#SlidingWindow

Question

The efficiency of a 16 Mbps ethernet is 3/4.Cable length is 100 km and frame size is 1000 bytes.If the network is using sliding window protocol,what is the window size?

Answer 1

Hope this helps ,correct me if am wrong

Comments

check the transmission time once  u have considered window size also in that  .

where transmission time will be calculated for one frame only  then it will be multiplied with required no of frames or window size .

and throughput = window size*frame size / TT + 2PT   where TT will be transmission  time of one frame only

check it once and tell me if i am wrong anywhere .

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But efficiency of ethernet is 1/(1+6.44a). For N window size it should be N/(1+6.44a). Isn't it?

Answer 2

efficiency =3/4

bandwidth=16 Mbps

length =100 KM

frame size=1000 Bytes

efficiency =useful time/total time

here useful time will be the time taken in sending required no of frames (window size) .

useful time =window size *Transmission time

total time =TT + 2PT

let window size=N

so efficiency=N*TT /TT+2PT

                =N/(1+2a) where a=PT/TT

now TT=data size / bandwidth

           =1000 *8  / (16*10^6) =500*10^-6 sec

      PT=length/ speed

          =100*10^3 / 3*10^8   where speed is speed of light (by default)

          =1/3 * 10^-3 sec  

now,  efficiency = N / 1 +2a = 3/4 (given)

                        =>3+6a =4N  

                           3+6*10/15=4N      [where a=pt/tt = 1/3 * 10^-3 *(500 *10^-6 ) =10/15)

                    N=7/4=1.75

              that says we can send 1.75 frames at one time ..that means  1 frame at one time

          so window size =1

correct me if i am wrong
 

Comments

why in useful time u are sending only 1 frame and actually in useful time it must be WS*frame size because overall data which is being sended will be counted as useful time
TT = WS* frame size / BW

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yes i was wrong i am correcting it