Question

Suppose two hosts A and B are directly connected. Length of the link is 10,000 km and the transmission rate is 10 Mbps. The propagation speed of the link in 0.5 * 108 m/s. Based on this information answer the following questions. (i) Suppose A sends a file of 100 MB size. How long does it take to send the file? Assume the file is sent continuously. (ii) Suppose the original file is broken up into 100 packets where size of each packet is 1 MB. B sends an ACK for each packet and A cannot send a packet until the previous packet is acknowledged. Transmission time of an ACK packet is negligible. How long does it take to send the file?
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Answer 1

Answer:

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Explanation:

he distance (Distance) between two hosts A and B = 20,000 km

a)

The distance (Distance) between two hosts A and B = 20,000 km

                                                                                   =2\times 10^{7} meters \left ( since 1km=10^{3}m \right )

Trasmission rate(R) of the direct link between A and B =2Mbps

                                                                                       =2\times 10^{6}bps \left ( 1Mbps =10^{6}bps \right )

Propagation Speed(S) of the link between A and B =2.5\times 10^{8} meters/sec

Calculate the propagation delay:

                            d_{prog}=\frac{Distance}{Speed} =\frac{2\times 10^{7}}{2.5\times10^{8} } =0.08sec

Calculate the band-width delay product:

                           R \times d_{prog} =2\times 10^{6}\times 0.08 =16\times 10^{4}bits

Therefore, band-with delay product is 160000bits

b)

Size of the file =800000 bits =8\times 10^{5}bits

Trasmission rate(R) of the direct link between A and B =2Mbps

                                                                                       =2\times 10^{6}bps \left ( 1Mbps =10^{6}bps \right )

The band-width delay product:

                           R \times d_{prog} =2\times 10^{6}\times 0.08 =16\times 10^{4}bits

Therefore, the maximum number of bits at a given time will be 160000bits.

c)

The product of band-windth delay is equal to the maximum number of bits on the transmission line.

d)

Trasmission rate(R) of the direct link between A and B =2Mbps

                                                                                       =2\times 10^{6}bps \left ( 1Mbps =10^{6}bps \right )

Propagation Speed(S) of the link between A and B =2.5\times 10^{8} meters/sec

Formula to calculate the length of 1 bit on the transmission line =\frac{Speed(S)}{Transmission rate(R)}

Length of 1bit = \frac{Speed(S)}{Transmission rate(R)} =\frac{2.5\times 10^{8}}{2\times 10^{6}} =125m/bit

Therefore, it is longer than a football field.

e)

 A general expression for the width = (Transmission rate(R) * Speed(s) )/ length of the link (m)