- ahmad alhayek
الرئيسية صورة وعبرة كلام الناس خواطر

 HTTP Performance. Here, we consider the performance of HTTP, comparing non-persistent HTTP with persistent HTTP Suppose the page your browser wants to download is 100K bits long, and contains 10 embedded images (with file names img01.jpg, img02.jpg, . img10.jpg), each of which is also 100K bits long. The page and the 10 images are stored on the same server, which has a 300 msec roundtrip time (RTT) from your browser. We will abstract the network path between your browser and the Web server as a 100 Mbps link. You can assum 20 points Save Answ the time it takes to transmit a GET message into th transmit the base file and the embedded objects into the link. This means that the server-to-client link has both a 150 msec one-way propagation delay, as well as a transmission delay associated with it. (Review page 39 in the textbook if you are uncertain about the difference between transmission delay and propagation delay.) In your answer, be sure to account for the time needed to set up a TCP connection (1 RTT) e link is zero, but you should account for the time it takes to a. Assuming non-persistent HTTP (and assuming no parallel connections are open between the browser and the server), how long is the response time the time from when the user requests the URL to the time when the page and its embedded images are displayed? Be sure to describe the various components that contribute to this delay b. Again, assume non-persistent HTTP, but now assume that the browser can open as many parallel TCP connections to the server as it wants. What is the response time in this case c. Now assume persistent HTTP (HTTP1.1). What is the response time, assuming no pipelining? Clearly mark your answers for each sub-question. (Example Ans. a. This is my answer to part a Ans. b. This is my answer to part b and so on...) Show your work d. Now suppose persistent HTTP with pipelining is used. What is the response time? T T T Arial

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  • Anonymous's Avatar
    Anonymous answered this
    7,497 answers

    Solution:

    a)

    The transmission time of 100k bits page is

    = L/R= 100*10^3/100*10^6= 1 ms

    the transmission time of each 100 kbits image= 100*10^3/100*10^6= 1ms

    delay associated with the given scenario

    = 2*RTT*6 + transmission time for file + transmission time for images

    = 2 * 300 * 6 + 1ms + (10*1) ms= 3611 ms

    = 3.611 s

    b)

    The delay in this scenario is

    = 2 * RTT for first object + 2 * RTT of all images + transmission time for file + transmission time for images

    = 600 + 600 + 1 + 1

    = 1.202 s

    c)

    2 RTT for base file + RTT * 10(for images) + transmission time for file + transmission time for images

    = 600 + 10 * 300 + 10 + 10= 3620 ms

    = 36.11 ms

    d)

    2 * RTT for base file + RTT for all images (parallel) + transmission time for file + transmission time for images

    = 300 + 600 + 10 + 10= 920 ms

    = 0.92 s

    Please, please

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ahmad alhayek تصميم ahmad alhayek جميع الحقوق محفوظة 2016

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