prelab4_solution

# prelab4_solution - (128 hosts, last 8 digits: bbbbbbb)...

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Now, we have 223.1.17.0/24 in hand, which means that the last 8 digits can be utilized while the beginning 24 bits cannot be modified. One solution of this division could be: Subnet 1: 223.1.17.0 ~ 223.1.17.127
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Unformatted text preview: (128 hosts, last 8 digits: bbbbbbb) Subnet 2: 223.1.17.128~223.1.17.191 (64 hosts, last 8 digits: 10 bbbbbb) Subnet 3: 223.1.17.192~223.1.17.255 (64 hosts, last 8 digits: 11 bbbbbb) To write it in CIDR form, we present the first avaliable address of a subnet (also known as NetID), and indicate after slash the # of beginning bits that are fixed. This goes to: Subnet 1: 223.1.17.0/25 Subnet 2: 223.1.17.128/26 Subnet 3: 223.1.17.192/26 Problem 16 The maximum size of data field in each fragment = 480 (20 bytes IP header). Thus the number of required fragments 7 480 20 3000 = = Each fragment will have Identification number 422. Each fragment except the last one will be of size 500 bytes (including IP header). The last datagram will be of size 120 bytes (including IP header). The offsets of the 7 fragments will be 0, 60, 120, 180, 240, 300, 360. Each of the first 6 fragments will have flag=1; the last fragment will have flag=0....
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## This note was uploaded on 11/09/2010 for the course ECE 441 taught by Professor Davidstarobinski during the Fall '09 term at BU.

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prelab4_solution - (128 hosts, last 8 digits: bbbbbbb)...

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