Computer Networks Test 1

Concept:

Stateless Protocols are the type of network protocols in which the client sends the request to the server and server response back according to the current state.

It does not require the server to retain session information or status about each communicating partner for multiple requests.

Explanation:

Application Layer protocol: HTTP and DNS

Transport Layer protocol: TCP and UDP

Stateless protocol: HTTP, UDP, DNS

Therefore, UDP is a stateless transport layer protocol

Important Point:

UDP is an unreliable connectionless-transport layer Protocol used for its simplicity and efficiency in applications where error can be provided by the application layer

UDP provided process-to-process communication

Data:

Bandwidth = 20 Mbps = 20 × 10⁶ bps

Data network can pass on average = 60,000 frames in 120 seconds.

1 frame = 5,000 bits

Calculation:

Therefore, data network can pass in 120 seconds = 60,000 × 5,000 = 300 × 10⁶ bits

Throughput is data network can pass in 1 second

Throughput = \(\frac{300 \times 10^6}{120}\)= 2.5 Mbps.

The throughput of this network in is 2.5 Mbps 

Explanation:

Physical layer: Concerned with transmitting raw bits over a communication channel

Physical layer is the only layer of OSI network model which actually deals with the physical connectivity of two different stations. This layer defines the hardware equipment, cabling, wiring, frequencies, pulses used to represent binary signals etc.

Transport layer: True end-to-end layer from source to destination

It is termed as an end-to-end layer because it provides a point-to-point connection rather than hop-to- hop, between the source host and destination host to deliver the services reliably.

Session layer: Provide token management service

The session layer of OSI model is responsible for establishing, managing, synchronizing and terminating sessions between end-user application processes. Through token, this layer prevents the two users to simultaneously attempt the same critical operation.

Presentation layer: Concerned with the syntax and semantics of the information transmitted

It is responsible for data encryption and decryption of sensitive data before they are transmitted over common channels.

Data:

Baud rate = 2 × 107 b/s

Formula:

For Manchester encoding:

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

Calculation:

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

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

\(\therefore bit\;rate = \frac{{{1.5\times{10}^7}}}{2} = 7.5 \times {10^6}\;b/s\)

The value of k is 7.5

Data:

Length of cable(D) = 3km

Bandwidth (BW) = 30 MB/s

Speed = V = 2.5 × 108 m/s = 2.5 × 105 km/s

Frame size = Lmin 

Formula:

For CSMA/CD

​\({T_t} \ge 2 \times {T_p}\)​

Calculation:

\({T_t} \ge 2 \times {T_p}\)

\(L \ge 2 \times {T_p} \times BW\)

\({L} \ge 2 \times \frac{D}{V} \times BW\)

\({L} \ge 2 \times \frac{3}{2.5\times10^5} \times 30 \times10^6 \;bytes\)​​

If one node sends the data, then other two should not send to avoid collisions

Probability = N1 OR N2 OR N3 s

= N1 × (1 – N2) × (1 – N3) + (N1 – 1) × (N2) × (1 – N3) + (N1 – 1) × (1 – N2) × (N3)

= 0.2 × 0.6 × 0.4 + 0.8 × 0.4 × 0.4 + 0.8 × 0.6 × 0.6

\(= \frac{{48}}{{1000}}\; + \frac{{128}}{{1000}} + \frac{{288}}{{100}}\)

\(= \frac{{464}}{{1000}}\)

OSI model​

• Layers of OSI model: Physical Layer, Data Link Layer, Network Layer, Transport Layer, Session Layer, Presentation Layer, Application Layer
• Packet segmentation is the process of dividing a data packet into smaller units for transmission over the network; Packet segmentation happens at the transport layer
• Data compression (bit-rate reduction) involves encoding information using fewer bits than the original representation; Data compression is done at Presentation Layer
• Dialog control and Token management is done at the session layer

Data:

Length of packet = L = 10,000 bytes = 8 × 10⁴ bit

Bandwidth = BW = 10⁷ b/s

Distance = d= 5000 km

Propagation speed = v =2 × 10⁵ km/s

Transmission delay = T_t

Propagation delay = T_p

Formula:

\({T_t} = \frac{L}{{BW}}\)

\({T_p} = \frac{d}{v}\)

Calculation:

\({T_t} = \frac{{8 \times {{10}^4}}}{{{{10}^7}}} = 8\;ms\)

\({T_p} = \frac{{5000}}{{2 \times {{10}^5}}} = 25\;ms\)

Difference = T_p - T_t = 25 - 8 = 17 ms

Reading MAC address form left to right:

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

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

1^st byte = 2A = 0010 1010

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

1B : 3A : 23 : 7F : 32 : 1F

1^st byte = 1B = 0001 1011

C9 : EA : B2 : 1D : 64 : 15

1^st byte = C9 = 1100 1001

Broadcast MAC address: All bits are 1

 FF : FF : FF : FF : FF : FF

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

Note:

Every Broadcast MAC address is also a Multicast MAC address