Jarkom2013 Jarkom2013 TCPIP Network Design
Chapter 8: TCP/IP Network
Design
Network design with Classfull IP
Addressing
Classless Addressing and
Variable-Length Subnet Masks
Quiz:
Untuk alamat IP dibawah ini (classful IP), sebutkan :
Representasi binernya
Kelasnya
Network ID, rentang IP Host, dan IP Broadcast nya
a. 175.175.240.7
b. 192.168.59.95
c. 100.20.20.1
IP Addressing
1. Classfull
Conventional/ Default/ Standard Net Mask
SubNetting
2. Classless
Variable Length Subnet Mask (VLSM)
Classless Inter Domain Routing (CIDR)
TCP/IP Network Design
1. Network design with Classfull IP
Addressing
2. Classless Addressing and VariableLength Subnet Masks
1. Classful IP Addressing
There are three basic classes of addresses
known as class A, B, or C networks
Classful IP Addressing
Classful addresses are broken apart on octet
boundaries.
The first few bits of each segment address is
used to denote the address class of the
segment.
The class ID plus network ID portions of the
IP address are known as the network prefix,
the net-work number, or the major network
Private IPv4 address spaces
IP kelas A : 127.x.x.x loopback IP localhost
Public Versus Private IP Addresses
Some IP services require what’s called a secure endto-end connection—IP traffic must be able to move in
encrypted form between the sender and receiver
without intermediate translation
Most organizations need public IP addresses only for
two classes of equipment:
Devices that permit organizations to attach networks to the
Internet
Servers that are designed to be accessible to the Internet
8
Subnetting
When IP address classes were established, networks
were composed of a relatively small number of
relatively expensive computers.
As time went on and the PC exploded into LAN’s, the
strict boundaries of the classful addressing address
classes became restrictive and forced an inefficient
allocation of addresses.
Class C address with its limit of 254 (28-2) hosts per
network is too small for most organizations, while a
Class B address with its limit of 65,534 (28 * 28 – 2)
hosts per subnet is too large.
Subnetting
Networks grew and needed to be divided or
segmented in order to improve traffic flow.
Routers join two separate net-works.
Networks that are separated by routers must
have different network IDs so that the router
can distinguish between them.
This accelerated the depletion of IP
addresses.
RFC 950
RFC 950 gave users a way to subnet, or
provide a third layer of organization or
hierarchy between the existing network ID
and the existing host ID.
Since the network IDs could not be altered,
the only choice was to “borrow” some of the
host ID bits.
These “borrowed” bits constitute the subnet
portion of the address.
A subnet mask identifies which bits are used
for the subnet ID.
Extended Network Prefix
The extended network prefix is the classful network
prefix (/16 in the case of a Class B address) plus the
number of bits borrowed from the host ID
This figure illustrates the network prefix, extended
network prefix, subnet mask in binary and decimal,
network ID, subnet ID, and host ID
Subnet Masks
Alternative Subnet masks for a Class B
network
Contoh 1: subnetting
Bank XYZ memiliki alamat Jaringan IP 131.179.0.0
Saat ini bank tersebut memiliki 50 kantor cabang.
Dalam 5 tahun ke depan diperkirakan jumlah
tersebut akan meningkat menjadi 150 kantor
cabang.
Setiap kantor cabang memiliki sebuah subnet, yang
maksimal akan terdiri dari 50 host.
Tentukan subnet mask yang sesuai untuk
bank tersebut !
Latihan: Langkah-langkah penyelesaian (2)
1. Tentukan kelas IP Address: 131.179.0.0
•
•
Kelas B
Netmask (default) = 255.255.0.0
2. Hitung jumlah subnet yang dibutuhkan
•
•
150 kantor cabang = 150 subnet, butuh minimum 8 bit
255.255.255.0 (terjawab)
3. Periksa apakah sisa bit memenuhi kebutuhan host
•
•
Tiap subnet butuh hingga 50 host
Sisa untuk porsi host ada 8 bit maksimum 254 host
memenuhi !
Contoh 2: Subnetting
• Sebuah jaringan toko eceran, bernama ABC, saat ini
memiliki 80 toko. Diperkirakan perkembangannya adalah
20 toko/ tahun, selama 8 tahun ke depan.
