A group of N stations share 128Kbps line in Abramson's Aloha mode.
Each station transmits a 2000bit frame on an average of once every 50
seconds even if the previous one has not yet been sent. What is the
maximun value for N?

Hi anshul2579!!


In ALOHA mode, the maximum efficiency (or fraction of usable bandwidth) is given
by:
   1/2.e = 0.184

"An analysis [Abramson, 1973b] of the random access method of
transmitting packets in a pure ALOHA channel shows that the normalized
theoretical capacity of such a channel is 1/2e=0.184. Thus the average
data rate which can be supported is about one sixth the data rate
which could be supported if we were able to synchronize the packets
from each user in order to fill up the channel completely."
http://research.microsoft.com/users/GBell/Computer_Structures_Principles_and_Examples/csp0433.htm

This means that the maximum channel utilization is 0.184 (or 18.4%).

For this problem we have:

  maxBandwidth = 0.184 * 128 kbps = 23.55 kbps = 23,550 bps


The required bandwidth for each station is:

  reqBandwidth = 2000 bit / 50 sec = 40 bps


The maximum value of N is

  N = maxBandwidth / reqBandwidth = 23,550 bps  / 40bps = 588 stations



I hope that this helps you.

Best regards.
livioflores-ga