05

TD 5

Exerice 1

1

1)

La machine est de classe C

2)

255.255.255.0

3)

193.55.28.0

4)

193.55.28.255

5)

\(2^{8} - 2 = 254\)

2

1)

Classe C

2)

\(32\ /\ 256 = 0.125\)

3)

Classe B \(32\ /\ 65536 = 4.8828125e-4\)

Exercice 2

1

1)

Classe B

2)

On a 16 bits pour les machines, on en retire 5 (30 machines \(< 32 = 2^{5}\)) donc il reste \(2^{11} - 2\) machines disponibles

3)

Masque /21

4)

On regarde les bits 17 à 21: 75 = 0b01001011 et 0b01001 = 9

Exercice 3

1

1)

\(66 + 1 = 67\) bien contingues on crop: /23 et \(2^{9}\) adresses IP

2)

Total: \(2^{9} - 2\)
Séparés: \(\left( 2^{8} - 2 \right) \times 2\)

3)

Diffusion = broadcast 195.162.67.255

Exercice 4

1

1)

A: 000, B: 001, C: 010, D: 011, E: 100

2)

\(\mathcal{U}_{A} = \frac{16}{32},\mathcal{U}_{B} = \frac{30}{32},\mathcal{U}_{C} = \frac{4}{32},\mathcal{U}_{D} = \frac{9}{32},\mathcal{U}_{D} = \frac{30}{32},\)

3)

B: 000/27, E: 001/27, A: 0100/28, D: 0101/28, C: 011100/30
\({\hat{\mathcal{U}}}_{A} = \frac{16}{16},{\hat{\mathcal{U}}}_{B} = \frac{30}{32},{\hat{\mathcal{U}}}_{C} = \frac{4}{4},{\hat{\mathcal{U}}}_{D} = \frac{9}{16},{\hat{\mathcal{U}}}_{D} = \frac{30}{32},\)

2

1)

Après la séparation on a \(2^{5} - 2 = 30\) adresses disponibles par réseau

2)

A: 0/25, B: 10/26, C: 11/26

3)

A: 126, B: 62, C: 62

Exercice 5

1

1)

11111111.11111111.11111000.00000000

2)

/21

3)

10000000.11010011.10101111.11111111
ie 128.211.175.255

4)

addresses utilisables: de 10000000.11010011.10101000.00000001 à 10000000.11010011.10101111.11111110

5)

A: 00/23, B: 01/23, C: 1000/25, D: 1001/25, E: 01010/26 Première: A: 128.211.168.1 B: 128.211.170.1 C: 128.211.172.1 D: 128.211.172.129 E: 128.211.171.1

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