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# Reference models
## 1
Which layer packs bits into bytes, bytes into frames, uses hardware addressing and implements error detection? Which device (network hardware) operates on this layer in principle?
- 2nd layer: Data link layer; Switches work on this layer
## 2
At which layer does the assignment of addresses and the forwarding of data packets from one end of the network to another take place? Which network hardware operates on this layer?
- Later 3: Network layer; Router
## 3
Which layer is responsible for establishing, running and terminating a process communication between 2 systems?
- Layer 5: Session layer
## 4
At which layer does data stream segmentation take place for communication between endpoints, is a mechanism for establishing/disestablishing virtual connections provided, and does detection and elimination of transport errors as well as flow control or congestion avoidance take place?
- Layer 4: Transport layer
## 5
Which layer determines signal levels, transmission speeds, and connector pinouts, and transmits bits in a transmission channel? Which network hardware operates on this layer?
- Layer 1: Physical layer; Network interface card
## 6
What boundary does the transport layer form between layers 5-7 and 1-3?
- The transport layer forms the boundary between the lower transporting layers, which are primarilly concerned with network tasks, and the application oriented layers
## 7
Contrast the OSI layer model with the widely used TCP/IP layer model. Which layers correspond to each other?
-
## 8
What are the addressing names of layers 2, 3, 4 and 7?
- Layer 2: MAC-adresses
- Layer 3: IP-adresses
- Layer 4: Port number
- Layer 7: Application specific adresses such as E-Mail or URL
## 9
Can collisions occur in point-to-point networks?
- No, the switch sends the data out only at the reciver port. With original Ethernet 10 Mbit/s with CSMA/CD collisions can still happen between switch and connected host
## 10
How are computer networks classified according to their extent?
- ![[2023-04-18_11-08.png]]
# Tools

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# 1 Data transmission
## 1
To calculate the time it will take to transfer a 129 MB mp4-video file over a 300 Kbit/s communication channel, we first need to convert the file size and the data rate to a common unit.
File size: 129 MB (megabytes)
1 byte = 8 bits
129 MB * 1024 KB/MB * 1024 B/KB * 8 bits/byte = 1,076,633,600 bits
Data rate: 300 Kbit/s (kilobits per second)
300 Kbit/s * 1,000 bits/Kbit = 300,000 bits/s
Now, divide the file size by the data rate to find the time it takes to transfer the file:
Time (seconds) = File size (bits) / Data rate (bits/s)
Time (seconds) = 1,076,633,600 bits / 300,000 bits/s
Time (seconds) ≈ 3,588.78 seconds
To convert this into minutes, divide by 60:
Time (minutes) = 3,588.78 seconds / 60
Time (minutes) ≈ 59.81 minutes
It will take approximately 59.81 minutes to transfer the 129 MB mp4-video file over a 300 Kbit/s communication channel without any interferences or noises on the carrier.
## 2
According to the Nyquist theorem, the maximum data transfer rate (also known as channel capacity) can be determined using the following formula:
C = 2 * B * log2(M)
Where:
C = Channel capacity (data transfer rate) in bits per second (bps)
B = Bandwidth in Hertz (Hz)
M = Number of signal levels (distinct symbols)
In this case, the bandwidth (B) is 4 MHz, and there are 8 signal levels (M).
C = 2 * 4,000,000 Hz * log2(8)
C = 2 * 4,000,000 Hz * 3
C = 24,000,000 bps
The data transfer rate according to the Nyquist theorem at a bandwidth of 4 MHz and 8 signal levels is 24 Mbps (megabits per second).
## 3
To calculate the maximum data transmission rate of an ADSL-carrier, we can use the Shannon-Hartley theorem, which is given by the following formula:
C = B * log2(1 + SNR)
Where:
C = Channel capacity (maximum data transmission rate) in bits per second (bps)
B = Bandwidth in Hertz (Hz)
SNR = Signal-to-Noise Ratio (dimensionless)
First, we need to convert the SNR from decibels (dB) to a linear scale. To do this, we use the following formula:
SNR_linear = 10^(SNR_dB / 10)
In this case, SNR_dB = 20 dB and B = 1 MHz.
SNR_linear = 10^(20 / 10)
SNR_linear = 10^2
SNR_linear = 100
Now, we can plug these values into the Shannon-Hartley theorem:
C = 1,000,000 Hz * log2(1 + 100)
C ≈ 1,000,000 Hz * log2(101)
C ≈ 1,000,000 Hz * 6.658211
C ≈ 6,658,211 bps
The maximum data transmission rate of the ADSL-carrier with the given parameters (SNR = 20 dB and B = 1 MHz) is approximately 6.66 Mbps (megabits per second).
