Sistem Pengalamatan

 Sistem pengalamatan

Mencari network dan broadcast !

Soal:

a.       192.168.1.100/24

b.      172.71.17.17/29

c.       202.125.100.200/26

d.      10.10.10.10/30

e.       191.168.100.160/27

f.       220.200.20.30/28

g.      180.60.50.150/25

 Jawab:

a.       192.168.1.100/24

llllllll.llllllll.llllllll.00000000

Subnet mask    : 255.255.255.0

Network          : 192.168.1.0

Broadcast        : 192.168.1.255

b.      172.71.17.17/29

llllllll.llllllll.llllllll.lllll000

Subnet mask    : 255.255.255.248

Network          : 2^x = 2⁵ = 32

Host                :2^y  = 2³ – 2 = 8 – 2 = 6

Kelipatan         : 256 – 248 = 8

( 0,8,16,24,32,40,48,56,64,72,80 ……)

Netwok	172.71.17.16	172.71.17.32
Ip Awal	172.71.17.17	172.71.17.33
Ip akhir	172.71.17.30	172.71.17.38
Broadcast	172.71.17.31	172.71.17.39

c.       202.125.100.200/26

llllllll.llllllll.llllllll.ll000000

Subnet mask    : 255.255.255.192

Network          : 2^x  = 2² = 4

Host                : 2^y  - 2 = 2⁶ – 2 = 64 – 2 = 62

Kelipatan         : 256 – 192 = 64

(0,64,128,192,256)

Network	202.125.100.128	202.125.100.192
Ip awal	202.125.100.129	202.125.100.193
Ip akhir	202.125.100.190	202.125.100.254
broadcast	202.125.100.191	202.125.100.255

d.      10.10.10.10/30

llllllll.llllllll.llllllll.llllll00

Subnet mask    : 255.255.255.252

Network          : 2^x = 2⁶ = 64

Host                : 2^y – 2 = 2² – 2 = 4-2=2

Kelipatan         : 256 – 252 = 4

(0,4,8,12,16,20,24,28,32,36,40,…)

Network	10.10.10.8	10.10.10.12
Ip awal	10.10.10.9	10.10.10.13
Ip akhir	10.10.10.10	10.10.10.14
Broadcast	10.10.10.11	10.10.10.15

e.       191.168.100.160/27

llllllll.llllllll.llllllll.lll00000

subnet mask    : 255.255.255.224

Network          : 2^x = 2³ = 8

Host                :2^y-2= 2⁵ – 2 = 32 – 2 = 30

Kelipatan         : 256 – 224 = 32

(0,32,64,96,128,160,192,…)

Network	191.168.100.160	191.168.100.192
Ip awal	191.168.100.161	191.168.100.193
Ip akhir	191.168.100.190	191.168.100.222
Broadcast	191.168.100.191	191.168.100.223

f.       220.200.20.30/28

llllllll.llllllll.llllllll.llll0000

Subnet mask    : 255.255.255.240

Network          : 2^x = 2⁴ = 16

Host                : 2y – 2 = 2⁴ – 2 = 16 – 2 = 14

Kelipatan         : 256 – 255 = 16

(0,16,32,48,64,80,96, …)

Network	220.200.20.16	220.200.20.32
Ip awal	220.200.20.17	220.200.20.33
Ip akhir	220.200.20.30	220.200.20.46
Broadcast	220.200.20.31	220.200.20.47

g.      180.60.50.150/25

llllllll.llllllll.llllllll.l0000000

subnet mask    : 255.255.255.128

Network          :2^x = 2¹ = 2

Host                :2^y – 2 = 2⁷ – 2 = 128 – 2 = 126

Kelipatan         : 256 – 128 = 128

(0,128,256)

Network	180.60.50.0	180.60.50.128
Ip awal	180.60.50.1	180.60.50.129
Ip akhir	180.60.50.126	180.60.50.254
Broadcast	180.60.50.127	180.60.50.255