  Question 1

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In ethernet, when Manchester encoding is used, baud rate is 106 b/s. Find the bit rate?

SOLUTION

For Manchester encoding:

$$Baud\;rate = 2 \times bit\;rate$$

$$bit\;rate = \frac{{Baud\;rate}}{2}$$

$$\therefore bit\;rate = \frac{{{{10}^6}}}{2} = 5 \times {10^5}\;b/s$$

Question 2

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Which is/are the Application layer protocol?

SOLUTION

Application layer protocols: FTP, SMTP and IMAP

1. Simple Mail Transfer Protocol (SMTP) is the standard protocol for sending emails across the Internet.
2. IMAP is the Internet mail protocols used for retrieving emails.

Open Shortest Path First (OSPF) is a routing protocol for Internet Protocol (IP) networks. It uses a link state routing (LSR) algorithm and falls into the group of interior gateways.

Question 3

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The address of a class A host is to be split into subnets with a 14-bit subnet number.  The remaining bits in host id is x and the maximum number of hosts in each subnet is y. What is the value of x + y?

SOLUTION

In class A, 8 bits are reserved for Network id

14 bit more is added due to subnetting

Total Network Id = 8 + 14 = 22

Number of bit in host id = x =32 - 22 = 10

The number of available hosts = y = 2h − 2,

= 210 - 2

= 1024 - 2

1022

x + y = 10 + 1022 = 1032

Tips and Tricks:

Since first address and last address cannot be assigned to a Host

Question 4

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A computer uses the following sequence numbers

0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,0,1,2,3,4……

What are the senders and receivers window size is possible in SR protocol?

SOLUTION

In selective repeat protocol,

window size of sender = window size of receiver

Also, there are 16 distinct sequence number

∴ Ws + Wr   =  16

Ws = Wr

∴ Ws + Ws   =  16

∴ W=  8

Also Wr =  8

Question 5

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Let the size of congestion window of a TCP connection be 64 KB when a timeout occurs. The round-trip time is 50 milliseconds and the maximum segment size is 2 KB. The time taken (in milliseconds) by the TCP connection to get 64 KB congestion window is

SOLUTION

Number of segments $$= \frac{{64}}{2} = 32$$

Threshold $$= \frac{{32}}{2} = 16$$

 Transmission Segment (1 segment = 2 MSS) 1 1 2 2 3 4 4 8 5 16 6 17 7 18 8 19 9 20 10 21 11 22 12 23 13 24 14 25 15 26 16 27 17 28 18 29 19 30 20 31 21 32

Total Round Trip Time (RTT) = 21 - 1 = 20

1 RTT = 50 millisecond

Total time = 20 × 50 = 1000 milliseconds

Tips and Trick:

Slow start phase (number of RTT) = ⌈Log216⌉ = 4

Congestion avoidance (number of RTT) = 32(last segment) – 16 (segment till slow start)

Total Round trip time (RTT) = 4 + 16 = 20

Also note that it is not compulsory to divide it into segments

1 RTT = 50 millisecond

Total time = 20 × 50 = 1000 milliseconds

Question 6

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Which of the following statement is correct about HTTP?

SOLUTION

HTTP (Hypertext transfer protocol) is used mainly to access data on world wide web. It functions as combination of FTP and SMTP. It is like FTP because it transfers files and similar to SMTP because data transferred between client and server look like SMTP messages. Explanation: Some points about HTTP are:
• HTTP uses the services of TCP on well known port 80
• HTTP is a stateless protocol.
• A client can directly connect to a server using TELNET, which logs into port 80.
• HTTP prior to version 1.1 specified a non-persistent connection while a persistent connection is the default in version 1.1.
• In a non –persistent connection, one TCP connection is made for each request/response.
• In persistent connection, serves leaves the connection open for more requests after sending a response.
• HTTP supports proxy servers.

Question 7

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Consider the following destination address of standard Ethernet

43:10:1B:2D:06:EE

What is the type of the above given destination Ethernet address?

SOLUTION

Multicast MAC address: Least significant bit (MSB) of 1st byte is 1

43:10:1B:2D:06:EE

1st byte = 43 = 0100 0011

Examples of other types of MAC address:

Unicast MAC address: Least significant bit (LSB) of 1st byte is 0

2A : 1F : 41 : 4C : 11:19

1st byte = 2A = 0100 0011

FF : FF : FF : FF : FF : FF

1st byte = FF = 1111 1111 (also multicast MAC address)

Question 8

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A pure ALOHA network transmits 400-bit frames on a shared channel of 400 kbps. What is the maximum throughput(in frames) if the system (all stations together) produces 500 frames per second?