## 4
To increase the maximum data transmission rate to 16 Mbit/s, the provider can explore two possibilities: increasing the bandwidth (B) or improving the signal-to-noise ratio (SNR).
1. Increasing the Bandwidth (B):
By increasing the available bandwidth, the provider can achieve higher data transmission rates. To determine the required bandwidth for a target rate of 16 Mbit/s, we can use the Shannon-Hartley theorem:
C = B * log2(1 + SNR)
We already know that the SNR_linear is 100. Rearrange the formula to solve for the required bandwidth (B):
B = C / log2(1 + SNR_linear)
B = 16,000,000 bps / log2(1 + 100)
B ≈ 16,000,000 bps / 6.658211
B ≈ 2,402,907 Hz
The provider would need to increase the bandwidth to approximately 2.4 MHz to achieve a maximum data transmission rate of 16 Mbit/s.
2. Improving the Signal-to-Noise Ratio (SNR):
Alternatively, the provider can improve the signal-to-noise ratio (SNR) by reducing the noise or increasing the signal strength. This can be achieved through various techniques, such as better cables, advanced error correction, or signal amplification. To calculate the required SNR, we can rearrange the Shannon-Hartley theorem formula:
SNR_required = 2^(C / B) - 1
In this case, C = 16 Mbit/s and B = 1 MHz.
SNR_required = 2^(16,000,000 bps / 1,000,000 Hz) - 1
SNR_required = 2^16 - 1
SNR_required ≈ 65,535
To convert this linear value to decibels (dB), use the following formula:
SNR_dB = 10 * log10(SNR_linear)
SNR_dB = 10 * log10(65,535)
SNR_dB ≈ 48.17 dB
The provider would need to improve the SNR to approximately 48.17 dB to achieve a maximum data transmission rate of 16 Mbit/s with the existing 1 MHz bandwidth.
In summary, the provider can either increase the bandwidth to around 2.4 MHz or improve the SNR to around 48.17 dB to achieve the desired maximum data transmission rate of 16 Mbit/s.
# 2 encoding
## 5
Unmanaged carriers typically refer to communication channels that lack the proper infrastructure, organization, or management needed to ensure optimal data transmission. Some main problems when transmitting data over an unmanaged carrier include:
1. Interference and Noise:
Unmanaged carriers often experience higher levels of interference from other signals, as well as increased background noise. These factors can lead to signal degradation, reduced data transmission rates, and higher error rates.
2. Security and Privacy:
Unmanaged carriers may not have robust security measures in place to protect data from unauthorized access, tampering, or eavesdropping. This can result in data breaches, data corruption, and privacy violations.
3. Congestion and Bandwidth Limitations:
Due to a lack of proper resource allocation and management, unmanaged carriers are more prone to network congestion and bandwidth limitations. This can lead to increased latency, dropped connections, and slower data transmission rates.
4. Reliability and Stability:
Unmanaged carriers may not have the necessary infrastructure or failover mechanisms in place to ensure reliable and stable connections. This can result in frequent connection drops, interruptions, or inconsistent data transmission quality.
5. Scalability:
With an unmanaged carrier, it may be difficult to scale the network and accommodate more users or devices without facing significant performance degradation or other challenges.
6. Lack of Quality of Service (QoS) and Prioritization:
Unmanaged carriers typically do not provide Quality of Service (QoS) or traffic prioritization features. This means that critical or time-sensitive data may not receive the necessary priority, leading to reduced performance for applications and services that depend on low-latency or high-reliability connections.
7. Difficulty in Troubleshooting and Monitoring:
Without proper management and monitoring tools, diagnosing and resolving issues on an unmanaged carrier can be challenging. This can result in prolonged periods of downtime, reduced network performance, and increased costs for troubleshooting and maintenance.
## 6
The main purpose of line encoding is to convert digital data into a format suitable for transmission over a physical communication channel, such as a wire, fiber-optic cable, or wireless link. Line encoding involves transforming the digital bit stream into a series of electrical, optical, or electromagnetic signals that can be transmitted, received, and decoded by the receiver.
