Kurose and Ross Edition 6 Exercise 1 Question P27 (Page No 76)

Question

Suppose two hosts, A and B, are separated by 20,000 kilometers and are connected by a direct link of R = 1Gbps. Suppose the propagation speed over the link is $ 2.5×10^8$ meters/sec

1. Calculate the bandwidth-delay product, $R \times d _{prop}$.

2. Consider sending a file of 800,000 bits from Host A to Host B. Suppose the file is sent continuously as one big message. What is the maximum number of bits that will be in the link at any given time?
3.  What is the width (in meters) of a bit in the link?

 

Answer 1

 

1. R x dprop = (1x10^9) X {2x10^7/2.5x10^8}  = 8x10^7 bits
2. 800,000 bits since Maximum = min(8x10^7 , 8x10^5) 
3. Width of a bit = length of link/bandwidth-delay

             Width of a bit = m/(R · tprop)

             = m/(R · m/s) = s/R =2.5x10^8/ 10^9 = 0.25 meters