• Setiap hari diperlukan upload data penjualan dari masingmasing toko ke kantor pusat. Dengan begitu tiap toko
memerlukan sebuah router dan sebuah komputer yang
terhubung ke router tersebut (tiap toko merupakan
subnet yg butuh 2 host ID).
• Alamat Jaringan IP yang dimiliki jaringan toko tersebut
adalah 165.32.0.0
• Tentukan solusi masalah tsb dg subnetting?
Alamat IP kelas apakah yang dimiliki jaringan
toko ABC? Tuliskan representasi binernya.
165.32.0.0
10100101.00100000.00000000.00000000
IP kelas B, sehingga network mask nya
adalah 255.255.0.0 (default, classfull).
Hitung kebutuhan maksimal subnet jaringan tersebut,
hingga 8 tahun ke depan. Kemudian putuskan
subnetting yang paling sesuai (classfull).
80+(8x20) = 240 , jadi toko
membutuhkan minimal 240 subnet.
Untuk memenuhi kebutuhan tersebut
maka digunakan 8 (28) bit subnet mask
255.255.255.0, agar mampu
menampung hingga 254 subnet ID.
Berapa jumlah host ID minimum yang diperlukan tiap
toko. Apakah Jumlah tersebut dapat dipenuhi dengan
subnet mask hasil perhitungan sebelumnya ? Jelaskan.
Minimum butuh 2 host ID per toko,
atau 2 host ID per subnet.
Karena tersisa 8 bit untuk host ID per
subnet, berarti jumlah ini sangat
memadai bila hanya diperlukan 2 host
ID per subnet.
Tuliskan solusi subnet mask tersebut dalam representasi
biner. Tunjukkan bagian yang menandakan network ID,
subnet ID, dan host ID. Tunjukkan pula bagian yang
disebut sebagai extended network prefix.
Gambarkan topologi jaringan toko ABC hasil
rancangan anda tersebut berikut detil alamat IP
nya.
Misal, 1 router kantor pusat beserta
servernya, dan beberapa router toko beserta
PC nya. Cantumkan alamat IP yang relevan.
165.32.0.0/ 24
165.32.1.1
165.32.2.1
165.32.3.1
....
165.32.4.1
(R), 165.32.1.2 (H) pusat
(R), 165.32.2.2 (H) toko 1
(R), 165.32.3.2 (H) toko 2
(R), 165.32.4.2 (H) toko 240
Limitations of Classful Addressing and
Fixed-Length Subnet Masks
1. Wasted addresses. Only one subnet mask
can be used for a net-work prefix
2. A shrinking pool of available IPv4 addresses
Limitations of Classful Addressing and
Fixed-Length Subnet Masks (2)
Routing protocols such as RIP are unable to
transmit subnet masks or extended network
prefix information along with network IDs and
IP addresses
All routers, servers, and workstations within a
given network all have the same subnet
mask,
A fixed subnet mask implies a fixed subnet
size for all subnets of a given network ID.
Limitations of Classful Addressing and
Fixed-Length Subnet Masks (3)
Subnets must be sized to accommodate the
largest required subnet within a given
network ID, resulting in wasted host
addresses that cannot be recovered or used
by other subnets.
We need solutions....
Classless addressing and Variable-length subnet
masks !
TCP/IP Network Design
1. Network design with Classfull IP
Addressing
2. Classless Addressing and
Variable-Length Subnet Masks
Variable-Length Subnet Masks
Variable-Length Subnet Masks
RFP 1009
RFP1009 specifies how a single
network ID can have different subnet
masks among its subnets.
Minimizes the wasted IP addresses
forced by a single subnet mask per
network ID as defined by the original
RFC 950 Subnetting technique
Implementing VLSM
The routers on the network where VLSM is
implemented must be able to share sub-net masks
and/or extended network prefixes with each router
advertisement.
Routing protocols such as OSPF and IS-IS do this,
whereas RIP and IGRP do not.
RIPv2 (RFC 1388), added support for VLSM to RIPv1.
All routers supporting VLSM must support a longest
match routing algorithm.
