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diberikan Class C network 204.24.93.0/24, ingin di subnet dengan kebutuhan berdasarkan jumlah host: neta=14 hosts, netb=28 hosts, netc=2 hosts, netd=7 hosts, nete=28 hosts. Secara keseluruhan terlihat untuk melakukan hal tersebut di butuhkan 5 bit host(2^5-2=30 hosts) dan 27 bit net, sehingga:
Neta (14 hosts): 204.24.93.0/27 => ada 30 hosts; tidak terpakai 16 hosts
netb (28 hosts): 204.24.93.32/27 => ada 30 hosts; tidak terpakai 2 hosts
netc ( 2 hosts): 204.24.93.64/27 => ada 30 hosts; tidak terpakai 28 hosts
netd ( 7 hosts): 204.24.93.96/27 => ada 30 hosts; tidak terpakai 23 hosts
nete (28 hosts): 204.24.93.128/27 => ada 30 hosts; tidak terpakai 2 hosts
Dengan demikian terlihat adanya ip address yang tidak terpakai dalam jumlah yang cukup besar. Hal ini mungkin tidak akan menjadi masalah pada ip private akan tetapi jika ini di alokasikan pada ip public(seperti contoh ini) maka terjadi pemborosan dalam pengalokasian ip public tersebut.
Untuk mengatasi hal ini (efisiensi) dapat digunakan metoda VLSM, yaitu dengan cara sebagai berikut:
1. Buat urutan berdasarkan penggunaan jumlah host terbanyak (14,28,2,7,28 menjadi 28,28,14,7,2).
2. Tentukan blok subnet berdasarkan kebutuhan host:
28 hosts + 1 network + 1 broadcast = 30 --> menjadi 32 ip ( /27 )
14 hosts + 1 network + 1 broadcast = 16 --> menjadi 16 ip ( /28 )
7 hosts + 1 network + 1 broadcast = 9 --> menjadi 16 ip ( /28 )
2 hosts + 1 network + 1 broadcast = 4 --> menjadi 4 ip ( /30 )
Sehingga blok subnet-nya menjadi:
netb (28 hosts): 204.24.93.0/27 => ada 30 hosts; tidak terpakai 2 hosts
nete (28 hosts): 204.24.93.32/27 => ada 30 hosts; tidak terpakai 2 hosts
neta (14 hosts): 204.24.93.64/28 => ada 14 hosts; tidak terpakai 0 hosts
netd ( 7 hosts): 204.24.93.80/28 => ada 14 hosts; tidak terpakai 7 hosts
netc ( 2 hosts): 204.24.93.96/30 => ada 2 hosts; tidak terpakai 0 hosts
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Mencari IP Network, Host min, Host max, Broadcast, Hosts/net Dalam Format Decimal
bit IP : 0-7. 8-15 . 16-23 . 24-31
net prefix: 8 . 16 . 24 . 32
Classful IP Address:
class A : 1.0.0.0 s/d 126.255.255.255
class B : 128.0.0.0 s/d 191.255.255.255
class C : 192.0.0.0 s/d 223.255.255.255
Netmask default:
class A : /8(netprefix) atau 255.0.0.0
class B : /16(netprefix) atau 255.255.0.0
class C : /24(netprefix) atau 255.255.255.0
Rumusan:
IP = oct1 . Oct2 . Oct3 . Oct4
Netmask = mask_oct1 . Mask_oct2 . Mask_oct3. Mask_oct4
Wildcard = 255.255.255.255 - Netmask
net1 = int(oct1:(256-mask_oct1)x(256-mask_oct1)
net2 = int(oct2:(256-mask_oct2)x(256-mask_oct2)
net3 = int(oct3:(256-mask_oct3)x(256-mask_oct3)
net4 = int(oct4:(256-mask_oct4)x(256-mask_oct4)
Network = net1.net2.net3.net4
Broadcast = Network + Wildcard
Host min = Network + 0.0.0.1
Host max = Broadcast - 0.0.0.1
Host/net = (2^(32-netprefix))-2
Catatan:
Network: nilai oct4 harus genap
Host min: nilai oct4 harus ganjil
Host max: nilai oct4 harus genap
Broadcast: nilai oct4 harus ganjil
Contoh:
1) Nilai netprefix berada dalam range: /24 < netprefix < /32
A) IP = 118.98.176.214/30
32-30=2 -> 2^2=4 -> 256-4=252
