Patterson Chap 1 Foundtion Q 2

Question

 Calculate the total time required to transfer a 1000-KB file in the
following cases, assuming an RTT of 50 ms, a packet size of 1 KB
data, and an initial 2 × RTT of “handshaking” before data is sent:
(a) The bandwidth is 1.5 Mbps, and data packets can be sent
continuously.
(b) The bandwidth is 1.5 Mbps, but after we finish sending each
data packet we must wait one RTT before sending the next.
(c) The bandwidth is “infinite,” meaning that we take transmit
time to be zero, and up to 20 packets can be sent per RTT.
(d) The bandwidth is infinite, and during the first RTT we can
send one packet (21−1), during the second RTT we can send
two packets (22−1), during the third we can send four (23−1),
and so on. (A justification for such an exponential increase
will be given in Chapter 6.)
 

Comments

plz adjust font properly..

Answer 1

part a) 

      Initial Handshaking = 2 * R.T.T. = 2*50 ms = 100 ms

      we can send continuously : 

              to send 1000KB file, & BandWidth is 1.5 Mbps..

              So, time to transmit is   1000KB / 1.5 Mbps  = 5333.33 ms

              Total Time  = T.T. + Initial Handshaking time

                                  =  5333.33 + 100 = 5433.33 ms 

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part b)

   Initial Handshaking = 2 * R.T.T. = 2*50 ms = 100 ms

   time to send 1 packet  = 1KB /  1.5 Mbps = 16 / 3  ms

   Inter packet gap = 1 R.T.T. = 50 ms

   Total Time  = Initial Handshaking time + 1000 packets T.T.

                                     + 999 * R.T.T.(waiting time)

                       =  100 + 1000(16/3) + 999(50)

                       =  55383.33 ms = 55.38 sec.

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part c)

   Initial Handshaking = 2 * R.T.T. = 2*50 ms = 100 ms

   packet size is 1 KB , so 1000 packets will be transmitted ,only 20 per R.T.T.

   So, it takes 1000 / 20 = 50 R.T.T. 

   but in last R.T.T , transmission time is '0' , so we will not consider it

                              => 49 R.T.T. = 2450 ms

   Total Time  = Initial Handshaking time + packets T.T.

                       = 100 + 2450 = 2550 ms

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part d)

   Initial Handshaking = 2 * R.T.T. = 2*50 ms = 100 ms

   1000 packets to be transmitted ,

      In 1st R.T.T   = 1 packet

      In 2nd R.T.T. = 2 packet

      In 3rd R.T.T. = 4 packets  & so on 

    like this we will take 10 R.T.T. 

    But again in last R.T.T time taken will be '0' , so will not consider this

                      => 9 R.T.T = 450 ms

     Total Time  = Initial Handshaking time + packets T.T.

                         = 100 + 450 = 550 ms

     

Comments

in ans d why is it not taking last rtt...also in option b... ??/

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reason for part d) :  Bandwidth is  $\infty$  , so packets transmission will take no time , Hence considered '0' {mentioned in question..}

part b) as we have to wait for 1 r.t.t  after each packet transmission , so after transmitting last packet we will not have to wait , transmitting 1000 packets , and waiting 999 r.t.t in between (after each packet)

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In part b ,why is the propagation time of the last packet tp=rtt/2 not taken into consideration. Considering the fact tt +tp is less than rtt ,so its not taken in other packets,but the last packet will be sent after n-1 rtt and so its total time should be taken .Please clarify sir