SHARE
SPREAD
HELP

The Tradition of Sharing

Help your friends and juniors by posting answers to the questions that you know. Also post questions that are not available.

To start with, Sr2Jr’s first step is to reduce the expenses related to education. To achieve this goal Sr2Jr organized the textbook’s question and answers. Sr2Jr is community based and need your support to fill the question and answers. The question and answers posted will be available free of cost to all.

The pleasure of all reading is doubled when one lives with another who shares the same. -Katherine Mansfield

 

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

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.

TextbookTextbookTextbookTextbookTextbookTextbookTextbookTextbookTextbookTextbookTextbookTextbookTextbookTextbookTextbookTextbookTextbook

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)

 

1 0

Discussions

Post the discussion to improve the above solution.