Tài liệu tổng hợp về địa chỉ IP.pdf

7/30/2019 Ti liu tng hp v a ch IP.pdf

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Ti liu tng hp va ch IP

Ti liu v a ch IP

Ngi tng hp: Nng Dng Hng

Lp 9LTCD-IT11

Trng H KD&CN H Ni

Kha hc: 2012 2014

Cu trc IPGii thiu chung:

Nh chng ta bit Internet l mt mng my tnh ton cu , do hng nghn mng my tnh tkhp mi ni ni li to nn. Khc vi cch t chc theo cc cp: ni ht, lin tnh, quc t camt mng vin thng nh mng thoi chng hn, mng Internet tchc ch c mt cp, ccmng my tnh d nh, d to khi ni vo Internet u bnh ng vi nhau. Do cch t chc nhvy nn trn Internet c cu trc a ch, cch nh a ch c bit, trong khi cch nh a ch
http://khanh.com.vn/post/2010/09/17/Cau-truc-dia-chi-IP.aspxhttp://khanh.com.vn/post/2010/09/17/Cau-truc-dia-chi-IP.aspx


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i vi mng vin thng li n gin hn nhiu.i vi mng vin thng nh mng thoi chng hn, khch hng cc vng khc nhau honton c th c cng s in thoi, phn bit vi nhau bng m vng, m tnh hay m quc t. ivi mng Internet , do cch t chc ch c mt cp nn mi mt khch hng hay mt my ch (Host ) hoc Router u c mt a ch internet duy nht m khng c php trng vi bt k ai.

Do vy m a ch trn Internet thc s l mt ti nguyn.Hng chc triu my ch trn hng trm nghn mng. a ch khng ctrng nhau cn phic cu trc a ch c bit qun l thng nht v mt T chc ca Internet gi l Trung tmthng tin mng Internet -Network Information Center ( NIC ) ch tr phn phi, NIC ch phna ch mng ( Net ID ) cn a ch my ch trn mng ( Host ID ) do cc T chc qun lInternet ca tng quc gia mt t phn phi. (Trong thc t c th nh tuyn (routing ) trnmng Internet ngoi a ch IP cn cn n tn ring ca cc my ch (Host) - Domain Name ).Cc phn tip theo chng ta hy nghin cu cu trc c bit ca a ch Internet.

a/ Thnh phn v hnh dng ca a ch IP:a ch IP ang c s dng hin ti (IPv4) c 32 bit chia thnh 4 Octet ( mi Octet c 8 bit,

tng ng 1 byte ) cch m u t tri qua phi bt 1 cho n bt 32, cc Octet tch bit nhaubng du chm (.), bao gm c 3 thnh phn chnh.

Bit 1............................................................................ 32

Bit nhn dng lp ( Class bit ) a ch ca mng ( Net ID ) a ch ca my ch ( Host ID )

Ghi ch: Tn l a ch my ch nhng thc t khng ch c my ch m tt c cc my con(Workstation), cc cng truy nhp v.v..u cn c a ch.

Bit nhn dng lp (Class bit) phn bit a ch lp no.1/ - a ch Internet biu hin dng bit nh phn:

x y x y x y x y. x y x y x y x y. x y x y x y x y. x y x y x y x yx, y = 0 hoc 1.

V d:

0 0 1 0 1 1 0 0. 0 1 1 1 1 0 1 1. 0 1 1 0 1 1 1 0. 1 1 1 0 0 0 0 0

bit nh phn Octet 1 Octet 2 Octet 3 Octet 4

2/ - a ch Internet biu hin dng thp phn: xxx.xxx.xxx.xxxx l s thp phn t 0 n 9V d: 146. 123. 110. 224


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Dng vit y ca a ch IP l 3 con s trong tng Octet. V d: a ch IP thng thy trnthc t c th l 53.143.10.2 nhng dng y l 053.143.010.002.

b / Cc lp a ch IPa ch IP chia ra 5 lp A,B,C, D, E. Hin ti dng ht lp A,B vgn ht lp C, cn lp D v

E T chc internet ang dnh cho mc ch khc khng phn, nn chng ta ch nghin cu 3lp u.

Qua cu trc cc lp a ch IP chng ta c nhn xt sau:

Bit nhn dng l nhng bit u tin - ca lp A l 0, ca lp B l 10, ca lp C l 110. Lp D c 4 bit u tin nhn dng l 1110, cn lp E c 5 bt u tin nhn dng l 11110. a ch lp A: a ch mng t v a ch my ch trn tng mng nhiu. a ch lp B: a ch mng va phi v a ch my ch trn tng mng va phi. a ch lp C: a ch mng nhiu, a ch my ch trn tng mng t.


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Nh vy nu chng ta thy 1 a ch IP c 4 nhm s cch nhau bng du chm, nu thy nhms th nht nh hn 126 bit a ch ny lp A, nm trong khong 128 n 191 bit a ch ny lp B v t 192 n 223 bit a ch ny lp C.

Ghi nh:a ch thc t khng phn trong trng hptt c cc bit trong mt hay nhiuOctet s dng cho a ch mng hay a ch my ch u bng 0 hay u bng 1. iu ny ngcho tt c cc lp a ch.

I - a ch lp A:

Tng qut chung:

Bit th nht l bit nhn dng lp A = 0. 7 bit cn li trong Octet th nht dnh cho a ch mng. 3 Octet cn li c 24 bit dnh cho a ch ca my Ch.

- net id: 126 mng- host id:16.777.214 my ch trn mt mng

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a/ a ch mng (Net ID)1/ Kh nng phn a ch

Khi m s bit chng ta m t tri qua phi, nhng khi tnh gi tr thp phn 2n ca bit li tnht phi qua tri, bt u t bit 0. Octet th nht dnh cho a ch mng, bit 7 = 0 l bit nhn dnglp A. 7 bit cn li t bit 0 n bit 6 dnh cho a ch mng ( 2 7 ) = 128. Nhng trn thc t a

ch khi tt c cc bit bng 0 hoc bng 1 u khng phn cho mng. Khi gi tr cc bit u bng0, gi tr thp phn 0 l khng c ngha, cn a ch l 127 khi cc bit u bng 1 dng thngbo ni b, nn trn thc t cn li 126 mng.

Cch tnh a ch mng lp A.S th t Bit (n)- tnh t phi qua tri: 6 5 4 3 2 1 0Gi tr nh phn (0 hay 1) ca Bit: x x x x x x xGi tr thp phn tng ng khi gi tr bit = 1 s l 2 nGi tr thp phn tng ng khi gi tr bit = 0 khng tnh.Gi tr thp phn ln nht khi gi tr ca 7 bit u bng 1 l 127.Xin xem bng tnh trn vn gi tr ca tt c cc Bit

Nh vy kh nng phn a ch ca lp A cho 126 mng

2/ Biu hiu a ch trn thc t: T 001 n 126

B / a ch ca cc my ch trn mt mng1/ Kh nng phn a chBa Octet sau gm 24 bit c tnh t bit 0 n bit 23 dnh cho a ch my ch trn tng mng.

Vi cch tnh nh trn, c tng s my ch trn mt mng ta c.


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a ch khi cc bit u bng 0 hay bng 1 b ra. Trn thc t cn li 224-2 = 16 777 214Nh vy kh nng phn a ch cho 16 777 214 my ch.

2/ Biu hin a ch trn thc t

Octet 2 Octet 3 Octet 4


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Tng qut li ti a ch ca mt mng, khi ln lt thay i cc gi tr ca cc Octet 2, 3, 4.ta sc 16 777 216 kh nng thay i m cc con s khng trng lp nhau ( Combinations ) c nghi l 16 777 216 a ch ca my ch trn mng, nhng thc t phn ch l

(256 x 256 x 256) - 2 =16 777 214Biu hin trn thc t l ba s thp phn trong 3 Octet cch nhau du.T 000. 000. 0001 n 255. 255. 254

Kt lun:a ch lp A c th phn cho 126 mng v mi mt mng c 16 777 214 my ch.Ni cch khc a ch thc t s t 001.000.000.001 n 126.255.255.254

V d: Mt a ch y ca lp A: 124. 234. 200. 254. Trong :a ch mng: 124a ch my ch: 234.200.254

II - a ch lp B:

Tng qut chung:

2 bit u tin nhn dng lp B l 1 v 0. 14 bit cn li trong 2 Octet u tin dnh cho a ch mng. 2 Octet cn li gm 16 bit dnh cho a ch my Ch.


