Computer Networks Test 2

For Manchester encoding:

\(Baud\;rate = 2 \times bit\;rate\)

\(bit\;rate = \frac{{Baud\;rate}}{2}\)

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

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.

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

In selective repeat protocol,

window size of sender = window size of receiver

Also, there are 16 distinct sequence number

∴ W_s + W_r   =  16

W_s = W_r

∴ W_s + W_s   =  16

∴ Ws =  8

Also W_r =  8

Number of segments \(= \frac{{64}}{2} = 32\)

Threshold \(= \frac{{32}}{2} = 16\)

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

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.

Reading MAC address form left to right:

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

43:10:1B:2D:06:EE 

1^st byte = 43 = 0100 0011

Examples of other types of MAC address:

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

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

1st byte = 2A = 0100 0011

Broadcast MAC address: All bits are 1

 FF : FF : FF : FF : FF : FF

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

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

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 devices: simplex, 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

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

Concept

Explanation

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

T_n = 4 + (n - 1) (3)

31 = 1 + 3n

∴ n = 10

10 + 31 = 41

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.

Subnet mask: 255.255.255.254→ 255.255.255.1111 1110

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

Subnet Mask & X’s IP address →  200.20.15.2

∴ Subnet Id = 200.20.15.2

Y’s IP address: 200.20.15.1 →       200.20.15.0000 0001

Subnet Mask & Y’s IP address →  200.20.15. 0

∴ Subnet Id = 200.20.15. 0

Z’s IP address: 200.20.15.2 →      200.20.15.0000 0010

Subnet Mask & Z’s IP address → 200.20.15.2

∴ 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 AND

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:

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

\(P = \frac{{600}}{{3 \times {{10}^5}}} = 2 \times {10^{ - 3}}\;s = 2\;ms\)

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

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

0010 = (0001) ⊕ → EXOR (0011)

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

(0000111) ⊕ (1100110)

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 \(= \left( {12 - 1} \right) \times \frac{{1456}}{8} = 2002\) 

• 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 (T_p) is more than efficiency is less. Therefore, it is not suitable for high propagation delay network like satellite network

Address is: 128.100.40.36

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

Consider the subnet mask 255.255.255.128

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

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

It represents router R2.

File size (data) = 2200 byte

Header size = 110 byte

Bandwidth (BW) = 10⁶ byte/second

Data per packet = 2200 ÷ 20 = 110B

Header size = 110 B

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

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

A → R1 → R2 → R3 → R4 → R5 → B

6 hops are needed

∴ time needed for 1^st 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

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.