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A datagram network allows routers to drop packets whenever they need to. The
probability of a router discarding a packet is p. Consider the case of a source host
connected to the source router, which is connected to the destination router, and then
to the destination host. If either of the routers discards a packet, the source host eventually
times out and tries again. If both host-router and router-router lines are counted
as hops, what is the mean number of
(a) hops a packet makes per transmission?
(b) transmissions a packet makes?
(c) hops required per received packet?
in Computer Networks by Boss (10.5k points) | 49 views

1 Answer

+1 vote
(a) Each  packet  may make 1, 2 or 3 hops.  For 1 hop,  the  first
     router  drops it and the  probability  is p.  For 2 hops,  it goes  through
     first router but not the second and the probability is (1-p)p.  For 3 hops,
     it goes through both routers and the  probability is (1-p)(1-p).  Mean hops
     per transmission is given by

     1 x p + 2 x (1-p)p + 3 x (1-p)(1-p) which simplifies to p**2 - 3p +3
    
     
     (b)  The probability of successful transmission all the way is (1-p)**2
     Let us denote it by w. The average number of transmissions per packet is
     given by
     
      w + 2w(1-w) + 3w(1-w)**2 +   +  nw(1-w)**(n-1) + ....
      
      which reduces to  1/w, that is,  1/(1-p)**2
      
     
     (c) mean hops  per  packet  = mean  hops  per  transmission  x mean  number
      of transmissions which is ( pxp - 3p + 3 ) / ((1-p) x (1-p))
by Boss (10.2k points)

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