DHCP ACK
B. DHCP DISCOVER
C. DHCP OFFER
D. DHCP REQUEST
E. Client receives IP configuration
AA, B, C, D, E
BB, C, D, A, E
CC, B, A, E, D
DD, A, E, C, B
Answer:(B) B, C, D, A, E
Explanation
DHCP follows the DORA exchange: Discover, Offer, Request, Acknowledge, then the client applies the received configuration. This is B, C, D, A, E. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.
Identify the correct statements about multiple access techniques.
A. ALOHA is used for multiple access on a shared medium.
B. TDMA uses Walsh tables for time-slot allocation.
C. CSMA/CD was invented for wireless networks.
D. CDMA codes are sequences generated by orthogonal codes.
AA and B only
BB and C only
CC and D only
DA and D only
Answer:(D) A and D only
Explanation
ALOHA is a shared-medium multiple-access scheme and CDMA uses orthogonal spreading codes. Walsh codes are associated with CDMA, while CSMA/CD was designed for wired Ethernet, not wireless networks.
I. Retrieves email from a mail server II. Remote login over a network III. Maps IP addresses to MAC addresses on a LAN IV. Accesses email on the server without downloading
AA-II, B-I, C-III, D-IV
BA-III, B-IV, C-I, D-II
CA-IV, B-III, C-I, D-II
DA-IV, B-III, C-II, D-I
Answer:(C) A-IV, B-III, C-I, D-II
Explanation
IMAP accesses email on the server without downloading it; ARP maps IP addresses to MAC addresses; POP retrieves email; TELNET provides remote login. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.
Sequence the following stages of the sliding-window protocol for flow control.
A. Sender moves the window to the next set of frames.
B. Sender sends a window of frames.
C. Receiver acknowledges the received frames.
D. If acknowledgement is not received, sender retransmits the frame.
AB, A, C, D
BB, C, D, A
CB, D, C, A
DB, C, A, D
Answer:(D) B, C, A, D
Explanation
The sender sends a window, the receiver acknowledges it, and the sender advances its window. Retransmission is the fallback when an acknowledgement is not received, giving B, C, A, D. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.
Consider a noiseless channel with bandwidth 5000 Hz transmitting a signal with two signal levels. The maximum bit rate is
A2500 bps
B10000 bps
C5000 bps
D20000 bps
Answer:(B) 10000 bps
Explanation
By Nyquist’s formula, maximum bit rate = 2B log₂L = 2 × 5000 × log₂2 = 10000 bps. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.
A CSMA/CD network has bandwidth 20 Mbps. If maximum propagation time is 25 μs, what is the minimum frame size?
A500 bits
B1000 bits
C1500 bits
D2000 bits
Answer:(B) 1000 bits
Explanation
For collision detection, a frame must last at least the round-trip propagation time: 2 × 25 μs. At 20 Mbps this is 20×10⁶ × 50×10⁻⁶ = 1000 bits. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.
Which of the following are key features of Mobile Ad hoc Networks (MANETs)?
A. Self-organizing and decentralized
B. High mobility of nodes
C. Requires fixed infrastructure for operation
D. Network topology is dynamic and constantly changing
E. Requires a central controller to manage network traffic
AA, B and E only
BB, C, D and E only
CA, B and D only
DA, B and C only
Answer:(C) A, B and D only
Explanation
MANETs are infrastructure-free, decentralized and self-organizing, with mobile nodes and changing topology. They do not require a fixed infrastructure or a central controller. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.
In a token ring network the transmission speed is 10⁷ bps and the propagation speed is 200 m/μs. The 1-bit delay in this network is equivalent to ________________
A20 m
B30 m
C50 m
D40 m
Answer:(A) 20 m
Explanation
The transmission speed is 10⁷ bps, so one bit occupies 1/10⁷ s = 0.1 μs on the wire. In that time the signal advances 0.1 μs × 200 m/μs = 20 m. The 1-bit delay is therefore equivalent to 20 metres of cable, so option A is correct.
Consider a CSMA/CD network that transmits
data at the rate of 100 Mbps over a 1 Kilometre
cable with no repeater. If the minimum frame size
required for this network is 1250 bytes, what is
the signal speed (km/sec) in the cable.
A8000
B16000
C12000
D20000
Answer:(D) 20000
Explanation
The minimum frame of 1250 bytes is 10,000 bits, which at 100 Mbps takes 10,000/10⁸ = 10⁻⁴ s to transmit. Collision detection requires that time to cover the round trip of 2 × 1 km. The signal speed is therefore 2 km / 10⁻⁴ s = 20,000 km/s — option D.
Host A (on TCP/IPv4 Networks) send an IP
Datagram D to host B (also on TCP/IPv4 network).
Assume that no error occurred during
transmission of D, when D reaches B. Which of
following header field may differ from that of
original datagram D ?
A. TTL
B. Checksum
C. Fragment offset
D. Source IP
E. Destination IP
A(A) and (B) Only
B(A), (B) and (C) Only
C(A), (B), (C) and (D) Only
D(A), (B), (C) and (E) Only
Answer:(B) (A), (B) and (C) Only
Explanation
TTL is decremented by every router, so (A) changes. The header checksum covers TTL and must be recomputed at each hop, so (B) changes. Fragment offset changes whenever the datagram is fragmented in transit, so (C) may differ. Source and destination IP are never rewritten by ordinary routing, so (D) and (E) stay fixed. Only (A), (B) and (C) — option B.
Arrange the following steps in a proper sequence
for the typical process of a DNS query :
A. Query authoritative DNS Server
B. Query local DNS Server
C. Local DNS Server Check Cache
D. Query root DNS Server
E. Query TLD DNS Server
A(A), (C), (E), (B), (D)
B(E), (A), (C), (B), (D)
C(A), (B), (C), (D), (E)
D(B), (C), (D), (E), (A)
Answer:(D) (B), (C), (D), (E), (A)
Explanation
Resolution starts at the local recursive resolver (B), which first checks its cache (C). On a miss it queries a root server (D), which refers it to the TLD server (E), which in turn refers it to the authoritative server (A) holding the record. That order is option D.
A class B network on the Internet has a subnet
mask of 255.255.248.0, what is the maximum
number of hosts per subnet?
A1022
B1023
C2046
D2047
Answer:(C) 2046
Explanation
The mask 255.255.248.0 is 11111111.11111111.11111000.00000000, leaving 11 zero bits for the host field. That gives 2¹¹ = 2048 addresses per subnet, less the network and broadcast addresses, so 2046 usable hosts — option C.
Consider the three way hand shaking process
followed during TCP connection establishment
between two hosts A and B . Let S and R be two
random 32 -bit starting sequence numbers
chosen by A and B, respectively. Suppose A
sends a TCP segment having SYN bit =1, SEQ
number = S and ACK bit =0 and B accepts the
connection request.
Which one of the following choices represents
the information present in the TCP segment
header that is sent by B to A ?
ASYN bit =1, SEQ number =S+1, ACK bit =0, ACK number =R, FiN bit =0
BSYN bit =0, SEQ number =R, ACK bit =1, ACK number =S+1, FiN bit =0
CSYN bit =1, SEQ number =R, ACK bit =1, ACK number =S+1, FiN bit =0
DSYN bit =1, SEQ number = R, ACK bit =1, ACK number =S, FiN bit =0
Answer:(C) SYN bit =1, SEQ number =R, ACK bit =1, ACK number =S+1, FiN bit =0
Explanation
B’s reply is a SYN-ACK. It carries B’s own initial sequence number R, sets SYN = 1 because B is opening its own direction of the connection, sets ACK = 1, and acknowledges A’s stream with ACK number S + 1. FIN stays 0. Option C matches all five fields.
Two computers A and B are configured as follows
: A has IP address 203.197.2.53 and subnet mask
255.255.128.0, B has IP address 203.197.75.201
and subnet mask 255.255.192.0. What one of the
following statement is true?
AA and B both assume they are on the same network
BB assumes, A is on same network, A assumes B is on different network
CA assumes, B is on same network, but B assumes, A is on a different network
DA and B both assume they are on different network
Answer:(C) A assumes, B is on same network, but B assumes, A is on a different network
Explanation
A applies its own mask to both addresses: 203.197.2.53 and 203.197.75.201 both reduce to 203.197.0.0, so A concludes B is local. B applies its own mask: its address reduces to 203.197.64.0 but A’s reduces to 203.197.0.0, so B concludes A is remote. Option C describes exactly this asymmetry.
A. In IPv4 addressing, a block of address can be
define as x.y.z.t/ n in which x.y.z.t define one of
the address and / n define the mask
B. The first address in the block can be found by
setting the rightmost 32-n bits to 0s
C. Address of class C is used for multicast
communication
D. There are five classes in IPv4 address
E. Supernetting combine several networks into
one large network
A(A), (B) and (C) Only
B(A), (B), (D) and (E) Only
C(A), (C), (D) and (E) Only
D(C), (D) and (E) Only
Answer:(B) (A), (B), (D) and (E) Only
Explanation
(A) is true — x.y.z.t/n names a block by one address and a prefix length. (B) is true — zeroing the rightmost 32 − n bits yields the first address of the block. (D) is true — IPv4 defines classes A through E. (E) is true — supernetting aggregates contiguous blocks into one larger route. (C) is false: multicast uses class D, not class C. So (A), (B), (D) and (E) — option B.
Has functions are used to produce the message digests which are then encrypted with a private key to get:
APublic key
BDigital signature
CCipher text
DData Encryption Standard
Answer:(B) Digital signature
Explanation
The key idea is stated in the problem: Has functions are used to produce the message digests which are then encrypted with a private key to get:. The correct answer is B: Digital signature. It follows the governing definition, algorithm, or calculation in this data communication and computer networks item; the other choices fail to satisfy one or more stated conditions.
A multiplexes combines for 100 Kbps channels using a time slot of 2 bits. What is the bit rate ?