Netmask: 255.255.255.252
Wildcard = 255.255.255.255 - 255.255.255.252 = 0.0.0.3
Network:
net1 = int(118:(256-255))x(256-255) = 118
net2 = int(98:(256-255))x(256-255) = 98
net3 = int(176:(256-255))x(256-255) = 176
net4 = int(214:(256-252))x(256-252) = 212
=> = 118.98.176.212
Broadcast: 118.98.176.212 + 0.0.0.3 = 118.98.176.215
Host min: 118.98.176.212 + 0.0.0.1 = 118.98.176.213
Host max: 118.98.176.215 - 0.0.0.1 = 118.98.176.214
Hosts/net: (2^(32-30))-2 = 2
B) 118.98.176.214/29
32-29=3 -> 2^3=8 -> 256-8=248
Netmask: 255.255.255.248
Wildcard = 255.255.255.255 - 255.255.255.248 = 0.0.0.7
Network:
net1 = int(118:(256-255))x(256-255) = 118
net2 = int(98:(256-255))x(256-255) = 98
net3 = int(176:(256-255))x(256-255) = 176
net4 = int(214:(256-248))x(256-248) = 208
=> = 118.98.176.208
Broadcast: 118.98.176.208 + 0.0.0.7 = 118.98.176.215
Host min: 118.98.176.208 + 0.0.0.1 = 118.98.176.209
Host max: 118.98.176.215 - 0.0.0.1 = 118.98.176.214
Hosts/net: (2^(32-29))-2 = 6
2) Nilai netprefix berada dalam range: /16 < netprefix < /24
A) 118.98.176.214/22
24-22=2 -> 2^2=4 -> 256-4=252
Netmask: 255.255.252.0
Wildcard = 255.255.255.255 - 255.255.252.0 = 0.0.3.255
Network:
net1 = int(118:(256-255))x(256-255) = 118
net2 = int(98:(256-255))x(256-255) = 98
net3 = int(176:(256-252))x(256-252) = 176
net4 = int(214:(256-0))x(256-0) = 0
=> = 118.98.176.0
Broadcast: 118.98.176.0 + 0.0.3.255 = 118.98.179.255
Host min: 118.98.176.0 + 0.0.0.1 = 118.98.176.1
Host max: 118.98.179.255 - 0.0.0.1 = 118.98.179.254
Hosts/net: (2^(32-22))-2 = 1022
B) 118.98.176.214/21
24-21=3 -> 2^3=8 -> 256-8=248
Netmask: 255.255.248.0
Wildcard = 255.255.255.255 - 255.255.248.0 = 0.0.7.255
Network:
net1 = int(118:(256-255))x(256-255) = 118
net2 = int(98:(256-255))x(256-255) = 98
net3 = int(176:(256-248))x(256-248) = 176
net4 = int(176:(256-0))x(256-0) = 0
=> = 118.98.176.0
Broadcast: 118.98.176.0 + 0.0.7.255 = 118.98.183.255
Host min: 118.98.176.0 + 0.0.0.1 = 118.98.176.1
Host max: 118.98.183.255 - 0.0.0.1 = 118.98.183.254
Hosts/net: (2^(32-21))-2 = 2046
3) Nilai netprefix berada dalam range: /8 < netprefix < /16
A) 118.98.176.214/14
16-14=2 -> 2^2=4 -> 256-4=252
Netmask: 255.252.0.0
Wildcard = 255.255.255.255 - 255.252.0.0 = 0.3.255.255
Network:
net1 = int(118:(256-255))x(256-255) = 118
net2 = int(98:(256-252))x(256-252) = 96
net3 = int(176:(256-0))x(256-0) = 0
net4 = int(214:(256-0))x(256-0) = 0
=> = 118.96.0.0
Broadcast: 118.96.0.0 + 0.3.255.255 = 118.99.255.255
Host min: 118.96.0.0 + 0.0.0.1 = 118.96.0.1
Host max: 118.99.255.255 - 0.0.0.1 = 118.99.255.254
Hosts/net: (2^(32-14))-2 = 262142
B) 118.98.176.214/13
16-13=3 -> 2^3=8 -> 256-8=248
Netmask: 255.248.0.0
Wildcard = 255.255.255.255 - 255.248.0.0 = 0.7.255.255
Network:
net1 = int(118:(256-255))x(256-255) = 118
net2 = int(98:(256-248))x(256-248) = 96
net3 = int(176:(256-0))x(256-0) = 0
net4 = int(214:(256-0))x(256-0) = 0
=> = 118.96.0.0
Broadcast: 118.96.0.0 + 0.7.255.255 = 118.103.255.255
Host min: 118.96.0.0 + 0.0.0.1 = 118.96.0.1
Host max: 118.103.255.255 - 0.0.0.1 = 118.103.255.254
Hosts/net: (2^(32-13))-2 = 524286
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