CS 312 - Internet Concepts
Fall 2011: Tues/Thurs 3-4:15pm, Dragas 1117

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Quiz 2 - HTTP Connections, RTTs, and Delays

Thursday, Oct 13, 2011

Question 2 is an actual question from the 2010 mid-term exam.

Consider downloading the webpage of www.gomonarchs.com The base HTML page contains the following embedded objects:

  • http://www.gomonarchs.com/banner.jpg
  • http://www.odu-football.com/photos/coach.jpg
  • http://www.odu-basketball.com/ads/tickets.jpg

This question will consider the time it takes to download the entire main web page for www.gomonarchs.com from a remote web browser. The round trip time from the browser to any server can be represented by RTT. Transmission and queuing delays are ignored. You may assume that the IP addresses for all of the hostnames are already cached.

  1. If persistent HTTP connections are used, how many TCP connections are needed to download the entire web page (base page, plus each embedded object)?
  2. If persistent HTTP connections are used, how many RTTs would it take to download the entire web page? Explain the purpose of each RTT. Show your work.
  3. If each object (including the base page) has a size of 1000 bytes, the access link is 4 Mbps, and the RTT to each server is 50 ms, what is the total download time for the entire web page if using persistent HTTP connections? For this question, you must calculate transmission and propagation delays. Show your work.

Solutions

1) 3 TCP connections:

  1. gomonarchs.com
  2. odu-football.com
  3. odu-basketball.com

2) 7 RTTs:

  1. handshake gomonarchs.com
  2. request/response (req/rsp) base page
  3. req/rsp banner
  4. handshake odu-football.com
  5. req/rsp coach
  6. handshake odu-basketball.com
  7. req/rsp tickets

3) 358 ms

 dprop = number of RTTs * RTT 
 dprop = 7 * 50 ms = 350 ms

 dtrans for one object = L/R

 L = 1000 B = 8,000 b

 8,000 b *   1000 ms        8
           ----------   =  --- ms = 2 ms
           4,000,000 b      4 

 dtrans = number of objects * dtrans for one object
 dtrans = 4 * 2 ms = 8 ms

 de2e = dtrans  + dprop 
 de2e  = 8 ms + 350 ms = 358 ms