Q2.4. Suppose two hosts, A and B, are separated by 10 000 kilometres, and are co

Question

Q2.4. Suppose two hosts, A and B, are separated by 10 000 kilometres, and are connected by a direct link of R 1 Mbps. Suppose the propagation speed over the link is 2.5 10A8 metres/sec. Calculate the bandwidth-delay product, R Tprop. a. b. 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? 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. Q2.5. Suppose you click on a link within your web browser to obtain a web page. The IP address for the associated URL is not cached in your local host, so a DNS look-up is necessary to obtain the IP address. Suppose that n DNS servers are visited before your host receives the IP address from DNS: the successive visits incur an RTT of RTT1 RTTn. Further suppose that two images, sized 100 KB and 80 KB respectively. RTT2 are embedded in the web page associated with the link, while the size of the web page is 4 KB. Let RTTO denote the RTT between the local host and the server containing the object, and let R denote the average end-to-end transmission rate between your computer and the web server. How much time elapses from when the client clicks on the link until the client receives both the web page and the two images?

Explanation / Answer

Hi,

Below are the solutions to your problem:

Answer 1:

Answer:

a) 80,000 bits

Explaination:

10,000 kilometers is 10 million meters. The speed of the link is 250 million meters/second. That makes propagation delay of 0.08 seconds from end to end (10/250)

Pumping 1000kbps (1mbps) into a line that has 0.04 second delay, means you get 1000*.04=80kbits of data in "the pipeline", waiting for propagation to finish...

i.e 1Mbps* (10000*1000 / (2.5*10^8)) = 0.8 Mb

b) 80,000 bits

c) The bandwidth-delay product of a link is the maximum number of bits that can be in the link.

d) the width of a bit = length of link / bandwidth-delay product, so 1 bit is 125 meters long, which is longer than a football field

e) m/((m/s)*R) = s/R

Question 2:

Answer:

The total amount of time to get the IP address is RTT1 + RTT2+..... + RTTn .

Once the IP address is known, RTTO elapses to set up the TCP connection and another RTTO elapses to request and receive the small object. The total response time is 2RTT0+ RTT1 + RTT2 +....+ RTTn

Now further:

RTT1+ RTT2+..+ RTTn+ 2RTT0+2*2 RTT0= 2RTT0+ RTT1+ RTT2+...+ RTTn

Therefore using the above obtained equation:

Query the DNS once +setup connection once+ request 2 files

Hope that helps..HAPPY ANSWERING!!!!!!!!

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