TUGAS
JARINGAN DASAR
(JAWABAN)
1)
198.53.67.0/30 =
11111111.11111111.11111111.11111100
= 255.255.255.252
Jumlah subnet = 26
= 64 subnet
Jumlah host per subnet = 22-2= 2 host
Blok subnet =
256-252= 4,subnet lengkapnya adalah 0,4,8,12
Host dan broadcast yg valid?
|
Subnet
|
198.53.67.0
|
198.53.67.4
|
198.53.67.8
|
198.53.67.12
|
|
Host pertama
|
198.53.67.1
|
198.53.67.5
|
198.53.67.9
|
198.53.67.13
|
|
Host terakhir
|
198.53.67.2
|
198.53.67.66
|
198.53.67.130
|
198.53.67.194
|
|
Broadcast
|
198.53.67.3
|
198.53.67.67
|
198.53.67.131
|
198.53.67.195
|
a)
Subnet
mask=255.255.255.252
b)
Alamat subnet=(
198.53.67.0) (198.53.67.4) (198.53.67.8) (198.53.67.12)dst (198.53.67.252)
c) Alamat Broadcast=(198.53.67.3) (198.53.67.67)
(198.53.67.131) (198.53.67.195) dst (198.53.67.252)
d) Jumlah host yg dapat di gunakan =
Jumlah host x Jumlah subnet
= 2 x 64 = 128
e) Serta alamat subnet yg ke-3 =
198.53.67.8
2) 202.151.37.0/26 =
11111111.11111111.11111111.11000000
= 255.255.255.192
Jumlah
subnet =22=4 subnet
Jumlah
host per subnet = 26-2=62 host
Blok subnet=
256-192= 64 subnet lengkapnya adalah 0,64,128,192
Host dan
broadcast yg valid?
|
subnet
|
202.151.37.0
|
202.151.37.64
|
202.151.37.128
|
202.151.37.192
|
|
Host pertama
|
202.151.37.1
|
202.151.37.65
|
202.151.37.129
|
202.151.37.193
|
|
Host terakhir
|
202.151.37.62
|
202.151.37.126
|
202.151.37.190
|
202.151.37.254
|
|
Boadcast
|
202.151.37.63
|
202.151.37.127
|
202.151.37.191
|
202.151.37.255
|
a) Subnet mask = 255.255.255.192
b) Alamat subnet = (202.151.37.0) (202.151.37.64)
(202.151.37.128) (202.151.37.192)
c) Alamat Broadcast=(202.151.37.63) (202.151.37.127)
(202.151.37.191) (202.151.37.255)
d) Jumlah host yg di gunakan = 62 x 4
=248
e) Serta alamat subnet ke-3= 202.151.37.128
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