Suppose Host A wants to send a file of size F = 1500 bytes to Host B. The path f

Question

Suppose Host A wants to send a file of size F = 1500 bytes to Host B. The path from Host A to Host B has three links having rates R1 =500 kbps, R2 = 1 kbps, and R3 = 2 Mbps, respectively. Assume there is no other traffic on this network. Assume negligible processing delays. Assume the packets have a size of L = 520 bytes (20 bytes are for header - carry no data from the file - 1000 bytes for payload). Assume no other traffic in the network and negligible propagation delay. Answer the following question:

(a) What is the end-to-end delay from the time Host A starts transmitting the third packet to the time when the file is completely received at Host B? Motivate your calculation.

(b) How much queuing delay does the second packet experience in the case where R1 =1 kbps, R2 = 2 Mbps, and R3 = 500 kbps, respectively? Motivate your calculation.

Explanation / Answer

(a)

According to the question,

(A)-------R1---------(X1)------R2---------(X2)-------R3---------(B)     

Assume X1 and X2 are separator of the links R1, R2 and R3.

R1 = 500 Kbps , R2 = 1Kbps,    R3 = 2 Mbps                                          

pkt size = 520 B (including 20 B header)

file size = 1500 B

number of pkts = (1500/500) = 3

Assume transmission delay at A, X1, and X2 are T1, T2 and T3 respectively.

then T1 = 520*8 b/500Kbps = 8.32 ms

        T2 = 520*8 b/ 1Kbps = 4160 ms

        T3 = 520*8 b/ 2Mbps = 2.08 ms

Since pkt 1 reaches to X1 at 8.32 ms.

         pkt 2 reaches to X1 at 16.64 ms and pkt 1 is being transmitted at X1.

         pkt 3 reaches to X1 at 24.96 ms and pkt 2 is in queue at X1 and pkt 1 is being transmitted.

Then, At X1, queuing delay of pkt 2 = 4160 - 8.32 = 4151.68 ms

                   and T2 = 4160 ms

                   total delay of pkt 2 at X1= 8311.68 ms

       So, queuing delay of pkt 3 at X1 = 8311.68 - 16.64 = 8295.04 ms

             and T2 = 4160 ms

(b)

             total delay at X1 of pkt 3 = 8295.04 + 4160 = 12455.04 ms

      queuing delay at X2 of pkt 3 = 0 ms

      So, Total end-to-end delay of pkt 3 = 8.32 ms + 12455.04 ms + 2.08 ms = 12465.44 ms

(b) Given, R1 = 1 Kbps, R2 = 2Mbps, R3 = 500 Kbps

    then, T1 = 520 * 8b / 1Kbps = 4160 ms

             T2 = 520 * 8b / 2Mbps = 2.08 ms

             T3 = 520 *8b / 500Kbps = 8.32 ms

So, pkt 1 reaches to X1 at 4160 ms.

       pkt 2 reaches to X1 at 8320 ms and pkt 1 reaches to X2 at 4162.08 ms and pkt 1 reaches

               to B at 4170.40 ms.

      Since, there is no pkt at X1 when pkt 2 reaches to X1. Similarly, there is no pkt at X2 when pkt 2 reaches to X2. So, no queuing delay for pkt 2.

      So, total queuing delay = 0 ms

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