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- net id: 16.382 mng-host id: 65.534 my ch trn mt mng

a/ a ch mng1/ Kh nng phn a ch

Octet 1 Octet 2

Hai Octet u tin c 16 bit phn cho a ch mng, 2 bit ( bit 1 v bit 2 ) k t tri sang cgi tr ln lt l 1 v 0 dng nhn dng a chlp B. Nh vy cn li 14 bit cho Net ID -a ch mng.

Theo cch tnh nh ca a ch mng Lp A ta c


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Tng t nh a ch Lp A, cc bit u bng 0 v cc bit u bng 1 c b ra, nn thc t

gi tr thp phnch t 1 n 16 382 c ngha phn c cho 16 382 mng.

2/ Biu hin trn thc t.Biu hin a ch trn thc t th hin s thp phn trong 2 Octet cch nhau bng du chm (. ).Cch tnh s thp phn cho tng Octet mt.

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a ch mng ca Lp A t 001 n 126. ( khng phn 127 ). Nh vy a ch mng ca Lp B Octet th nht s t 128 cho n 191.Nh vy gi tr thp phn ca Octet 1 t 128 n 191.

Octet 2

Nh vy gi tr thp phn ca Octet 2 t 001 n 254. Nh vy: a ch mng lp B biu hin trn thc t gm 2 Octet t 128.001 cho n 191. 254 cngha phn c cho 16 382 mng ( 214 - 2 ).

b / a ch cc my ch trn mt mng1 / Kh nng phn a ch

Octet 3 v 4 gm 16 bit dnh cho a ch ca cc my ch trn tng mng.

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a ch ca cc bit bng 0 v bng 1 b ra, Kh nng thc t cn li 65534 a ch ( 216 - 2)phn cho cc my ch trn mt mng.

2/ Biu hin a ch trn thc t

Octet 3

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Biu hin a ch my ch trn thc t ca Lp B l t 000. 001 n 255. 254Kt lun: a ch Lp B c th phn cho 16 382 mng v mi mng c n 65 534 my ch. Nicch khc a ch phn trong thc t s t 128. 001. 000. 001 n 191. 254. 255. 254

V d: Mt a ch y ca lp B l 130.130.130.130. Trong :a ch mng: 130.130a ch my ch: 130.130

III - a ch Lp C:

Tng qut chung:

3 bit u tin nhn dng lp C l 1,1,0. 21 bit cn li trong 3 Octet u dnh cho a ch mng. Octet cui cng c 8 bit dnh cho a ch my ch.

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- net id: 2.097.150 mng- host id: 254 mych/1 mng

a / a ch mng1/ Kh nng phn a ch21 bit cn li ca 3 Octet u dnh cho a ch mng

Cc bit u bng 0 hay bng 1 khng phn, nn kh nng phn a ch cho mng lp C l 2 097150 hoc bng 221 - 2.

2/ Biu hin trn thc t

Octet 1

1


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Octet 2

1


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Octet 3

Kt lun:a ch dnh cho mng ca lp C c kh nng phn cho 2097150 mng, ni cch khctrn thc t s t 192. 000. 001 n 223. 255. 254

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b / a ch my ch trn tng mng1/ Kh nng phn a chOctet 4 c 8 bit phn a ch cho cc my ch trn mt mng.

Octet 4

Nh vy kh nng cho mych trn tng mng ca a ch lp C l 254 hay 28 - 2.

2/ Biu hin trn thc t: T 001 n 254.

Kt lun: a ch lp C c th phn cho 2 097 150 mng v mi mt mng c 254 my ch. Nicch khc s t 192. 000. 001. 001 n 223. 255. 254.254

V d mt a ch Internet lp C y : 198. 010. 122. 230. Trong : a ch mng: 198.010.122a ch my ch: 230

V d: Trung tm thng tin mng Internet vng Chu - Thi bnh dng ( APNIC ) phn choVDC 8 a ch ca lp C c th phn cho 8 mng t 203.162.0.0 cho n 203.162.7.0. Nhm sth nht l 203 cho bit y l nhng khi a ch lp C.

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a ch y ca mt khi a ch 203.162.0.0 phi l 203.162.000.000, chng ta c sdng trn vn octet cui cng c ngha l c 254 a ch my ch v u cui trn mt mng.V d mng 203.162.0 s c a ch u cui t 203.162.0.000 n 203.162.0. 255. Nh vy tngcng VDC c 8x254=2032 a ch l thuyt phn cho cc my ch v u cui trn 8 mng

203.162.0 ; 203.162.1;.....203.162.7 v.v..

Nh vy a ch mng l c nh, chng ta ch c quyn phn a ch cho my ch trnmng .

IV - a ch mng con ca Internet (IP subnetting):

a/ Nguyn nhnNh nu trn a ch trn Internet thc s l mt ti nguyn, mt mng khi gia nhp Internetc Trung tm thng tin mng Internet ( NIC) phn cho mt s a ch va dng vi yu culc , sau ny nu mng pht trin thm li phi xin NIC thm, l iu khng thun tin cho

cc nh khai thc mng.Hn na cc lp a ch ca Internet khng phi hon ton ph hp vi yu cu thc t, a chlp B chng hn, mi mt a ch mng c th cp cho 65534 my ch, Thc t c mng nh chc vi chc my ch th s lng ph rt nhiu a ch cn li m khng ai dng c . khcphc vn ny v tn dng ti a a ch c NIC phn, bt u t nm 1985 ngi ta nghn a ch mng con.Nh vy phn a ch mng con l m rng a ch cho nhiu mng trn c s mt a ch mngm NIC phn cho, ph hp vi s lng thct my ch c trn tng mng.

b/ Phng php phn chia a ch mng conTrc khi nghin cu phn ny chng ta cn phi hiu qua mt s khi nim lin quan ti vic

phn a ch cc mng con.1/ - Default Mask: (Gi tr trn a ch mng) c nh nghatrc cho tng lp a ch A,B,C.Thc cht l gi tr thp phn cao nht (khi tt c 8 bit u bng 1) trong cc Octet dnh cho ach mng - Net ID.

Default Mask:

Lp A 255.0.0.0

Lp B 255.255.0.0

Lp C 255.255.255.0

2/ - Subnet Mask:( gi tr trn ca tng mng con)

Subnet Mask l kt hp ca Default Mask vi gi tr thp phn cao nht ca cc bit ly t ccOctet ca a ch my ch sang phn a ch mng to a ch mng con.

Subnet Mask bao gi cng i km vi a ch mngtiu chun cho ngi c bit a ch

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mng tiu chun ny dng c cho 254 my ch hay chia ra thnh cc mng con. Mt khc n cngip Router trong vic nh tuyn cuc gi.

Nguyn tc chung:

Ly bt mt s bit ca phn a ch my ch to a chmng con.Ly i bao nhiu bit ph thuc vo s mng con cn thit (Subnet mask) m nh khai thc mngquyt nh s to ra.V a ch lp A v B u ht, hn na hin ti mng Internet ca Tng cng ty do VDC qunl ang c phn 8 a ch mng lpC nn chng ta s nghin cu k phn chia a ch mngcon lp C.

a/ a ch mng con ca a ch lp CClass c:

a ch lp C c 3 octet cho a ch mng v 1 octet cui cho a ch my ch v vy ch c 8 bitl thuyt to mng con, thc t nu dng 1 bit m mng con v 7 bit cho a ch my chth vn ch l mt mng v ngc li 7 bit cho mng v 1 bit cho a ch my ch th mtmng ch c mt my, nh vy khng logic, t nht phi dng 2 bit m rng a ch v 2 bit

cho a ch my ch trn tng mng. Do vy trn thc t ch dng nh bng sau. Default Mask ca lp C : 255.255.255.0

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Nh vy mt a ch mng lp C ch c 5 trng hp la chn trn (Hay 5 Subnet Mask khcnhau), tu tng trng hp c th quyt nh s mng con.1/ Trng hp 1 - Hai mng con

Subnet Mask 255.255.255.192.