The implemented network topology must match the
distribution of addresses and definition of subnets
Route Aggregation with VLSM
Dipecah jadi 254 subnet
Dipecah jadi 254 subnet
Dipecah jadi 6 subnet
Subnetting
Conventional vs VLSM
Contoh 1: VLSM
Alokasi network = 128.25.0.0/21
Tentukan alokasi subnetwork untuk :
a)
b)
c)
d)
e)
Network
Network
Network
Network
Network
A terdiri dari 31 PC
B terdiri dari 255 PC
C terdiri dari 100 PC
D terdiri dari 525 PC
E terdiri dari 10 PC
Analisis alamat jaringan IP
128.25.0.0/21
Range network : 128.25.0.0/21 - 128.25.7.255/21
128.25.0.0
= 1000 0000 . 0001 1001 . 0000 0000. 0000 0000
128.25.7.255 = 1000 0000 . 0001 1001 . 0000 0111 . 1111 1111
/21
= 1111 1111 . 1111 1111 . 1111 1000 . 0000 0000
Jumlah host maksimum: 7 x 256 – 2 = 1790 host ID
Tentukan kebutuhan subnet
Subnetwork A : 31 + 2 = 33 64 6 bit
Masking /26 atau 255.255.255.192
Subnetwork B : 255 + 2 = 257 512 9 bit
Masking /23 atau 255.255.254.0
Subnetwork C : 100 + 2 = 102 128 7 bit
Masking /25 atau 255.255.255.128
Subnetwork D : 525 + 2 = 527 1024 10 bit
Masking /22 atau 255.255.252.0
Subnetwork E : 10 + 2 = 12 16 4 bit
Masking /28 atau 255.255.255.240
Urutkan dari masking terbesar ke terkecil!
D, B, C, A, E
Range NW awal : 128.25.0.0/21 - 128.25.7.255/21
Subnetwork D (1024): 128.25.0.0/22 - 128.25.3.255/22
Subnetwork B (512) : 128.25.4.0/23 - 128.25.5.255/23
Subnetwork C (128) : 128.25.6.0/25 - 128.25.6.127/25
Subnetwork A (64) : 128.25.6.128/26 -128.25.6.191/26
Subnetwork E (16) : 128.25.6.192/28 -128.25.6.207/28
Sisa = 128.25.6.208 – 128.25.7.255 304 alamat IP
Contoh 2: VLSM
Alokasi network = 192.168.0.0/20
Tentukan alokasi subnetwork untuk :
a)
b)
c)
d)
e)
f)
Network
Network
Network
Network
Network
Network
A terdiri dari 1000 PC
B terdiri dari 555 PC
C terdiri dari 320 PC
D terdiri dari 130 PC
E terdiri dari 750 PC
F terdiri dari 10 PC
Tentukan kebutuhan subnet
Subnetwork A : 1000 + 2 = 1002 1024 10 bit
Masking /22 atau 255.255.252.0
Subnetwork B : 555 + 2 = 557 1024 10 bit
Masking /22 atau 255.255.255.0
Subnetwork C : 320 + 2 = 322 512 9 bit
Masking /23 atau 255.255.255.254
Subnetwork D : 130 + 2 = 132 256 8 bit
Masking /24 atau 255.255.255.0
Subnetwork E : 750 + 2 = 752 1024 10 bit
Masking /22 atau 255.255.252.0
Subnetwork F : 10 + 2 = 12 16 4 bit
Masking /28 atau 255.255.255.240
Urutkan dari masking terbesar ke terkecil!
A, E,B,C,D,F
Range NW awal : 192.168.0.0/20 - 192.168.15.255/20
Subnetwork A (1024):192.168.0.0/22 -192.168.3.255/22
Subnetwork E (1024):192.168.4.0/22 -192.168.7.255/22
Subnetwork B (1024):192.168.8.0/22 -192.168.11.255/22
Subnetwork C (512) : 192.168.12.0/23 -192.168.13.255/23
Subnetwork D (256) :192.168.14.0/24 -192.168.14.255/24
Subnetwork F (16) : 192.168.15.0/28 -192.168.15.255/28
Sisa= 192.168.15.16 -192.168.15.255 240 alamat IP
Classless Inter-Domain Routing
(CIDR)
CIDR was announced in September
1993 and is documented in RFCs 1517,
1518, 1519, and 1520.