Some key objectives of line encoding include:
1. Clock Recovery:
Line encoding helps facilitate clock recovery at the receiver by incorporating sufficient timing information within the encoded signal. This allows the receiver to synchronize its internal clock with the transmitter's clock, ensuring accurate data recovery.
2. Synchronization:
Line encoding schemes can provide synchronization between the transmitter and receiver, making it easier to identify the start and end of data frames or individual bits within the transmitted signal.
3. Minimizing Signal Degradation:
Line encoding can help minimize signal degradation due to factors such as noise, interference, and attenuation. Certain encoding schemes can reduce the impact of these factors on the transmitted signal, improving the overall reliability and quality of the communication link.
4. Power Spectral Density Control:
Line encoding can help control the power spectral density of the transmitted signal, which is important for minimizing interference with other signals and meeting regulatory requirements for signal emissions.
5. Error Detection and Correction:
Some line encoding schemes incorporate redundancy or error detection and correction mechanisms, which can help identify and correct errors introduced during transmission.
6. Bandwidth Efficiency:
Line encoding can help optimize the use of available bandwidth by using efficient encoding techniques that minimize the required bandwidth for a given data rate.
In summary, the main purpose of line encoding is to convert digital data into a format suitable for transmission over a physical communication channel, addressing challenges such as synchronization, clock recovery, signal degradation, and bandwidth efficiency.
## 7
For the bit sequence 1011011101, the resulting NRZ (Non-Return-to-Zero) and NRZI (Non-Return-to-Zero Inverted) codes can be illustrated as follows:
NRZ Code:
- The NRZ code uses a fixed voltage level to represent each binary bit. A high voltage level may represent a binary 1, while a low voltage level may represent a binary 0.
- For the given bit sequence, the NRZ code would be as follows:
1 0 1 1 0 1 1 1 0 1
↑ ↓ ↑ ↓ ↑ ↓ ↓ ↓ ↑ ↓
H L H L H L L L H L
NRZI Code:
- The NRZI code uses a change in voltage level to represent a binary 1, and no change in voltage level to represent a binary 0.
- For the given bit sequence, the NRZI code would be as follows:
1 0 1 1 0 1 1 1 0 1
↑ ↓ ↑ ↑ ↓ ↑ ↓ ↓ ↓ ↑
H L H H L H L L L H
Note: The first bit in both codes can be either a high or low voltage level, depending on the starting condition of the transmission line. In this example, we have assumed a high voltage level for the first bit.
## 8
4B5B encoding is a line coding scheme that converts groups of four bits into five-bit symbols for transmission over a communication channel. To decode the original bit-stream from the given 4B5B encoded signal, we need to reverse this process using a decoding table or algorithm.
The decoding table for 4B5B encoding is as follows:
4B5B | Decoded Bits
----------------------
0000 | 0000
0001 | 0001
0010 | 0010
0011 | 0011
0100 | 0100
0101 | 0101
0110 | 0110
0111 | 0111
1000 | 1000
1001 | 1001
1010 | 0000 (special character)
1011 | 0000 (special character)
1100 | 0000 (special character)
1101 | 0000 (special character)
1110 | 0000 (special character)
1111 | 0000 (special character)
To decode the given 4B5B encoded signal, we can simply look up each five-bit symbol in the decoding table and convert it back to its original four-bit value.
The given 4B5B encoded signal is:
111001011101110010101010010101101011111001001
The decoded bit-stream would be:
1110 0101 1101 0111 0010 1010 1001 0110 1011 1111 0010 01
The special characters (1010, 1011, 1100, 1101, 1110, 1111) do not correspond to any valid four-bit values, so they are discarded in the decoding process.
Therefore, the original bit-stream that was encoded using 4B5B is:
1101010110101100101001001010
Note that the original bit-stream has 28 bits, whereas the 4B5B encoded signal has 44 bits, which is due to the fact that each group of four bits is encoded into a five-bit symbol.
## 9
Fast Ethernet uses a combination of 4B5B encoding and MLT-3 encoding to transmit data over the physical communication channel. To convert the given bit sequence according to this scheme, we need to apply 4B5B encoding followed by MLT-3 encoding.
