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A network has a data transmission bandwidth of 20 x 106 bits per second. It uses CSMA/CD in the MAC layer. The maximum signal propagation time from one node to another node is 40 microseconds. The minimum size of a frame in the network is __________ bytes.
Correct answer is '200 Bytes'. Can you explain this answer?
Verified Answer
A network has a data transmission bandwidth of 20 x 106bits per second...
Since CSMA/CD
Transmission Delay = RTT
hence,
L=B RTT
L=B 2 Tpropagation delay
L=(20 106) 2 40 10-6
=20 2 40
=1600bits
=200bytes
Hence 200Bytes is the answer.
Transmission Delay = RTT
hence,
L=B RTT
L=B 2 Tpropagation delay
L=(20 106) 2 40 10-6
=20 2 40
=1600bits
=200bytes
Hence 200Bytes is the answer.
Most Upvoted Answer
A network has a data transmission bandwidth of 20 x 106bits per second...
Understanding CSMA/CD and Frame Size
In a CSMA/CD (Carrier Sense Multiple Access with Collision Detection) network, the minimum size of a frame is essential to ensure that the sender can detect collisions effectively. This is particularly crucial in networks where multiple devices share the same communication medium.
Key Factors Influencing Frame Size
- Bandwidth: The given bandwidth is 20 x 10^6 bits per second (bps).
- Propagation Time: The maximum signal propagation time is 40 microseconds (µs). This indicates the time it takes for a signal to travel from one node to another.
Calculating Round Trip Time
To determine the minimum frame size, we first calculate the round trip time (RTT):
- RTT = 2 * Propagation Time
- RTT = 2 * 40 µs = 80 µs
Data Transmission During RTT
Next, we calculate how much data can be transmitted during this round trip time:
- Data transmitted in 80 µs = Bandwidth * RTT
- Data transmitted = 20 x 10^6 bps * 80 x 10^-6 s = 1600 bits
Converting Bits to Bytes
To find the minimum frame size in bytes:
- Minimum frame size in bytes = Data transmitted (bits) / 8
- Minimum frame size = 1600 bits / 8 = 200 bytes
Conclusion
Thus, the minimum size of a frame in this network is 200 bytes. This size ensures that the sender can detect collisions in a timely manner, maintaining efficient network communication.
In a CSMA/CD (Carrier Sense Multiple Access with Collision Detection) network, the minimum size of a frame is essential to ensure that the sender can detect collisions effectively. This is particularly crucial in networks where multiple devices share the same communication medium.
Key Factors Influencing Frame Size
- Bandwidth: The given bandwidth is 20 x 10^6 bits per second (bps).
- Propagation Time: The maximum signal propagation time is 40 microseconds (µs). This indicates the time it takes for a signal to travel from one node to another.
Calculating Round Trip Time
To determine the minimum frame size, we first calculate the round trip time (RTT):
- RTT = 2 * Propagation Time
- RTT = 2 * 40 µs = 80 µs
Data Transmission During RTT
Next, we calculate how much data can be transmitted during this round trip time:
- Data transmitted in 80 µs = Bandwidth * RTT
- Data transmitted = 20 x 10^6 bps * 80 x 10^-6 s = 1600 bits
Converting Bits to Bytes
To find the minimum frame size in bytes:
- Minimum frame size in bytes = Data transmitted (bits) / 8
- Minimum frame size = 1600 bits / 8 = 200 bytes
Conclusion
Thus, the minimum size of a frame in this network is 200 bytes. This size ensures that the sender can detect collisions in a timely manner, maintaining efficient network communication.
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A network has a data transmission bandwidth of 20 x 106bits per second. It uses CSMA/CD in the MAC layer. The maximum signal propagation time from one node to another node is 40 microseconds. The minimum size of a frame in the network is __________ bytes.Correct answer is '200 Bytes'. Can you explain this answer? for Computer Science Engineering (CSE) 2025 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about A network has a data transmission bandwidth of 20 x 106bits per second. It uses CSMA/CD in the MAC layer. The maximum signal propagation time from one node to another node is 40 microseconds. The minimum size of a frame in the network is __________ bytes.Correct answer is '200 Bytes'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A network has a data transmission bandwidth of 20 x 106bits per second. It uses CSMA/CD in the MAC layer. The maximum signal propagation time from one node to another node is 40 microseconds. The minimum size of a frame in the network is __________ bytes.Correct answer is '200 Bytes'. Can you explain this answer?.
A network has a data transmission bandwidth of 20 x 106bits per second. It uses CSMA/CD in the MAC layer. The maximum signal propagation time from one node to another node is 40 microseconds. The minimum size of a frame in the network is __________ bytes.Correct answer is '200 Bytes'. Can you explain this answer? for Computer Science Engineering (CSE) 2025 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about A network has a data transmission bandwidth of 20 x 106bits per second. It uses CSMA/CD in the MAC layer. The maximum signal propagation time from one node to another node is 40 microseconds. The minimum size of a frame in the network is __________ bytes.Correct answer is '200 Bytes'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A network has a data transmission bandwidth of 20 x 106bits per second. It uses CSMA/CD in the MAC layer. The maximum signal propagation time from one node to another node is 40 microseconds. The minimum size of a frame in the network is __________ bytes.Correct answer is '200 Bytes'. Can you explain this answer?.
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