Solution
Problem 7
The total amount of time to get the IP address(traverse through n DNS servers to resolve the IP) is RTT1 + RTT2 +Λ + RTTn
Once the IP address is known, RTT0 elapses to set up the TCP connection and another RTT0 elapses to request and receive the small object. The total response time is 2RTT0 + RTT1 + RTT2 +Λ + RTTn
Problem 8
(a) Non-persistent HTTP with no parallel TCP connections, for 3 objects
(b) Non-persistent HTTP with parallel connections
(c) Persistent HTTP
Problem 9
a) 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:
The traffic intensity on the link is (10 requests/sec). Thus, the average access delay is (0.06 sec)/
(1 – 0.06)(10) = 0.15 seconds. The total average response time is therefore 0.15 sec + 2 sec = 2.15 sec.
b) The traffic intensity on the access link is reduced by 40% since the 40% of the requests are satisfied within the institutional network. Thus the average access delay,
Thus the average response time is reduced from 2.15 sec to 0.8316 sec
