Tanenbaum

Question

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?

Comments

(a) host - router ,then router - router,then router to destination

so,in total 3 hops for a packet.

(b) mean number of transmissions -

it might be possible that there is only 1 transmission i.e packet is nt dropped by any router.

or it might be the case that in first transmission,packet is dropped by a router,so again source will start transmission,hence 'p' this time

in this way 1+ p + p*p  +  p*p*p +........

so,mean number of transmissions are (1/1-p)

please correct me

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i think for c it is always 3 becouse  hops required  host - router ,then router - router,then router to destination

correct me if iam wrong.

Answer 1

B.)

A) THERE ARE THREE HOPS IN TOATAL

    LINKS                   HOPS count

     L1   -----------------   1

     L2   -----------------   (1-P) * 1

     L3   ----------------    (1-P)*(1-P)*1

   TOTAL                   1 +(1-P) +(1-P)^2  = 1+(1-P)(2-P)  mean no. of hops a packet makes per transmission.

C) TOTAL NO. OF HOPS PER RECEIVED PACKETS IS 3 .

Answer 2

mean number of transmissions -

p + (1-p)p +(1-p)p^2..........

so thee above is the probability but not the mean of no of transmissions mean of no of transmissions will be

3+6+9+12+..............................................

3(1+2+3+............................)

3(n(n+1)/2)

asit is mean we need to divide with no of transmissions so it whould be 

(3n+3)/2

according to me option A and option C are similar    hops a packet makes per transmission? (which means no of hops   for a succesful transmission )

(c) hops required per received packet this also mean the same that is 3

Answer 3

(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))

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