CIDR is also sometimes referred to as
supernetting.
Implementing CIDR
CIDR addresses are issued in blocks known
as CIDR blocks.
“The first octet” is meaningless in
determining how many subnets or host IDs
can be defined for a given CIDR block.
The only factor that determines the capacity
of a CIDR block is the network prefix
assigned by the Internet authorities when the
CIDR block is issued.
Implementing CIDR
The net-work prefix issued by the Internet
authorities indicates the number of bits used
for the major network ID.
These bits are reserved and cannot be used
by the end users for subnet IDs or host IDs.
For example, if a CIDR block were issued
with a network prefix of /18, that would imply
that the first 18 bits from left to right were
reserved for the network ID and leave 14 bits
(32–18) for subnet IDs and host IDs.
CIDR Block Capacity
Contoh 1: CIDR
Identifikasikan NW ID, rentang IP Host, dan BC
IP untuk alamat IP 192.168.235.1/20
Jawaban:
IP : 192.168.235.1/20
Alamat IP :
Subnetmask :
Maka :
NW ID
:
Rng Host ID :
BC IP
:
11000000.10101000.11101011.00000001 = 192.168.235.1
11111111.11111111.11110000.00000000 = 255.255.240.0
11000000.10101000.11100000.00000000
11000000.10101000.11100000.00000001
11000000.10101000.11100000.00000010
.
.
.
.
11000000.10101000.11101111.11111101
11000000.10101000.11101111.11111110
11000000.10101000.11101111.11111111
= 192.168.224.0
= 192.168.224.1/20
= 192.168.224.2/20
= 192.168.239.253/20
= 192.168.239.254/20
= 192.168.239.255
Review:
1. Apa tujuan VLSM?
2. Apa tujuan CIDR?
3. Jelaskan perbedaan classful routing protocol dengan
classles routing protocol.
Terimakasih
Sampai jumpa di pertemuan berikutnya
Design
Network design with Classfull IP
Addressing
Classless Addressing and
Variable-Length Subnet Masks
Quiz:
Untuk alamat IP dibawah ini (classful IP), sebutkan :
Representasi binernya
Kelasnya
Network ID, rentang IP Host, dan IP Broadcast nya
a. 175.175.240.7
b. 192.168.59.95
c. 100.20.20.1
IP Addressing
1. Classfull
Conventional/ Default/ Standard Net Mask
SubNetting
2. Classless
Variable Length Subnet Mask (VLSM)
Classless Inter Domain Routing (CIDR)
TCP/IP Network Design
1. Network design with Classfull IP
Addressing
2. Classless Addressing and VariableLength Subnet Masks
1. Classful IP Addressing
There are three basic classes of addresses
known as class A, B, or C networks
Classful IP Addressing
Classful addresses are broken apart on octet
boundaries.
The first few bits of each segment address is
used to denote the address class of the
segment.
The class ID plus network ID portions of the
IP address are known as the network prefix,
the net-work number, or the major network
Private IPv4 address spaces
IP kelas A : 127.x.x.x loopback IP localhost
Public Versus Private IP Addresses
Some IP services require what’s called a secure endto-end connection—IP traffic must be able to move in
encrypted form between the sender and receiver
without intermediate translation
Most organizations need public IP addresses only for
two classes of equipment:
Devices that permit organizations to attach networks to the
Internet
Servers that are designed to be accessible to the Internet
8
Subnetting
When IP address classes were established, networks
were composed of a relatively small number of
relatively expensive computers.
As time went on and the PC exploded into LAN’s, the
strict boundaries of the classful addressing address
classes became restrictive and forced an inefficient
allocation of addresses.
Class C address with its limit of 254 (28-2) hosts per
network is too small for most organizations, while a
Class B address with its limit of 65,534 (28 * 28 – 2)
hosts per subnet is too large.
Subnetting
Networks grew and needed to be divided or
segmented in order to improve traffic flow.
