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Computer Networking : A Top-Down Approach

Computer Networking : A Top-down Approach

Authors:
James F. Kurose, Keith W. Ross
Exercise:
Problems
Chapter:
Computer Networks And The Internet
Edition:
6
ISBN:
9780132856201
Question:
25

 

Question

Suppose two hosts, A and B, are separated by 20,000 kilometers and are connected by a direct link of R = 2 Mbps. Suppose the propagation speed over the link is 2.5 x 10^8 meters/sec.

a. Calculate the bandwidth-delay product, R _ dprop.

b. Consider sending a file of 800,000 bits from Host A to Host B. Suppose the file is sent continuously as one large message. What is the maximum number of bits that will be in the link at any given time?

c. Provide an interpretation of the bandwidth-delay product.

d. What is the width (in meters) of a bit in the link? Is it longer than a football field?

e. Derive a general expression for the width of a bit in terms of the propagation speed s, the transmission rate R, and the length of the link m.

Answer

The 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)

 

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