T mt a ch tiu chun to c a ch cho hai mng con, mi mt mng c 62 my ch.S dng hai bit (bit 7 v 6) ca phn a ch my ch to mng con. Nh vy cn li 6 bit

phn cho my ch.

a/ Tnh a ch mng

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a ch ca mng l gi tr ca bit 7 v 6 ln lt bng 0 v 1.Trong trng hp chia a chmng con khng bao gi c dng a ch khi cc bit u bng 0 hay bng 1. Do vy trnghp 2 mng con ni trn, a ch mng con s l:

Mng con 1: a ch mng xxx.xxx.xxx.64Mng con 2: a ch mng xxx.xxx.xxx.128

b/ Tnh a ch cho my ch cho mng con 1Chng ta ch cn 6 bit cho a ch my ch trn tng mng.

Mi mng cn li 62 a ch cho my ch.Mng 1: T xxx.xxx.xxx. 065 n xxx.xxx.xxx.126

c/ Tnh a ch cho my ch cho mng con 2Tng t nh cch tnh trn ta c

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Mng 2: a ch my ch trn mng 2.T xxx.xxx.xxx.129 n xxx.xxx.xxx.190.Tng qut li:

a/ Mng con th nht* / a ch mng con: xxx.xxx.xxx.064* / a ch cc my ch trn mng con ny t.

xxx.xxx.xxx. 065

xxx.xxx.xxx. 066

xxx.xxx.xxx. 067

..............

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n xxx.xxx.xxx. 126

b/ Mng con th 2

*/ a ch mng con: xxx.xxx.xxx. 128

*/ a ch cc my ch trn mng con ny t.

xxx.xxx.xxx. 129

xxx.xxx.xxx. 130

.............

n xxx.xxx.xxx. 190

a ch my ch t 1 n 62 v t 193 n 254 v 127 ; 191 b mt, ngha l mt 130 a ch. V d: a ch tiu chun lp C l 196. 200. 123Subnetmask 255.255.255.192

T a ch ny ta c 2 mng con l:

* Mng 1: a ch mng 196.200.123.064

a ch My ch trn mng ny.T 196.200.123.065 n 196. 200. 123. 126.

* Mng 2: a ch mng 196.200.123.128a ch my ch trn mng ny.T 196.200.123.129 n 196.200.123. 190

2/ Trng hp 2 - Su mng conSubnetmask: 255.255.255.224.

To c 6 mng con, mi mng con c 30 my cha/ Tnh a ch Mng con

Trng hp ny s dng 3 bit ( bit 7,6,5) ca a ch my ch (Octet 4) b sung cho a ch mngtiu chun to mng con.

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B trng hp cc bit u bng 0 hay 1,chng ta cn li a ch ca 6 mng con sau.

xxx.xxx.xxx.32 ; Mng con 1






b / Tnh a ch my ch cho mng con 1

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Nh vy a ch my ch ca mng con 1 s t 33 n 62.Tng t nh cch tnh nu trn chng ta c thtnh c cho tt c cc trng hp cn li(xem bng 1) v c tng hp li nh sau.1/ Trng hp 1: Subnetmask 255.255.255.192

2 mng con. 62 my ch mi mng.

2/ Trng hp 2: Subnetmask 255.255.255.224



14 mng con. 14 my ch mi mng



5/ Trng hp 5: Subnetmask 255.255.255.252.

62 mng con.

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2 my ch mi mng.

Xem bng tnh a ch cho cc trng hp trn

V d: a ch mng lp C m NIC phn cho VDC l 203.162.4.0. Trn a ch ny phn ra 2mng con th a ch s l.Mng 1: a ch mng 203.162.4.64.a ch my ch trn mng t 203.162.4.65 n 203.162.4.126 Mng 2: a ch mng 203.162.4.128.a ch my ch trn mng t 203.162.4.129 n 203.162.4.190

b/ a ch mng con t a ch lp B

Default Mask ca lp B l 255.255.0.0

Class B:

Net ID - Khi phn a ch mng con s dng Octet 3a ch lp B c 2 Octet th 3 v th 4 dnh cho a ch my ch nn v nguyn l c th lyc c 16 bit to a ch mng . Nu t mt a ch mng c NIC phn chng ta nh mrng ln 254 mng v mi mng s c 254 my ch. Trng hp ny s ly ht 8 bit ca octetth 3 b sung vo a ch mng v ch cn li 8 bit thc t cho a ch my ch, theo cch tnh sthp phn 2n gi tr ca 8 bit nh nu phn lp C, chng ta s c:

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Bng phn chia a ch mng con lp B

a ch lp B v l thuyt c 2 octet u cho a ch mng, khi chia mng con theo phng phps dng tt c 8 bit trong 3 octet cho a ch mng, trn thc tng ng vi lp C, nh vy v ach NIC phn l lp B nhng cch t chc a ch li lp C ( Xem Bng ph lc phn a chmng con lp B ).Trong bng ny cn ch ct 6 - khong cch a ch gia 2 mng con gii thiu cho chng tacch tnh a ch cc mng con, a ch cc my ch trn tng mng lin quan ti ct 7,8,9,10.

V d: Trng hp Subnetmask 255.255.240.0 l r nht.Chia c 14 mng con, mi mng con c 4094 my ch, khong cch a ch gia hai mngcon l 16.0 c ngha.

Mng con 1 c a ch l xxx.yyy.16.0 ; Mng con 2 sc a ch l xxx.yyy.16.0 + 16.0 =xxx.yyy.32.0 c tip tc nh vy ta stnh c a ch ca tng mng con v mng con 14 lxxx.yyy. 224.0.

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a ch my chu tin trn mng con 1 l xxx.yyy.16.1 ; a ch my chu tin trn mngcon 2 s l xxx.yyy.16.1 + 16.0 = xxx.yyy.32.1. Tip tc nh vy ta stnh a chc my chu tin ca mng con 14 l xxx.yyy.224.1 v.v..

Tng t chng ta bit c a ch cui cng ca cc my ch trn mt mng con.

Theo hng dn ny chng ta s tm c cc trng hp khc.Tm li chia a ch mng con cng phi theo mt quy lut nht nh ngoi mun ca chng ta,khi chia mng con cng b mt kh nhiu a ch, mt t hay nhiu tu thuc vo cc trng hpc th.

Bng phn chia a ch mng con lp B:

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Subnet mask v cch chia

biCu trc IPta bit s qua v khi nim v cu trc ca IPv4. Bi ny mnh s trnh byr v cch chia SubnetmaskSubnet l g?: Hiu n gin vy. Khi ta chia mt Network ra thnh nhiu Network nh hn th

cc Network nh ny c gi l Subnet.

V sao cn phi chia Subnet mask?

Nh ta bit mng Internet s dng a ch IPv4 32 bit v phn chia ra cc lp A,B,C,D , tuynhin, vi mt h thng a ch nh vy vic qun lvn rt kh khn . Nu nh mt mng c

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http://khanh.com.vn/post/2010/10/20/Subnet-mask-va-cach-chia.aspxhttp://khanh.com.vn/post/2010/09/17/Cau-truc-dia-chi-IP.aspxhttp://khanh.com.vn/post/2010/09/17/Cau-truc-dia-chi-IP.aspxhttp://khanh.com.vn/post/2010/09/17/Cau-truc-dia-chi-IP.aspxhttp://khanh.com.vn/post/2010/09/17/Cau-truc-dia-chi-IP.aspxhttp://khanh.com.vn/post/2010/10/20/Subnet-mask-va-cach-chia.aspx


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cp mt a ch lp A th c ngha n c th cha ti 16*1.048.576 a ch ( my tnh ) .Vi slng my tnh ln nh vy rt t cng ty hoc t chc dng ht c iu gy lng ph ach IP. trnh tnh trng cc nh nghin cu a ra mt phng php l s dng mt nmng con ( Subnet mask ) phn chia mng ra thnh nhng mng con gi l Subnet. Subnetmask l mt con s 32 bit bao gm n bit 1 ( thng l cc bit cao nht ) dng nh a ch

mng con v m bit 0 dng nh a ch my trong mng con vi n+m=32 . Subnet mask phi c cu hnh cho mi my tnh trong mng v phi c nh ngha cho migiao din Router. Nh vy, ta phi dng cng mt Subnet mask cho ton b mng vt l cngchung mt a ch Internet. Trong thc t, d dng cho hot ng qun l cc my trongmng, thng chia nh cc mng ln trong cc lp mng (A, B, C) thnh cc mng nh hn.Qu trnh ny c thc hin bng cch ly mt s bit phn nh danh host s dng chovic nh a ch mng. Tu theo cch s dng ca ngi qun tr mng ( s subnet v s hosttrn mi subnet ) m s lng bit ly phn host nhiu hay t. tch bit gia a ch mng v a ch host ngi ta dng netmask. tch bit gia Subnetaddress v a ch host ngi ta dng Subnet mask.

