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?

Question

Answer

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%.

Answer

Here , packet size is 2 kb or 2 * 1024 bytes or 2 * 1024 * 8 bits

Transmission time is defined as ( packet size / link speed )

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%

Similar Computer Science Engineering (CSE) Doubts

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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.

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