Assignment No. 3
Semester: Spring 2015
Computer Network- CS610 / Total Marks: 20
Assignment
Solution:
Binary numbers in the decimal dotted notations
(A) 10000010 10101000 00001010 00101000
10000010 = 1*27+0*26+0*25+0*24+0*23+0*22+1*21+0*20 =128+2 = 130
10101000 = 1*27+0*26+1*25+0*24+1*23+0*22+0*21+0*20 =128+32+8 = 168
00001010 = 0*27+0*26+0*25+0*24+1*23+0*22+1*21+0*20 = 8+2 =10
00101000 = 0*27+0*26+1*25+0*24+1*23+0*22+0*21+0*20 =32+8 = 40
130.168.10.40
(B) 00001010 01111000 00010100 00000000
00001010 = 0*27+0*26+0*25+0*24+1*23+0*22+1*21+0*20 = 8+2 =10
01111000 = 0*27+1*26+1*25+1*24+1*23+0*22+0*21+0*20 = 64+32+16+8 = 120
00010100 = 0*27+0*26+0*25+1*24+0*23+1*22+0*21+0*20 = 16+4 = 20
00000000 = 0*27+0*26+0*25+0*24+0*23+0*22+0*21+0*20 = 0
10.120.20.0
2) Classes of both ip addresses are
Ip class
130.168.10.40 B
10.120.20.0 A
3) prefix and suffix parts of both ip addresses are
Ip prefix suffix
130.168.10.40 130.168 10.40
10.120.20.0 10 120.20.0
4) addition of last two digit of V.U ID (25) in the first octet is
Before addition after addition
130.168.10.40 155.168.10.40
10.120.20.0 35.120.20.0
Ip address class
155.168.10.40 B
35.120.20.0 A