Add new subnetting exercise.

1/PAR/assignments/t5-subnetting-3.txt

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+Se necesitarán 8 redes: 8 = 2^3 => 3 + 24 = 27 bits de red => 5 bits de
+dispositivo. La máscara de red sería de 27 (ó 255.255.255.224). Esto permitiría
+unos 30 hosts (32 - 2). Las direcciones de cada red serían:
+
+Red:
+000 00000 = 0 => 197.142.68.0
+001 00000 = 32 => 197.142.68.32
+010 00000 = 64 => 197.142.68.64
+011 00000 = 96 => 197.142.68.96
+100 00000 = 128 => 197.142.68.128
+101 00000 = 160 => 197.142.68.160
+110 00000 = 192 => 197.142.68.192
+111 00000 = 224 => 197.142.68.224
+
+IPs válidas:
+197.142.68.1-30
+197.142.68.33-62
+197.142.68.65-94
+197.142.68.97-126
+197.142.68.129-158
+197.142.68.161-190
+197.142.68.193-222
+197.142.68.225-254
+
+Direcciones de difusión:
+197.142.68.31
+197.142.68.63
+197.142.68.95
+197.142.68.127
+197.142.68.159
+197.142.68.191
+197.142.68.223
+197.142.68.255