Routers join two separate net-works.
Networks that are separated by routers must
have different network IDs so that the router
can distinguish between them.
This accelerated the depletion of IP
addresses.
RFC 950
RFC 950 gave users a way to subnet, or
provide a third layer of organization or
hierarchy between the existing network ID
and the existing host ID.
Since the network IDs could not be altered,
the only choice was to “borrow” some of the
host ID bits.
These “borrowed” bits constitute the subnet
portion of the address.
A subnet mask identifies which bits are used
for the subnet ID.
Extended Network Prefix
The extended network prefix is the classful network
prefix (/16 in the case of a Class B address) plus the
number of bits borrowed from the host ID
This figure illustrates the network prefix, extended
network prefix, subnet mask in binary and decimal,
network ID, subnet ID, and host ID
Subnet Masks
Alternative Subnet masks for a Class B
network
Contoh 1: subnetting
Bank XYZ memiliki alamat Jaringan IP 131.179.0.0
Saat ini bank tersebut memiliki 50 kantor cabang.
Dalam 5 tahun ke depan diperkirakan jumlah
tersebut akan meningkat menjadi 150 kantor
cabang.
Setiap kantor cabang memiliki sebuah subnet, yang
maksimal akan terdiri dari 50 host.
Tentukan subnet mask yang sesuai untuk
bank tersebut !
Latihan: Langkah-langkah penyelesaian (2)
1. Tentukan kelas IP Address: 131.179.0.0
•
•
Kelas B
Netmask (default) = 255.255.0.0
2. Hitung jumlah subnet yang dibutuhkan
•
•
150 kantor cabang = 150 subnet, butuh minimum 8 bit
255.255.255.0 (terjawab)
3. Periksa apakah sisa bit memenuhi kebutuhan host
•
•
Tiap subnet butuh hingga 50 host
Sisa untuk porsi host ada 8 bit maksimum 254 host
memenuhi !
Contoh 2: Subnetting
• Sebuah jaringan toko eceran, bernama ABC, saat ini
memiliki 80 toko. Diperkirakan perkembangannya adalah
20 toko/ tahun, selama 8 tahun ke depan.
• Setiap hari diperlukan upload data penjualan dari masingmasing toko ke kantor pusat. Dengan begitu tiap toko
memerlukan sebuah router dan sebuah komputer yang
terhubung ke router tersebut (tiap toko merupakan
subnet yg butuh 2 host ID).
• Alamat Jaringan IP yang dimiliki jaringan toko tersebut
adalah 165.32.0.0
• Tentukan solusi masalah tsb dg subnetting?
Alamat IP kelas apakah yang dimiliki jaringan
toko ABC? Tuliskan representasi binernya.
165.32.0.0
10100101.00100000.00000000.00000000
IP kelas B, sehingga network mask nya
adalah 255.255.0.0 (default, classfull).
Hitung kebutuhan maksimal subnet jaringan tersebut,
hingga 8 tahun ke depan. Kemudian putuskan
subnetting yang paling sesuai (classfull).
80+(8x20) = 240 , jadi toko
membutuhkan minimal 240 subnet.
Untuk memenuhi kebutuhan tersebut
maka digunakan 8 (28) bit subnet mask
255.255.255.0, agar mampu
menampung hingga 254 subnet ID.
Berapa jumlah host ID minimum yang diperlukan tiap
toko. Apakah Jumlah tersebut dapat dipenuhi dengan
subnet mask hasil perhitungan sebelumnya ? Jelaskan.
Minimum butuh 2 host ID per toko,
atau 2 host ID per subnet.
Karena tersisa 8 bit untuk host ID per
subnet, berarti jumlah ini sangat
memadai bila hanya diperlukan 2 host
ID per subnet.
Tuliskan solusi subnet mask tersebut dalam representasi
biner. Tunjukkan bagian yang menandakan network ID,
subnet ID, dan host ID. Tunjukkan pula bagian yang
disebut sebagai extended network prefix.
Gambarkan topologi jaringan toko ABC hasil
rancangan anda tersebut berikut detil alamat IP
nya.
