*.*.*.*
8bits.8bits.8bits.8bits = 32bits
1octet = 8bits
81.180.231.16
81 = 64+16+1
180 = 128+32+16+4
231= 128+64+32+4+2+1
16=16
decimal bit 128.64.32.16.8.4.2.1
81 01010001
180= 10110100
231= 11100111
16 00010000
binary = 01010001.10110100.11100111.00010000
----------class A network structure------------
0nnnnnn. 11111111.11111111.11111111 (255.255.255)
^network number ^from here u may administrate
clasa A incepe intodeauna in primul bit cu 0 deci raman 7 pentru
identitatea retelei adica 127 posibile din 128.
exemple : 10.0.0.0
44.0.0.0
101.0.0.0
127.0.0.0
---------class B network structure----- ----- -------
10nnnnnn.nnnnnnnn. 11111111.11111111 (255.255)
^network number 2bout ^from here u may admin
^incep de la 128 pana la 191
137.55.0.0
129.33.0.0
190.254.0.0
150.0.0.0
168.30.0.0
128.0.0.0 and 191.255.0.0
-----------class C network structure----- ----- --------
110nnnnn.nnnnnnnn.nnnnnnnn. 11111111
^network number 3b out admin 255
incepe la 192 si se termina in 223
192.153.186.0
199.0.44.0
191.0.0.0
222.222.31.0
192.0.0.0 and 223.255.255.0
-------- ----- ------ ----- ----- -----------
deci avem :
Clasa A putine retele dar multe adrese
Clasa B 16.284 de retele si 65.536 de adrese
Clasa C multe retele si foarte putin adrese adika 256
-------- ----- ------ ----- ----- -----------
Assigning secondary addresses on cisco routers is done using IOS configuration commands.
Here is an
example of how to assign a primary IP address and two secondary IP addresses to an Ethernet interface:
interface ethernet 0
ip address 183.55.2.77 255.255.255.0
ip address 204.238.7.22 255.255.255.0 secondary
ip address 88.127.6.209 255.255.255.0 secondary
The router's Ethernet 0 interface now has addresses in the 183.55.0.0 network,
the 204.238.7.0 network,and the 88.0.0.0 network.
-------- ----- ------ ----- ----- -----------
decimal - 10 digits
binary - 2 digits
hexadecimal - 16 digits
----- ----- ----------------Subnetting
de ex: 1245 adika 1x1000 + 2x100 + 4x10 + 5x1 = decimal
2^7 2^6 2^5 2^4 2^3 2^2 2^1 2^0
128 64 32 16 8 4 2 1
-------- ----- ------ ----
153.88.0.0 = network address / a class B address adika:
10nnnnnn.nnnnnnnn. 11111111.11111111,
primi 16 bitsi sunt network address !
----- ----- --------- ----- --------
sa luam ca exemplu 153.88.240.0 subnet pentru 153.88.0.0 network
deci: administratorul tine 240^ - 8 bits pentru subnetare
rezulta ca o adresa IP pentru 153.88.0.0 network cu 153.88.240.0 subnet
ar fi : 153.88.240.22 >
atunci daca luam clasa A de ip observam ca masca de subnet naturala ar fi
pentru >> 0nnnnnn. 11111111.11111111.11111111
subnet 255 .0 .0 .0
deci masca de subnet exista in primi 8 biti
pentru clasa B ar fi
10nnnnn.nnnnnnnn. 11111111.11111111
255 .255 .0 .0
-------- ----- ------ ---
Ident de retea este 153.88 si ident de host este 4.240 in expresia IP 153.88.4.240
Deci
Fiind o adresa clasa B, inseamna ca primi 16 biti reprezinta network number
Iar ceilalti 16 din totalul de 32bitsi reprezinta host number
Subnetam 153.88.0.0 cu maska 255.255.255.0 si atunci inseamna ca am adugat un field de subnet.
Adika 153.88.0.0 reprezinta inca network number deci ident retea, dar cu o maska de 255.255.255.0 al treilea octet (255.255.255.0) este folosit pentru a indica unde este locat numarul de subnet - atunci numarul de subnet este 4 si in sfarsit numarul de host este 240
255. 255. 255. 0 pentru o retea clasa B IP
11111111.11111111.1111111.00000000
^Netw ^Netw ^Subn ^Host
Cand hotaram de cate subneturi avem nevoie, si stabilim sa zicem 73, transformam valoarea decimala in binar, de aici aflam ca ne ocupa 7 biti din 16 valabili, si ii alocam ca atare, ceea ce ramane.respectiv 9 biti reprezinta spatiul de alocare a adreselor IP (hosts)
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