Question

Consider the figure above where users in an institutional network need to download

a web page from a remote web server, with an average rate of 5 requests per second.
The rate of the LAN (or throughput between each two end systems) is 10Mbps,
while the rate of the access link is 1.5Mbps. The web page is an HTML file of
negligible size, with reference to 3 objects of 50 Kbits each, each one fitting in one
single packet. HTTP requests and TCP connection establishment messages have
negligible sizes.

a) [10 pts] Compute the average LAN utilization and the access link utilization.

c) [10 pts] Now consider that a web cache (proxy) is added to the institutional
network, and for 40% of the requests it can send the local copy of the web page
instead of requesting it to the remote web server. Re-compute the average LAN
utilization and the access link utilization for this new case.

d) [10 pts] If the RTT between two end systems in the institutional network is
RTTlan = 50 msec, compute the savings in the average time to retrieve the whole
web page between the case in a) and the case in c) in case of non-persistent, non-
parallel, non-pipelined HTTP (Assume the bottleneck link between users and
remote web server is the access link).

e) [10 pts] Re-compute the savings as in d) but in case of a persistent, non-parallel,
pipelined HTTP.

Answer 1

a) The average LAN utilization is 7.5% and the access link utilization is 5%. b) With the addition of a web cache, the access link utilization becomes 3%. c) The savings in average time to retrieve the whole web page between cases a) and c) is 50 msec. d) No calculation for parallel, persistent, and pipelined HTTP.

Step-by-step explanation:

a) To calculate the average LAN utilization, we need to find the rate at which packets are being sent over the LAN compared to its capacity. Since each object is 50 Kbits and can fit in a single packet, the size of each packet is 50 Kbits. The rate of the LAN is 10Mbps, which is equivalent to 10,000,000 bits per second. Therefore, the average LAN utilization is:

Average LAN utilization = (Rate at which packets are being sent over the LAN) / (Capacity of the LAN)

= (5 requests per second) * (3 objects) * (50 Kbits per object) / (10,000,000 bits per second)

= 0.075 or 7.5%

To calculate the access link utilization, we need to find the rate at which packets are being sent over the access link compared to its capacity. The rate of the access link is 1.5Mbps, which is equivalent to 1,500,000 bits per second. Therefore, the access link utilization is:

Access link utilization = (Rate at which packets are being sent over the access link) / (Capacity of the access link)

= (5 requests per second) * (3 objects) * (50 Kbits per object) / (1,500,000 bits per second)

= 0.05 or 5%

b) With the addition of a web cache that can send the local copy of the web page for 40% of the requests, the average LAN utilization remains the same as before, as the LAN is still being utilized for all requests. However, the access link utilization decreases, as 40% of the requests no longer need to go through the access link. Therefore, the new access link utilization is:

New access link utilization = (Rate at which packets are being sent over the access link) / (Capacity of the access link)

= (5 requests per second) * (3 objects) * (50 Kbits per object) * (0.6) / (1,500,000 bits per second)

= 0.03 or 3%

c) To calculate the savings in the average time to retrieve the whole web page between case a) and case c) for non-persistent, non-parallel, non-pipelined HTTP, we need to consider the number of round trips between the users and the web server. In case a), each object requires a round trip, so the total number of round trips is 3. In case c), with the web cache sending the local copy for 40% of the requests, 2 objects require a round trip and 1 object can be retrieved from the cache with no round trip required. Therefore, the average time to retrieve the whole web page in case a) is:

Average time in case a) = (Number of objects) * (RTT between two end systems)

= 3 * 50 msec

And the average time to retrieve the whole web page in case c) is:

Average time in case c) = (Number of round trips for objects requiring round trips) * (RTT between two end systems)

= 2 * 50 msec

The savings in the average time to retrieve the whole web page between case a) and case c) is the difference between the average time in case a) and case c):

Savings = (Average time in case a) - (Average time in case c))

= (3 * 50 msec) - (2 * 50 msec)

= 50 msec

d) To calculate the savings in the average time to retrieve the whole web page for parallel, persistent, and pipelined HTTP, we need to consider the number of parallel connections and the number of objects requested in parallel. In the case of parallel, persistent, and pipelined HTTP, multiple requests can be sent over a single TCP connection and multiple responses can be received in parallel. However, in this scenario, we do not have enough information to calculate this value.