Computer Networks Previous Year Questions

Explanation

Bit rate = number of bits per unit time; baud rate = number of signal units per unit time. Here each signal unit carries 6 bits, so baud rate = bit rate / 6 = 3000/6 = 500 baud/sec.

Explanation

The original ethernet IEEE 802.3 standard define the minimum ethernet frame size as 64 bytes and maximum as 1518 bytes. Where payload size in ethernet defined the minimum 46 bytes and maximum 1500 bytes.

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A bridge functions in the data link layer, which means it cannot read IP addresses; it has access to the physical (MAC) addresses of all stations connected to it.

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2-phase locking is a concurrency control method that guarantees serializability, but because transactions may block one another while holding locks, 2PL can lead to deadlocks. Timestamp-based (time phase) ordering is a non-lock concurrency method: it ensures conflict serializability and no deadlock can occur since no transaction ever waits. So the answer is (b) II only.

Explanation

The host-to-host layers main purpose is to shield the upper layer application from the complexity of network. So the answer is (a).

Explanation

At 1 Mbps the sender sends 1,000,000 bits per second. The receiver clock is 0.1% faster, so the extra bits received = 1,000,000 × 0.1/100 = 1000 bits, i.e. the receiver receives 1,001,000 bits per second. Extra bits per second = 1000 bps. Hence option (c) is correct.

Explanation

With n = 4 bits for sequence numbers, the maximum sender window size (Go-Back-N) is 2ⁿ − 1 = 2⁴ − 1 = 15. So the answer is 15.

Explanation

IPv4 is the fourth version in the development of the IP. and IPv4 use loose source for an IP option can be used for address translation and IPv4 also use strict source routing for the contrast with loose source routing, in every step of the route is decided in advance where the packet is send and IPv4 also time stamp. So the correct answer is (d).

Explanation

Correct match: A. Call control protocol – iii. Connection management; B. A-bis – i. Interface between BTS and BSC; C. BSMAP – iv. Works between MSC and BSS; D. CDMA – ii. Spread spectrum.

Explanation

The signal sends 2000 signal elements (baud) per second and 6000 bits per second, so each signal element carries 6000/2000 = 3 bits/baud.

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HDLC (High-level Data Link Control) is a bit-oriented, code-transparent, synchronous data link layer protocol from ISO. It can be used for point-to-point and point-to-multipoint links.

Explanation

Two phase locking is a pessimistic protocol (it assumes conflicts will happen) because it applies locks. Time stamping is an optimistic protocol because it uses no locking and handles conflicts later. So II and III are correct.

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In basic two-phase locking there is a chance of deadlock. The timestamp-ordering protocol ensures conflict serializability and, since no transaction waits, it is free from deadlock. So the answer is (2) only.

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Session initiation protocol is an application layer control protocol that can establish, modify & terminate multimedia sessions such as internet telephony calls.

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Total length 16 bit field define the entire packet size in IPv4, including header and data, in bytes. The minimum length packet is 20 bytes (20 byte header. + 0 byte data) and maximum is 65, 535 bytes the maximum value of a 16 bit word.

Explanation

Circuit switching transmission technique guarantees the data packets will be received by the receiver in the same order in which they were sent by the sender because path is fixed and all packet goes through that path in same order.

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Key LAYER gives column order L = 3, A = 1, Y = 5, E = 2, R = 4. Writing the message row-wise: C O M P U T E R N E T W O R K Reading columns in order 1 (A), 2 (E), 3 (L), 4 (R), 5 (Y): OEW PNR CTT UEK MRO. Encrypted message: OEWPNRCTTUEKMRO.

Explanation

FIN control fields in TCP header is used to specify whether the sender has no more data to transmit i.e., close the connection from sender side.

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The maximum payload of TCP segment is 65, 495 bytes. Because maximum IP packet is 65, 535 bytes out of which 20 is used for header. TCP header is 20 bytes. So, 2¹⁶ – 20 – 20 = 65535 – 40 = 65495 bytes left for TCP.

Explanation

Services provided by a wireless access point in 802.11 WLAN: (i) Integration (ii) Disassociation (iii) Association (iv) Authentication (v) Privacy (vi) Distribution (vii) MSDU delivery. Error correction is not provided by the wireless access point in 802.11 WLAN.

Explanation

Flow control is a function of the data link layer and transport layer, not a network layer congestion policy. Flow control manages the rate of data transmission between two nodes to prevent a fast sender from overwhelming a slow receiver. Network layer congestion policies are: (1) Virtual circuits vs datagram inside the subnet (2) Packet queueing and service policy (3) Packet discard policy (4) Routing algorithm (5) Packet lifetime management.

Explanation

Identification, Flags and Fragment offset are the IPv4 header fields used for fragmentation. Type of service is not related to fragmentation.

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A socket address is the combination of an IP address & a port number, much like one end of a telephone connection is the combination of a phone number and a particular extension.

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With eight clusters, each containing 9 regions of 10 routers, each router needs 10 entries for local routers, 8 entries for the other regions in its own cluster and 7 entries for the distant clusters: total 10 + 8 + 7 = 25.

