Contoh Soal Jaringan Komputer: Routing

-->

ROUTING

N1 = 50 (6 bit) N5 = 2 (2 bit)

N2 = 40 (6 bit) N6 = 2 (2 bit)

N3 = 60 (6 bit) N7 = 2 (2 bit)

N4 = 70 (7 bit)

Alokasi IP = 124.128.0.0/16

Jawaban :

Subnet Mask = 255.255.255.192

Alokasi IP

N1 = 124.128.0.1 - 124.128.0.126

N2 = 124.128.0.129 - 124.128.0.254

N3 = 124.128.1.1 - 124.128.1.126

N4 = 124.128.1.129 - 124.128.1.254

N5 = 124.128.2.1 - 124.128.2.126

N6 = 124.128.2.129 - 124.128.2.254

N7 = 124.128.3.1 - 124.128.3.126

R1

e0 124.128.2.1

e1 124.128.2.129

R2

e0 124.128.2.130

e1 124.128.3.1

R3

e0 124.128.2.2

e1 124.128.0.1

e2 124.128.0.129

R4

e0 124.128.3.2

e1 124.128.1.1

e2 124.128.1.129

Tabel Routing

R1

Destination Network	Gateway	Interface
124.128.2.0 /25	Direct	e0
124.128.2.128 /25	Direct	e1
124.128.0.0 /25	124.128.2.2	e2(R­3­­­­­)
124.128.0.128 /25	124.128.2.2	e2(R­3­­­­­)
124.128.3.0 /25	124.128.2.130	e0(R2)
124.128.1.0 /25	124.128.2.130	e0(R2)
124.128.1.128 /25	124.128.2.130	e0(R2)

R2

Destination Network	Gateway	Interface
124.128.2.128 /25	Direct	e1(R1)
124.128.3.0 /25	Direct	e0(R4)
124.128.1.0 /25	124.128.3.2	e0(R4)
124.128.1.128 /25	124.128.3.2	e0(R4)
124.128.2.0 /25	124.128.2.129	e1(R1)
124.128.0.0 /25	124.128.2.129	e1(R1)
124.128.0.128 /25	124.128.2.129	e1(R1)

R3

Destination Network	Gateway	Interface
124.128.2.0 /25	Direct	e0
124.128.0.0 /25	Direct	e1
124.128.0.128 /25	Direct	e2
124.128.2.128 /25	124.128.2.1	e0(R1)
124.128.3.0 /25	124.128.2.1	e0(R1)
124.128.1.0 /25	124.128.2.1	e0(R1)
124.128.1.128 /25	124.128.2.1	e0(R1)

R4

Destination Network	Gateway	Interface
124.128.3.0 /25	Direct	e0
124.128.1.0 /25	Direct	e1
124.128.1.128 /25	Direct	e2
124.128.2.128 /25	124.128.3.1	e0(R2)
124.128.2.0 /25	124.128.3.1	e0(R2)
124.128.0.0 /25	124.128.3.1	e0(R2)
124.128.0.128 /25	124.128.3.1	e0(R2)

Implementasi Dalam Paket Tracer

DOWNLOAD VERSI PDF | File Packet Tracer (pkt)