A100 Kbps
B200 Kbps
C40 Kbps
D1000 Kbps
Answer:(C) 40 Kbps
Explanation
The key idea is stated in the problem: A multiplexes combines for 100 Kbps channels using a time slot of 2 bits. What is the bit rate ?. The correct answer is C: 40 Kbps. It follows the governing definition, algorithm, or calculation in this data communication and computer networks item; the other choices fail to satisfy one or more stated conditions.
Which of the following is not a field in TCP header ?
ASequence Number
BChecksum
CFregmentation offset
DWindow size
Answer:(C) Fregmentation offset
Explanation
The key idea is stated in the problem: Which of the following is not a field in TCP header ?. The correct answer is C: Fregmentation offset. It follows the governing definition, algorithm, or calculation in this data communication and computer networks item; the other choices fail to satisfy one or more stated conditions.
In "bit stuffing", each frame begins and ends with a bit pattern in hexadecimal ?
AOx8c
BOx6F
COxEF
DOx7E
Answer:(D) Ox7E
Explanation
The key idea is stated in the problem: In "bit stuffing", each frame begins and ends with a bit pattern in hexadecimal ?. The correct answer is D: Ox7E. It follows the governing definition, algorithm, or calculation in this data communication and computer networks item; the other choices fail to satisfy one or more stated conditions.
Which of the following are example of CSMA channel sensing methods ?
A. 1-persistent
B. 2-persistent
C. p-persistent
D. o-persistent Choose the correct answer from the options given below :
The work done by UDP is/are :
A. Congestion control
B. Flow control
C. Retransmission
D. Segments transmission Choose the correct answer from the options given below : (Q)
A. and
D. Only
AA and D only
BC only
CD only
DB and C only
Answer:(C) D only
Explanation
The key idea is stated in the problem: The work done by UDP is/are :
A. Congestion control
B. Flow control
C. Retransmission
D. Segments transmission Choose the correct answer from the options given below : (Q)
A. and
D. Only. The correct answer is C: D only. It follows the governing definition, algorithm, or calculation in this data communication and computer networks item; the other choices fail to satisfy one or more stated conditions.
Which of the following statement/s is/are NOT CORRECT ?
A. OSPF is based on distance-vector routing protocol.
B. Both link-state and distance-vector routing are based on the least cost goal.
C. BGP4 is based on the path-vector algorithm.
D. The three-node instability can be avoided using split horizon combined with poison reverse.
E. _ RIP is based on link state algorithm. Choose the correct answer from the options given below :
The key idea is stated in the problem: Which of the following statement/s is/are NOT CORRECT ?
A. OSPF is based on distance-vector routing protocol.
B. Both link-state and distance-vector routing are based on the least cost goal.
C. BGP4 is based on the path-vector algorithm.
D. The three-node . The correct answer is A: (A), (D) and &) Only. It follows the governing definition, algorithm, or calculation in this data communication and computer networks item; the other choices fail to satisfy one or more stated conditions.
Which of the following statement/s are CORRECT ?
A. NRZ is a bipolar scheme in which the positive voltage define bit is 0 (zero).
B. NRZ-L and NRZ-I both have an average signal rate of N/2.
C. The idea of RZ and NRZ-L are combined into Manchester scheme.
D. NRZ-L and NRZ-I both have DC component problems.
E. The minimum bandwidth of Manchester and differential Manchester is 3 times that of NRZ. Choose the correct answer from the options given below :
Which of the following are commonly used parsing techniques in NLP (Natural Language Processing) for syntatic analysis.
A. Top down parsing
B. Bottom Up parsing
C. Dependency parsing
D. _ Statistical machine translation
E. Earley parsing Choose the correct answer from the options given below :
Evaluate these statements: Statement I: If H is non empty finite subset of a group G and abeH V a, beH, then H is also a group. Statement II: There is no homomorphism exist from (Z, +) to (Q, +); where Z is set of integers and Q is set of rational number. In the light of the above statements, choose the most appropriate answer from the options given below :
ABoth Statement I and Statement II are correct
BBoth Statement I and Statement II are incorrect
CStatement I is correct but Statement II is incorrect
DStatement I is incorrect but Statement II is correct
Answer:(C) Statement I is correct but Statement II is incorrect
Explanation
The key idea is stated in the problem: Evaluate these statements: Statement I: If H is non empty finite subset of a group G and abeH V a, beH, then H is also a group. Statement II: There is no homomorphism exist from (Z, +) to (Q, +); where Z is set of integers and Q is set of rational number. In t. The correct answer is C: Statement I is correct but Statement II is incorrect. It follows the governing definition, algorithm, or calculation in this data communication and computer networks item; the other choices fail to satisfy one or more stated conditions.
Evaluate these statements: Statement I: A thread is a dispatchable unit of work that does not executes sequentially and is not interruptible Statement II: It is not possible to alter the behaviour of a thread by altering its context when thread is suspended. In the light of the above statements, choose the most appropriate answer from the options given below :
ABoth Statement I and Statement II are correct
BBoth Statement I and Statement II are incorrect
CStatement I is correct but Statement II is incorrect
DStatement I is incorrect but Statement II is correct
Answer:(B) Both Statement I and Statement II are incorrect
Explanation
The key idea is stated in the problem: Evaluate these statements: Statement I: A thread is a dispatchable unit of work that does not executes sequentially and is not interruptible Statement II: It is not possible to alter the behaviour of a thread by altering its context when thread is suspended. I. The correct answer is B: Both Statement I and Statement II are incorrect. It follows the governing definition, algorithm, or calculation in this data communication and computer networks item; the other choices fail to satisfy one or more stated conditions.
Evaluate these statements: Statement I: In Reuse Oriented Model, Modification of the old system parts appropriate to the new requirements. Statement II: In Reuse Oriented Model, Integration of the modified parts are not possible into the new systems. In the light of the above statements, choose the most appropriate answer from the options given below :
ABoth Statement I and Statement II are correct
BBoth Statement I and Statement II are incorrect
CStatement I is correct but Statement II is incorrect
DStatement I is incorrect but Statement II is correct
Answer:(C) Statement I is correct but Statement II is incorrect
Explanation
The key idea is stated in the problem: Evaluate these statements: Statement I: In Reuse Oriented Model, Modification of the old system parts appropriate to the new requirements. Statement II: In Reuse Oriented Model, Integration of the modified parts are not possible into the new systems. In the li. The correct answer is C: Statement I is correct but Statement II is incorrect. It follows the governing definition, algorithm, or calculation in this data communication and computer networks item; the other choices fail to satisfy one or more stated conditions.
Evaluate these statements: Statement I: In datagram networks - routers hold state information about connections Statement II: In virtual circuit network- each virtual circuit requires router table space per connection In the light of the above statements, choose the most appropriate answer from the options given below :
ABoth Statement I and Statement II are correct
BBoth Statement I and Statement II are incorrect
CStatement I is correct but Statement II is incorrect
DStatement I is incorrect but Statement II is correct
Answer:(D) Statement I is incorrect but Statement II is correct
Explanation
The key idea is stated in the problem: Evaluate these statements: Statement I: In datagram networks - routers hold state information about connections Statement II: In virtual circuit network- each virtual circuit requires router table space per connection In the light of the above statements, choo. The correct answer is D: Statement I is incorrect but Statement II is correct. It follows the governing definition, algorithm, or calculation in this data communication and computer networks item; the other choices fail to satisfy one or more stated conditions.
Food X contains 6 units of Vitamin D per gram and 7 units of Vitamin E per gram and costs Rs. 12 per gram. Food Y contains 8 units of Vitamin D per gram and 12 units of Vitamin E per gram and costs Rs. 20 per gram. The daily minimum requirements of vitamin D and vitamin E are 100 units and 120 units respectively. Let x and y be the quantities in grams of Food X and Food Y.
AZ=6x+7y
BZ=8x+12y
CZ=12x +20y
DZ=20x+12y
Answer:(C) Z=12x +20y
Explanation
The key idea is stated in the problem: The cost funtion of total food is :. The correct answer is C: Z=12x +20y. It follows the governing definition, algorithm, or calculation in this data communication and computer networks item; the other choices fail to satisfy one or more stated conditions.
Number of tuples by applying right outer join on relation X and Y is/are :
X (S, Si, C)
Y (S, P, D)
(J, 1, M)
(B, 2, N)
(R, 3, H)
(T, 4, G)
(J, S₁, CA)
(B, P₁, AB)
(R, D₁, DC)
(A, H₁, MD)
A16
B5
C3
D4
Answer:(D) 4
Explanation
The key idea is stated in the problem: Number of tuples by applying right outer join on relation X and Y is/are :. The correct answer is D: 4. It follows the governing definition, algorithm, or calculation in this data communication and computer networks item; the other choices fail to satisfy one or more stated conditions.
A sports website has 20,000 concurrent users. Each user submits a request once every 2 minutes on average, and each transaction downloads a new article of average length 3 kilobytes. What is the throughput?
A8 megabits/s
B4 megabits/s
C6 megabits/s
D2 megabits/s
Answer:(B) 4 megabits/s
Explanation
Throughput here is the total data delivered per second across all users, found by first computing the request rate and then the resulting data rate.
A TCP server application is programmed to listen on port P on host S. A TCP client is connected to the TCP server over the network. While the TCP connection is active, the server crashes and reboots. Assume that the client does not use a TCP keepalive timer.
Statement I: The TCP application server on S can listen on port P after the reboot.
Statement II: If the client sends a packet after the server reboots, it will receive an RST segment.
ABoth Statement I and Statement II are true
BBoth Statement I and Statement II are false
CStatement I is true but Statement II is false
DStatement I is false but Statement II is true
Answer:(A) Both Statement I and Statement II are true
Explanation
A crash and reboot wipes out all of the server's in-memory TCP connection state, which changes how the OS and application respond afterward.
Statement I is true because rebooting clears any prior binding to port P, so once the server application restarts, it is free to bind and listen on P again.
Statement II is true because the rebooted server no longer has any record of the old connection, so when it receives a segment belonging to that now-unknown connection, TCP responds with an RST segment to tell the client the connection no longer exists.
