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Add new subnetting exercises.

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Nicolás A. Ortega Froysa
2022-01-14 18:53:27 +01:00
parent 21cce726d1
commit 2e50db192b
2 changed files with 96 additions and 0 deletions
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1/PAR/assignments/t5-ejercicio-subnetting-2.txt Normal file
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1)
215.83.25.0 => 255.255.255.X
6 ~= 2^3 => 3 bits de máscara subred => 8 - 3 = 5 bits de host
Máscara global de red: 2^7 + 2^6 + 2^5 = 224 => 255.255.255.224
Número de hosts: 2^5 - 2 = 30
Dir. Subred | Dir. Bcast | Rango IPs
---------------------------------|---------------|------------------
000 00000 = 0 => 215.83.25.0 | 215.83.25.31 | 215.83.25.1-30
001 00000 = 32 => 215.83.25.32 | 215.83.25.63 | 215.83.25.33-62
010 00000 = 64 => 215.83.25.64 | 215.83.25.95 | 215.83.25.65-94
011 00000 = 96 => 215.83.25.96 | 215.83.25.127 | 215.83.25.97-126
100 00000 = 128 => 215.83.25.128 | 215.83.25.159 | 215.83.25.129-158
101 00000 = 160 => 215.83.25.160 | 215.83.25.191 | 215.83.25.161-190
2)
116.0.0.0 => 255.X.X.X
15 ~= 2^4 => 4 bits de máscara subred => 24 - 4 = 20 bits de host
Máscara global de red: 255.240.0.0
Número de hosts: 2^20 - 2 = 1 048 574
Dir. Subred | Dir. Bcast | Rango IPs
-------------------------------|-----------------|------------------------------
0000 0000 = 0 => 116.0.0.0 | 116.15.255.255 | 116.0.0.1 - 116.15.255.254
0001 0000 = 16 => 116.16.0.0 | 116.31.255.255 | 116.16.0.1 - 116.31.255.254
0010 0000 = 32 => 116.32.0.0 | 116.47.255.255 | 116.32.0.1 - 116.47.255.254
0011 0000 = 48 => 116.48.0.0 | 116.63.255.255 | 116.48.0.1 - 116.63.255.254
0100 0000 = 64 => 116.64.0.0 | 116.79.255.255 | 116.64.0.1 - 116.79.255.254
0101 0000 = 80 => 116.80.0.0 | 116.95.255.255 | 116.80.0.1 - 116.95.255.254
0110 0000 = 96 => 116.96.0.0 | 116.111.255.255 | 116.96.0.1 - 116.111.255.254
0111 0000 = 112 => 116.112.0.0 | 116.127.255.255 | 116.112.0.1 - 116.127.255.254
1000 0000 = 128 => 116.128.0.0 | 116.143.255.255 | 116.128.0.1 - 116.143.255.254
1001 0000 = 144 => 116.144.0.0 | 116.159.255.255 | 116.144.0.1 - 116.159.255.254
1010 0000 = 160 => 116.160.0.0 | 116.175.255.255 | 116.160.0.1 - 116.175.255.254
1011 0000 = 176 => 116.176.0.0 | 116.191.255.255 | 116.176.0.1 - 116.191.255.254
1100 0000 = 192 => 116.192.0.0 | 116.207.255.255 | 116.192.0.1 - 116.207.255.254
1101 0000 = 208 => 116.208.0.0 | 116.223.255.255 | 116.208.0.1 - 116.223.255.254
1110 0000 = 224 => 116.224.0.0 | 116.239.255.255 | 116.224.0.1 - 116.239.255.254
3)
183.24.0.0 => 255.255.X.X
10 ~= 2^4 => 4 bits de máscara subred => 16 - 4 = 12 bits de host
Máscara global de red: 255.255.240.0
Número de hosts: 2^12 - 2 = 4094
Dir. Subred | Dir. Bcast | Rango IPs
--------------------------------|-----------------|------------------------------
0000 0000 = 0 => 183.24.0.0 | 183.24.15.255 | 183.24.0.1 - 183.24.15.254
0001 0000 = 16 => 183.24.16.0 | 183.24.31.255 | 183.24.16.1 - 183.24.31.254
0010 0000 = 32 => 183.24.32.0 | 183.24.47.255 | 183.24.32.1 - 183.24.47.254
0011 0000 = 48 => 183.24.48.0 | 183.24.63.255 | 183.24.48.1 - 183.24.63.254
0100 0000 = 64 => 183.24.64.0 | 183.24.79.255 | 183.24.64.1 - 183.24.79.254
0101 0000 = 80 => 183.24.80.0 | 183.24.95.255 | 183.24.80.1 - 183.24.95.254
0110 0000 = 96 => 183.24.96.0 | 183.24.111.255 | 183.24.96.1 - 183.24.111.254
0111 0000 = 112 => 183.24.112.0 | 183.24.127.255 | 183.24.112.1 - 183.24.127.254
1000 0000 = 128 => 183.24.128.0 | 183.24.143.255 | 183.24.128.1 - 183.24.143.254
1001 0000 = 144 => 183.24.144.0 | 183.24.159.255 | 183.24.144.1 - 183.24.159.254

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1)
a. 10 bits de máscara, 8 son del primer 113, lo cual 10 - 8 = 2 bits => 2^2 = 4 subredes.
b. 2^(32-10) - 2 = 4 194 302 hosts
c. 2^7 + 2^6 = 192 => 113.192.0.0
d. 113.191.255.255
e.
Direcciones de subred 1: 113.0.0.1 - 113.63.255.254
Dirección de subred 1: 113.0.0.0
Dirección de broadcast de subred 1: 113.63.255.255
f. Sí, pertenece a la subred 1.
2)
a. 16 primeros usados por 176.46, lo cual 20 - 16 = 4 bits => 2^4 = 16 subredes.
b. 2^(32-20) - 2 = 4094 hosts
c. 2^6 + 2^5 = 96 => 176.46.96.0
d. No existe.
e.
Direcciones de subred 5: 176.46.80.1 - 176.46.95.254
Dirección de subred 5: 176.46.80.0
Dirección de broadcast de subred 5: 176.46.95.255
f. Sí, pertenece a la subred 15
3)
a. 255.255.255.248
b. 29-24 = 5 => 2^5 = 32 subredes
c. 2^(32-29) - 2 = 8 hosts
d. 2^3 * (21 - 1) = 235.171.39.160
e. 2^3 * (23 - 1) - 1 = 235.171.39.176
f.
Direcciones de subred 23 = 235.171.39.177-182
Dirección de subred 23 = 235.171.39.176
Dirección de broadcast de subred 23 = 235.171.39.183
g. No lo son. 235.171.39.248 es la dirección de red de la subred 32, y la 235.171.39.247 es el broadcast de la subred 31.
h. 2^5 = 32 => sería mejor una máscara de (32 - 5) = 27 y sobrarían (30 - 17) = 13 direcciones.