Question

Consider the following institutional network that is connected to the Internet. Suppose that the average object size is 240,000 bits and that the average request rate from the institution’s browsers to the origin servers is 62 requests per second. Also suppose that the amount of time it take from when the router on the Internet side of the access link forwards an HTTP request until it receives the response is 1.5 seconds on average (see Section 2.2.5). Model the total average response time as the sum of the average access delay (that is, the delay from Internet router to institution router) and the average Internet delay. For the average access delay, use α/(1- αλ), where α is the average time required to send an object over the access link and λ is the arrival rate of objects to the access link. a. Find the total average response time. b. Now suppose a cache is installed in the institutional LAN. Suppose the hit rate is 0.3. Find the total average response time.

Answer 1

a). The time taken to transmit an object size L=Object size / Access Link Capacity

= (240000 bits) / (15,000,000 bits/sec) =0.016 seconds

The traffic intensity on the link = (62 requests/sec) * (0.016 sec/request) = 0.992.

The average access delay = (0.016 sec) / (1 - 0.992) = 2 seconds

The total average response time = 2 sec + 1.5 sec = 3.5 seconds

b).  The average access delay = Time taken to transmit an object / 1-(hit rate * traffic intensity)

=(0.016 sec)/[1 – (0.3)*(0.992)] = 0.022 seconds

The response time is approximately zero if the request is satisfied by the cache.

Average response time for cache misses = average access delay+time taken by http protocol to response to router

=0.022 sec + 1.5 sec = 1.522 seconds

So the average response time =hit rate * response time taken by cache to hit+miss rate * response time taken by cache to miss

=(0.7)(0 sec) + (0.3)(1.522 sec) = 1.24 seconds.

Thus the average response time is reduced from 3.5 seconds to 1.24 seconds.