Theo quy c, cc a ch IP c chia ra lm ba lp nh sau:Class Subnet mask trong dng nh phn Subnet maskLp A 11111111 00000000 00000000 00000000 255.0.0.0Lp B 1111111111111111 00000000 00000000 255.255.0.0Lp C 111111111111111111111111 00000000 255.255.255.0

Nh ta bit, lp A s dng 1 octet u tin lm Network ID. S dng 8 bit u c set gitr thnh 1, v 24 bit sau set gi tr 0 => c Subnet Mask 255.0.0.0. Tng t vi cc lp kia.V d IP: 192.168.1.0/24y l a ch thuc lp C. V con s 24 c ngha l ta s dng 24 bit cho phn Network ID, vcn li 8 bit cho Host ID.

Chia Subnet Mask nh th no? y, mnh s trnh by cch ngn gn gip bn cth tnh nhm c. Ly v d c th nhsau:

Cng ty thu mt ng IP l 192.168.1.0. By gi ng gim c yu cu phn lm chia lm 3mng con cho ba phng ban trong cng ty. Hy thc hin vic chia subnet ny.

Trc ht ta phn tch cu trc ca a ch: 192.168.1.0 nh sau: + a ch NetMask: 255.255.255.0+ Network ID: 11111111.11111111.11111111

+ HostID: 00000000

Trong v d ny ta cn chia lm 3 mng con (3 subnet) nn ta cn s dng 2 bit phn Host ID thm vo Network ID. Lm sao bit c s bit cn mn thm? Ta c cng thc :2^n>=m (vi m l s subnet cn chia, n l s bit cn mn). y 2^2>=3.Sau khi mn 2 bit, ta c cu trc mi dng nh phn l (bit mn ta set gi tr bng 1 nh):+ a ch NetMask:: 11111111.11111111.11111111.11000000+ Network ID: 11111111.11111111.11111111.11

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+ Host ID: 000000=> dng thp phn l: 255.255.255.192

a ch IP mi lc ny l: 192.168.1.0/26 (con s 26 l 24 + 2 bits mn).Ta xc nh "bc nhy" cho cc subnet:

Bc nhy k=256-192=64=> Ta c cc mng con sau:Ip: 192.168.1.0 Netmask: 255.255.255.192

Ip: 192.168.1.64 Netmask: 255.255.255.192Ip: 192.168.1.128 Netmask: 255.255.255.192Ip: 192.168.1.192 Netmask: 255.255.255.192

Nh vy s my trn mi mng bng bao nhiu?S bits ca Host ID cn li sau khi b Network ID mn: x = 32-26 = 6=> S my trn mi mng: 2^n-2 = 2^6-2 = 62 my

Cc vn lin quan ti vic chia mng concc bn c th hi p / chia s ti dinn vi link sauhttp://kythuatmang.vn/forums/6-MANG-CAN-BAN.html

Cch chia Subnet nhanh

Do trc mnh c vit biSubnet mask v cch chiac trnh by v cch chia Subnet cn

bn. Cch tnh v nh phng ph chia Subnet c nhiu cch, bi ny trnh by cchtnh rt nhanh, tham kho nh . VIA t blog ca Anh L Cng.

Khi tnh ton IP, chia subnet th chng ta thng p dng cng thc tnh ton.

Cng thc tnh l 2^n v 2^h - 2 ( 2^m -2 ),

tnh tng s subnet c c sau khi chia ta dng cng thc 2^n, trong n l s bitmn chia subnet trong octet (mn lm network id).

tnh tng s host/subnet ta dng cng thc 2^h-2, trong h l tng s bit cn lidng lm host sau khi mn . Ta phi tr 2 v cn b a ch subnet id v broadcast.

Ni s s qua cch tnh truyn thng nh vy thi, gi chng ta tm hiu cch nhm nhanh bngcch m lng tay nh!

PHNG PHP CHIA SUBNET BNG CCH M LNG NGN TAY

u tin cc bn xe bn tay tri ra v m theo hnh:

3
http://kythuatmang.vn/forums/6-MANG-CAN-BAN.htmlhttp://kythuatmang.vn/forums/6-MANG-CAN-BAN.htmlhttp://kythuatmang.vn/forums/6-MANG-CAN-BAN.htmlhttp://www.khanh.com.vn/post/2011/01/07/Cach-chia-Subnet-nhanh.aspxhttp://khanh.com.vn/post/2010/10/20/Subnet-mask-va-cach-chia.aspxhttp://khanh.com.vn/post/2010/10/20/Subnet-mask-va-cach-chia.aspxhttp://khanh.com.vn/post/2010/10/20/Subnet-mask-va-cach-chia.aspxhttp://khanh.com.vn/post/2010/10/20/Subnet-mask-va-cach-chia.aspxhttp://www.khanh.com.vn/post/2011/01/07/Cach-chia-Subnet-nhanh.aspxhttp://kythuatmang.vn/forums/6-MANG-CAN-BAN.html


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Cc bn bn tay chng ta c tt c 14 lng tay, mi lng tay tng trng cho 1 bit nh! ^^

m 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384.

m i m li cho thuc i nh cc bn.

- tnh tng s lng Subnet idc c sau khi chia, ta m s bit mn lm subnet id trongoctet l ra. Mn 3 bit th m 2 4 8, mn 4 bit th m 2 4 8 16, gi tr ca bit m sau cngchnh l tng s subnet id sau khi c chia ra.

V d: 10.10.0.0 /13 ---> mn 5 bit ---> m 2 4 8 16 32. Vy mng ny c 32 subnet.

- tnh bc nhy trong mi subnet id. Ta m s bit cn li dng lm host trong ringoctet . Gi tr ca bit c m sau cng cng l gi tr ca bc nhy trong octet .

V d: 172.35.0.0/19. Tc l a ch lp B s mn 3 bit octet th 3 lm subnet id. Dngphng php m ta c 2 4 8, 3 bit mn ri, vy tng s subnet id l 8. Ta bit trong octetth 3 sau khi cho mn 3 bit lm net id th cn li 5 bit lm host, vy ta m 2 4 8 16 32, 5 bitri, gi tr l 32, v cng chnh l bc nhy ca subnet id., th xem no:

-172.35.0.0/19

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-172.35.32.0/19

-172.35.64.0/19

-172.35.96.0/19

-172.35.128.0/19

-172.35.160.0/19

-172.35.192.0/19

-172.35.224.0/19

Ta c tng cng 8 subnet id, vi bc nhy l 32.

- tnh a ch broadcast ca mt subnet idta ly subnet id k tip gim 1. V d, tnhbroadcast ca subnet id 172.35.64.0/19, ta ly subnet id k tip l 172.35.96.0/19 gim 1 ==172.35.95.255/19 y chnh l broadcast ca subnet id 172.35.64.0

- tnh s host trong mt subnet, ta m ton b s bit host cn li trong subnet v ly gi trbit sau cng -2, Lu l khng phn bit octet. Nhc li, ta ly gi tr ca bit c m saucng - 2 ta c s host trong subnet id c th xi.

Trong v d subnet 172.35.64.0/19, ta nhn bit ton b s bit dng lm host cn li l 13. Tam 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192, 13 bit ri, ok, s host trong mng sl 8192 -2 = 8190. V sao -2, v ta phi tr b a ch subnet id v broadcast. Hay n gin hn

c th nhn thy l s host c th xi c trong dy:

172.35.64.1/19 ----> 172.35.95.254/19

V ng thi n cng lt gia 2 subnet id v broadcast.

Lu : Phng php m t 2 khng c dng tnh tng s gi tr ca 1 octet chy t 0->255. Hay ni cch khc l khng c dng tnh tng gi tr ca 1 dy bit nh 10101101. tnh tng s gi tr ca dy trn ta phi m t 1, cng cc gi tr c bit 1 vi nhau.

Ngoi ra, yu cu cc bn cn nh v thuc:

1xxxxxxx =128

11xxxxxx =192

111xxxxx =224

1111xxxx =240

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11111xxx =248

111111xx =252

1111111x =254

11111111 =255

v

2^0 = 1

2^1 = 2

2^2 = 4

2^3 = 8

2^4 = 16

2^5 = 32

2^6 = 64

2^7 = 128

2^8 = 256

Cc bn cng c th dng bn tay phi ghi nh cc gi tr trn , dng nhm nhanh subnetmask ca mng.

