home work

Question

A system uses the Sliding Window Protocol is having a bandwidth of 10Mbps with a window size of 100. What is the size of data if the distance between the sender and receiver is 72000km and the propagation speed is 3 x 10^8 m/sec? Given utilization is 0.5

Comments

$24120$ bits?

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24120 bits is the correct answer..

Answer 1

$\eta = N/(1+2a)$

And $\eta$ is given as $0.5 =  1/2 $

and Given,

                  $N = 100$

                  $B = 10Mbps = 10*10^6 bps $

                 $L = 72000 * 10^3 mtr $

                 $Propagation Speed = 3*10^8 m/sec$

        Means Travel $3*10^8 m ------> 1 sec$

                            $ 1 m ------> 1/3*10^8  sec$

                           $72 * 10^6 m ------> 0.24 sec $

Now $Propagation Delay (Tp)= 0.24 sec$

$\eta = 1/2$

$100/1+2a = 1/2$

$a=99.5$

    Where ${a=(Propagation time)/(Transmission Time) = Tp* B/L}$

$L =  Tp * B / 99.5$

$L=(0.24  * 10*10^6 )/99.5 == 24120b$

divide this by $8$ to get ans in Byte

$L=24120/8=3015 Bytes$

Done!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Comments

thank you , but i have 4 options to select answer , your answer dont match any one

2048 bytes

3015 bytes

4096 bytes

3072 bytes

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@Samaa.qudah i did some modification in ans..... check in last. 

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thank you