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A network using CSMA/CD has a bandwidth 10 Mbps. If the maximum propagation time (including delays in the devices and ignoring the time needed to send a jamming signal) is 25.6 microseconds. What is the minimum size of the frame?
- a)512 bytes
- b)1024 bytes
- c)64 bytes
- d)32 bytes
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
A network using CSMA/CD has a bandwidth 10 Mbps. If the maximum propag...
CSMA/CD and Bandwidth
CSMA/CD (Carrier Sense Multiple Access with Collision Detection) is a protocol used in Ethernet networks to avoid data collisions. In this protocol, each device on the network listens for a carrier signal before transmitting data onto the network. If two devices transmit data at the same time, a collision occurs and both devices stop transmitting and wait for a random amount of time before trying again.
Bandwidth is the maximum amount of data that can be transmitted over a network in a given time period. It is measured in bits per second (bps). In this question, the bandwidth of the network is given as 10 Mbps (10 million bits per second).
Propagation Time
Propagation time is the time it takes for a signal to travel from one end of the network to the other. It includes delays in the devices on the network and the time it takes for the signal to travel over the physical medium (copper wire, fiber optic cable, etc.). In this question, the maximum propagation time is given as 25.6 microseconds.
Minimum Frame Size
The minimum frame size for a CSMA/CD network is calculated using the following formula:
Minimum Frame Size = 2 x Propagation Time x Bandwidth
In this question, the maximum propagation time is given as 25.6 microseconds and the bandwidth is given as 10 Mbps. Substituting these values in the formula, we get:
Minimum Frame Size = 2 x 25.6 microseconds x 10 Mbps
Minimum Frame Size = 512 bits
Since 1 byte = 8 bits, the minimum frame size is:
Minimum Frame Size = 512 bits / 8 bits per byte
Minimum Frame Size = 64 bytes
Therefore, the correct answer is option C, which is 64 bytes.
CSMA/CD (Carrier Sense Multiple Access with Collision Detection) is a protocol used in Ethernet networks to avoid data collisions. In this protocol, each device on the network listens for a carrier signal before transmitting data onto the network. If two devices transmit data at the same time, a collision occurs and both devices stop transmitting and wait for a random amount of time before trying again.
Bandwidth is the maximum amount of data that can be transmitted over a network in a given time period. It is measured in bits per second (bps). In this question, the bandwidth of the network is given as 10 Mbps (10 million bits per second).
Propagation Time
Propagation time is the time it takes for a signal to travel from one end of the network to the other. It includes delays in the devices on the network and the time it takes for the signal to travel over the physical medium (copper wire, fiber optic cable, etc.). In this question, the maximum propagation time is given as 25.6 microseconds.
Minimum Frame Size
The minimum frame size for a CSMA/CD network is calculated using the following formula:
Minimum Frame Size = 2 x Propagation Time x Bandwidth
In this question, the maximum propagation time is given as 25.6 microseconds and the bandwidth is given as 10 Mbps. Substituting these values in the formula, we get:
Minimum Frame Size = 2 x 25.6 microseconds x 10 Mbps
Minimum Frame Size = 512 bits
Since 1 byte = 8 bits, the minimum frame size is:
Minimum Frame Size = 512 bits / 8 bits per byte
Minimum Frame Size = 64 bytes
Therefore, the correct answer is option C, which is 64 bytes.
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Community Answer
A network using CSMA/CD has a bandwidth 10 Mbps. If the maximum propag...
We know that,
L= Size of frame
BW→ Bandwidth = 10 Mbps = 10 × 106 b/s
Tp → Propagation time = 25.6μs = 25.6 × 10−6 s
Tt → Transmission time
As we know,
For CSMA/CD,
Tt ≥ 2 × Tp
Tt = L/BW
Tt = L/BW
Now,
L/BW ≥ 2 × Tp
L/BW ≥ 2 × Tp
L ≥ 2 × Tp × BW
L ≥ 2 × 25.6 × 10−6 × 10 × 106
L ≥ 512 bits
L ≥ 64 × 8 bits (8 Bit = 1 byte)
Lmin = 64 bytes
Therefore, the minimum size of a frame in the network is 64 bytes.
Hence, the correct option is (C).
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A network using CSMA/CD has a bandwidth 10 Mbps. If the maximum propagation time (including delays in the devices and ignoring the time needed to send a jamming signal) is 25.6 microseconds. What is the minimum size of the frame?a)512 bytesb)1024 bytesc)64 bytesd)32 bytesCorrect answer is option 'C'. Can you explain this answer?
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