Add new subnetting exercises.
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1/PAR/assignments/t5-ejercicio-subnetting-2.txt
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1/PAR/assignments/t5-ejercicio-subnetting-2.txt
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1)
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215.83.25.0 => 255.255.255.X
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6 ~= 2^3 => 3 bits de máscara subred => 8 - 3 = 5 bits de host
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Máscara global de red: 2^7 + 2^6 + 2^5 = 224 => 255.255.255.224
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Número de hosts: 2^5 - 2 = 30
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Dir. Subred | Dir. Bcast | Rango IPs
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---------------------------------|---------------|------------------
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000 00000 = 0 => 215.83.25.0 | 215.83.25.31 | 215.83.25.1-30
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001 00000 = 32 => 215.83.25.32 | 215.83.25.63 | 215.83.25.33-62
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010 00000 = 64 => 215.83.25.64 | 215.83.25.95 | 215.83.25.65-94
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011 00000 = 96 => 215.83.25.96 | 215.83.25.127 | 215.83.25.97-126
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100 00000 = 128 => 215.83.25.128 | 215.83.25.159 | 215.83.25.129-158
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101 00000 = 160 => 215.83.25.160 | 215.83.25.191 | 215.83.25.161-190
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2)
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116.0.0.0 => 255.X.X.X
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15 ~= 2^4 => 4 bits de máscara subred => 24 - 4 = 20 bits de host
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Máscara global de red: 255.240.0.0
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Número de hosts: 2^20 - 2 = 1 048 574
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Dir. Subred | Dir. Bcast | Rango IPs
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-------------------------------|-----------------|------------------------------
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0000 0000 = 0 => 116.0.0.0 | 116.15.255.255 | 116.0.0.1 - 116.15.255.254
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0001 0000 = 16 => 116.16.0.0 | 116.31.255.255 | 116.16.0.1 - 116.31.255.254
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0010 0000 = 32 => 116.32.0.0 | 116.47.255.255 | 116.32.0.1 - 116.47.255.254
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0011 0000 = 48 => 116.48.0.0 | 116.63.255.255 | 116.48.0.1 - 116.63.255.254
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0100 0000 = 64 => 116.64.0.0 | 116.79.255.255 | 116.64.0.1 - 116.79.255.254
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0101 0000 = 80 => 116.80.0.0 | 116.95.255.255 | 116.80.0.1 - 116.95.255.254
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0110 0000 = 96 => 116.96.0.0 | 116.111.255.255 | 116.96.0.1 - 116.111.255.254
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0111 0000 = 112 => 116.112.0.0 | 116.127.255.255 | 116.112.0.1 - 116.127.255.254
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1000 0000 = 128 => 116.128.0.0 | 116.143.255.255 | 116.128.0.1 - 116.143.255.254
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1001 0000 = 144 => 116.144.0.0 | 116.159.255.255 | 116.144.0.1 - 116.159.255.254
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1010 0000 = 160 => 116.160.0.0 | 116.175.255.255 | 116.160.0.1 - 116.175.255.254
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1011 0000 = 176 => 116.176.0.0 | 116.191.255.255 | 116.176.0.1 - 116.191.255.254
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1100 0000 = 192 => 116.192.0.0 | 116.207.255.255 | 116.192.0.1 - 116.207.255.254
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1101 0000 = 208 => 116.208.0.0 | 116.223.255.255 | 116.208.0.1 - 116.223.255.254
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1110 0000 = 224 => 116.224.0.0 | 116.239.255.255 | 116.224.0.1 - 116.239.255.254
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3)
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183.24.0.0 => 255.255.X.X
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10 ~= 2^4 => 4 bits de máscara subred => 16 - 4 = 12 bits de host
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Máscara global de red: 255.255.240.0
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Número de hosts: 2^12 - 2 = 4094
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Dir. Subred | Dir. Bcast | Rango IPs
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--------------------------------|-----------------|------------------------------
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0000 0000 = 0 => 183.24.0.0 | 183.24.15.255 | 183.24.0.1 - 183.24.15.254
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0001 0000 = 16 => 183.24.16.0 | 183.24.31.255 | 183.24.16.1 - 183.24.31.254
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0010 0000 = 32 => 183.24.32.0 | 183.24.47.255 | 183.24.32.1 - 183.24.47.254
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0011 0000 = 48 => 183.24.48.0 | 183.24.63.255 | 183.24.48.1 - 183.24.63.254
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0100 0000 = 64 => 183.24.64.0 | 183.24.79.255 | 183.24.64.1 - 183.24.79.254
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0101 0000 = 80 => 183.24.80.0 | 183.24.95.255 | 183.24.80.1 - 183.24.95.254
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0110 0000 = 96 => 183.24.96.0 | 183.24.111.255 | 183.24.96.1 - 183.24.111.254
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0111 0000 = 112 => 183.24.112.0 | 183.24.127.255 | 183.24.112.1 - 183.24.127.254
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1000 0000 = 128 => 183.24.128.0 | 183.24.143.255 | 183.24.128.1 - 183.24.143.254
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1001 0000 = 144 => 183.24.144.0 | 183.24.159.255 | 183.24.144.1 - 183.24.159.254
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34
1/PAR/assignments/t5-ejercicio-subnetting-3.txt
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1/PAR/assignments/t5-ejercicio-subnetting-3.txt
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1)
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a. 10 bits de máscara, 8 son del primer 113, lo cual 10 - 8 = 2 bits => 2^2 = 4 subredes.
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b. 2^(32-10) - 2 = 4 194 302 hosts
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c. 2^7 + 2^6 = 192 => 113.192.0.0
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d. 113.191.255.255
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e.
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Direcciones de subred 1: 113.0.0.1 - 113.63.255.254
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Dirección de subred 1: 113.0.0.0
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Dirección de broadcast de subred 1: 113.63.255.255
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f. Sí, pertenece a la subred 1.
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2)
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a. 16 primeros usados por 176.46, lo cual 20 - 16 = 4 bits => 2^4 = 16 subredes.
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b. 2^(32-20) - 2 = 4094 hosts
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c. 2^6 + 2^5 = 96 => 176.46.96.0
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d. No existe.
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e.
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Direcciones de subred 5: 176.46.80.1 - 176.46.95.254
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Dirección de subred 5: 176.46.80.0
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Dirección de broadcast de subred 5: 176.46.95.255
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f. Sí, pertenece a la subred 15
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3)
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a. 255.255.255.248
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b. 29-24 = 5 => 2^5 = 32 subredes
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c. 2^(32-29) - 2 = 8 hosts
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d. 2^3 * (21 - 1) = 235.171.39.160
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e. 2^3 * (23 - 1) - 1 = 235.171.39.176
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f.
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Direcciones de subred 23 = 235.171.39.177-182
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Dirección de subred 23 = 235.171.39.176
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Dirección de broadcast de subred 23 = 235.171.39.183
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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.
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h. 2^5 = 32 => sería mejor una máscara de (32 - 5) = 27 y sobrarían (30 - 17) = 13 direcciones.
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