Question

[Analytical Question] Caching. We have an enterprise network connected to the Internet. Assume the average web request rate from this network’s browsers is 22 requests per second, and average object size transferred is 1,000,000 bits. Also assume the capacity of the institutional link providing access of this enterprise to the internet is 20Mbps. Model and estimate the average response time for this enterprise with and without a proxy server assuming a hit rate of 60% for the proxy server.

Group of answer choices

1.

La/R = 22 x 200,000 / 10,000,000 < 1 --> Access Delay < Secs

Internet Delay (can be considered constant in order of mSec) + Access Delay (over the order of seconds and increasing) + LAN Delay (negligible)

60% hit rate

0.6 x 22,000,000 = 7,200,000 --> 7,200,000 / 20,000,000 = 0.66 -> La/R << 1 --> Access Delay < Secs

2.

La/R = 22 x 1,000,000 / 20,000,000 > 1 --> Increasing delay

Internet Delay (can be considered constant in order of mSec) + Access Delay (over the order of seconds and increasing) + LAN Delay (negligible)

40% of requests still should be sent to original servers

0.4 x 22,000,000 = 8,800,000 --> 8,800,000 / 20,000,000 = 0.44 -> La/R << 1 --> Access Delay < Secs

3.

La/R = 22 x 1,000,000 / 20,000,000 > 1 --> Increasing delay

Internet Delay (can be considered constant in order of mSec) + Access Delay (over the order of seconds and increasing) + LAN Delay (negligible)

hit rate = 60%

0.6x 22,000,000 = 7,200,000 --> 7,200,000 / 20,000,000 = 0.36 -> La/R << 1 --> Access Delay < Secs

4.

L/R = 1,000,000 / 10,000,000 << 1 --> Access Delay < Secs

Internet Delay (can be considered constant in order of mSec) + Access Delay (over the order of seconds and increasing) + LAN Delay (negligible)

40% of requests still should be sent to original servers

0.4 x 22,000,000 = 880,000 --> 880,000 / 20,000,000 = 0.044 -> La/L << 1 --> Access Delay < Secs

Answer 1

Option 2 is the right configuration:

That is the following option is right:

2.

La/R = 22 x 1,000,000 / 20,000,000 > 1 --> Increasing delay

Internet Delay (can be considered constant in order of mSec) + Access Delay (over the order of seconds and increasing) + LAN Delay (negligible)

40% of requests still should be sent to original servers

0.4 x 22,000,000 = 8,800,000 --> 8,800,000 / 20,000,000 = 0.44 -> La/R << 1 --> Access Delay < Secs

Explanation:

The Traffic Intensity (queuing delay) is La/R, where

R=link bandwidth (bps)

L=packet length (bits) that is 22 packets x 1,000,000 (1 packet size)

a=average packet arrival rate

La/R = 22 x 1,000,000 / 20,000,000 > 1 more data arriving than can be serviced

Further , since the hit ration 60%, 40% of the requests must be satisfied from the web server.

Thus access delay = 0.4 x 22,000,000 = 8,800,000 --> 8,800,000 / 20,000,000 = 0.44