3. EXECUTIVE SUMMARY
1.
In the Next Generation Telecommunication (LTE), all
communication devices will use IP as the identity.
2.
IP will be used to define routing from one node to others and
vice versa. IP is the 3rd layer in OSI standard. Layer 3 (Network
layer) is used to bring packet from one node to other using
logical address (Packet Forwarder/routing).
3.
Allocation of IPV4 is limited. Therefore plan is needed and we
can also divide IP network into some IP sub-network
(subnetting)
4.
IPv6 contain 128 bits (IPv4 only 32 bits). As the limitation of IP
allocation, IPv6 will replace IPv4.
2
4. OBJECTIVES
After this presentation, participants will know:
OSI layer applied in LTE
How to do Subnetting
Grouping BTS/NodeB in a VLAN
IP Configuration in BSC/RNC/LTE
3
6. OSI LAYER
Application: End User Interface (http, ftp, telnet, dns, etc)
Data format sent: ASCII, binary, JPEG, other compression, etc.
Open, maintain and terminate communication session: SQL, netbios, RPC, etc
How to deliver data reliable or unreliable, connectionless (UDP) or connection
oriented (TCP). It has function: error & flow control, sequence number,
acknowledgement.
Bring packet from one node to other using logical address (Packet Forwarder
/Routing)
Communication data between one node and others using Hardware Address
(MAC, LLC, etc). It identify the topology used (PTP, PTM – FR/ATM, BUS, Token
Rng, etc). Also function for error control and flow control.
Change data from Data link BITS
5
7. OSI LAYER IN LTE
User Plane & Control Plane
User Plane Protocol Stack
Control Plane Protocol Stack
6
10. *
* Downlink Physical Layer Procedures
* For E-UTRA, the following downlink physical layer procedures are especially
important:
Cell search and synchronization:
Scheduling:
Link Adaptation:
Hybrid ARQ (Automatic Repeat Request)
11. *
* Uplink Physical Layer Procedures
* For E-UTRA, the following uplink physical layer procedures are especially important:
Random access
Uplink scheduling
Uplink link adaptation
Uplink timing control
Hybrid ARQ
16. The three sublayers are
Medium access Control(MAC)
Radio Link Control(RLC)
Packet Data Convergence Protocol(PDCP)
*
[Source: E-UTRAN Architecture(3GPP TR 25.012 ]
23. Internet Protocol Overview
From Wikipedia:
The Internet Protocol (IP) is the principal communications protocol in the Internet
protocol suite for relaying datagrams across network boundaries. This function
of ROUTING enables internetworking, and essentially establishes the Internet.
Internet Protocol sends data packets with unreliable/connectionless (no warranty
success or not)
The responsibility is handled in upper layer.
22
24. IP V4
Communication between TCP/IP network needs identity known as IP
address.
- IP address contain 32 bits.
- IP address divided into Network ID and Host ID
- 32 bits IP divided into 4 parts, each part has 8 bits.
- Every 8 bits can be converted to decimal 0 to 255.
Dec:
xxx
.
xxx .
xxx
. xxx
Bit : xxxxxxxx.xxxxxxxx.xxxxxxxx.xxxxxxxx
Note: IPv6 has 128 bit
23
25. IP V4 Classification
xxxxxxxx . xxxxxxxx . xxxxxxxx . xxxxxxxx
IP divided into 3 class
Class A:
Network ID (8bit)
0xxxxxxx
Host ID (24 bit)
xxxxxxxx.xxxxxxxx.xxxxxxxx
Class B:
Network ID (16 bit)
10xxxxxx.xxxxxxxx
Host ID (16 bit)
xxxxxxxx.xxxxxxxx
Class C:
Network ID (8bit)
110xxxxx.xxxxxxxx.xxxxxxxx
Host ID (24 bit)
xxxxxxxx
Note: IPv6 has 128 bit
24
26. IP V4 Classification
Class
Bit awal
Jumlah
Jaringan
Jumlah Host
Private IP address by International
Assigned Number Authority (IANA)
A
1 - 126
126
16 777 214
10.0.0.0 sampai 10.255.255.255
B
128 - 191
16 384
65 534
172.16.0.0 sampai 172.31.255.255
C
192 - 223
2 097 152
254
192.168.0.0 sampai 192.168.255.255
Note: 127.0.0.0 is used for loopback address
25
27. IP Netmask
To separate Network ID and Host ID, NETMASK is used with definition:
- Network ID use binary 1
- Host ID use binary 0
Netmask natural:
11111111 00000000 00000000 00000000 = 255.0.0.0
11111111 11111111 00000000 00000000 = 255.255.0.0
11111111 11111111 11111111 00000000 = 255.255.255.0
Netmask bit
Netmask Dec
1111 1111
255
1111 1110
254
1111 1100
252
1111 1000
248
1111 0000
240
1110 0000
224
1100 0000
192
1000 0000
128
26
28. Broadcast Address & Network Address
Broadcast address is needed in a network.
Function of Broadcast Address:
- To give information to the network for an existing service
- Finding information in a network
192.168.1.2
192.168.1.4
192.168.1.3
192.168.1.0
192.168.1.1
Local broadcast: 255.255.255.255
Directed broadcast: 192.168.1.255
27
Network ID:
1st Host:
:
:
Last host:
IP broadcast:
192.168.1.0
192.168.1.1
:
:
192.168.1.254
192.168.1.255
29. Exercise
Jaringan
Class
First IP
Last IP
Broadcast IP
10.0.0.0
A
10.0.0.1
10.255.255.254
10.255.255.255
128.3.0.0
B
128.3.0.1
128.3.255.254
128.3.255.255
172.16.0.0
B
172.16.0.1
172.16.255.254
172.16.255.255
192.168.16.0
C
192.168.16.1
192.168.16.254
192.168.16.255
191.254.0.0
B
191.254.0.1
191.254.255.254
191.254.255.255
224.19.2.0
C
224.19.2.1
224.19.2.254
224.19.2.255
223.253.25.0
C
223.253.25.1
223.253.25.254
223.253.25.255
126.0.0.0
A
126.0.0.1
126.255.255.254
126.255.255.255
30. SUBNETTING
Subnetting diperlukan untuk membangun SUB-Jaringan dari Jaringan yang ada.