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Mn 1 bit : 128

Mn 2 bit : 192

Mn 3 bit : 224

Mn 4 bit : 240

Mn 5 bit : 248

Mn 6 bit : 252

Mn 7 bit : 254

Mn 8 bit : 255

V d : 10.10.0.0 /13 --mn 5 bit ---> S/M: 255.248.0.0

155.55.3.32 /28 -- mn 12 bit = 8 +4 ----> S/M: 255.255.255.240

Nu cc bn nhun nhuyn cch tnh ny, ti tin rng cc bn s tnh ton, chia subnet rtnhanh!!

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Subnet Masks

Khi ta chia mt Network ra thnh nhiu Network nh hn, cc Network nh ny c gi lSubnet. Theo quy c, cc a ch IP c chia ra lm ba Class (lp) nh sau:

Address Class Subnet mask trong dng nh phn Subnet maskClass A 11111111 00000000 00000000 00000000 255.0.0.0

Class B 11111111 11111111 00000000 00000000 255.255.0.0Class C 11111111 11111111 11111111 00000000 255.255.255.0

Subnet Mask ca Class A bng 255.0.0.0 c ngha rng ta dng 8 bits, tnh t tri qua phi (ccbits c set thnh 1), ca a ch IP phn bit cc NetworkID ca Class A. Trong khi , ccbits cn st li (trong trng hp Class A l 24 bits uc reset thnh 0) c dng biu dincomputers, gi l HostID. Nu bn cha quen cch dng s nh phn hy c qua bi H thngs nh phn.

Subnetting

Hy xt n mt a ch IP class B, 139.12.0.0, vi subnet mask l 255.255.0.0 (c th vit l:139.12.0.0/16, y s 16 c ngha l 16 bits c dng cho NetworkID). Mt Network vi ach th ny c th cha 65,534 nodes hay computers (65,534 = ( 2^16) 2 ) . y l mt con squ ln, trn mng s c y broadcast traffic.

Gi t chng ta chia ci Network ny ra lm bn Subnet. Cng vic s bao gm ba bc: 1) Xc nh ci Subnet mask2) Lit k ID ca cc Subnet mi3) Cho bit IP address range ca cc HostID trong mi Subnet

Bc 1: Xc nh ci Subnet mask m cho n 4 trong h thng nh phn (cho 4 Subnet) ta cn 2 bits. Cng thc tng qut l: Y = 2^Xm Y = con s Subnets (= 4)X = s bits cn thm (= 2)

Do ci Subnet mask s cn 16 (bits trc y) +2 (bits mi) = 18 bitsa ch IP mi s l 139.12.0.0/18 ( con s 18 thay v 16 nh trc y). Con s hosts ti ac trong mi Subnet s l: ((2^14)2) = 16,382. V tng s cc hosts trong 4 Subnets l: 16382

* 4 = 65,528 hosts.

Bc 2: Lit k ID ca cc Subnet miTrong a ch IP mi (139.12.0.0/18) con s 18 ni n vic ta dng 18 bits, m t bn tri, ca32 bit IP address biu din a ch IP ca mt Subnet. Subnet mask trong dng nh phn Subnet mask11111111 11111111 11000000 00000000 255.255.192.0

Nh th NetworkID ca bn Subnets mi c l:

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Subnet Subnet ID trong dng nh phn Subnet ID1 10001011.00001100.00000000.00000000 139.12.0.0/182 10001011.00001100.01000000.00000000 139.12.64.0/18

3 10001011.00001100.10000000.00000000 139.12.128.0/18

4 10001011.00001100.11000000.00000000 139.12.192.0/18

Bc 3: Cho bit IP address range ca cc HostID trong mi Subnet V Subnet ID dng ht 18 bits nn s bits cn li (32-18= 14) c dng cho HostID.

Nh ci lut dngcho Host ID l tt c mi bits khng th u l 0 hay 1.Subnet HostID IP address trong dng nh phn HostID IP address Range 1 10001011.00001100.00000000.0000000110001011.00001100.00111111.11111110 139.12.0.1/18

-139.12.63.254/182 10001011.00001100.01000000.00000001

10001011.00001100.01111111.11111110 139.12.64.1/18-139.12.127.254/18

3 10001011.00001100.10000000.0000000110001011.00001100.10111111.11111110 139.12.128.1/18

-139.12.191.254/184 10001011.00001100.11000000.00000001

10001011.00001100.11111111.11111110 139.12.192.0/18139.12.255.254

Bn c thy trong mi Subnet, ci range ca HostID t con s nh nht (mu xanh) n cons ln nht u y ht nhau khng?

By gi ta th t cho mnh mt bi tp vi cu hi:Bn c th dng Class B IP address cho mt mng gm 4000 computers c khng? Cu tr lil C. Ch cn lm mt bi ton nh.

Gi t ci IP address l 192.168.1.1. Thay v bt u vi Subnet mask, trc ht chng ta tnhxem mnh cn bao nhiu bits cho 4000 hosts.Con s hosts tac th c trong mt network c tnh bng cng thc:Y = (2^X 2)

Nh ci lut dng cho Host ID l tt c mi bits khng th u l 0 hay 1.4094 = (2^12 2)

X = 12 , ta cn 12 bits cho HostIDs, do Subnet mask s chim 20 (=32-12) bits.Qu trnh tnh ton ni trn ny mang tn l Variable Length Subnet Mask(VLSM). SupernettingGi t ta mng ca ta c 3 Subnets:Accounting: gm 200 hostsFinance : gm 400 hostsMarketing: gm 200 hostsBn ha mng vi Internet v c Internet Service Provider (ISP) cho 4 Class C IP addresses

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nh sau:192.250.9.0/24

192.250.10.0/24192.250.11.0/24

192.250.12.0/24

Bn c 3 segments v bn mun mi segment cha mt Network.

By gi bn lm sao?a ch IP trong Class C vi default subnet mask 24 cho ta con s Hosts tia trong miNetwork l [(2^X) 2] = (2^8) 2 = 254. Nh th segments Accounting v Marketing khng btr ngi no c.Nhng ta thy Segment Finance cn thm 1 bit mi . Ta lm nh sau:Bc 1: Lit k Network IP addresses trong dng nh phn 192.250.9.0/24 11000000 11111010 00001001 00000000 (1)

192.250.10.0/24 11000000 11111010 00001010 00000000 (2)192.250.11.0/24 11000000 11111010 00001011 00000000 (3)

192.250.12.0/24 11000000 11111010 00001100 00000000 (4)Bc 2: Nhn din network prefix notation23 bits u (t tri qua phi) ca 2 network IP address (2) and (3) u ging nhau. Nu chng ta thu Subnet mask t 24 xung 23 cho (2) v (3) ta s c mt Subnet c th cung cp508 hosts.IP address ca mi segment tr thnh:Accounting: gm 200 hosts: 192.250.9.0/24Finance: gm 400 hosts: 192.250.10.0/23Marketing: gm 200 hosts: 192.250.12.0/24By gi IP address 192.250.11.0 tr thnh mt HostID tm thng trong Subnet192.250.10.0/23.

Qu trnh ta lm va qua bng cch bt s bits trong Subnet mask khi gom hai hay bn (v.v..)subnets li vi nhau tng con s HostID ti a trong mt Subnet, c gi lSUPERNETTING.

Supernetting uc dng trong router b xung CIDR (Classless Interdomain Routing v VLSM(Variable Length Subnet Mask).

V lun lun nh rng trong internetwork, NETWORK ID phi l a ch c o (unique).

Phng php chia mng con (Subnet)theo VLSMQua qu trnh ging dy cc sinh vin, c bit mt s bn vn cn bng vi cchchia a ch mng con theo VLSM, phng php ny s gip chng ta kim sot c smng mi sinh ra, s mng dng, s mng d tha cn li, sau y ti s hng dncc bn thc hin vic ny mt cch d dng bng v d minh ha. Trc ht, chng ta

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phi hiu r cu trc ca a ch IP v4 v ngha ca mt s khi nim: v d cc lp ach IP v4, Net_id, host_id, Subnet Mask, gii a ch kh dng, a ch mng, chia thnh tho, chng ta cn nm r mt s khi nim v cng thc sau y:- Bit, byte.

- Khi nim v s nh phn, thp phn.

