The figure below shows an institutional network connected to the Internet. Hosts

Question

The figure below shows an institutional network connected to the Internet. Hosts in this institutional network can download files from both remote origin servers and a local web cache onigin publik Mbpá access link lweb Assume the following: An average file size = 100,000 bits. The average request rate for files from institution's browsers = 40 The propagation delay from the institutional router to any origin server and back to router i3 2 seC requests/sec. The average time to download a file is the sum of the average access delay and the (round-trip) propagation delay from the server where the file resides in. The average access delay at a link can be calculated by c, where T is the average time required to send a file over the link (ie., the transmission delay), u is the utilization at the link and is defined as (average bits transmitted per second)/(link rate).

Explanation / Answer

3 a,b)

3a)

utilization(u)=average bits transmitted per second/link rate(R)

average bits transmitted per second = average requests per second * bits per request(average file size) = 40 *100000 bits = 4Mbps

0.8 = 4 Mbps/R

or R=4/.8=5Mbps

b)Average delay for downloading a file= average access delay + average propogation delay= T/(1-u) + 2 seconds

T= transmission delay=average time to send the file over the access link = file size/link speed= 100000bits/5 Mbps=100000/5000000=0.02 seconds

Average access delay=T/(1-u) = 0.02/0.2=0.1 seconds

Average delay for downloading file = 0.1 seconds + 2 seconds = 2.1 seconds

3 c,d,e)

Now the access link is 2Mbps

T= transmission delay=average time to send the file over the access link = file size/link speed= 100000bits/2 Mbps=100000/2000000=0.05 seconds

As the proxy server propogation delay is not mention for proxy server to client, we will consider 0.2 ms as the total 2 way propogation delay.

The installation of the local proxy server will help in reducing the propogation delay. For all the files cached by it, there will be no need to access original server and total propogation delay will be the propogation delay of cache server(0.2 ms or 0.0002 seconds). If the file is not cached, the propogation delay will be the propogation delay of cache server+propogation of the original server (2 seconds+0.2 milli seconds)=2.0002 seconds

Average delay for downloading a file= average access delay + average propogation delay

3 c) Average delay for a file from the original server(uncached file) =0.05 + 2.0002=2.0502 seconds

3 d) Average delay for a file cached with local proxy server = 0.05 + 0.0002 = 0.0502 seconds

3 e) On an average 60% files will be taken from cache server and 40% from original server.

Hence Average delay for downloading any file in this case = 2.0502 * 0.4 + 0.0502 * 0.6 = 0.8502 seconds

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