Since neither the port binding nor the missing connection state blocks these behaviors, both statements hold.
In the standard Ethernet with transmission rate of 10 Mbps, assume that the length of the medium is 2500 m and size of a frame is 512 bytes. The propagation speed of a signal in a cable is normally 2 × 10⁸ m/s. The transmission delay and propagation delay are
A25.25 μs and 51.2 μs
B51.2 μs and 12.5 μs
C10.24 μs and 50.12 μs
D12.5 μs and 51.2 μs
Answer:(B) 51.2 μs and 12.5 μs
Explanation
Propagation delay is the medium's length divided by the signal's propagation speed:
Propagation delay = 2500 m / (2 × 10⁸ m/s) = 1.25 × 10⁻⁵ s = 12.5 μs
Transmission delay is the frame size divided by the link's bit rate. This is the standard textbook Ethernet example built around the 64-byte minimum Ethernet frame, which equals 512 bits, so the "512" in the question is a bit count rather than a byte count:
Transmission delay = 512 bits / (10 × 10⁶ bps) = 5.12 × 10⁻⁵ s = 51.2 μs
Together this gives a transmission delay of 51.2 μs and a propagation delay of 12.5 μs, in that order.
A. 25.25 μs and 51.2 μs — neither value matches the computed transmission or propagation delay.
B. 51.2 μs and 12.5 μs — correct, matches the transmission delay followed by the propagation delay.
C. 10.24 μs and 50.12 μs — neither value matches the computed delays.
D. 12.5 μs and 51.2 μs — has the correct two values but in reversed order (propagation before transmission), while the question asks for transmission delay first.
Based on this IP datagram, what is its transmission efficiency, calculated as payload size divided by total datagram size?
An IPv4 datagram header, shown in hexadecimal, begins: 45 00 00 54 00 03 00 00 20 06 .....
The first byte (45) splits into a version of 4 and an IHL of 5, giving a header length of 5 x 4 = 20 bytes. The next two bytes (00 54) give the total datagram length as 0x0054 = 84 bytes. The flags and fragment offset bytes (00 00) show no fragmentation. The Time to Live byte (20) equals 0x20 = 32. The protocol byte (06) equals 6, which identifies TCP.
A76.19%
B80.50%
C82.24%
D85.45%
Answer:(A) 76.19%
Explanation
Decoding the header gives a total datagram length of 84 bytes and a header length of 5 x 4 = 20 bytes, so the payload carried inside the datagram is 84 - 20 = 64 bytes.
Efficiency = payload size / total size = 64 / 84 = 0.7619, or 76.19%.
Which transport-layer protocol is being carried as the payload of this IP datagram?
An IPv4 datagram header, shown in hexadecimal, begins: 45 00 00 54 00 03 00 00 20 06 .....
The first byte (45) splits into a version of 4 and an IHL of 5, giving a header length of 5 x 4 = 20 bytes. The next two bytes (00 54) give the total datagram length as 0x0054 = 84 bytes. The flags and fragment offset bytes (00 00) show no fragmentation. The Time to Live byte (20) equals 0x20 = 32. The protocol byte (06) equals 6, which identifies TCP.
AICMP
BSCTP
CTCP protocol
DIGMP
Answer:(C) TCP protocol
Explanation
The protocol field, the tenth byte of the header, holds the value 06 in hexadecimal, which is 6 in decimal. In the IANA protocol number assignments used by IPv4, protocol number 6 identifies TCP, while 1 is ICMP, 2 is IGMP, and 132 is SCTP. So the payload inside this datagram is a TCP segment.
An IPv4 datagram header, shown in hexadecimal, begins: 45 00 00 54 00 03 00 00 20 06 .....
The first byte (45) splits into a version of 4 and an IHL of 5, giving a header length of 5 x 4 = 20 bytes. The next two bytes (00 54) give the total datagram length as 0x0054 = 84 bytes. The flags and fragment offset bytes (00 00) show no fragmentation. The Time to Live byte (20) equals 0x20 = 32. The protocol byte (06) equals 6, which identifies TCP.
A64 bytes
B74 bytes
C84 bytes
D104 bytes
Answer:(C) 84 bytes
Explanation
The total length field occupies the third and fourth bytes of the header, 00 54 in hexadecimal, which equals 0x0054 = 84 in decimal. This total length field already reports the full size of the IP datagram, header plus payload, so the datagram is 84 bytes.
Based on the Time to Live value carried in this IP datagram, through how many more routers can the packet still travel before being discarded?
An IPv4 datagram header, shown in hexadecimal, begins: 45 00 00 54 00 03 00 00 20 06 .....
The first byte (45) splits into a version of 4 and an IHL of 5, giving a header length of 5 x 4 = 20 bytes. The next two bytes (00 54) give the total datagram length as 0x0054 = 84 bytes. The flags and fragment offset bytes (00 00) show no fragmentation. The Time to Live byte (20) equals 0x20 = 32. The protocol byte (06) equals 6, which identifies TCP.
A22
B26
C30
D32
Answer:(D) 32
Explanation
The Time to Live byte in the header is 20 in hexadecimal, which equals 32 in decimal. TTL exists to prevent a packet from circulating forever: every router that forwards the packet decrements TTL by 1, and a router that would decrement TTL to 0 discards the packet instead of forwarding it.
Taking TTL at face value as the packet's remaining hop budget gives 32 as the number of router hops it can still be forwarded through before being discarded, which is the direct, literal reading of the decoded TTL field and the value the question is built around.
An IPv4 datagram header, shown in hexadecimal, begins: 45 00 00 54 00 03 00 00 20 06 .....
The first byte (45) splits into a version of 4 and an IHL of 5, giving a header length of 5 x 4 = 20 bytes. The next two bytes (00 54) give the total datagram length as 0x0054 = 84 bytes. The flags and fragment offset bytes (00 00) show no fragmentation. The Time to Live byte (20) equals 0x20 = 32. The protocol byte (06) equals 6, which identifies TCP.
A10 bytes
B20 bytes
C30 bytes
D40 bytes
Answer:(B) 20 bytes
Explanation
The header length is derived from the IHL field, the second half of the first header byte. The byte 45 in hex splits into version = 4 and IHL = 5. IHL is measured in units of 4-byte words, so header size = 5 x 4 = 20 bytes.
Consider two hosts P and Q connected through router R. The MTU of link P-R is 1500 bytes, and the MTU of link R-Q is 820 bytes. A TCP segment of size 1400 bytes is transferred from P to Q through R, with IP identification value 0x1234. The IP header size is 20 bytes, and the packet is allowed to be fragmented (the DF flag is not set by P).
A. Two fragments are created at R, and the IP datagram size carrying the second fragment is 620 bytes
B. If the second fragment is lost, then R resends the fragment with IP identification value 0x1234
C. If the second fragment is lost, then P is required to resend the entire TCP segment
D. The TCP destination port can be determined by analyzing the second fragment only
Which of the above statements are correct?
AA, B and C only
BA and C only
CC and D only
DB and D only
Answer:(B) A and C only
Explanation
At P, the IP datagram carrying the 1400-byte TCP segment plus a 20-byte IP header totals 1420 bytes, which fits within the P-R link's 1500-byte MTU, so no fragmentation happens at P.
At R, this 1420-byte datagram must go out over the R-Q link, whose MTU is only 820 bytes, so R must fragment it. Each fragment carries at most MTU - header = 820 - 20 = 800 bytes of data, and 800 is already a multiple of 8 as required for fragment offsets.
Fragment 1 carries 800 bytes of data, total size 800 + 20 = 820 bytes, offset 0.
Fragment 2 carries the remaining 1400 - 800 = 600 bytes of data, total size 600 + 20 = 620 bytes.
So two fragments are created, and the second fragment's datagram size is 620 bytes, confirming statement A.
Fragment loss recovery in IP/TCP happens end to end, not hop by hop. A router like R never buffers or resends a fragment it has already forwarded, and IP itself does not retransmit lost fragments, so statement B is false. Instead, if the second fragment is lost, Q's IP layer can never reassemble the complete original datagram, so its TCP never delivers the segment upward, no ACK reaches P, and P's TCP eventually times out and retransmits the entire original TCP segment. This confirms statement C.
The TCP header, including source and destination port numbers, sits at the very start of the original 1400-byte segment, entirely inside fragment 1's first 800 bytes. Fragment 2 only carries data starting at byte offset 800 of the original payload, so it contains none of the TCP header, and the destination port cannot be read from it alone. This makes statement D false.
Only A and C are true, matching the option A and C only.
A Computer on a 10 Mbps network is regulated by a token bucket. The token bucket is filled at a rate of 2 Mbps. It is initially filled to capacity with 16 megabits. What is the maximum duration for which the computer can transmit at the full 10 Mbps?
A1.6 seconds
B2 seconds
C5 seconds
D8 seconds
Answer:(B) 2 seconds
Explanation
While transmitting at 10 Mbps, the bucket keeps refilling at 2 Mbps, so the net drain rate is 10 − 2 = 8 Mbps.
The bucket starts with 16 megabits stored, and it empties when those 16 megabits are consumed at the net drain rate.
In the standard Ethernet with transmission rate of 10Mbps, assume that the length of the medium is 2500m and size of a frame is 512 bits. The propagation speed of a signal in a cable is normally 2 × 10⁸ m/s. Calculate Transmission delay and propagation delay.
A25.25μs and 51.2μs
B51.2μs and 12.5μs
C10.24μs and 50.12μs
D12.5μs and 51.2μs
Answer:(B) 51.2μs and 12.5μs
Explanation
Transmission delay is frame size divided by the link's transmission rate.
Transmission delay = 512 bits / 10 Mbps = 51.2 μs
Propagation delay is the cable length divided by the signal's propagation speed.
Consider two hosts P and Q connected through a router R. The maximum transmit unit (MTU) value of the link between P and R is 1500 bytes and between R and Q is 820 bytes. A TCP segment of size of 1400 bytes is transferred from P to Q through R with IP identification value of 0x1234. Assume that IP header size is 20 bytes. Further the packet is allowed to be fragmented, that is Don't Fragment (DF) flag in the IP Header is not set by P.
