Question
Consider Figure 2.12, 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 servers is 16 requests 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 (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 Δ/(1 – Δ
), where Δ is the average time required to send an object over the access link and is the arrival rate of objects to the access link.
a. Find the total average response time.
b. Now suppose a cache installed in the institutional LAN, Suppose the miss rate is 0.4. Find the total response time.
Answer
- The time to transmit an object of size L over a link or rate R is L/R. The average time is the average size of the object divided by R:
= (850,000 bits)/(15,000,000 bits/sec) = .0567 sec
The traffic intensity on the link is given by 
= (16 requests/sec)(.0567 sec/request) = 0.907. Thus, 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.
- The traffic intensity on the access link is reduced by 60% since the 60% of the requests are satisfied within the institutional network. Thus the average access delay is (.0567 sec)/[1 – (.4)(.907)] = .089 seconds. The response time is approximately zero if the request is satisfied by the cache (which happens with probability .6); the average response time is .089 sec + 3 sec = 3.089 sec for cache misses (which happens 40% of the time). So the average response time is (.6)(0 sec) + (.4)(3.089 sec) = 1.24 seconds. Thus the average response time is reduced from 3.6 sec to 1.24 sec.