- Php ton AND- Cc bin i t nh phn sang thp phn, t thp phn sang nh phn.- Cu trc a ch IP, gii hn ca cc lp IP- Khi nim v default mask, mask, subnet, subneting ....!- Cc a ch ringLu :- a ch mng (subnet) : tt c cc bit dnh cho phn host bng 0- a ch broadcast: tt c cc bit dnh cho phn host bng 1.- a ch u tin hp l: l a ch lin sau a ch mng (subnet)- a ch cui cng hp l: l a ch lin trc a ch broadcast=> phi hiu r v phn bit khi nim n v m l g p dng cng thc cho ng

Cng thc:

+ S subnet c to ra: 2^m (m: s bit mn ca phn Host ID) (Ch : ng l cng thc ny phi l 2^m 2 v phi loi tr i 2 mng u tin subnet zero vmng cui cng subnet broadcast, nhng vi cc dng Router hin nay ca Cisco h tr lnh Router(config)# ip subnet-zero do ta vn c th s dng 2 mng m khng phi loi tr b i)+ S host / subnet: 2^n 2 (n: s bit cn li ca phn Host ID sau khi b mn m bit)+ Subnet Mask mi = Subnet Mask c + m (l s bit va b mn) + a ch kh dng l cc a ch IP c th gn cho mi host, thit b (Lu : c nhiu cch hoc th thut tnh a ch mng con , nhng cch no cngphi da vo nn tng gc l s thay i cc bit mn sinh ra mng con mi, do tt nht chng ta nn tham kho theo phng php VLSM)

V D MU 1:Cho gii a ch 172.35.0.0/16, hy Subnet cp cho cc mng con :A: 320 hostB: 115 host

C: 80 hostD: 30 host

E: 2 hostF: 2 host

G: 2 host

theo phng php VLSM?

Hng dn gii mu:-Theo u bi cho a ch ban u l X: 172.35.0.0/16=> i ra h nh phn ta c:10101100.00100011.00000000.0000000011111111.11111111.00000000.00000000

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(Phn gch chn chnh l phn bit host, vic chia t a ch trn thnh nhiu Subnet chnhl vic bin ihay gi l mn cc bit phn host_id chuyn thnh cc bit Net_id; Nhnvo s bit 1 ca a ch Subnet Mask ta s phn bit c danh gii: cc bit bn trn bit 1chnh l Net_id, cc bit bn trn bit 0 l host_id)

- B1: Theo VLSMth ta s phi chia X cho cc mng theo chiu gim dn, tc l chia

cho mng c s host cao nht ri thp nht cui cng-> sp xp li ta c:+A: 320+B:115

+C:80+D:30

+E:2+F:2

+G:2- B2: +Thc hin chia Xcho mng A u tin, p dng cng thc: 2^n - 2 320 => n=9(chnh l s bit cn li cha b mn) => s bit mn l m= 32 (l tng s bit ca 1a ch IP v4)16 (s bit thuc phn Net_id ca a ch cho)9 ( s bit cn li) = 7

=> SM (Subnet Mask mi) = SM (Subnet Mask c) + m = 16 + 7 = 23 ( vit tt l /23)& s Subnet ( mng con ) c to ra l: 2^m = 27 = 128vi SM thay it /16 thnh /23 (cc bit trong khong ny ca X chuyn sang Octetth 3) nn ta c101100.00100011.00000000.00000000172.16. 0. 0

(bit mu en khng in m & b gch chn chnh l 7 bit va mn, vic sinh ra ccSubnet con chnh l da vo vic thay i v tr v gi tr t 0 thnh 1 ca nhng bit ny)Vy cc mng con c sinh ra t X l:Mng X1:10101100.00100011.00000000.00000000 -> 172.35.0.0/23Mng X2:10101100.00100011.00000010.00000000 -> 172.35.2.0/23Mng X3:10101100.00100011.00000100.00000000 -> 172.35.4.0/23 vn vn .Mng X127: 10101100.00100011.11111100.00000000 -> 172.35.252.0/23Mng X128: 10101100.00100011.11111110.00000000 -> 172.35.254.0/23(ch , ta thy ch cn tnh n mng th 3 tr i l ta c th tm c bc nhygia 2 mng lin k l 2: ly octet tng ng ca mng sau tr octet mng trc)=> ly mng con u tin X1: 172.35.0.0/23 cp cho mng A: 320 host+ Tip theo,ly mng X2 (l a ch mng ln nht tip theo) chia cho mng B:115 host.Tng t trn, theo cng thc: 2^n - 2 115 => n=7 => m = 32-23-7 = 2=> SM (Subnet Mask mi) = SM (Subnet Mask c) + m = 23 +2 = 25 ( vit tt l /25)& s Subnet ( mng con ) c to ra l: 2m = 22 = 4vi SM thay i t /23 thnh /25 (cc bit trong khong ny ca X2 lin quan n c Octet3 v Octet th 4) nn ta c cc mng con mi sinh ra t X2:Mng X21:10101100.00100011.00000010.00000000 -> 172.35.2.0/25Mng X22:10101100.00100011.00000010.10000000 -> 172.35.2.128/25Mng X23:10101100.00100011.00000011.00000000 -> 172.35.3.0/25Mng X24:10101100.00100011.00000011.10000000 -> 172.35.3.128/25=>ly mng X21: 172.35.2.0/25 cp cho mng B: 115 host+ Tip theo, ta s dng mng con X22: 172.35.2.128/25 chia cho mng C: 80 hos t

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Tng t trn, theo cng thc: 2^n - 2 80 => n=7 => m = 32-25-7 = 0 (Va p, khi gitr m=0 iu ny chng t l mng ang chia ch c th cp va hoc tha mt s IPcho mng c s host ang yu cu, y l 115 host- ch : khi s dng VLSM th m skhng bao gi nhn gi tr m)=> cp lun X22 cho mng C: 80 host

+ Lc nycn phi dng n gii a ch X23 chia cho mng D: 30 hostTng t trn, theo cng thc: 2^n - 2 30 => n=5 => m = 32-25-5 = 2=> SM (Subnet Mask mi) = SM (Subnet Mask c) + m = 25 +2 = 27 ( vit tt l /27)& s Subnet ( mng con ) c to ra l: 2m = 22 = 4vi SM thay i t /25 thnh /27 (cc bit trong khong ny ca X23 lin quan n Octetth 4) nn ta c cc mng con mi sinh ra t X23:

Mng X231:10101100.00100011.00000011.00000000 -> 172.35.3.0/27Mng X232:10101100.00100011.00000011.00100000 -> 172.35.3.32/27Mng X233:10101100.00100011.00000011.01000000 -> 172.35.3.64/27Mng X234:10101100.00100011.00000011.01100000 -> 172.35.3.96/27

=> ly X231: 172.35.3.0/27cp cho mng D: 30 host+ Ly X232 chia cho cc mng E: 2 host, F: 2 host, G: 2 hostTng t trn, theo cng thc: 2^n - 2 2 => n=2 => m = 32-27-2 = 3=> SM (Subnet Mask mi) = SM (Subnet Mask c) + m = 27 +3 = 30 ( vit tt l /30)& s Subnet ( mng con ) c to ra l: 2^m = 23 = 8vi SM thay i t /27 thnh /30 (cc bit trong khong ny ca X232 lin quan n Octetth 4) nn ta c cc mng con mi sinh ra t X232 l:Mng X2321:10101100.00100011.00000011.00100000 -> 172.35.3.32/30MngX2322:10101100.00100011.00000011.00100100 -> 172.35.3.36/30Mng X2323:10101100.00100011.00000011.00101000 -> 172.35.3.40/30 vn vn .Mng X2327:10101100.00100011.00000011.00111000 -> 172.35.3.56/30Mng X2328:10101100.00100011.00000011.00111100 -> 172.35.3.60/30=> ly mng Mng X2321: 172.35.3.32/30 cp cho mng E: 2 host=> ly mng Mng X2322: 172.35.3.36/30 cp cho mng F: 2 host=> ly mng Mng X2323: 172.35.3.40/30 cp cho mng G: 2 hostKT LUN- Sau khi cp cc a ch mng con cho cc mng A, B, C, D, E, F, G s cn d ccmng cha c s dng ( ginh khi cn ta c th s dng cp pht hoc chianh tip). Phng php VLSM ny s gip ta kim sot c phn a ch d thacha c s dng.(C th vo kim tra xem kt qu c trng, ng hay khng)Xong ri !!!