Subnetting diperlukan untuk lebih mengefisiensikan/utilize alokasi IP address yang ada.
Tujuan Subnetting:
-
Memadukan teknologi jaringan yang berbeda
Menghindari limitasi jumlah simpul dalam satu segmen
Mereduksi traffic yang disebabkan oleh broadcast atau pun collision
Jaringan di bawah ini bisa kita bagi menjadi beberapa sub-jaringan dengan menggunakan router.
192.168.1.2
192.168.1.3
192.168.1.4
192.168.1.0
192.168.1.1
31. SUBNETTING
Dari gambar sebelumnya kita akan
membagi IP jaringan 192.168.1.0 menjadi 4
buah sub-jaringan.
192.168.1.0 mempunyai
Network ID = 192.168.1.0
Broadcast ID = 192.168 1.255
Host ID = 192.168.1.1-254
192.168.1.0
192.168.1.128
192.168.1.64
192.168.1.192
Karena ada 4 subjaringan maka langkah
selanjutnya adalah memecah IP tersebut
menjadi 4 bagian.
192.168.1.0 =
11000000.10101000.00000001.00000000
Karena 4 subnet = 22 maka jumlah bit untuk
subnet = 2
11000000.10101000.00000001.00000000
Simplenya 256/4 ~= 64
Sehingga didapat IP jaringan 4 subjaringan:
- 192.168.1.0
- 192.168.1.64
- 192.168.1.128
- 192.168.1.192
32. SUBNETTING
Sub Network 1
Network ID = 192.168.1.0 =
11000000.10101000.00000001.00000000
Broadcast ID = 192.168.1.63
Host ID = 192.168.1.1-62
192.168.1.0
192.168.1.128
192.168.1.64
192.168.1.192
Sub Network 2
Network ID = 192.168.1.64 =
11000000.10101000.00000001.01000000
Broadcast ID = 192.168.1.127
Host ID = 192.168.1.65-126
Sub Network 3
Network ID = 192.168.1.128 =
11000000.10101000.00000001.10000000
Broadcast ID = 192.168.1.191
Host ID = 192.168.1.129-190
Sub Network 4
Network ID = 192.168.1.192 =
11000000.10101000.00000001.11000000
Broadcast ID = 192.168.1.255
Host ID = 192.168.1.193-254
33. Exercise
1. Pada jaringan Class B & C dibutuhkan 50 subnet dengan masing2 dapat mempunyai 4 hosts.
Berapa subnet bits yang dibutuhkan? Bisakah?
Class B (172.16.0.0 = 10110000.00010000.00000000.00000000)
Karena 50 ~ 64 = 26 maka bit subnet yang dibutuhkan adalah 6 bit.
Sisa 10 bits HOST
10110000.00010000.00000000.00000000
Class C (192.168.1.0 = 11000000.10101000.00000001.00000000)
Karena 50 ~ 64 = 26 maka bit subnet yang dibutuhkan adalah 6 bit.
Sisa 2 bits HOST
11000000.10101000.00000001.00000000
2. Dari data di atas berapa subnet-mask nya?
Ingat!!
- Network ID use binary 1
- Host ID use binary 0
Maka subnet-mask adalah:
Class B: 11111111.11111111.11111100.00000000 = 255.255.252.0
Class C: 11111111.11111111.11111111.11111100 = 255.255.255.252
34. Exercise
3. Tentukan IP subnet/sub-jaringan dari Class B & C tersebut?
Class B:
172.16.0.0 = 10110000.00010000.00000000.00000000
172.16.4.0 = 10110000.00010000.00000100.00000000
172.16.8.0 = 10110000.00010000.00001000.00000000
172.16.12.0 = 10110000.00010000.00001100.00000000
:
:
:
172.16.252.0 = 10110000.00010000.11111100.00000000
38. IP ROUTING IN BSC/RNC
- IP ROUTING (IPRT) should be created from BSC/RNC to other nodes to establish connection.
- For checking whether connection is open between BSC/RNC with other nodes we can use PING command.
- If no RTO found then we can create IP routing and UP/CP connection
DSTIP = IP Subnet
DSTMASK = Subnetmask
NEXTHOP = IP router
39. INTERFACE IP
We can know how many BTS grouped in a VLAN from IPRT command.
For example, Abis IP is set from 10.48.x.x to 10.54.x.x
From the CFGMML (LST IPRT) we got:
For IP Subnet=10.48.0.32, DSTMASK=255.255.255.224 (11111111.11111111.11111111.11100000) #HOST = 25 -2 = 32
IP HOST/BTSIP = 10.48.0.32.0-62
IP BROADCAST=10.48.0.63
Other example for RNC CFGMML:
For IP Subnet=10.176.2.0, DSTMASK=255.255.255.0 (11111111.11111111.11111111.00000000) #HOST = 28 -2 = 254
IP NODEBIP = 10.176.2.1-254
IP BROADCAST=10.176.2.255
43. IPV6
IPV6 compared to IPv4 has some advantages:
- Larger address space (contain 128 bits), means 2^128 = 3.4 x 10^38 IP address can be defined
- Multicasting