SOLUTION

For a pure ALOHA

Efficiency(η) = G × e-2G

If η is maximum then  G = 1/2

η = 1/(2e) = 0.1839

Throughput = number of frames × η = 500× 0.1839 = 91.96 frames

This implies only 91 frames will survive.

Note:

Number of frames is always less then 92

Question 9

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Consider the following set of statements: Which of the above statements is/are true?

SOLUTION

Physical Layer

• It is concerned with the connection of the devices to the media. In a point to point configuration, two devices are connected through a dedicated link. In a multipoint configuration, the sender link is shared among devices.
• It defines the direction of transmission between two devicessimplex, half-duplex or full-duplex.
• In simplex mode, the only one device can send; the other can only receive. In half-duplex, two devices can send and receive but not at the same time. In a full-duplex mode, two devices can send and receive at the same time.

The network layer

• It is responsible for packet forwarding including routing through intermediate routers.

The transport layer

It can be either connectionless or connection-oriented. A connectionless transport layer treats each segment as an independent packet and delivers it to the transport layer at the destination machine. A connection-oriented transport layer makes a connection with the transport layer at the destination machine first before delivering the packet. When the data is transferred, the connection is terminated.

Therefore option 1, 2 and 4 are correct

Question 10

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If bandwidth of the link is 2Gbps and frame size is 1000 bits. Distance between two nodes is 6km and velocity of signal in medium is 3 × 108 m/sec and it is given that channel is error free and pure sliding window protocol is used, then what is the sender window size?

SOLUTION

Data:

Frame size = 1000 bits

Bandwidth = 2 Gbps

Distance = 6 km

Velocity = 3 * 108 m/sec

Calculation:

$${\rm{Transmission\;time}}\left( {{{\rm{T}}_{\rm{t}}}} \right) = {\rm{}}\frac{{{\rm{frame\;size}}}}{{{\rm{bandwidth}}}} = {\rm{}}\frac{{1000}}{{2\; \times \;{{10}^9}}} = 0.5{\rm{\;\mu sec}}$$

$${\rm{Propagation\;delay\;}}\left( {{{\rm{T}}_{\rm{d}}}} \right) = {\rm{}}\frac{{6\; \times \;1000}}{{3\; \times \;{{10}^8}}} = 20{\rm{\mu sec}}$$

As, we know that a = $$\frac{{{\rm{Propagation\;delay}}}}{{{\rm{Transmission\;delay}}}} = {\rm{}}\frac{{20}}{{0.5}} = 40$$

Maximum window size that is possible in case of error free sliding window protocol = 1 + 2 × a

= 1 + 2 × 40 = 81

Question 11

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Consider a sender sends data using stop and wait protocol to the receiver. If the receiver wants 31 packets and every 4th packet is lost. What is the total number of packets sent by the sender?

SOLUTION

Concept

In stop and wait protocol, packet lost during transmission is the only packet retransmitted.

Explanation

 1 2 3 4 4 5 6 7 7 8 9 10 10 11 12 13 13 14 15 16 16 17 18 19 19 20 21 22 22 23 24 25 25 26 27 28 28 29 30 31 31

Lost packets: 4, 7, 10, 13, 16, 19, 22, 25, 28, 31

Total packets = 31 + 10 = 41

Tips and Tricks:

Lost of packet happens in Arithmetic progression 4, 7, 10... (difference of 3)

Total packet lost = n

Tn = 4 + (n - 1) (3)

31 = 1 + 3n

∴ n = 10

10 + 31 = 41

Question 12

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Which of the following statements are true?

(a) Three broad categories of Networks are:

(i) Circuit Switched Networks

(ii) Packet Switched Networks

(iii) Message Switched Networks

(b) Circuit Switched Network resources need not be reserved during the set up phase.

(c) In packet switching there is no resource allocation for packets.

SOLUTION

Concept:

A network consists of two kind of elements:

1) Network nodes known as actual computing end system

2) connection between these along which data transmission travel.

There must be some method to switch data flows between these network nodes.

Explanation:

There are three basic type of switching.