A. Two fragments are created at R and IP datagram size carrying the second fragment is 620 bytes
B. If the second fragment is lost then R resend the fragment with IP identification value of 0x1234
C. If the second fragment lost then P required to resend the entire TCP segment
D. TCP destination port can be determined by analyzing the second fragment only
Which of the following statement is/are true. Choose the correct answer from the options given below:
AA, B and C only
BA and C only
CC and D Only
DB and D Only
Answer:(B) A and C only
Explanation
The 1400-byte TCP segment plus its 20-byte IP header forms a 1420-byte datagram, which fits under the 1500-byte P-R MTU, so R receives it whole and must refragment its 1400 bytes of payload to fit the 820-byte R-Q MTU.
Each fragment can carry at most 820-20=800 bytes of data, which is already a multiple of 8 so no further rounding is needed.
First fragment: 800 bytes data + 20-byte header = 820 bytes total.
Second fragment: 1400-800=600 bytes data + 20-byte header = 620 bytes total.
This gives exactly two fragments, with the second one totalling 620 bytes, matching statement A.
Routers never retransmit lost IP fragments themselves, that responsibility belongs to the original TCP sender, so statement B's claim that R resends the fragment is false.
Because TCP operates on whole segments and IP has no fragment-level retransmission, P is the one that must resend the entire TCP segment if any fragment is lost, matching statement C.
Only the first fragment, at offset zero, carries the TCP header with the port numbers, so a later fragment alone cannot reveal the destination port, making statement D false.
The Solution to Silly Window Syndrome problem is/are:
A. Nagle's Algorithm
B. Clark's Algorithm
C. Jacobson's Algorithm
D. Piggy backing Algorithm
Choose the correct answer from the options given below:
AA and B only
BA and C only
CC and D only
DB and D only
Answer:(A) A and B only
Explanation
Silly Window Syndrome has two named textbook fixes, one for each end of the connection.
Nagle's algorithm addresses the sender side by delaying small outgoing writes until there is enough data or an acknowledgment arrives, and Clark's algorithm addresses the receiver side by not advertising small window increases until a full segment's worth of space opens up.
Jacobson's algorithm is about round-trip-time estimation for retransmission timeouts, and piggybacking is about combining acknowledgments with outgoing data, so neither one specifically solves Silly Window Syndrome.
I. It provides mobile internet connection with faster data transfer rates
B. GSM
II. It allows users to connect to a network or to other devices over wireless channel
C. UMTS
III. Accessing mechanism for multiple transmitters over a single channel
D. WiFi
IV. It is cellular technology employs hybrid of FDMA and TDMA
List I and List II for the matching question
AA-III, B-IV, C-II, D-I
BA-III, B-IV, C-I, D-II
CA-II, B-III, C-IV, D-I
DA-II, B-I, C-IV, D-III
Answer:(B) A-III, B-IV, C-I, D-II
Explanation
Match each wireless or mobile technology to what it is best known for.
A-III, B-IV, C-I, D-II
CDMA is an accessing mechanism that lets multiple transmitters share one channel using distinct codes, matching III. GSM is a cellular technology built on a hybrid of FDMA and TDMA, matching IV. UMTS is the 3G technology that provides mobile internet access with faster data transfer rates, matching I. WiFi is what lets a device connect to a network or other devices over a wireless channel, matching II.
Rearrange the following sequence in the context of OSI Layers:
A. Transforming the raw bits in the form of frame for transmission
B. Transmission of raw bits over communication channel
C. Handling user interfaces
D. Control and monitoring of subnet
E. Transmission data through connection oriented or connection less using datagrams
Choose the correct answer from the options given below:
AA-B-C-D-E
BB-C-D-E-A
CB-A-D-E-C
DA-B-D-E-C
Answer:(C) B-A-D-E-C
Explanation
Match each description to its OSI layer, then order the layers from physical up to application.
Transmitting raw bits over the channel is the Physical layer (B). Framing those raw bits for transmission is the Data Link layer (A). Controlling and monitoring the subnet is the Network layer (D). Delivering data connection-oriented or connectionless using datagrams is the Transport layer (E). Handling user interfaces is the Application layer (C).
Ordering these from the bottom of the stack to the top gives B, A, D, E, C.
A TCP Server application is programmed to listen on port P on Host S. A TCP Client is connected to the TCP Server over the network. Consider that while the TCP Connection is active the server is crashed and rebooted. Assume that the client does not use a TCP keepalive timer. Which of the following behaviours is/are possible?
Statement I: If client is waiting to receive a packet, it may wait indefinitely
Statement II: If the client sends a packet after the server reboot, it will receive the FIN segment
In the light of the above statements, choose the correct answer from the options given below
ABoth Statement I and Statement II are true
BBoth Statement I and Statement II are false
CStatement I is true but Statement II is false
DStatement I is false but Statement II is true
Answer:(C) Statement I is true but Statement II is false
Explanation
A client that is passively blocked waiting for data has no built-in way to detect a silent server crash, since without a keepalive timer nothing prompts it to check whether the connection is still alive, so Statement I is true, it may wait indefinitely.
After rebooting, the server's TCP stack has no memory of the old connection at all, so any packet the client sends for that connection is unrecognized. A host that receives a segment for a connection it has no record of replies with a RST, not a FIN, because FIN is reserved for gracefully closing a connection the host still knows about, so Statement II is false.
An organization is granted the block 130.56.0.0/16. The administrator wants to create 1024 subnets.
What will be the prefix length of each subnet?
A/24
B/25
C/26
D/27
Answer:(C) /26
Explanation
Creating 1024 equal subnets from a /16 block requires borrowing enough host bits to distinguish 1024 subnets.
1024 = 2^10, so 10 bits must be borrowed from the host portion.
New prefix length = 16 + 10 = /26
The official answer key for this exact sitting lists a different option here, but that would require a /24 prefix (256 addresses per subnet), which directly contradicts the same key's own confirmed answer to the "addresses per subnet" question in this set (64 addresses, which only a /26 prefix produces). Because those two answers cannot both be correct for the same subnetting scenario, /26 is kept here as the internally consistent answer, and this conflict is noted for a human to double-check against the original key.
An organization is granted the block 130.56.0.0/16. The administrator wants to create 1024 subnets.
Find first and last addresses of the last subnet.
A130.56.0.0 130.56.254.255
B130.56.255.0 130.56.255.255
C130.56.0.192 130.255.255.255
D130.56.255.192 130.56.255.255
Answer:(D) 130.56.255.192 130.56.255.255
Explanation
The last of 1024 subnets, each holding 64 addresses, starts 64 addresses before the very end of the /16 block.
The block's last address is 130.56.255.255, so the last subnet's first address is 65536 − 64 = 65472 addresses from the start, which works out to 130.56.255.192.
The last subnet spans 130.56.255.192 through 130.56.255.255.
An organization is granted the block 130.56.0.0/16. The administrator wants to create 1024 subnets.
Find the subnet mask.
A255.255.255.0
B255.255.255.128
C255.255.255.192
D255.255.255.224
Answer:(C) 255.255.255.192
Explanation
A /26 prefix sets the first 26 bits of the address to 1, which spans all of the first three octets plus the top two bits of the fourth octet.
The fourth octet's top two bits set gives 11000000 in binary, which is 192 in decimal.
The subnet mask is 255.255.255.192.
NTA's official final answer key for this exact sitting marked this question as dropped (full marks awarded to all candidates). The mask 255.255.255.192 is the direct, unambiguous consequence of the confirmed /26 prefix used elsewhere in this same question set, so the drop does not appear to reflect any ambiguity in the mask itself.
Which protocol is used to acquire the MAC Address of a host whose IP address is known?
ADynamic Host control protocol
BAddress Resolution Protocol
CReverse Address Resolution Protocol
DUser Datagram Protocol
Answer:(B) Address Resolution Protocol
Explanation
The Address Resolution Protocol (ARP) broadcasts a request containing a known IP address and receives back the corresponding MAC address of the host that owns it, which is exactly the direction of lookup described here.
Reverse ARP does the opposite lookup, finding an IP address from a known MAC address, DHCP assigns IP configuration to hosts, and UDP is a transport-layer protocol unrelated to address resolution, so ARP is the only option that matches IP-to-MAC resolution.
In Go-back-N protocol, the values of S_f (Send window, the first outstanding packet) = 3, S_n (Send window, the next packet to be send) = 7, after receiving the packet with AckNo.6, the value of S_n is _______.
A6
B7
C5
D4
Answer:(B) 7
Explanation
In Go-Back-N, S_f marks the oldest unacknowledged packet and S_n marks the next new packet the sender will transmit, and these two pointers are updated by different events.
Receiving a cumulative acknowledgment such as AckNo.6 slides S_f forward past all packets up to and including packet 6, but it does not by itself cause the sender to transmit further packets.
S_n only advances when the sender actually sends a new packet, and no new packet transmission is described here, so S_n stays at its given value of 7.
What is the minimum Hamming Distance in a block code if we want to be able to detect up to S errors?
AS
BS+1
C1
D0
Answer:(B) S+1
Explanation
To reliably detect up to S bit errors in a codeword, every pair of valid codewords must differ in at least S+1 bit positions, since an error pattern of exactly S flipped bits could otherwise transform one valid codeword into another undetected.
A minimum Hamming distance of only S would allow an S-bit error to land exactly on another valid codeword, which defeats detection, so S+1 is the smallest distance that guarantees every possible error of up to S bits is caught.
Consider a computer network using the distance vector routing algorithm in its network layer. The objective is to find the shortest cost path from router R to routers P and Q. Assume that R does not initially know the shortest routes to P and Q. Assume that R has three neighboring routers, namely X, Y and Z. During one iteration, R measures its distance as 3, 2, and 5 from its neighbors X, Y and Z, respectively. Router R gets routing vectors from its neighbors that indicate a distance of 7, 6 and 5 from routers X, Y and Z to router P, respectively. The routing vectors also indicate that the distance to router Q from routers X, Y and Z are 4, 6, 8 respectively.
