computer networks

Question

Suppose within your Web browser you click on a link to obtain a Web page. The IP address for the associated URL is not cached in your local host, so a DNS lookup is necessary to obtain the IP address. Suppose that two DNS servers are visited before your host receives the IP address from DNS. The first DNS server visited is the local DNS cache, with an RTT delay of RTT0 = 5 msecs. The second DNS server contacted has an RTT of 7 msecs. Initially, let's suppose that the Web page associated with the link contains exactly one object, consisting of a small amount of HTML text. Suppose the RTT between the local host and the Web server containing the object is RTTHTTP = 46 msecs.

Answer 1

Based on the information provided, let's break down the steps involved and calculate the total time it takes to obtain the web page.

1. DNS Lookup:

- RTT delay to the local DNS cache (RTT0) = 5 ms

- RTT delay to the second DNS server (RTT1) = 7 ms

- Total DNS lookup time = RTT0 + RTT1

2. Retrieving the web page:

- RTT delay between the local host and the web server (RTTHTTP) = 46 ms

Now, let's calculate the total time:

Total Time = DNS Lookup Time + Retrieval Time

1. DNS Lookup Time:

- RTT0 = 5 ms

- RTT1 = 7 ms

- DNS Lookup Time = RTT0 + RTT1 = 5 ms + 7 ms = 12 ms

2. Retrieval Time:

- RTTHTTP = 46 ms

Total Time = DNS Lookup Time + Retrieval Time

= 12 ms + 46 ms

= 58 ms

Therefore, the total time it takes to obtain the web page, including DNS lookup and retrieval, is 58 milliseconds.