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A network has a data transmission bandwidth of 20 × 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.
- a)1600
- b)200
- c)800
- d)100
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A network has a data transmission bandwidth of 20 × 106 bits per...
Calculation:
- The minimum size of a frame can be calculated using the formula:
- Frame Size = 2 × Propagation Delay × Bandwidth
- Given Propagation Delay = 40 microseconds and Bandwidth = 20 × 10^6 bits per second
Calculation Steps:
1. Convert Propagation Delay to seconds: 40 microseconds = 40 × 10^-6 seconds
2. Calculate Frame Size:
Frame Size = 2 × 40 × 10^-6 seconds × 20 × 10^6 bits per second
= 2 × 40 × 20 = 1600 bits
= 1600 / 8 bytes (1 byte = 8 bits)
= 200 bytes
Conclusion:
- The minimum size of a frame in the network is 200 bytes.
- The minimum size of a frame can be calculated using the formula:
- Frame Size = 2 × Propagation Delay × Bandwidth
- Given Propagation Delay = 40 microseconds and Bandwidth = 20 × 10^6 bits per second
Calculation Steps:
1. Convert Propagation Delay to seconds: 40 microseconds = 40 × 10^-6 seconds
2. Calculate Frame Size:
Frame Size = 2 × 40 × 10^-6 seconds × 20 × 10^6 bits per second
= 2 × 40 × 20 = 1600 bits
= 1600 / 8 bytes (1 byte = 8 bits)
= 200 bytes
Conclusion:
- The minimum size of a frame in the network is 200 bytes.
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Community Answer
A network has a data transmission bandwidth of 20 × 106 bits per...
Data:
L = size of frame
BW → bandwidth = 20 × 106 b/s
Tp → propagation time = 40 μs = 40 × 10-6 s
Tt → transmission time
Formula:
For CSMA/CD,
Tt ≥ 2 × Tp
Calculation:
L ≥ 2 × 40 × 10-6 × 20 × 106
L ≥ 1600 bits
L ≥ 200 bytes
Lmin = 200 bytes
Therefore, the minimum size of a frame in the network is 200 bytes.
L = size of frame
BW → bandwidth = 20 × 106 b/s
Tp → propagation time = 40 μs = 40 × 10-6 s
Tt → transmission time
Formula:
For CSMA/CD,
Tt ≥ 2 × Tp
Calculation:
L ≥ 2 × 40 × 10-6 × 20 × 106
L ≥ 1600 bits
L ≥ 200 bytes
Lmin = 200 bytes
Therefore, the minimum size of a frame in the network is 200 bytes.
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A network has a data transmission bandwidth of 20 × 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.a)1600b)200c)800d)100Correct answer is option 'B'. Can you explain this answer?
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