Which of the following statements is/are true with respect to the new routing table at router R after updating during this iteration?
A. The distance from R to P will be stored as 10
B. The distance from R to Q will be stored as 7
C. The next hop router for a packet from R to P is Y
D. The next hop router for a packet from R to Q is Z
Choose the correct answer from the options given below:
AB and C only
BA and C only
CC and D only
DB and D only
Answer:(A) B and C only
Explanation
The distance vector algorithm updates R's distance to a destination as the minimum, over every neighbor, of R's own link cost to that neighbor plus the neighbor's reported distance to the destination.
Distance to P via each neighbor: via X = 3+7 = 10, via Y = 2+6 = 8, via Z = 5+5 = 10
The minimum is 8 via Y, so R stores 8 (not 10) as its distance to P, and Y becomes the next hop for P.
Distance to Q via each neighbor: via X = 3+4 = 7, via Y = 2+6 = 8, via Z = 5+8 = 13
The minimum is 7 via X, so R stores 7 as its distance to Q, with X (not Z) as the next hop.
This confirms statement B (distance to Q is 7) and statement C (next hop to P is Y) are correct, while A (distance to P as 10) and D (next hop to Q as Z) are both wrong.
Consider the routing table of an organization's router.
Which of the following prefixes in CIDR notation can be collectively used to correctly aggregate all of the subnets in the routing table?
A. 12.20.164.0/20
B. 12.20.164.0/22
C. 12.20.164.0/21
D. 12.20.168.0/22
Choose the correct answer:
Subnet Number
Subnet Mask
Next Hop
12.20.164.0
255.255.252.0
R1
12.20.170.0
255.255.254.0
R2
12.20.168.0
255.255.254.0
Interface 0
12.20.166.0
255.255.254.0
Interface 1
Default
—
R3
Organization router's subnet table
AA and B Only
BA and C Only
CC and D Only
DB and D Only
Answer:(D) B and D Only
Explanation
Converting each mask to its block of covered third-octet values shows the routing table actually covers the continuous range 164 through 171, once the overlap between rows is accounted for.
12.20.164.0/255.255.252.0 is a /22, an aligned 4-value block covering 164-167. 12.20.170.0/255.255.254.0 is a /23, a 2-value block covering 170-171. 12.20.168.0/255.255.254.0 is a /23 covering 168-169. 12.20.166.0/255.255.254.0 is a /23 covering 166-167, which is already fully contained inside the first row's 164-167 block.
The true union of all four rows is therefore the range 164-171, eight consecutive third-octet values. This range is not itself a single power-of-two-aligned block starting at 164, since a true /21 (8-value) block starting at 164 would need 164 to be a multiple of 8, which it is not.
Instead, this exact 8-value range splits cleanly into two valid /22 blocks: 12.20.164.0/22 (164-167, aligned since 164 is a multiple of 4) and 12.20.168.0/22 (168-171, aligned since 168 is a multiple of 4). Together these two prefixes exactly cover 164-171 with no wasted address space, matching options B and D, while option A's /20 wrongly stretches the range down to 160 and up to 175, and option C's /21 is not properly aligned to start at 164.
Consider two hosts P and Q are connected through a router R. The maximum transfer unit (MTU) value of the link between P and R is 1500 bytes and between R and Q is 820 bytes. A TCP segment of size 1400 bytes is transferred from P to Q through R with IP identification value of 0x1234. Assume that IP header size is 20 bytes. The packet is allowed to be fragmented, that is, the Don't Fragment (DF) flag in the IP header is not set by P.
Which of the following statement is/are not true?
A. Two fragments are created at R and IP datagram size carrying the second fragment is 620 bytes
B. If the second fragment is lost then R resends the fragment with IP Identification value of 0x1234
C. If the second fragment is lost then P is required to resend the entire TCP segment
D. TCP destination port can be determined by analyzing the second fragment only
Choose the correct answer from the options given below:
AA and B Only
BA and C Only
CC and D Only
DB and D Only
Answer:(D) B and D Only
Explanation
From P to R, the full 1420-byte datagram (1400-byte segment plus 20-byte header) fits under the 1500-byte MTU, so it travels as one piece to R.
At R, forwarding onto the 820-byte MTU link forces fragmentation, since 820 minus the 20-byte header leaves 800 bytes of payload per fragment, itself a clean multiple of 8 as required by the fragment offset field. The first fragment carries 800 bytes of data (820 bytes total), and the second fragment carries the remaining 1400-800=600 bytes of data plus its own 20-byte header, for 620 bytes total, matching statement A exactly, so A is true.
IP itself has no per-fragment retransmission mechanism. If any fragment of a datagram is lost, the router does not resend just that fragment. The destination simply cannot reassemble the whole datagram and eventually discards what it has, so statement B is false.
Since reassembly at Q fails, Q never acknowledges the corresponding TCP data, and it is the original sender P's transport layer that eventually times out and retransmits the entire TCP segment, not just a fragment, making statement C true.
The TCP header, containing the source and destination ports, occupies the very start of the segment's data and therefore lands entirely inside the first fragment. The second fragment carries only a later slice of the payload with no TCP header at all, so the destination port cannot be read from the second fragment alone, making statement D false.
Since B and D are the false statements, the correct option is B and D Only.
Arrange the following in the correct order with respect to key generation for the RSA algorithm.
A. Choose an integer e such that 1 < e < λ(n)
B. Compute Key length n = p*q
C. Find d which is equal to 1/e
D. Choose two large prime numbers p and q
E. Compute λ(n)
Choose the correct answer from the options given below:
AA-B-C-D-E
BD-C-B-E-A
CD-B-E-A-C
DA-B-D-E-C
Answer:(C) D-B-E-A-C
Explanation
RSA key generation follows a fixed dependency chain, where each step needs a value computed in the step before it.
First, two large primes p and q must be chosen (D), since everything else depends on them. Their product gives the modulus n = p*q (B). From p and q, λ(n) (the totient-like function used for key generation) is computed (E). Only once λ(n) is known can an integer e be chosen satisfying 1 < e < λ(n) (A). Finally, d, the modular inverse of e with respect to λ(n), is computed last (C), since it needs e to already be fixed.
Which mode is a block cipher implementation as a self synchronizing stream cipher?
ACipher Block Chaining Mode
BCipher Feedback Mode
CElectronic Codebook Mode
DOutput Feedback Mode
Answer:(B) Cipher Feedback Mode
Explanation
In Cipher Feedback (CFB) mode, previously transmitted ciphertext is fed back into the encryption shift register to generate the next keystream, so the decryption side automatically resynchronizes once enough correct ciphertext bits have shifted through, which is why CFB is called self-synchronizing. Output Feedback (OFB) mode instead feeds back the cipher's own output rather than the ciphertext, so a lost bit is never recovered and OFB is a synchronous, not self-synchronizing, stream cipher.
Which one is a connectionless transport – layer protocol that belongs to the Internet protocol family?
ATransmission Control Protocol (TCP)
BUser Datagram Protocol (UDP)
CRouting Protocol (RP)
DDatagram Control Protocol (DCP)
Answer:(B) User Datagram Protocol (UDP)
Explanation
UDP sends independent datagrams without establishing a connection or tracking session state, making it the connectionless transport-layer protocol in the TCP/IP suite. TCP is connection-oriented, and the other two listed names are not real Internet protocol family transport-layer protocols.
Consider an error free 64 kbps satellite channel used to send 512 byte data frames in one direction with very short acknowledgements coming back the other way. What is the maximum throughput for window size of 15?
A32 kbps
B48 kbps
C64 kbps
D70 kbps
Answer:(C) 64 kbps
Explanation
This is a sliding-window pipelining problem using the standard textbook values for a geostationary satellite link, a one-way propagation delay of 270 ms.
Frame transmission time Tf = (512 × 8 bits) / 64,000 bps = 64 ms
Round-trip time = 2 × 270 ms = 540 ms
A window of size w keeps the channel fully busy once w × Tf ≥ Tf + RTT, that is once w ≥ (64 + 540) / 64 ≈ 9.4
Since the given window of 15 already exceeds this threshold, the sender never has to stall waiting for acknowledgements, so the link runs at its full rated capacity.
Which layer divides each message into packets at the source and re-assembles them at the destination?
ANetwork layer
BTransport layer
CData link layer
DPhysical layer
Answer:(A) Network layer
Explanation
The network layer is responsible for routing individual packets end to end, and in the classic layer-function description it is the layer that divides an outgoing message into packets at the source and reassembles the incoming packets back into the original message at the destination. The data link layer instead frames and error-checks data over one physical hop, and the physical layer only converts bits into electrical or optical signals.
Match List I with List II regarding cryptographic key sizes.
List-I
List-II
A. DES
I. Key size - 256
B. AES
II. Key size - 1024
C. 3 DES
III. Key size - 56
D. RSA
IV. Key size - 168
A(A)-(I), (B)-(II), (C)-(IV), (D)-(III)
B(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
C(A)-(III), (B)-(IV), (C)-(II), (D)-(I)
D(A)-(IV), (B)-(II), (C)-(III), (D)-(I)
Answer:(B) (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
Explanation
DES uses a 56-bit key.
AES, as referenced here, uses a 256-bit key.
Triple DES (3DES) has an effective key size of 168 bits, three 56-bit DES keys applied in sequence.
RSA is commonly cited in this context with a 1024-bit key.
Match List I with List II regarding socket programming calls.
List-I
List-II
A. BIND
I. Block the caller until a connection attempt arrives
B. LISTEN
II. Give a local address to a socket
C. ACCEPT
III. Show willingness to accept connections
D. SOCKET
IV. Create a new point
A(A)-(I), (B)-(III), (C)-(II), (D)-(IV)
B(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
C(A)-(III), (B)-(II), (C)-(I), (D)-(IV)
D(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
Answer:(B) (A)-(II), (B)-(III), (C)-(I), (D)-(IV)
Explanation
BIND assigns a local address to a socket.