Misal, 1 router kantor pusat beserta
servernya, dan beberapa router toko beserta
PC nya. Cantumkan alamat IP yang relevan.
165.32.0.0/ 24
165.32.1.1
165.32.2.1
165.32.3.1
....
165.32.4.1
(R), 165.32.1.2 (H) pusat
(R), 165.32.2.2 (H) toko 1
(R), 165.32.3.2 (H) toko 2
(R), 165.32.4.2 (H) toko 240
Limitations of Classful Addressing and
Fixed-Length Subnet Masks
1. Wasted addresses. Only one subnet mask
can be used for a net-work prefix
2. A shrinking pool of available IPv4 addresses
Limitations of Classful Addressing and
Fixed-Length Subnet Masks (2)
Routing protocols such as RIP are unable to
transmit subnet masks or extended network
prefix information along with network IDs and
IP addresses
All routers, servers, and workstations within a
given network all have the same subnet
mask,
A fixed subnet mask implies a fixed subnet
size for all subnets of a given network ID.
Limitations of Classful Addressing and
Fixed-Length Subnet Masks (3)
Subnets must be sized to accommodate the
largest required subnet within a given
network ID, resulting in wasted host
addresses that cannot be recovered or used
by other subnets.
We need solutions....
Classless addressing and Variable-length subnet
masks !
TCP/IP Network Design
1. Network design with Classfull IP
Addressing
2. Classless Addressing and
Variable-Length Subnet Masks
Variable-Length Subnet Masks
Variable-Length Subnet Masks
RFP 1009
RFP1009 specifies how a single
network ID can have different subnet
masks among its subnets.
Minimizes the wasted IP addresses
forced by a single subnet mask per
network ID as defined by the original
RFC 950 Subnetting technique
Implementing VLSM
The routers on the network where VLSM is
implemented must be able to share sub-net masks
and/or extended network prefixes with each router
advertisement.
Routing protocols such as OSPF and IS-IS do this,
whereas RIP and IGRP do not.
RIPv2 (RFC 1388), added support for VLSM to RIPv1.
All routers supporting VLSM must support a longest
match routing algorithm.
The implemented network topology must match the
distribution of addresses and definition of subnets
Route Aggregation with VLSM
Dipecah jadi 254 subnet
Dipecah jadi 254 subnet
Dipecah jadi 6 subnet
Subnetting
Conventional vs VLSM
Contoh 1: VLSM
Alokasi network = 128.25.0.0/21
Tentukan alokasi subnetwork untuk :
a)
b)
c)
d)
e)
Network
Network
Network
Network
Network
A terdiri dari 31 PC
B terdiri dari 255 PC
C terdiri dari 100 PC
D terdiri dari 525 PC
E terdiri dari 10 PC
Analisis alamat jaringan IP
128.25.0.0/21
Range network : 128.25.0.0/21 - 128.25.7.255/21
128.25.0.0
= 1000 0000 . 0001 1001 . 0000 0000. 0000 0000
128.25.7.255 = 1000 0000 . 0001 1001 . 0000 0111 . 1111 1111
/21
= 1111 1111 . 1111 1111 . 1111 1000 . 0000 0000
Jumlah host maksimum: 7 x 256 – 2 = 1790 host ID
Tentukan kebutuhan subnet
Subnetwork A : 31 + 2 = 33 64 6 bit
Masking /26 atau 255.255.255.192
Subnetwork B : 255 + 2 = 257 512 9 bit
Masking /23 atau 255.255.254.0
Subnetwork C : 100 + 2 = 102 128 7 bit
Masking /25 atau 255.255.255.128
Subnetwork D : 525 + 2 = 527 1024 10 bit
Masking /22 atau 255.255.252.0
Subnetwork E : 10 + 2 = 12 16 4 bit
Masking /28 atau 255.255.255.240
Urutkan dari masking terbesar ke terkecil!
