361-Midterm-solution

361-Midterm-solution - 1. (a) (b) COMP361 Midterm Exam...

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COMP361 Midterm Exam Suggested Solution 1. (a) Width of a bit = 1/(1*10 6 )*2.5*10 8 =250m. R * d prop =1*10 6 *10*10 3 /(2.5*10 8 )=40. (b) The maximal bits on the link = 10*10 3 /(2.5*10 8 )*10 6 =40. Time to transmit the packet = 40000/(1*10 6 ) + 10*10 3 /(2.5*10 8 ) = 40.04 ms 2. Upper bound: ( M=N-1) N + NM + NM 2 + … + NM (K-1) = N(1 + M + M 2 + … + M (K-1) ) = N(1-M K )/(1-M) = N[(N-1) K - 1]/(N-2) . Lower bound: 0 since there may be no neighbor nodes connected to Alice. 3. (a) UDP for DNS, TCP for HTTP. (b) The total amount of time to get the IP address is 12 n RRT RRT RRT + ++ " . Once the IP address is known, O RRT elapses to set up the TCP connection and another O RRT elapses to request and receive the small object. The total response time is 2 on RRT RRT RRT RRT +++ + " (c) Non-persistent HTTP: 1. At most one object is sent over a TCP connection; 2. Each TCP connection is closed after the server sends the object; Persistent HTTP: Multiple objects can be sent over single TCP connection between client and server. The total amount of time to get the IP address is n RRT RRT RRT + " .
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361-Midterm-solution - 1. (a) (b) COMP361 Midterm Exam...

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