LISTEN marks a socket as willing to accept incoming connections.
ACCEPT blocks the caller until a connection attempt actually arrives.
SOCKET creates a brand new communication endpoint.
Match List I with List II regarding OSI layer functions.
List-I
List-II
A. Physical layer
I. Routing of the signals divide the outgoing message into packets, to act as network controller for routing data
B. Data link layer
II. Make and break connections, define voltages and data rates, convert data bits into electrical signal
C. Network layer
III. Synchronization, error detection and correction. To assemble outgoing message into frames
D. Presentation layer
IV. It works as a translating layer
A(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
B(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
C(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
D(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
Answer:(D) (A)-(II), (B)-(III), (C)-(I), (D)-(IV)
Explanation
The physical layer makes and breaks connections, defines voltages and data rates, and converts bits into electrical signals.
The data link layer handles synchronization and error detection and correction, assembling the outgoing message into frames.
The network layer routes signals and divides the outgoing message into packets.
The presentation layer works as a translating layer between applications and the network format.
Based on the passage, the transmission and propagation delays are respectively
A 3000 km long trunk operates at 1.536 mbps and is used to transmit 64 bytes frames and uses sliding window protocol. The propagation speed is 6 μ sec/km.
Based on the passage, the minimum number of bits required in the sequence number field of the packet is
A 3000 km long trunk operates at 1.536 mbps and is used to transmit 64 bytes frames and uses sliding window protocol. The propagation speed is 6 μ sec/km.
A6 bits
B7 bits
C5 bits
D4 bits
Answer:(B) 7 bits
Explanation
The bandwidth-delay ratio a = propagation delay / transmission delay = 18000 / 333.33 = 54.
The window size needed for maximum efficiency is 1 + 2a = 1 + 108 = 109.
A sequence number field needs enough distinct values to cover this window, and 2^6 = 64 is too small while 2^7 = 128 comfortably covers 109, so the minimum sequence number field width is 7 bits.
Based on the passage, the sender window size to get the maximum efficiency is
A 3000 km long trunk operates at 1.536 mbps and is used to transmit 64 bytes frames and uses sliding window protocol. The propagation speed is 6 μ sec/km.
A108
B109
C55
D56
Answer:(B) 109
Explanation
With a = propagation delay / transmission delay = 18000 / 333.33 = 54, the sender window size that keeps the link continuously busy is
W = 1 + 2a = 1 + 2(54) = 109
This is the smallest window size at which the sender never has to stall waiting for an acknowledgement.
Based on the passage, if only 6 bits are reserved for sequence number field, then the efficiency of the system is:
A 3000 km long trunk operates at 1.536 mbps and is used to transmit 64 bytes frames and uses sliding window protocol. The propagation speed is 6 μ sec/km.
A0.587
B0.875
C0.578
D0.50
Answer:(A) 0.587
Explanation
A 6-bit sequence number field allows a maximum window of 2^6 = 64, which is smaller than the ideal window of 109 found earlier, so the link cannot be kept continuously busy.
Based on the passage, the maximum achievable throughput is
A 3000 km long trunk operates at 1.536 mbps and is used to transmit 64 bytes frames and uses sliding window protocol. The propagation speed is 6 μ sec/km.
A0.768
B0.678
C0.901
D0.887
Answer:(C) 0.901
Explanation
Using the 6-bit sequence number scenario, the link's efficiency was 64/109 ≈ 0.587.
The data link layer uses piggybacking to attach acknowledgements to outgoing frames (IV). The network layer handles routing and can produce billing/accounting information about packets (III). The transport layer provides genuine end-to-end (source-to-destination) delivery (I). The session layer provides token management for dialog control (II).
Statement I: Telnet, FTP and HTTP are application-layer protocols.
Statement II: The Iridium project was planned to launch 66 low-orbit satellites.
In light of the above statements, choose the correct answer from the options given below:
ABoth Statement I and Statement II are true
BBoth Statement I and Statement II are false
CStatement I is true but Statement II is false
DStatement I is false but Statement II is true
Answer:(A) Both Statement I and Statement II are true
Explanation
Telnet, FTP and HTTP all run at the application layer of the TCP/IP model (Statement I true). The Iridium satellite-phone constellation was designed around 66 low-Earth-orbit satellites (Statement II true).
APacket switching leads to better utilization of bandwidth resources than circuit switching
BPacket switching results in less variation in the delay than circuit switching
CPacket switching sender and receiver can use any bit rate, format or framing method unlike circuit switching
DPacket switching can lead to reordering unlike circuit switching
Answer:(B) Packet switching results in less variation in the delay than circuit switching
Explanation
Packet switching statistically shares links (better utilization), lets endpoints differ in rate and format, and can deliver packets out of order. But because packets queue independently at each hop, packet switching has MORE delay variation (jitter) than a dedicated circuit, so statement B is false.
In Ethernet using Manchester coding, the bit rate is ______.
AHalf the baud rate
BTwice the baud rate
CThrice the baud rate
DSame as the baud rate
Answer:(A) Half the baud rate
Explanation
Manchester encoding places a mid-bit transition, so each data bit uses two signal elements. The data (bit) rate is therefore half the signalling (baud) rate.
A Class B address is subnetted using a 6-bit subnet number. What is the maximum number of subnets and the maximum number of hosts per subnet?
A62 subnets and 262142 hosts
B64 subnets and 1024 hosts
C64 subnets and 262142 hosts
D62 subnets and 1022 hosts
Answer:(D) 62 subnets and 1022 hosts
Explanation
Class B has 16 host bits. Borrowing 6 for the subnet gives 2^6 - 2 = 62 usable subnets (excluding all-zeros and all-ones) and 16 - 6 = 10 host bits, i.e. 2^10 - 2 = 1022 hosts per subnet.
In electronic mail, which protocol allows the transfer of multimedia content?
AIMAP
BSMTP
CPOP3
DMIME
Answer:(D) MIME
Explanation
MIME (Multipurpose Internet Mail Extensions) extends email to carry non-ASCII text, images, audio, video and attachments by encoding them for transport over SMTP. IMAP and POP3 only retrieve messages; SMTP alone carries plain 7-bit text.
A message is sent using public-key cryptography. Which statement is true?
AThe sender encrypts using the receiver's public key
BThe sender encrypts using their own public key
CThe receiver decrypts using the sender's public key
DThe receiver decrypts using their own public key
Answer:(A) The sender encrypts using the receiver's public key
Explanation
For confidentiality the sender encrypts with the receiver's public key, and only the receiver, holding the matching private key, can decrypt. Encrypting with the sender's key or decrypting with any public key would not keep the message secret.
A. Frequency-division multiplexing can be handled by digital circuits.
B. Time-division multiplexing can be handled by analog circuits.
C. Wavelength-division multiplexing is used with optical fibre to combine signals.
D. Frequency-division multiplexing can be applied when the bandwidth of a link is greater than the combined bandwidth of the signals to be transmitted.
Choose the correct answer from the options given below:
AA and B only
BC and D only
CA and D only
DB and D only
Answer:(B) C and D only
Explanation
WDM combines optical signals of different wavelengths on one fibre (C true), and FDM needs a link whose bandwidth exceeds the sum of the signal bandwidths so each gets its own frequency band (D true). FDM is an analog technique (not A) and TDM is digital (not B).
A. X.25 is a connection-oriented network.
B. X.25 does not support switched virtual circuits.
C. Frame relay service provides acknowledgements.
D. Frame relay service provides detection of transmission errors.
Choose the correct answer from the options given below:
AA and D only
BB and D only
CC and D only
DB and C only
Answer:(A) A and D only
Explanation
X.25 is connection-oriented, setting up a virtual circuit before data transfer (A true). Frame relay carries a CRC and discards frames that fail the check, so it detects transmission errors (D true) - but it does not correct them or send acknowledgements (C false), and X.25 does support switched virtual circuits (B false).
Let G(x) be the generator polynomial used for CRC checking. The condition that G(x) must satisfy to catch all errors consisting of an odd number of inverted bits is:
A(x + 1) is a factor of G(x)
B(x - 1) is a factor of G(x)
C(x^2 + 1) is a factor of G(x)
D(1 - x^2) is a factor of G(x)
Answer:(A) (x + 1) is a factor of G(x)
Explanation
An error pattern E(x) with an odd number of 1-terms evaluates to 1 at x = 1, so it is not divisible by (x + 1). If (x + 1) divides G(x), then G(x) cannot divide such an E(x), so the CRC remainder is non-zero and every odd-bit error is detected.
Answer:(A) Worker node and master node respectively
Explanation
DataNodes store and serve HDFS data blocks on worker machines. The NameNode maintains filesystem metadata and coordinates the cluster, so it is the master service.
A code has valid words 0000000000, 0000011111, 1111100000, and 1111111111. If the received word is 0000000111, which word was sent?
A0000011111
B0000000000
C1111100000
D1111111111
Answer:(A) 0000011111
Explanation
Decode by minimum Hamming distance. The received word differs from 0000011111 in two positions but from 0000000000 in three positions; the two words beginning with 11111 are farther away.
Using RSA with p = 3, q = 11, and d = 7, find e and encrypt 19.
A20, 19
B33, 11
C3, 28
D77, 28
Answer:(C) 3, 28
Explanation
Here φ(n) = (3 − 1)(11 − 1) = 20. Since 7e ≡ 1 (mod 20), e = 3; with n = 33, encryption gives 19³ mod 33 = 28. Therefore option C is the only pair that has both the required inverse and ciphertext.
Protocols in which the sender sends one frame and waits for an acknowledgement before sending the next frame are called
ASimplex protocols
BUnrestricted simplex protocols
CSimplex stop-and-wait protocols
DRestricted simplex protocols
Answer:(C) Simplex stop-and-wait protocols
Explanation
Stop-and-wait allows only one outstanding frame, so the sender must wait for its acknowledgement. Sliding-window protocols, unlike stop-and-wait, permit multiple frames in transit.