Bi tp chia mng con

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C nhiu bn s mn Mng my tnh, nht l chia mng con.

V thmnh s gip cc bn lm sao lm tt bi tp chia mng con, cng nh l cch chia mng con

nhanh nht cc bn khng cn phi s n na.

I/ Mt cht thng tin cn bit:

A a ch IP

Hin nay hu ht chng ta ang xi a chIPv4 (IP version 4) v n sp ht.

Cu trc ca mt a ch IPv4 c 32 bit (4bytes) chia lm 4 octec.

N s c dng: xxxxxxxx.yyyyyyyy.zzzzzzzz.tttttttt (vi x y z t l bit 0 hoc 1).

ddng hn trong vic hiu, ng ta thng c di dng s thp phn

VD 192.168.4.2 thay v 11000000.10101000.00000100.00000010.

a ch IP gm c 2 phn. Phn Net ID v Host ID.

Chng ta chquan tm n a ch lp A,B,C

a ch mc nh lp A c octec u tin lm NET ID, 3 octet cn li lm Host ID

a ch mc nh lp B c 2 octec u tin lm NET ID, 2 octet cn li lm Host ID

a ch mc nh lp C c 3 octec u tin lm NET ID, 1 octet cn li lm Host ID

Cch nhn bit a chno thuc Lp A, lp B, hay lp C, ta da vo Octet u tin

Gi sa ch IP c dng x.y.z.t

Nu x t1 n 126 -> a ch lp A

Nu x t128 n 191 -> da ch lp B

Nu x t 192 -> 224 -> a ch lp C

Ring 127.0.0.1 l a chloop back. a chny chnh l a chi din chi my chng ta (Vo menu Run

g ping 127.0.0.1 th bit lin)

B- Subnet Mask.

C nhiu bn vn kh hnh dung c subnet mask l ci g.

Theo mnh th Subnet Mask (mt n mng) l mt chui 32bit tng tnh a chIP, c iu n gip

cho my tnh bit, phn no l Net ID, phn no l Host ID bng cch bt cc bit Phn Net ID ln 1, v cc

bit phn Host ID l 0.

V d ta c mt a ch mng l 10.0.0.0

Khi ta c th bit c mng ny thuc lp A (Octec u tin < 126).

Subnet Mask ca mng s l 255.0.0.0 (tc l 11111111.00000000.00000000.0000000)

Khi vo su hn bi tp mnh sni k hn v Subnet mask.

Tm li cc bn c hiu Subnet Mask nh mt ci mt n, ch no cc bn che i (Bt bit 1 ln) th ch

l Net ID, ch no cc bn khng che (Bt bit 0 ln) th ch l Host ID

II/ Bi tp

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1. Trong a ch IP, c 5 lp A, B, C, D, E. Lp B l lp c dy a ch:

a. 192.0.0.0 ti 223.255.255.255 b. 240.0.0.0 ti 255.255.255.255

c. 128.0.0.0 ti 191.255.255.255 d. 224.0.0.0 ti 239.255.255.255

Lp B l lp m octet u tin nm khong t128 n 191.

Vy dy a ch ca lp ny sl p n C

2. Byte u tin ca mt a ch IP c dng: 11011011. Vy n thuc lp no:

a. Lp A b. Lp B c. Lp C d. Lp D

cc bn c thi nh phn sang thp phn bit n l lp my.

Ring mnh th mnh da vo cc bit Octet u tin

0 l lp A, 10 l lp B, 110 l lp C, 1110 l lp D.

3 Subnet Mask no sau y l hp l:

a. 0.255.255.255 b. 0.0.0.255 c. 255.0.0.255 d. 255.255.255.0

Subnet mask th cc bit 1 lun ng trc cc bit 0 => Cu ng l cu D

NAT v PAT

Network Address Translation: NAT l k thut thay i cc a ch mng (Network Address)trong mt gi tin (packet) gy nh hng trong qu trnh nh hng i ca packet cho mtmc ch c th. a ch mng y mun ni n a ch IP WAN (Internet Protocol) layer 3,

ngoi ra cn c th thay i s port layer 4 theo m hnh phn lp OSI. Bn cnh a chmng cn c phn bit a ch ngun (source) v a ch ch (destination). Ty theo mc chdng NAT m ta thay i mt s hoc tt c cc loi a ch trn, trn cng mt packet.

y l loi hnh NAT ph bin nht, ta cn hay gi l Des NAT, hay NAT Inbound. NAT theokiu ny ch thay i a ch IP ch ca gi tin m khng ng n cc thnh phn khc. ycng l kiu c s dng mc nh trong cc ROUTER ( thit b nh tuyn ni chung ) theocch sau:

Khi client gi packet request i ti ROUTER, i s c dest IP l Virtual IP ca ROUTER(141.149.65.3), source IP l IP ca client (188.1.1.10). Do ROUTER i din cho tt c nhng

server thc ng sau, nn IP ca ROUTER cng l a ch i din, client s ch lin h viROUTER m khng bit c a ch tht ca cc server l g ( trc khi bit VIP ca router nphi nh DNS server dch t tn min sang a ch IP ). a ch IP i din ca ROUTER cn la ch IP o (Virtual IP - VIP) ROUTER s gi repquest i n server x l nn n thc hin thay dest IP trong i thnh IPca server (10.10.10.20), source IP vn gi nguyn (188.1.1.10). Thao tc ny gi l DestinationNAT ( dch chuyn IP ch ). Nh c dest IP l IP ca server nn request i s c nh tuyn tip n server x l.

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Khi server tr li, packet reply s i qua li ROUTER. Ti y packet c un-NAT, ngha lthay li a ch IP ca server (lc ny tr thnh source IP) bng VIP ca ROUTER.Reverse NAT

Trong c ch ca ROUTER, cc server tht ch c cp a ch IP private (v nhiu l do, trong

c s hn ch v s lng IP public v tnh bo mt, ngoi tr mt s trng hp c bit choserver kt ni thng IP public), ngha l khng th i ra Internet c. ROUTER c publish raInternet nhn request cho cc server bn trong, nn n thng c mt, hoc nhiu a ch IPpublic cc client kt ni t bn ngoi vo, v a ch ny cng l VIP. Khi client mun sdng dch v ca server bn trong, ROUTER thc hin vic chuyn i a ch dest IP t publicsang private n c server, l kiu NAT ta xt trn. Trong trng hp ngc li, khiserver bn trong mun khi to kt ni vi bn ngoi Internet, ROUTER phi thc hin vicchuyn a ch source IP ca packet i ra t private sang public c th lu thng trn Internet.Do cch hot ng nh vy nn kiu NAT ny cn c gi l SourceNAT, hay NAT outbound.

Port-Address Translation (PAT)

Hai trng hp trn ta xem xt hai kiu thay i a ch IP, trng hp th ba ny cn cgi l NAT port, ngha l s port TCP/UDP trong packet s b thay i ( y ta khng xtn cc protocol khc cng c dng port). PAT cng l mt phn tt yu trong cc chc nng

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chnh ca ROUTER. Cch hot ng ca n cng rt n gin: Khi ta lin kt (bind) port 80 caVIP trn ROUTER n port 1000 trn server tht, load balancer s thc hin vic chuyn i vy ton b yu cu port 80 (dest port) n port 1000 trn server.

Nhng li ch khi ta thc hin PAT:

- Bo mt l ch li u tin m ta thy ngay. Bng cch khng m nhng cng mc nh trnserver, ta c th gy kh khn hn cho vic tn cng c . Chng hn, ta c th chy mt Webserver trn port 4000, v lin kt port 80 ca VIP trn ROUTER n port 4000 ca server tht.Lc by gi, k tn cng khng th khai thc trc tip ln port 80 ca server tht c, v nkhng c m.- Kh nng co dn (scalability) PAT cho php ta chy cng mt ng dng trn nhiu port. Tytheo cch thit k ng dng, c th vic chy nhiu bn sao ca n s lm tng hiu sut phc vln. Chng hn ta c th chy my ch web IIS trn cc port 80, 81, 82 ca mi server tht. Sau ch cn lin kt port80 ca VIP vi mi port chy IIS ca server tht. Load balancer s phnb lu thng khng ch cho cc server m cn gia cc port trn tng server.

