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In a network,system packet size is 2kb,propagation time is30 msec and channel capacity is 10^6 bits/sec. What will be the transmission time?? What is the channel utilization of sender in%ge?
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In a network,system packet size is 2kb,propagation time is30 msec and ...
Answer
Transmission Time Calculation
- System packet size = 2KB = 2 x 1024 = 2048 bytes
- Channel capacity = 10^6 bits/sec = 1,000,000 bits/sec
- Transmission time = packet size / channel capacity
- Transmission time = (2048 x 8) / 1,000,000
- Transmission time = 0.016384 seconds
Channel Utilization Calculation
- Propagation time = 30 msec = 0.03 seconds
- Total time = Transmission time + Propagation time
- Total time = 0.016384 + 0.03
- Total time = 0.046384 seconds
- Channel utilization = Transmission time / Total time x 100%
- Channel utilization = 0.016384 / 0.046384 x 100%
- Channel utilization = 35.34%
Explanation
In this problem, we are given the system packet size, propagation time, and channel capacity. We need to calculate the transmission time and channel utilization of the sender.
To calculate the transmission time, we first convert the packet size from KB to bytes and then divide it by the channel capacity to get the time taken to transmit one packet.
To calculate the channel utilization, we add the transmission time to the propagation time to get the total time taken for one packet and then divide the transmission time by the total time and multiply by 100 to get the percentage of channel utilization.
Therefore, the transmission time for the given packet size and channel capacity is 0.016384 seconds, and the channel utilization of the sender is 35.34%.
Community Answer
In a network,system packet size is 2kb,propagation time is30 msec and ...
Here , packet size is 2 kb or 2 * 1024 bytes or 2 * 1024 * 8 bits
Link speed or channel capacity are the same only ; here link speed is 10^6 bits / sec
Trans. time = ( 2 * 1024 * 8 bits ) / ( 1 M-bits / sec ) ) = 16.3 milli sec
Now , channel utilisation is expressed as
1 / ( 1 + 2 * a ) where a is ( propagation delay / trans. time )
=> 1 /( 1 + 2 * ( 30 ms / 16.3 ) )
=> nearly 21.3 % or 21%
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In a network,system packet size is 2kb,propagation time is30 msec and channel capacity is 10^6 bits/sec. What will be the transmission time?? What is the channel utilization of sender in%ge?
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