The period of a signal is 100 ms. What is its frequency in kilohertz?
A10
B10⁷
C10⁻⁷
D10⁻²
Answer:(D) 10⁻²
Explanation
The signal period is 100 ms, and frequency is its reciprocal. Because 100 ms = 0.1 seconds, its frequency is 1/0.1 = 10 Hz. Dividing by 1,000 converts this to 0.01 kHz = 10⁻² kHz; therefore the correct option is D.
Which statements about hash functions are true?
A. A hash function maps an arbitrary-length message to a fixed-length code.
B. It maps a fixed-length message to a variable-length code.
C. Distinct messages can have the same hash value.
A(A) only
B(B) and (C) only
C(A) and (C) only
D(B) only
Answer:(C) (A) and (C) only
Explanation
A fixed-size digest is produced from inputs of arbitrary size. Since the input space is much larger than the digest space, collisions—different inputs with the same digest—are possible.
For Post Office Protocol (POP), which statements are true?
A. POP has delete and keep modes.
B. In delete mode, mail is deleted from the mailbox after retrieval.
C. In delete mode, mail is deleted before retrieval.
D. In keep mode, mail is deleted before retrieval.
E. In keep mode, mail remains in the mailbox after retrieval.
A(A) and (B) only
B(A), (D), and (E) only
C(A), (B), (C), and (D) only
D(A), (B), and (E) only
Answer:(D) (A), (B), and (E) only
Explanation
POP can either retain server copies or delete them after successful retrieval. Thus A, B, and E are true; deletion before a client retrieves a message would defeat the protocol.
A firewall can filter packets using which fields?
A. Source IP address
B. Destination IP address
C. TCP source port
D. UDP source port
E. TCP destination port
A(A), (B), and (C) only
B(B) and (E) only
C(A), (C), and (D) only
D(A), (B), (C), (D), and (E)
Answer:(D) (A), (B), (C), (D), and (E)
Explanation
Packet filters can inspect network-layer source and destination addresses and transport-layer source and destination ports for TCP or UDP. Every listed field is therefore usable in a firewall rule.
Match List I with List II
List I List II
A. Serial Line IP (SLIP) (Application layer
B. Border Gateway Protocol (BGP) (I Transport layer
C. User Data Protocol (UDP) (I) Data Link layer
D. Simple Network Management (IV) Network layer
Protocol
List I
List II
Items are listed in the question stem.
Match each item to its stated description or complexity.
SLIP operates at the data-link layer, BGP at the network layer, UDP at the transport layer, and SNMP at the application layer; option B is the correct matching.
The laws of nature put two fundamental limits on data rate of a channel. The
H.Nyquist limit restricts the number of independent samples per second to
twice the band-width in a Noiseless channel.
Shannon's major result about noised channel is that maximum data rate of a
channel whose band width is H Hz, and whose signal-to-noise ratio is S/N is
given by:
Maximum number of bits/sec = H log2(1 + =)
In the light of the above statements. choose the correct answer from the options given below
ABoth Statement I and Statement II are true
BBoth Statement I and Statement II are false
CStatement I is correct but Statement II is false
DStatement I is incorrect but Statement II is true.
Answer:(A) Both Statement I and Statement II are true
Explanation
Nyquist limits a noiseless H-Hz channel to 2H independent samples per second. Shannon’s capacity formula is H log₂(1 + S/N), so both statements are true.
Given below are two statements:
Statement I: In Ceasar Cipher each letter of Plain text is replaced by another letter for
encryption.
Statement II: Diffie-Hellman algorithm is used for exchange of secret key.
In the light of the above statements, choose the correct answer from the options given below
(dd) Both Statement I and Statement II are true
AOCR review required
BBoth Statement I and Statement II are false
CStatement I is correct but Statement IT is false
DStatement I is incorrect but Statement II is true
Answer:(A) OCR review required
Explanation
A Caesar cipher substitutes each plaintext letter by a shifted letter. Diffie–Hellman establishes a shared secret over an insecure channel, so both statements are true.
Consider a subnet with 720 routers. If a three-level hierarchy is chosen, with eight clusters, each containing 9 regions of 10 routers, then total number of entries in hierarchical table of each router is:
A25
B27
C53
D72
Answer:(A) 25
Explanation
In hierarchical routing with three levels (routers, regions, clusters):
Entries for local routers within the router's own region = 10 entries.
Entries for other peer regions within the same cluster = 9 - 1 = 8 entries.
Entries for other peer clusters within the network = 8 - 1 = 7 entries.
Total table entries for each router:
Total = 10 + 8 + 7 = 25 entries.
Consider the following statements with respect to network security:
A. Message confidentiality means that the sender and the receiver expect privacy.
B. Message integrity means that the data must arrive at the receiver exactly as they were sent.
C. Message authentication means the receiver is ensured that the message is coming from the intended sender.
Which of the statements is (are) correct?
AOnly A and B
BOnly A and C
COnly B and C
DA, B and C
Answer:(D) A, B and C
Explanation
Evaluating the foundational pillars of network security:
A. Message confidentiality — correct. Ensures unauthorized parties cannot intercept or read transmission contents, guaranteeing privacy between sender and receiver.
B. Message integrity — correct. Ensures transmitted data has not been altered, modified, or corrupted in transit.
C. Message authentication — correct. Verifies the sender's identity, ensuring the message originated from the claimed source.
AInternet Corporation for Assigned Names and Numbers
BInternet Corporation for Assigned Numbers and Names
CInstitute of Cooperation for Assigned Names and Numbers
DInternet Connection for Assigned Names and Numbers
Answer:(A) Internet Corporation for Assigned Names and Numbers
Explanation
ICANN stands for Internet Corporation for Assigned Names and Numbers. Established in 1998, it is the non-profit coordinating body responsible for maintaining and managing the core namespaces of the Internet, including IP address allocation, protocol parameter registries, and top-level domain systems.
A. Fiber optic cable is much lighter than copper cable.
B. Fiber optic cable is not affected by power surges or electromagnetic interference.
C. Optical transmission is inherently bidirectional.
A. Fiber optic cable is much lighter than copper cable — correct. Glass or plastic optical fibers have substantially less physical weight and bulk than copper wiring.
B. Fiber optic cable is not affected by power surges or electromagnetic interference — correct. Because signals travel as light pulses rather than electric currents, optical cables are immune to electromagnetic and radio frequency interference.
C. Optical transmission is inherently bidirectional — false. Light propagation along a single fiber strand is unidirectional. Bidirectional transmission requires two separate strands or wavelength-division multiplexing.
A. Windows Azure is a cloud-based operating system.
B. Google App Engine is an integrated set of online services for consumers to communicate and share with others.
C. Amazon Cloud Front is a web service for content delivery.
Which of the statements is (are) correct?
AOnly A and B
BOnly A and C
COnly B and C
DA, B and C
Answer:(B) Only A and C
Explanation
Evaluating cloud computing services and classifications:
A. Windows Azure is a cloud-based operating system — correct. Microsoft introduced Windows Azure as a cloud operating system providing compute and storage infrastructure.
B. Google App Engine is an integrated set of online services for consumers to communicate and share with others — false. Google App Engine is a Platform as a Service (PaaS) framework for deploying web applications, not a consumer social sharing platform.
C. Amazon Cloud Front is a web service for content delivery — correct. Amazon CloudFront is a globally distributed content delivery network (CDN) accelerating static and dynamic web content delivery.
Piconet is a basic unit of a bluetooth system consisting of __________ master node and up to __________ active slave nodes.
Aone, five
Bone, seven
Ctwo, eight
Done, eight
Answer:(B) one, seven
Explanation
In Bluetooth network architecture, a piconet is an ad-hoc network consisting of exactly one master node and up to seven active slave nodes communicating over shared frequency-hopping channels. Additional slave devices can remain synchronized in a low-power parked state.
A network with bandwidth of 10 Mbps can pass only an average of 12,000 frames per minute with each frame carrying an average of 10,000 bits. What is the throughput of this network?
A1,000,000 bps
B2,000,000 bps
C12,000,000 bps
D120,000,000 bps
Answer:(B) 2,000,000 bps
Explanation
Throughput measures the actual volume of data successfully transmitted per unit of time:
Total bits transmitted per minute:
Total = 12000 frames/min × 10000 bits/frame = 120000000 bits/min
Convert bits per minute to bits per second:
Throughput = 120000000 bits / 60 seconds = 2000000 bps
A fully connected network topology is a topology in which there is a direct link between all pairs of nodes. Given a fully connected network with n nodes, the number of direct links as a function of n can be expressed as
An(n+1)/2
B(n+1)/2
Cn/2
Dn(n-1)/2
Answer:(D) n(n-1)/2
Explanation
In a fully connected (mesh) network topology, every node has an undirected point-to-point link to every other node without self-connections.
The number of unique pairs of nodes that can be chosen from n distinct nodes is given by the combination formula:
C(n, 2) = n(n - 1) / 2.
A. n(n+1)/2 — incorrect because it includes n self-loops.
B. (n+1)/2 — incorrect linear formula.
C. n/2 — incorrect count representing only n/2 disconnected pairs.
In the TCP/IP model, encryption and decryption are functions of ______ layer.
Adata link
Bnetwork
Ctransport
Dapplication
Answer:(D) application
Explanation
In the standard TCP/IP protocol suite, the Application layer absorbs the responsibilities of the OSI Session and Presentation layers.
Data encryption, decryption, compression, and format conversions are handled either directly within application protocols or through security sublayers like TLS and SSL operating at the application level.
The RSA encryption algorithm also works in reverse, that is, you can encrypt a message with the private key and decrypt it using the public key. This property is used in
Aintrusion detection systems
Bdigital signatures
Cdata compression
Dcertification
Answer:(B) digital signatures
Explanation
In asymmetric cryptography using RSA, encrypting a message digest with the sender's private key creates a digital signature.
