This preview shows page 1. Sign up to view the full content.
Unformatted text preview: )
+ 10*(200/150+Tp + 100,000/150+ Tp )
=7351 + 24*Tp (seconds)
Assuming the speed of light is 300*106 m/sec, then Tp=10/(300*106)=0.03 microsec. Tp
is therefore negligible compared with transmission delay.
Thus, we see that persistent HTTP is not significantly faster (less than 1 percent) than the non persistent case with parallel download. Problem 5: (20 Points) In this problem, we consider the performance of HTTP, comparing non persistent HTTP with persistent HTTP. Suppose the page your browser wants to download is 500K bits long, and contains 5 embedded images (with file names img01.jpg, img02.jpg, …, img05.jpg), each of which is also 100K bits in length. The page and the 5 images are all stored on the same server, which has a 250 msec RTT from the ECE374: Homework 2...
View
Full
Document
This homework help was uploaded on 03/06/2014 for the course ECE 374 taught by Professor Ganz during the Spring '08 term at UMass (Amherst).
 Spring '08
 GANZ

Click to edit the document details