- Kh nng qun tr (manageability) chng hn khi host nhiu website trn mt b cc servertht, ta c th ch cn dng mt VIP i din cho tt c cc domain ca cc website. Lc nyROUTER s nhn tt c cc request n port 80 cng mt VIP. Web server ca ta c th chymi domain trn mt port khc nhau, chng hn www.abc.com trn port 81, www.xyz.com trnport 82. ROUTER c th gi lu thng n port ph hp da trn domain trong URL ca miHTTP request.

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Full NAT

Nh vy ta xt k thut NAT thay i ln lt a ch ch, a ch ngun, ri thay i portkhi nh tuyn. Mi s thay i l mt loi hnh NAT c ng dng trong tng trng hp ring.

Kt hp cc kiu thay i ny li, ta c mt kiu NAT khc phc tp hn l Full NAT. KiuNAT ny c tn gi nh vy v n bao gm cc thay i sau y trn gi tin request: a ch IP ngun (source IP) a ch IP ch (dest IP) Port ngun (source port)Lu source port y l port ca client, cn dest port l port c request trn server, chnghn port 80 v d trn.

Kiu NAT ny khc vi nhng kiu NAT trn ch, packet t server reply c th b quaROUTER m i thng n client ngoi Internet. Vn ch a ch IP ca server vn lprivate IP, do ng nhin packet reply mang a ch source l IP ni b s chng bao gi n

c client.

Vy th bng cch no buc server phi tr li thng qua ROUTER c NAT a ch IP ira Internet? Cch n gin nht l ta c th khai bo cho ROUTER l default gateway ca ccserver. Nhng cch ny yu cu ROUTER phi cng subnet vi cc server (cng Layer 2broadcast domain). Nu khng th nm cng subnet th sao? y l ch Full NAT khc bit

FULL NAT c s dng.

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Khi c thit lp thc hin Full NAT, ROUTER s thay source IP ca tt c cc requestpacket bng mt a ch c khai botrn ROUTER, xem nh l source IP, ri thay dest IP (lcny ang l VIP) thnh IP ca server (10.10.10.20), trc khi gi n server . Source IP ny cth ging hoc khc VIP, ty vo tng sn phm ROUTER. Nh vy tng t nh proxy, lc

ny server thts xem ROUTER nh l client yu cu mnh, v khng quan tm n client thcs na. V th server s reply li cho ROUTER v ROUTER s i li dest IP thnh IP ca clientthc s (188.1.1.100) gi i.

Nh vy source port c thay i ch no? Mi ln thay i mt source IP ca client thnhsource IP ca ROUTER, gi l mt session, th ROUTER thc hin lu li nhng thng tin caclient trong session bng cch i source port trong cng packet (lc ny ang l source portca client).

Source port lc ny c ngha nh l mt session ID khng hn khng km. Khi server reply vcho ROUTER, source port cng c gi tr v theo packet reply. Da vo source port ny,

ROUTER xc nh c session ca client trong bng lu thay li source IP, source port caclient nh c.

u im ca kiu NAT ny l cho php bn thc hin vic thay i a ch thng qua ROUTERtrn mi topology mng. Nhc im l khng ly c cc thng tin v IP, port t pha client.Nhng ng dng nh Web c s dng thng tin t source IP ca client th khng nn dng mhnh ny. a s cc sn phm ROUTER u cung cp chc nng log v report source IP ca ccrequest.

Enhanced NAT

Nhng k thut NAT va trnh by trn u xoay quanh vic thay i a ch IP, cng nh porttrong packet header. Tuy nhin c nhng protocol c bit cha thng tin a ch hay port nhngtrong packet payload, cng cn phi c thay i cng vi packet header.

iu ny i hi ROUTER phi hiu bit theo tng protocol c th. Khi nim enhanced NATni n kiu NAT phc tp c thc hin vi nhng hiu bit theo tng protocol c th lmcho nhng protocol hot ng c vi vic nh tuyn gi tin.

Trong s cc protocol c bit , thng dng nht l cc protocol streaming media (v d RTSP- Real Time Streaming Protocol). y cng l cc protocol s dng cn bng ti ph bin nht,v chng cc k ngn ti nguyn mng v tnh ton khi phi phc v ng thi cho hng trmn hng ngn ngi s dng.

Cc protocol streaming thng gm c hai kt ni, mt kt ni iu khin xy dng trn TCP vmt kt ni d liu da trn UDP. khi u, client khi to mt knh iu khin n mtwell-known port trn server. Client v server s tho thun cc iu khon cho knh iu khin.S tho thun gm c IP ca server v s port ca server m client s gi d liu n trn kt nid liu.

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Nu cc server c a ch IP private, ROUTER s thc hin Destination NAT cho kt ni iukhin. Nhng ng thi ROUTER cng phi xem cc thng tin tho thun v thay i mi thngtin v a ch IP hay port m server v client trao i sao cho client s gi d liu n VIP publicch khng phi IP private ca server (nhng thng tin ny nm trong payload ca packet).

Hn na, dest port c chn trong qu trnh tho thun li khng bit trc c nn phi x lrequest ngay c khi port cha c lin kt n bt k server no.

Tuy nhin, nhiu doanh nghip li c nhng chnh sch bo mt trn tng la lm cho nhngkt ni d liu trn nn tng UDP c th khng thnh cng. Do nhiu h thng streamingmedia cho php stream trn nn HTTP, ngha l ton b dng d liu s c gi i bng kt nic thit lp bi giao tip HTTP. iu ny lm cho vic NAT tr nn nh nhng hn.

Direct Server Return (DSR)

Xt li 2 kiu NAT trn, ta xem hai cch x l cng vi cc u nhc im ca chng l gn

default gateway hay dng k thut Full NAT p cc reply ca server khng phi i quaROUTER.

Nhng trng hp ta mun server tr li trc tip cho client m khng thng qua ROUTER thsao? y khng phi l cu hi v l, v trong trng hp nng lc x l ca ROUTER b giihn, th vic tch dng lu thng reply i trc tip m khng qua ROUTER s gip ROUTERtp trung vo x l cc lu thng request hiu qu hn, trnh tc nghn, nng cao hiu sut.

Nhng vi a ch IP private, cc server lm sao tr li trc tip qua Internet n client c?Ch vi mt cht ph php xoay quanh cc a ch IP l vn s c gii quyt: khi c thitlp thc hin DSR, ROUTER khng chuyn dest IP thnh IP ca server m vn gi nguyn lVIP (public IP). ROUTER ch i dest MAC thnh MAC ca server packet c th n cserver. Nh vy gii hn ca DSR l ROUTER v cc server phi nm cng subnet.

Vn cn li l lm sao khi server nhn request packet t ROUTER chuyn n s khng tchi, v dest IP khng phi l IP ca server, m l VIP! Mt cch n gin, ta cu hnh VIP la ch ca loopback interface trn mi server. ROUTER dng cch nyv li dng nhng tnhcht th v sau y ca loopback interface:

C th gn bt k a ch IP no, khng bt buc phi bt u bng 127. V loopback interface khng phi l mt thit b tht, n khng c a ch MAC, nn h thngs khng tr licho cc request ARP. Do s khng c h thng bn ngoi no bit c ach IP ca loop back interface. Tuy nhin, h thng vn nhn request n IP ca loop backinterface v tr li nh l cc interface khc.Ta c th thit lp mt a ch IP public cho loop back interface trn Linux nh sau:ifconfig lo 141.149.65.3 netmask 255.255.255.0 up.

Nh vy, bng mt cht th thut vi cc a ch, server tht khng cn a ch IP public vn cth nhn request v tr li trc tip cho client.

DSR rt hu ch cho nhng ng dng tn bng thng nh FTP, streaming media, khi m kch

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thc packet reply l rt ln so vi kch thc packet request. K thut ny cng c ng dngcho nhng protocol i hi phc tp khi thc hin NAT hay khng c ROUTER h tr.

Chng hn nhng giao thc streaming media nh trong phn Enhanced NAT c cp, th ta cth dng DSR thay v NAT. Ngi ta cng cn nhc DSR khi trin khai trn m hnh mng, khi

m lu thng reply t server khng m bo l s i ng ch.Ktlun

NAT trong ROUTER rt a dng, ty thuc vo tng tnh hung m ta p dng. Cng i su, tacng thy ci hay trong vic x l thng tin ca cc packet, nhng i tng lun l giiquyt cc vn vt l th v nh th no. Nm vng cc k thut NAT gip ta c th trin khaicc h thng nh tuyn, cn bng ti mt cch an ton v hiu qu nht

Ngun:Knh gii php VN
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