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A host number of 0 denotes the network address of a network. Addresses with a network number of 0 refer to the current network, allowing a host to send to a neighbour without needing to know the network number of its local network (IP packets with destination network address 0 are not routed).

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Multipurpose Internet Mail Extensions (MIME) is an internet standard that extends the format of email to support text in character sets other than ASCII, non-text attachments (audio, video, images, application programs etc.) and message bodies with multiple parts.

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Since, each in mesh network total number of cables are required ⁿC₂ = ¹⁰C₂ = 45 Each device needs to be connected to 9 other devices. So, each device needs to have 9 ports.

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If the angle of incidence is equal to the critical angle, the light beam bends along the interface. When a beam of light travels through media of two different densities, if the angle of incidence is greater than the critical angle, reflection occurs.

Explanation

(A) Line coding is the process of converting digital data into a digital signal – (iii). (B) Block coding provides redundancy to ensure synchronization and inherent error detection – (iv). (C) Scrambling provides synchronization without increasing the number of bits – (ii). (D) Pulse code modulation (PCM) is a technique to change an analog signal to digital data – (i).

Explanation

KILL (SIGKILL) cannot be blocked, caught or ignored, so the statement "KILL is a blockable signal" is not correct.

Explanation

The channel capacity of a discrete memoryless channel is defined by C = max over p(x) of I(x; y) = max over p(x) of [H(y) – H(y|x)] = max over p(x) of [H(x) – H(x|y)]. max H(x|y) does not define the channel capacity.

Explanation

Transmission rate of a circuit = frame rate × number of bits in a slot = 4000 × 8 bps = 32 Kbps. So, option (c) is correct.

Explanation

Multiplexing is of three types: (1) Frequency division multiplexing (FDM) – analog signals. (2) Wavelength division multiplexing (WDM) – analog signals (used in optical fiber). (3) Time division multiplexing (TDM) – digital signals; a single frame is made with interleaved digital signals from n devices. So (1), (2) and (4) are true and (3) is false.

Explanation

Throughput of pure ALOHA is S = G × e⁻²ᴳ, where G is the load (frames per frame-transmission time). Frame transmission time = 200 bits / 200 kbps = 1 ms. The system produces 500 frames per second = 0.5 frames per ms, so G = 1/2. S = 0.5 × e⁻¹ ≈ 0.184 (the maximum throughput of pure ALOHA). So the answer is (b).

Explanation

P1 arrives at time 1 and runs. At time 3, P2 arrives with a shorter remaining time (10 < 18), so there is a switch to P2 (1). P2 runs to completion at time 13 since no shorter job arrives, then there is a switch to P1 (2). P1 (remaining 18) runs to completion since its remaining time is less than P3's burst time (30), then there is a switch to P3 (3), which runs to completion. So 3 context switches are required.

Explanation

Fast Ethernet:- Cabling, 100 Base- Tx uses twisted pair cable and maximum segment size is 100 metres.

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Throughput = (12,000 × 10,000) / 60 bits per second = 2,000,000 bps = 2 Mbps. So answer is option (b).

Explanation

IMAP : The basic idea behind IMAP is for all email server to maintain a central repository that can be accessed from any machine

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Each route (hop) takes 8000/32000 = 0.25 second. For all 8 routes the total delay = 8 × 8000/32000 = 2 seconds.

Explanation

Statement I– True, High speed Ethernet can utilize various media, including optical fiber. It's not limited to just copper cables. Statement II– True, the definition of PPPoE is to encapsulate PPP frames within Ethernet frames. Statement III– False Since high speed Ethernet can work on optic fiber, this statement in incorrect. Statement IV– False, The correct encapsulation is PPP frames inside Ethernet frames.

Explanation

In Artificial Intelligence, a semantic network is a graph-based method of knowledge representation where nodes represent concepts, instances or events and arcs (links) represent relationships between them, such as is-a, part-of and kind-of.

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The correct match: A. Application layer – iv. HTTP; B. Transport layer – i. TCP; C. Network layer – ii. IP; D. Datalink layer – iii. PPP.

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Default gateway serves an access point or IP router that a networked uses to send information to a computer in another network or the internet. Default simply means that this gateway is used by default, unless an application specifies another gateway.

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Multilayer perceptron network and support vector machines are examples of supervised learning, while self organizing feature maps and Hopfield networks are examples of unsupervised learning.

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Wait & signal operations are the two atomic operations performed on a semaphore.

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Non-adaptive (static) routing algorithms do not base their routing decisions on measurements or estimates of the current traffic and topology. Instead, the route from one node to another is computed in advance.

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The transport layer is responsible for end-to-end communication over a network; it provides logical communication between application processes running on different hosts.

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Switch takes data sent from one network device & forwards it to the destination node based on MAC address. A switch is a multi port bridge with a buffer and a design that can boost its efficiency and performance. Switch is data link layer device. It can perform error checking before forwarding data.

Explanation

Gigabit Ethernet uses the standard Ethernet frame, whose destination and source address (MAC address) fields are each 48 bits long.