More exercises and examples

1) Dapatkan subnet mask yang digunakan....
i) Host = 2^5
= 32 - 2
= 30
ii) Network = 1110 0000 - 224
= 3 Network
iii) Senarai IP untuk semua segment
| 1. | 0 | 1 - 30 | 31 |
--------------------------------
| 2. | 32 | 33 - 62 | 63 |
--------------------------------
| 3. | 64 | 65 - 94 | 95 |
--------------------------------
| 4. | 96 | 97 - 122 |123 |
--------------------------------
| 5. | 124 | 125 - 158 |159 |
-----------------------------------
| 6. | 160 | 161 - 190 |191 |
-----------------------------------
| 7. |192 | 193 - 222 | 223 |
-----------------------------------
| 8. | 224 | 225 - 254 | 255 |

Cara mengira bilangan host dan network

1) 10 . 0 . 0 . 0 / 15
Class A = 255 . 0 . 0 . 0
i) Bilangan Host = 255 . 0 . 0 . 0
1111 1111 . 1111 1110 . 0000 0000 . 0000 0000
= 2^17
= 131072 - 2
= 131070
ii) Bilangan Network = 2^7
= 128

2) 128 . 3 . 0 . 0 / 30
30 = 1111 1100
i) Bilangan Host = 2^2 ii) Bilangan Network = 2^6
= 4 - 2 = 64
= 2

3) 192 . 168 . 1 . 0 / 27
27 = 1110 0000
i) Bilangan Host = 2^5 ii) Bilangan Network = 2^3
= 32 - 2 = 8
= 30

4) 11 . 0 . 0 . 0 / 12
Class A = 255 . 0 . 0 . 0
i) Bilangan Host = 2^20 ii) Bilangan Network = 2^4
= 1048576 - 2 = 16
= 1048574

5) 13 . 0 . 0 . 0 / 30
i) Bilangan Host = 2^2 ii) Bilangan Network = 2^6
= 4 - 2 = 64
= 2

6) 130 . 11 . 0 . 0 / 22
Class B = 255 . 255 . 0 . 0
i) Bilangan Host = 2^10 ii) Bilangan Network = 2^6
= 1024 - 2 = 64
= 1022

7) 132 . 9 . 0 . 0 / 18
Class B = 255 . 255 . 0 . 0
i) Bilangan Host = 2^14 ii) Bilangan Network = 2^2
= 16384 - 2 = 4
= 16382

8) 6 . 0 . 0 . 0 / 9
Class A = 255 . 0 . 0 . 0
i) Bilangan Host = 2^23 ii) Bilangan Network = 2^1
= 8388608 - 2 = 2
= 8388606

Valid / Invalid And specify their class

1) 126 . 100 . 1 . 1 = Valid (Class A)

2) 192 . 168 . 254 . 0 = Invalid (Class C)

3) 204 . 204 . 7 . 0 = Invalid (Class C)

4) 201 . 100 . 11 . 5 = Valid (Class C)

5) 192 . 5 . 5 . 5 = Valid (Class C)

6) 205 . 7 . 256 . 11 = Invalid (Class C)

7) 300 . 10 . 100 . 13 = Invalid

8) 0 . 1 . 2 . 3 = Invalid

9) 255 . 223 . 100 . 101 = Invalid (Class E)

10) 253 . 252 . 254 . 255 = Invalid (Class E)

11) 1 . 1 . 1 . 1 = Valid (Class A)

12) 160 . 1 . 1 . 258 = Invalid (Class B)

Contoh Pengiraan Dalam Networking

Soalan-soalan :

1) 192 . 168 . 1 . 12/29
29 = 1111 1000
i) Bilangan Host = 2^3 ii) Bilangan network = 2^5
= 8 - 2 = 32
= 6
iii) Subnet mask = 1111 1000
= 128 + 64 + 32 + 16 + 8
= 248
iv) Network Address = 192 . 168 . 1 . 8
Broadcast Address = 192 . 168 . 1 . 15

2) 192 . 168 . 1 . 250/26
26 = 1100 0000
i) Cari bilangan host = 2^6 ii) Cari bilangan network = 2^2
= 64 - 2 = 4
= 62
iii) Subnet mask = 1100 0000
= 128 + 64
= 192
iv) Network Address = 192 . 168 . 1 . 192
Broadcast Address = 192 . 168 . 1 . 255

3) 192 . 168 . 1 . 87/27
27 = 1110 0000
i) Cari bilangan host = 2^5 ii) Cari bilangan network = 2^3
= 32 - 2 = 8
= 30
iii) Cari subnet mask = 1110 0000
= 128 + 64 + 32
= 224
iv) Network Address = 192 . 168 . 1 . 64
Broadcast Address = 192 . 168 . 1 . 95

Cara mengira guna rumus anding

Rumus Anding

0 -- 0 = 0
1 -- 1 = 1
0 -- 1 = 0
1 -- 0 = 0

Contoh :
1) Diberi nombor-nombor yang berikut :
192 . 168 . 1 . 64
255 . 255 . 255 . 192
Penyelesaiannya : -

Langkah 1 :
192 . 168 . 1 . 64 (Tukar ke dalam bentuk binary spt dibawah)
1100 0000 . 1010 1000 . 0000 0001 . 0100 0000

255 . 255 . 255 . 192 (Tukar ke dalam bentuk binary spt dibawah)
1111 1111 . 1111 1111 . 1111 1111 . 1100 0000

Langkah 2 :
Tambahkan kedua-dua nombor yang sudah ditukar kepada binary tadi..

1100 0000 1010 1000 0000 0001 0100 0000
11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 +
-------------------------------------------------------------------
1100 0000 1010 1000 0000 0001 0100 0000
-------------------------------------------------------------------

Langkah 3 :

Tukar jawapan yang ditambah tadi kepada decimal :
1100 0000 . 1010 1000 . 0000 0001 . 0100 0000 (Jawapan tadi)
192 . 168 . 1 . 64

Jawapannya : 192 . 168 . 1 . 64

Subnetting

Subnetting

1. Subnetting is the process of breaking down an IP network into smaller sub-network called 'subnets'.

2. Network devided into small sub-network is for :
- easy to handle it
- high performance
- low broadcast
- high network
- low data collision

network example : 192.168.1.66/26 = IP
255.255.255.192 = Subnet mask

3. Each subnet is non-physical description ( ID ) for a physical sub-network( usually a switched network of host containing a single router in a multi-router network).

Soalan :

i) Cari bilangan host :
= 192.168.1.66/26 = 1100 0000 ( nilai 26 melebihi 24, jadi 24 + 2 = 26. Yang
menjadikan yang hidup iaitu 1 sebanyak 2, dan yang selebihnya 0 sebanyak 6 )
= 2^6 ( 2 perlu dikuasakan dengan berapa kosong yang ada iaitu hostnya )
= 64 – 2 ( ditolak 2 untuk network & broadcast )
= 62

ii) Cari bilangan network
= 2^2 ( 2 perlu dikuasakan dengan berapa 1 yang ada iaitu networknya )
= 4 network

TCP/IP

TCP/IP

• stands for Transmission Control Protocol / Internet Protocol
• it is the communication protocol for the Internet
• defines how electronic devices (like computer) should be connected to the Intenet, and how data should be transmiited between them
• inside the TCP/IP standart there are several protocols for handling data communication:

- TCP (Transmission Control Protocol) communication between applications
- UDP (User Datagram Protocol) simple communication between application
- IP (Internet Protocol) communication between computers
- ICMP (Internet Protocol Message Protocol) for errors and statistics
- DHCP (Dynamic Host Configuration Protocol) for dynamic addressing