Consider the Figure 2 for which there is an institutional network connected to t

Question

Consider the Figure 2 for which there is an institutional network connected to the internet. Suppose that the average object size is 850,000 bits and that the average request rate from the institution’s browsers to the origin server is 16 request per second. Also suppose that the amount of time it takes from when the router on the internet side of the access link forwards an HTTP request until it receives the response is three seconds on average. Model the total average response time as the sum of the average access delay (that is, the delay from the Internet router to the institution router) and the average internet delay. For the average access delay, use ?/(1???), where ? is the average time required to send and object over the access link and ? is the arrival rate of objects to the access link.

Find the total average response time.

Now suppose a cache is installed in the institutional LAN. Suppose the miss rate is 0.4. Find the total response time.

origin serverS public Internet 5.4 Mbps access link institutional network 1 Gbps LAN igure 2: Institutional network connected to the publicI

Explanation / Answer

Average object size = 850,000 bits

Link rate is = 150,00000 bits/sec

Average request rate from the institution’s browsers to the origin server = 16 request / second

So Average Response time is = 850,000 bits/150,00000 bits/sec = -.567 sec

The traffic intensity on the link is given by ß=(16 requests/sec)(.0567 sec/request) = 0.907.

the Average access delay is (.0567 sec)/(1-.907) = .6 seconds.

The total average response time is therefore .6 sec + 3 sec = 3.6 sec.

B) Average access delay is = .0567 sec

miss rate = 0.4

so (.0567 sec)/[1–(.4)(.907)] = .089 seconds.

the average response time is .089sec + 3 sec = 3.089 sec

So the average response time is (.6)(0 sec) + (.4)(3.089 sec) = 1.24 seconds.

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