4-03
4-04
4-06 Calculate the data in Table 4-2 according to the IP address requirements
A：
B
C:
4-07
4-09
(2) Connect 6 hosts. 255.255.255 is 3 8 1,248 converted into binary, (3
) 2 to the 12th power minus 2
(4) Not a valid subnet mask
(5) 12 16+2 16^ 0=192+2=194
2 16+15=47
14 16+4=20
8*16+1=129
194.48.20.129
(6) Yes. Each type of network has practical significance. Although the divided subnets are very Less, but still meaningful.
4-10
Each ip address only looks at the first paragraph The
first paragraph
0-127: Class A.
128-191: Class B
129-223: Class C
4-15
6 times. Passed through several routers +1.
4-18
Counting from front to back is the same.
212.56.100000100.0
212.56.100000101.0
212.56.100000110.0
212.56.100000111.0
212.56.132/22
4-19
The former address contains the latter address.
4-20
Choose A
4-21
2 means 00000010
0 means
the first 4 digits of 00000000 are the same
Select (1)
4-22
The first paragraph 152 is the same as
7 is 00000111
and 31 is 00011111
, so choose the fourth
4-23
(1)192-128 =
(2)240
(3)