Circuit switching:

• It occurs when a physical link is made between two network nodes for duration of session along which data can flow under the control of end systems.
• In circuit switching, a route and its associated bandwidth is reserved from source to destination, making circuit switching relatively inefficient since capacity is reserved whether or not the connection is in continuous use.

Message switching:

• It is a network switching technique does not establish a physical connection. It relies on the network to deliver a message successfully once sent.
• Message is routed entirety from the source node to the destination node, one hope at a time.
• During message routing, every intermediate switch in the network stores the whole message. If the entire network's resources are engaged or the network becomes blocked, the message-switched network stores and delays the message until ample resources become available for effective transmission of the message.
•  It is low cost network.

Packet switching:

• Individual message is divided into packets each of which travels independently through the network.
• Packets are dispatched from source and reassembles at the point of exit for the destination.
• There is no resource allocation for packets in packet switching. With this technique, faster transmission rate is achieved.

Question 13

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Consider three machines X, Y, and Z with IP address 200.20.15.3, 200.20.15.1, and 200.20.15.2 respectively. The subnet mask is set to 255.255.255.254 for all the three machines. Which one of the following is/are true?

SOLUTION

X’s IP address: 200.20.15.3 →      200.20.15. 0000 0011

∴ Subnet Id = 200.20.15.2

Y’s IP address: 200.20.15.1 →       200.20.15.0000 0001

∴ Subnet Id = 200.20.15. 0

Z’s IP address: 200.20.15.2 →      200.20.15.0000 0010

∴ Subnet Id = 200.20.15.2

X and Z belongs to same subnet 200.20.15.2

Y and Z belongs to different subnet 200.20.15.2

Tips and Tricks:

& → bitwise ANDSince Network Mask is 255.255.255.254, not have to check first 3-byte because subnet id will same as first 3 bytes of IP address. Only 4th byte need to bitwise AND with given IP address

Question 14

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Consider two hosts S and R connected directly by a link. A link has a transmission speed of 107 bits/sec. It uses data packets of size 400 bytes. Data needed to transmitted is 400 bytes. Length of the link is 600 km and the propagation speed are 3 × 108 meter per second. T is the transmission delay and P is the propagation delay. The propagation delay in millisecond and transmission delay in microsecond respectively are

SOLUTION

Data:

Packet size = L = 400 bytes = 3200 bits

Bandwidth = BW = 107 b/s

Distance = d= 600 km

Propagation speed = v = 3 × 108 m/s = 3 × 105 km/s

Transmission delay = T

Propagation delay = P

Formula:

$$T = \frac{L}{{BW}}$$

$$P = \frac{d}{v}$$

Calculation:

$$T = \frac{{3200}}{{{{10}^7}}} = 320 \times {10^{ - 6}}\;s = 320\;\mu s\;$$

$$P = \frac{{600}}{{3 \times {{10}^5}}} = 2 \times {10^{ - 3}}\;s = 2\;ms$$propagation delay = 2 ms and transmission delay = 320 μs.

Question 15

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There are 5 stations in a slotted LAN. Each station attempts to transmit with a probability 0.6 in each time slot. What is the probability that only one station transmits in a given time slot?

SOLUTION

Number of stations = n = 5

Probability = p = 0.6

Only one station transmits in a given slot

Using Bernoulli Expression:

5C1p(1 - p)5-1 = 5C1(0.6)(1 – 0.6)5-1

= 5 × (0.6) × (1 – 0.6)4 = 5 × (0.6) × (0.4)4= 0.0768

Question 16

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A linear Hamming code is used to map 4-bit messages to 7-bit code words. The encoder mapping is linear. If the message 0001 is mapped to the code word 0000111, and the message 0011 is mapped to the code word 1100110, then the message 0010 is mapped to

SOLUTION

• A linear code is an error correcting code.
• Any linear combination of codewords is also a code word.

 Message Codeword 000100110010 00001111100110?

0010 = (0001) ⊕ → EXOR (0011)

To get the code word, Exor the two code words.

(0000111) ⊕ (1100110)= 1100001

Question 17

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Consider host A sends a UDP datagram containing 16060 bytes of user data to host B over an Ethernet LAN. Ethernet MTU is 1500 byte. Size of the IP header is 40 byte and the size of the UDP header is 8 bytes. Which of the following is/are true about the number of fragments transmitted and contents of the offset field in the last fragment?

