HW2 Solutions - Problem 7 The total amount of time to get...

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Problem 7 The total amount of time to get the IP address is n RTT RTT RTT 2 1 . Once the IP address is known, O RTT elapses to set up the TCP connection and another O RTT elapses to request and receive the small object. The total response time is n o RTT RTT RTT RTT 2 1 2 Problem 8 a) o o n RTT RTT RTT RTT 2 3 2 1 n o RTT RTT RTT 1 8 . b) o o n RTT RTT RTT RTT 2 2 1 n o RTT RTT RTT 1 4 . c) o o n RTT RTT RTT RTT 2 1 n o RTT RTT RTT 1 3 . 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 : = (900,000 bits)/(15,000,000 bits/sec) = .06 sec The traffic intensity on the link is (15 requests/sec)(.06 msec/request) = .9. Thus, the average access delay is (.06 sec)/(1 - .9) = .6 seconds. The total average response time is therefore .6 sec + 2 sec = 2.6 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 is (.06 sec)/[1 – (.6)(.9)] = .12 seconds. The response time is approximately zero if the request is satisfied by
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This note was uploaded on 06/15/2009 for the course ECE 3076 taught by Professor Copeland during the Spring '08 term at Georgia Institute of Technology.

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HW2 Solutions - Problem 7 The total amount of time to get...

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