Because the corresponding public key is accessible to anyone, any receiver can decrypt the signature to verify authentication and message integrity, while proving non-repudiation since only the private key owner could produce that ciphertext.
You need 500 subnets, each with about 100 usable host addresses per subnet. What network mask will you assign using a class B network address?
A255-255-255-252
B255-255-255-128
C255-255-255-0
D255-255-254-0
Answer:(B) 255-255-255-128
Explanation
A Class B network provides 16 host bits with a default /16 mask.
Determining required bit allocations:
500 subnets require s subnet bits where 2^s >= 500, giving s = 9 bits (2^9 = 512 subnets).
100 usable hosts require h host bits where 2^h - 2 >= 100, giving h = 7 bits (2^7 - 2 = 126 hosts).
Total bits s + h = 9 + 7 = 16 bits.
Adding 9 subnet bits to the 16 network bits produces a /25 mask:
First three octets = 255.255.255
Fourth octet = 10000000 in binary = 128.
Resulting mask is 255-255-255-128.
You are designing a link layer protocol for a link with bandwidth of 1 Gbps (10⁹ bits/second) over a fiber link with length of 800 km. Assume the speed of light in this medium is 200000 km/second. What is the propagation delay in this link?
A1 millisecond
B2 milliseconds
C3 milliseconds
D4 milliseconds
Answer:(D) 4 milliseconds
Explanation
Propagation delay measures the time required for a signal bit to travel across the physical medium:
Distance d = 800 km.
Propagation speed v = 200,000 km/second.
Which of the following statements is/are true with regard to various layers in the Internet stack?
P: At the link layer, a packet of transmitted information is called a frame.
Q: At the network layer, a packet of transmitted information is called a segment.
AP only
BQ only
CP and Q
DNeither P nor Q
Answer:(A) P only
Explanation
Packet data units are named according to their architectural layer:
P is true because data link layer packets encapsulated with MAC headers and trailers are called frames.
Q is false because packets at the network layer are called datagrams (or IP packets). Segments are protocol data units formed at the transport layer (e.g. TCP segments).
Consider the following two statements with respect to IPv4 in computer networking:
P: The loopback (IP) address is a member of class B network.
Q: The loopback (IP) address is used to send a packet from host to itself.
What can you say about the statements P and Q?
AP-True; Q-False
BP-False; Q-True
CP-True; Q-True
DP-False; Q-False
Answer:(B) P-False; Q-True
Explanation
Evaluating IPv4 loopback network properties:
P is false because the entire 127.0.0.0/8 address block is designated for loopback testing. In classful addressing, first-octet values from 0 through 127 fall under Class A, not Class B (Class B ranges from 128 to 191).
Q is true because loopback packets short-circuit within the host network stack, allowing a machine to transmit IP packets directly to its own services without placing frames onto physical network media.
Suppose that everyone in a group of N people wants to communicate secretly with (N – 1) other people using symmetric key cryptographic system. The communication between any two persons should not be decodable by the others in the group. The number of keys required in the system as a whole to satisfy the confidentiality requirement is
AN(N – 1)
BN(N – 1)/2
C(N – 1)²
D2N
Answer:(B) N(N – 1)/2
Explanation
In a symmetric key cryptosystem, each pair of users must share a unique secret key to prevent other group members from deciphering their messages.
The number of distinct unordered pairs among N people is given by the combination formula C(N, 2).
C(N, 2) = N(N - 1) / 2.
Option A accounts for directed communication pairs rather than symmetric shared keys.
Option C and Option D do not reflect the pairwise combination formula.
Identify the correct sequence in which the following packets are transmitted on the network by a host when a browser requests a webpage from a remote server, assuming that the host has just been restarted.
(1) HTTP GET request, DNS query, TCP SYN
(2) DNS query, HTTP GET request, TCP SYN
(3) TCP SYN, DNS query, HTTP GET request
(4) DNS query, TCP SYN, HTTP GET request
AHTTP GET request, DNS query, TCP SYN
BDNS query, HTTP GET request, TCP SYN
CTCP SYN, DNS query, HTTP GET request
DDNS query, TCP SYN, HTTP GET request
Answer:(D) DNS query, TCP SYN, HTTP GET request
Explanation
Network communication requires resolving the server IP address before initiating a transport connection and transmitting application data.
DNS query: The newly restarted host has an empty DNS cache and must first resolve the domain name to an IP address.
TCP SYN: Once the IP address is known, the host initiates the TCP three-way handshake to establish a connection.
HTTP GET request: After the TCP connection is established, the application layer sends the HTTP GET request.
Therefore, the correct transmission order is DNS query, TCP SYN, HTTP GET request (Option D).
(i) Firewalls can screen traffic going into or out of an organization.
(ii) Virtual private networks can simulate an old leased network to provide certain desirable properties.
Choose the correct answer from the code given below :
Code :
A(i) only
BNeither (i) nor (ii)
CBoth (i) and (ii)
D(ii) only
Answer:(C) Both (i) and (ii)
Explanation
Both statements describe standard security and architecture principles in computer networking.
Statement (i) is true because network firewalls filter both ingress (incoming) and egress (outgoing) traffic according to defined access-control rules.
Statement (ii) is true because a Virtual Private Network creates secure, encrypted overlays across public networks, simulating the dedicated security and privacy of an old leased line.
Therefore, both statements (i) and (ii) are true (Option C).
A host is connected to a department network which is 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
Athe university network
Bthe Internet
Cthe subnet to which the host belongs
Dthe department network
Answer:(B) the Internet
Explanation
An Ethernet MAC address is a 48-bit globally unique hardware identifier administered by IEEE through Organizationally Unique Identifiers (OUIs).
Manufacturers assign each network interface controller a globally unique MAC address so that no two network adapters worldwide share the same physical address.
Because uniqueness is globally preserved across all interconnected devices, the largest network containing this host where its Ethernet address is unique is the entire Internet (Option B).
Options A, C, and D are smaller subnetworks within the global address space.
S1 : TCP handles both congestion and flow control
S2 : UDP handles congestion but not flow control
Which of the following options is correct with respect to the above statements (S1) and S2) ?
Choose the correct answer from the code given below :
Code :
AS1 is correct but S2 is not correct
BBoth, S1 and S2 are correct
CS1 is not correct but S2 is correct
DNeither S1 nor S2 is correct
Answer:(A) S1 is correct but S2 is not correct
Explanation
Statement S1 is correct and Statement S2 is incorrect based on transport-layer protocol capabilities.
TCP provides flow control using the receive window and congestion control using algorithms such as slow start, congestion avoidance, fast retransmit, and fast recovery.
UDP is a connectionless, best-effort protocol that provides neither flow control nor congestion control.
Because Statement S2 falsely attributes congestion control to UDP, Option A is the correct answer.
Consider ISO-OSI network architecture reference model. Session layer of this model offers dialog control, token management and __________ as services.
AFlow Control
BSyncronization
CErrors
DAsyncronization
Answer:(B) Syncronization
Explanation
In the OSI reference model, the primary services provided by the Session layer are dialog control, token management, and synchronization.
Synchronization allows the session layer to insert checkpoints into long data streams so that only data transmitted after the most recent checkpoint is retransmitted following a network crash.
Dialog control regulates communication turns in half-duplex or full-duplex transmission.
Token management prevents conflicting attempts to execute critical operations simultaneously.
Flow control (Option A) and error control (Option C) are managed by the data link and transport layers.
An Internet Service Provider (ISP) has following chunk of CIDR-based IP addresses available with it : 245.248.128.0/20. The ISP wants 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 ?
A245.248.136.0/21 and 245.248.128.0/22
B245.248.128.0/21 and 245.248.128.0/22
C245.248.132.0/22 and 245.248.132.0/21
D245.248.136.0/24 and 245.248.132.0/21
Answer:(A) 245.248.136.0/21 and 245.248.128.0/22
Explanation
Subdividing a /20 block assigns a /21 prefix for half the address space and a /22 prefix for a quarter without address collision.
The initial block is 245.248.128.0/20 (spanning third-octet values 128 through 143).
Giving half to organization A requires a /21 mask (covering 8 third-octet values):
Half 1: 245.248.128.0/21 (octets 128 to 135)
Half 2: 245.248.136.0/21 (octets 136 to 143)
Giving a quarter to organization B requires a /22 mask (covering 4 third-octet values) from the unused half:
From Half 1: 245.248.128.0/22 (octets 128 to 131) and 245.248.132.0/22 (octets 132 to 135).
In Option A, organization A receives 245.248.136.0/21 and organization B receives 245.248.128.0/22.
These two allocations are mutually disjoint subsets of the parent block, leaving 245.248.132.0/22 (the remaining quarter) for the ISP.
Option B produces overlapping allocations where B is inside A.
Option C and Option D feature invalid prefix lengths or overlapping segments.
Suppose a cloud contains software stack such as Operating systems, Application softwares, etc. This model is referred as ________ model.
AMaaS
BSaaS
CPaaS
DIaaS
Answer:(B) SaaS
Explanation
In cloud computing service taxonomies, delivering a full software stack including applications directly to end-users defines SaaS.
Software as a Service (SaaS) provides ready-to-use software applications managed entirely by the cloud provider, covering underlying operating systems, runtimes, and databases.
Platform as a Service (PaaS) provides development frameworks, tools, and execution runtimes for building custom applications.
Infrastructure as a Service (IaaS) provides virtualized raw computing resources, storage, and networking.
Therefore, delivering the complete application software stack corresponds to the SaaS model (Option B).
3G mobile communications standardized under IMT-2000 rely on Wideband CDMA (UMTS) and CDMA2000 to deliver multi-megabit mobile broadband data and multimedia streaming.
First generation (1G) used analog AMPS (Option D).
Second generation (2G) used narrowband digital systems including GSM, D-AMPS (Option C), and IS-95 narrowband CDMA (Option A).
Therefore, 3G systems are characterized by Broadband / Wideband CDMA (Option B).
This code corresponds to Option B: (iii) (iv) (ii) (i).
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