SOLUTION

 Application layer 16060(Data) Transport layer (UDP) 8(Header) 16060(Data) Network layer 40(Header) 16068(Data)
Data = 16068 Byte MTU = 1500 byte Data + header = 1500 Data = 1500 – 40  Data = 1460 Number of fragments = $$\frac{{16068}}{{1460}} = 12$$ Since initial fragment offset is not given assume it to be 0. 1460 in not divisible by remove 4 byte and add to last fragment Fragment offset of 2nd fragment = $$\frac{{1456}}{8} = 182$$ Fragment offset of 3rd fragment = $$182 + \frac{{1456}}{8} = 364$$ Fragment offset of 3rd fragment = $$182 + \frac{{1456}}{8} = 546$$ : : Fragment offset of 12th fragment = 2002 Therefore option 2 and 3 are correct. Tips and Tricks: Since there are 12 fragments Fragment offset of 12th fragment $$= \left( {12 - 1} \right) \times \frac{{1456}}{8} = 2002$$

Question 18

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Which of the following statements is/are true?

I. Bridge reduces collision domain

III. CSMA/CD is suitable for a high propagation delay network like satellite network

SOLUTION

• Bridge is a 2-layer device i.e. work at physical layer and data link layer. There is no collision within a bridge. Hence collision domain is reduced in bridge.
• Ethernet is based on CSMA/CD, efficiency is given by $$\eta = \frac{1}{{1 + \frac{{{T_p}}}{{{T_t}}}}}$$ if propagation delay (Tp) is more than efficiency is less. Therefore, it is not suitable for high propagation delay network like satellite network

Question 19

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Consider the routing table shown below. A router can deliver the packets directly over interface 0 and 1 or it can forward to router R2, R3 and R4.

 Subnet number Subnet mask Next Hop 128.100.40.128 255.255.255.128 Interface 0 128.96.40.0 255.255.255.128 Interface 1 128.100.40.0 255.255.255.128 R2 128.100.39.36 255.255.255.192 R3 Default R4

What does the router with packet addressed to 128.100.40.36?

SOLUTION

Hint : We have to do the AND of this IP address with the subnet mask and if it matches with the subnet number, then it is delivered to the corresponding router or interface otherwise it will go to router having default.

Explanation:

Binary equivalent to: 128.100.40.36

= 10000000. 01100100. 00101000. 00100100

In binary form, subnet mask = 11111111. 11111111. 11111111. 10000000

AND of subnet mask and IP address = 10000000. 01100100. 00101000. 00000000

It represents router R2.

Question 20

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Let host A sends a file a size of 2200 byte to host B through routers R1, R2, R3, R4 and R5 respectively. Each packet contains 110 bytes of header information along with the user data. Assume that the bandwidth of each link is 106 B/s. File is split into 20 equal parts, and these packets are transmitted from A to B. Consider only transmission time and ignore processing, queuing and propagation delays also no error in transmission. What is the time taken by host A to send the complete file to host B?

SOLUTION

File size (data) = 2200 byte

Bandwidth (BW) = 106 byte/second

Data per packet = 2200 ÷ 20 = 110B

$$Transmission\;time\left( {{T_t}} \right) = \frac{{data\; + \;header}}{{BW}}\;$$

$${T_t} = \frac{{110 + 110}}{{{{10}^6}}} = 0.22\;ms$$

A → R1 → R2 → R3 → R4 → R5 → B

6 hops are needed

∴ time needed for 1st packet = 6 × 0.22 ms = 1.32 ms

And time needed for remaining 19 packet = 19 × 0.22 ms = 4.18 ms

∴ total time = 1.32 + 4.18 = 5.5 ms

Question 21

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Consider a long-lived TCP session with an end-to-end bandwidth 2 GB/s. The session starts with a sequence number of 3465. The minimum time (in seconds correct up to two decimal places) before this sequence number can be used again is _____

SOLUTION

2 GB → 1 second

$$1B \to \frac{1}{{2 \times {{10}^9}\;}}\;seconds$$

Sequence number = 32 bits

$${2^{32}}B \to \frac{{{2^{32}}}}{{2 \times {{10}^9}\;}}\;seconds$$

$${2^{32}}B \to 2.147\;seconds$$

The minimum time before this sequence number can be used again is 2.15 seconds. ##### Current Test Score
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