Authors:

James F. Kurose, Keith W. Ross

Chapter:

Computer Networks And The Internet

Exercise:

Problems

Question:25 | ISBN:9780132856201 | Edition: 6

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 *_ *d*prop.

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*.

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

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

Propagation Speed(S) of the link between A and B

Calculate the propagation delay:

Calculate the band-width delay product:

Therefore, band-with delay product is 160000bits

b)

Size of the file =800000 bits

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

The band-width delay product:

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

Propagation Speed(S) of the link between A and B

Formula to calculate the length of 1 bit on the transmission line

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)*