1. 4B5B Encoding:
The first step is to apply 4B5B encoding to the bit sequence. 4B5B encoding converts groups of four bits into five-bit symbols, as per the following table:
4B5B | Bits
-------------
0000 | 0000
0001 | 0001
0010 | 0010
0011 | 0011
0100 | 0100
0101 | 0101
0110 | 0110
0111 | 0111
1000 | 1000
1001 | 1001
1010 | 1010
1011 | 1011
1100 | 1100
1101 | 1101
1110 | 1110
1111 | 1111
The given bit sequence is:
11010010000101101011
To apply 4B5B encoding, we need to split the bit sequence into groups of four bits and encode each group using the 4B5B table:
1101 0010 0001 0110 1011
1101 0001 0011 0101 1011
The encoded bit sequence using 4B5B is:
11010001001101011011
2. MLT-3 Encoding:
The second step is to apply MLT-3 encoding to the 4B5B encoded bit sequence. MLT-3 encoding uses a three-level voltage signal to represent the encoded bits, with positive, negative, and zero voltage levels.
To encode the 4B5B sequence using MLT-3, we need to assign a voltage level to each bit as follows:
0 | 0
1 | 1
V | Alternate between +V and -V for consecutive zeros
The encoded bit sequence using MLT-3 is obtained by applying this rule to each bit of the 4B5B encoded sequence:
11010001001101011011
+++---+++--+-+---+-
The resulting MLT-3 encoded signal is:
+ + + - - - + + + - + - + - - + - +
Therefore, the converted sequence according to the Fast Ethernet scheme is:
+++---+++--+-+---+-
# 3 Tools
## 10
# 4 Hardware
## 11
The main problem with a hub sending all network data to all connected network hosts is that it can result in a significant amount of unnecessary network traffic, which can reduce network performance, increase network congestion, and result in higher error rates.
Specifically, some of the main problems of this type of network traffic steering include:
1. Increased Network Congestion:
Since a hub sends all network data to all connected hosts, each host has to process all the traffic that is sent through the hub, even if it is not intended for them. This can result in increased network congestion and a reduction in overall network performance.
2. Higher Error Rates:
With a hub, every packet that is transmitted on the network is sent to every connected host. This can increase the likelihood of packet collisions and result in higher error rates due to the potential for packets to be corrupted or lost during transmission.
3. Security Risks:
Because all network data is sent to every connected host, there is a higher risk of unauthorized access to sensitive information. This can make it easier for attackers to intercept and analyze network traffic, potentially leading to security breaches or data theft.
4. Lack of Quality of Service (QoS):
With a hub, there is no way to prioritize or allocate bandwidth for different types of network traffic. This can result in delays, latency, and a lack of Quality of Service (QoS) for applications and services that require low-latency or high-bandwidth connections.
5. Inefficient Use of Bandwidth:
Sending all network data to all connected hosts can result in an inefficient use of available bandwidth. This is because network resources are consumed by transmitting data that may not be needed by some or all of the connected hosts.
6. Limited Scalability:
Hubs have limited scalability, as they can only support a limited number of connected devices before network performance and reliability begin to degrade. As a result, hubs are typically only used in small networks with a limited number of connected devices.
In summary, the main problem with a hub sending all network data to all connected hosts is that it can result in network congestion, higher error rates, security risks, a lack of QoS, inefficient use of bandwidth, and limited scalability.
## 12
The problem of unnecessary network traffic caused by a hub sending all network data to all connected network hosts can be addressed by using a switch instead of a hub. Switches are designed to selectively forward network data only to the connected hosts that need to receive it, based on the destination MAC (Media Access Control) address in the data packet.
Some ways switches solve this problem include:
1. Reduced Network Congestion:
Since a switch only forwards data to the host that needs to receive it, it reduces unnecessary network traffic and improves network performance. This reduces network congestion and can result in better overall network efficiency.
2. Lower Error Rates:
Switches use a store-and-forward mechanism to process packets, which can help reduce the likelihood of packet collisions and result in lower error rates.
3. Increased Security:
Switches can be configured with features such as port security, VLANs, and access control lists (ACLs) to improve network security and prevent unauthorized access to network resources.
4. Improved Quality of Service (QoS):
Switches can prioritize network traffic based on the type of application or service being used, allowing for more efficient use of network bandwidth and improving overall network performance.
5. Better Use of Available Bandwidth:
By selectively forwarding network data only to the connected hosts that need to receive it, switches make more efficient use of available network bandwidth.
6. Scalability:
Switches can support a larger number of connected devices than hubs, making them more scalable for larger networks.
In summary, using a switch instead of a hub can solve the problem of unnecessary network traffic caused by a hub sending all network data to all connected network hosts, by selectively forwarding data only to the connected hosts that need to receive it, improving network performance, security, QoS, bandwidth utilization, and scalability.