Network Assignment # 2

Question 1: The well-known port numbers for the simple file transfer protocol (SFTP) are as follows:- SFTP 115/tcp SFTP 115/udp For the network news transfer protocol (NNTP) are as follows:- NNTP 119/tcp NNTP 119/udp Question 2: Application layer protocols: DNS and HTTP. Transport layer protocols: UDP for DNS; TCP for HTTP. Question 3: The total amount of time to get the IP address is RTT1+RTT2+...+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+...+RTTn. Question 4: (a) As there are 4 HTTP calls. Each one needs to be set up and then retrieve an object so each requires 2RTT0. (b) It takes 2RTT0 to retrieve the html file. The HTTP connection then closes and 3 new connections are then set up, with all three of the objects being retrieved in parallel. These new connections require 2RTT0 as well. (c) After retrieving the original html file the connection is still kept open and it only takes one RTT to retrieve all three of the objects (since they were all requested at the same time). Question 5: 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: so ∆= (900,000 bits) / (1,500,000 bits/sec) = .6 sec The traffic intensity on the link is (1.5 requests/sec)(.6 msec / request) = .9. so, the average access delay is (.6 sec)/(1 - .9) = 6 seconds. The total average response time is therefore 6 sec + 2 sec = 8 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. As the average access delay is (.6 sec)/[1 – (.6)(.9)] = 1.2 seconds. The response time is approximately zero if the request is satisfied by the cache (which happens with probability .4); the average response time is 1.2 sec + 2 sec = 3.2 sec for cache misses (which happens 60% of the time). So the average response time is (.4)(0 sec) + (.6)(3.2 sec) = 1.92 seconds. So the average response time is reduced from 8 sec to 1.92 sec. Question 6: Alice sends her query to at most N neighbors. Each of these neighbors forwards the query to at most M = N-1 neighbors. Each of those neighbors forwards the query to at most M neighbors. So the maximum number of query messages is N + NM + NM2 + … + NM(K-1) = N(1 + M + M2 + … + M(K-1) ) = N(1-MK)/(1-M) = N[(N-1)K- 1]/(N-2) .