hw2b_answers_F11

hw2b_answers_F11 - CS 176A - Introduction to Computer...

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CS 176A -- Introduction to Computer Communication Networks Homework #2B--November 18, 2011 (due at 11:59pm) (1)How many valid host addresses are in the network prefix 128.0/14? 2^(32-14) -2 = 262142 (2) How many valid host addresses are in the network prefix 128.108.0/20? 2^(32-20) -2 = 4,094 (3) Is 130.207.52/22 a valid network prefix? Why or why not? 1000 0010 . 1100 1111 . 0011 0100 . 0000 0000 1111 1111 . 1111 1111 . 1111 1100 . 0000 0000 It is valid because all of the network bits are masked. (4) What range of addresses is covered by 130.216.35.16/28? 1000 0010.1101 1000.0010 0011.0001 0000 1000 0010.1101 1000.0010 0011.0001 1111 130.216.35.16 - 130.216.35.31 (5) What range of addresses is covered by 38.128/9? 0010 0110 . 1000 0000 . 0000 0000 . 0000 0000 1111 1111 . 1000 0000 . 0000 0000 . 0000 0000 38.128.0.0 - 38.255.255.255 (6) If 175.111/16 where split into 8 equal network prefixes, how many host addresses would be in each of the 8 new networks? The mask length of each prefix would be 19: 2^(32-19) – 2 = 8,190 hosts. (Not required by the answer, but the four prefixes would be: 175.111.0/19, 175.111.32/19, 175.111.64/19, and 175.111.96/19.) (7) If 175.111/16 where split into 8 equal network prefixes, what would be the network mask length of the resulting 8 equal network prefixes? What would this network mask be using dotted quad notation? Network mask length is 19 bits.
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hw2b_answers_F11 - CS 176A - Introduction to Computer...

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