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Start Datkom 5
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tblut committed Jun 8, 2015
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5.1)
a)
Es bleiben 32 - 28 = 4 Bits übrig für den Host-Anteil.
Also ist die Host-Range: 137.226.12.177 bis 137.226.12.190.
Damit liegt (1) 137.226.12.190 im Netz und (2) 137.226.12.193 nicht.

b)
Netz 1 (61 Hosts):
Netz-Adresse: 135.13.32.0/26 = 135.13.[00100000].[00|000000]
Broadcast-Adresse: 135.13.32.63
Host-Range: 135.13.32.1 - 135.13.32.62

Netz 2 (292 Hosts):
Netz-Adresse: 135.13.34.0/23 = 135.13.[0010001|0].[00000000]
Broadcast-Adresse: 135.13.35.255
Host-Range: 135.13.34.1 - 135.13.35.254

Netz 3 (170 Hosts):
Netz-Adresse: 135.13.33.0/24 = 135.13.[00100001].[|00000000]
Broadcast-Adresse: 135.13.33.255
Host-Range: 135.13.33.1 - 135.13.33.254

Netz 4 (120 Hosts):
Netz-Adresse: 135.13.32.128/25 = 135.13.[00100000].[1|0000000]
Broadcast-Adresse: 135.13.32.255
Host-Range: 135.13.32.129 - 135.13.32.254

Netz 5:
Netz-Adresse: 135.13.32.64/26 = 135.13.[00100000].[01|000000]
Broadcast-Adresse: 135.13.32.127
Host-Range: 135.13.32.65 = 135.13.32.126
Also 62 Rechner

c)
135.46.64.17: 4
135.46.61.56: 4
135.46.56.17: 1
135.46.55.39: 5

d)
137.226.96.0/19
Da selbe Anzahl an Hosts (genau genommen 6 mehr).

e)
Nein, da pro Subnetz nur maximal 1022 Rechner adressiert werden können, was je nach Anwendung zu wenig ist.

5.2)
Prot. | IP-Adresse lokal | Port lokal | IP-Adresse global | Port global | IP-Adresse Ziel | Ziel
-------------------------------------------------------------------------------------------------
TCP | 10.1.0.37 | 6397 | 134.130.17.12 | 6398 | 137.226.75.113 | 80
TCP | 10.1.0.5 | 8539 | 134.130.17.12 | 8540 | 137.226.55.29 | 80
TCP | 10.1.0.16 | 4939 | 134.130.17.12 | 4939 | 137.226.6.2 | 80

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