direccionamiento ip y subredes ejercicios resueltos 1194346207489436 2.pdf
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Muestre aquí su forma de proceder para el Problema 6:
Number of
Subnets
16
256 128 64 32
-
192 . 70 . 10 .
2
4
8 16
32
128 64 32 16
0
0
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
(Invalid range) 0
1
1
1
1
1
1
1
(Invalid range) 1
1
1
1
1
0
0
0
0
1
1
1
1
8
0
8
4
2
-
64 128 256
4
0
2
0
1
0
- Binary values
192.70.10.0
192.70.10.16
192.70.10.32
192.70.10.48
192.70.10.64
192.70.10.80
192.70.10.96
192.70.10.112
192.70.10.128
192.70.10.144
192.70.10.160
192.70.10.176
192.70.10.192
192.70.10.208
192.70.10.224
192.70.10.240
128
+64
240
Number of
Hosts
16
-2
14
16
-2
14
to
to
to
to
to
to
to
to
to
to
to
to
to
to
to
to
192.70.10.15
192.70.10.31
192.70.10.47
192.70.10.63
192.70.10.79
192.70.10.95
192.70.10.111
192.70.10.127
192.70.10.143
192.70.10.159
192.70.10.175
192.70.10.191
192.70.10.0207
192.70.10.223
192.70.10.239
192.70.10.255
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