D, B, C, A, E
Range NW awal : 128.25.0.0/21 - 128.25.7.255/21
Subnetwork D (1024): 128.25.0.0/22 - 128.25.3.255/22
Subnetwork B (512) : 128.25.4.0/23 - 128.25.5.255/23
Subnetwork C (128) : 128.25.6.0/25 - 128.25.6.127/25
Subnetwork A (64) : 128.25.6.128/26 -128.25.6.191/26
Subnetwork E (16) : 128.25.6.192/28 -128.25.6.207/28
Sisa = 128.25.6.208 – 128.25.7.255 304 alamat IP
Contoh 2: VLSM
Alokasi network = 192.168.0.0/20
Tentukan alokasi subnetwork untuk :
a)
b)
c)
d)
e)
f)
Network
Network
Network
Network
Network
Network
A terdiri dari 1000 PC
B terdiri dari 555 PC
C terdiri dari 320 PC
D terdiri dari 130 PC
E terdiri dari 750 PC
F terdiri dari 10 PC
Tentukan kebutuhan subnet
Subnetwork A : 1000 + 2 = 1002 1024 10 bit
Masking /22 atau 255.255.252.0
Subnetwork B : 555 + 2 = 557 1024 10 bit
Masking /22 atau 255.255.255.0
Subnetwork C : 320 + 2 = 322 512 9 bit
Masking /23 atau 255.255.255.254
Subnetwork D : 130 + 2 = 132 256 8 bit
Masking /24 atau 255.255.255.0
Subnetwork E : 750 + 2 = 752 1024 10 bit
Masking /22 atau 255.255.252.0
Subnetwork F : 10 + 2 = 12 16 4 bit
Masking /28 atau 255.255.255.240
Urutkan dari masking terbesar ke terkecil!
A, E,B,C,D,F
Range NW awal : 192.168.0.0/20 - 192.168.15.255/20
Subnetwork A (1024):192.168.0.0/22 -192.168.3.255/22
Subnetwork E (1024):192.168.4.0/22 -192.168.7.255/22
Subnetwork B (1024):192.168.8.0/22 -192.168.11.255/22
Subnetwork C (512) : 192.168.12.0/23 -192.168.13.255/23
Subnetwork D (256) :192.168.14.0/24 -192.168.14.255/24
Subnetwork F (16) : 192.168.15.0/28 -192.168.15.255/28
Sisa= 192.168.15.16 -192.168.15.255 240 alamat IP
Classless Inter-Domain Routing
(CIDR)
CIDR was announced in September
1993 and is documented in RFCs 1517,
1518, 1519, and 1520.
CIDR is also sometimes referred to as
supernetting.
Implementing CIDR
CIDR addresses are issued in blocks known
as CIDR blocks.
“The first octet” is meaningless in
determining how many subnets or host IDs
can be defined for a given CIDR block.
The only factor that determines the capacity
of a CIDR block is the network prefix
assigned by the Internet authorities when the
CIDR block is issued.
Implementing CIDR
The net-work prefix issued by the Internet
authorities indicates the number of bits used
for the major network ID.
These bits are reserved and cannot be used
by the end users for subnet IDs or host IDs.
For example, if a CIDR block were issued
with a network prefix of /18, that would imply
that the first 18 bits from left to right were
reserved for the network ID and leave 14 bits
(32–18) for subnet IDs and host IDs.
CIDR Block Capacity
Contoh 1: CIDR
Identifikasikan NW ID, rentang IP Host, dan BC
IP untuk alamat IP 192.168.235.1/20
Jawaban:
IP : 192.168.235.1/20
Alamat IP :
Subnetmask :
Maka :
NW ID
:
Rng Host ID :
BC IP
:
11000000.10101000.11101011.00000001 = 192.168.235.1
11111111.11111111.11110000.00000000 = 255.255.240.0
11000000.10101000.11100000.00000000
11000000.10101000.11100000.00000001
11000000.10101000.11100000.00000010
.
.
.
.
11000000.10101000.11101111.11111101
11000000.10101000.11101111.11111110
11000000.10101000.11101111.11111111
= 192.168.224.0
= 192.168.224.1/20
= 192.168.224.2/20
= 192.168.239.253/20
= 192.168.239.254/20
= 192.168.239.255
Review:
1. Apa tujuan VLSM?
2. Apa tujuan CIDR?
3. Jelaskan perbedaan classful routing protocol dengan
classles routing protocol.
Terimakasih
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