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Question

Consider 802.3 LAN with 500 stations attached to five 500 - meter segments. The data rate is 10Mbps and the slot time is 51.2μs. If all stations transmit with equal probability, what is the channel utilization using a frame size of 512 bytes(in %)? Assume no. of contention slots are 1.716

Answer 1

Contention Period:- It is a minimum time a host must transmit such that it can be sure that no other host packets has been transmitting .

Given , Contention slot time = $51.2$ $\mu s$

Number of contention slots =$1.716 $

So to send a packet we need $51.2*1.716=87.8592 $ $\mu s$

Transmission time =$\frac{512*8}{10*10^{6}}=409.6 \mu s$

So utilization = useful time / total time = $\frac{409.6}{409.6+87.8592}=0.82 =82$% .

REf: http://www.piya.ee.engr.tu.ac.th/course/le517/lecture_10.pdf

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