P.1 Trong cng ty c 3 phng ban: Sales, Manager, Accouting. Hin ti ang s dng ng mng 192.168.1.0/24. Policy ca cng ty qui nh: nhn vin phng ban no ch c truy xut ti nguyn ca phng ban . gii quyt c policy ny chng ta cn thc hin chia subnet cho network 192.168.1.0/24 Cu trc ca a ch 192.168.1.0/255.255.255.0 Phn tch a ch net mash 255.255.255.0 Network ID: 11111111.11111111.11111111 HostID: 00000000

S subnet trong network s bng s 2^n vi n l s bit ta mn thm phn host ID Ta cn phi mn 2 bit phn host ID v h thng cn 3 subnet Vi 2^2=4 >3 nn ta phi s dng 2 bit Mt n mng mi cho subnet h nhi phn s l tt c cc bit 1 Network ID: 11111111.11111111.11111111.11 HostID: 000000 H thp phn s l 255.255.255.192 i vi s bit mn, khi mn n bit th mt n mng mi s theo qui c sau N -> Subnet ,mask 1 -> 128 2 -> 192 3 -> 224 4 -> 240 5 -> 248 6 -> 252

7 -> 254 8 ->255 Ta cn tm bc nhy cho cc subnet Bc nhy k= 256-192=64 Cc subnet s l: 192.168.1.0/26 192.168.1.0 255.255.255.192 192.168.1.64/26 192.168.1.64 255.255.255.192 192.168.1.128/26 192.168.1.128 255.255.255.192 192.168.1.192/26 192.168.1.192 255.255.255.192 Gi n l s bit cn li ca phn host ID n = 32-network ID=32-26=6 S host ca mi subnet s bng 2n-2 =26-2= 62 Phn tch c th tng subnet 192.168.1.0/26 o Network Address: 192.168.1.0 o Subnet mask: 255.255.255.192 o Broadcast: 192.168.1.63 o Host 192.168.1.1 192.168.1.62 192.168.1.64/26 o Network Address: 192.168.1.64 o Subnet mask: 255.255.255.192 o Broadcast: 192.168.1.127 o Host: 192.168.1.65 192.168.1.126 192.168.1.128/26 o Network Address: 192.168.1.128 o Subnet mask: 255.255.255.192 o Broadcast: 192.168.1.191 o Host: 192.168.1.129 192.168.1.190 192.168.1.192/26 o Network Address: 192.168.1.192 o Subnet mask: 255.255.255.192

o Broadcast: 192.168.1.255 o Host: 192.168.1.193 192.168.1.254

P.2 - Ba trc gh qua mt s din n thy mt s bn thc mc v cch chia subnetmask nn hm nay mnh vit bi ny gip cc bn mi cn thc mc v vn ny. cc bn d hiu hn mnh s i lm cc bi tp v chia subnet t cc bn t hiu nha - Cng thc tnh: + x: s bt mn lm net + y: s bt cn li lm host => S lng net c c sau khi chia: 2^x - 2 S lng host c trong 1 net: 2^y - 2 BC NHY: b= 2^(8 - x) Bc nhy ny chng ta thun tin lm cc tnh ton v sau. Lit k: + Net th i = i * b (Ci ny gii thch hi lng nhng cht na cc bn xem vd s hiu) + host min: Net +1 + host max: Net (k) - 2 + Broadcast: Net (k) -1 Nhn vo my ci cng thc ny chc cc bn cm thy lng nhng kh hiu qu ng khng? Sau y mnh s i gii quyt 3 bi tp tiu biu chia subnet 3 class A B C (xem xong l hiu 100%) VD: Cho 1, 10.100.100.10/ 12 2, 172.17.100.100/ 19 3, 192.168.100.10/ 27

Yu cu: Tnh s subnet, host c c. Tm cc net, host min, host max v broadcast ca net. Gii: 1, y l ip thuc lp A (net.host.host.host) ngha l c 8 bit lm net v 24 bit lm host 10.100.100.10 / 12 10.100.100.10 / 255.240.0.0 vy mn 4 bit ca host lm net. Vy nn: + x = 4 => y = 24 - 4 = 20 + S subnet: 2^4 -2 = 14 + S host trong 1 net : 2^20 - 2 = .... Bc nhy: b = 2 ^ (8-x) = 16 + Lit k (ci ny quan trng n) - Net 1: 10.16.0.0 (net i = i *b - mnh ch quan tm ti Octet 2 v mn bit Octet ny) - Net 2: 10.32.0.0 => host min: 10.16.0.1 (Net +1) host max: 10.31.255.254 (Net (k) - 2) Broadcast: 10.31.255.255 (Net (k) -1) (Cc bn i chiu vi cng thc nh! Cc net cn li lm tng t) 2, 172.17.100.100 / 19 - Phn tch tng t cu 1 => x =3, y = 13 + S subnet: 2^3 -2 = 6 + S host trong 1 net : 2^13 - 2 = .... Bc nhy: b = 2 ^ (8-x) = 32 + Lit k - Net 1: 172.17.32.0 (net i = i *b - mnh ch quan tm ti Octet 3 v mn bit Octet ny) - Net 2: 172.17.64.0 => host min: 172.17.32.1 host max: 172.17.63.254 Broadcast: 172.17.63.255 3, Cu ny cc bn t lm cho hiu nh!

Cc bn c th dng nettool kim tra kt qu ng hay sai. link nettool 5 http://hotfile.com/dl/118291361/400e....0.70.zip.html CHC CC BN HC TT - bn no thy hay thanks cho mnh ly tinh thn vi nha!

P.3 Chia subnet l mt trong nhng gii php hu dng xy dng mng ni b, va bo mt, ngn chn broadcast, va tit kim ti nguyn trong vic phn pht a ch IP cho tng my trm. 1. Quy hoch a ch IP Bn cn xy dng mt mng ni b (mng LAN) cho mt vn phng, mt cng ty quy m va v nh vi khong 7 n 8 phng, mi phng 30 my. Vn cn t ra l lm th no cho hiu qu, va bo mt c h thng mng, phng trnh cc ri ro c th xy ra, m cn c th tit kim c ti nguyn mng.

Mt m hnh mng LAN. Mt trong nhng phng php xy dng hiu qu l quy hoch a ch IP bng cch chia subnet. Vi phng n ny, bn s ch cung cp va s a ch IP cho cc my tnh s dng, va chia ra thnh nhiu mng con, phng trnh c hin tng broadcast v nu xy ra s c th ch b trong cc b mt nhnh mng con. Mt a ch IP gm 4 octet, mi octet l 1 byte cha 8 bit, tng cng l 32 bit. a ch IP s c nhn din lp A, B hay C thng qua a ch subnet mask. Vi bi ton nh trn, chng ta s s dng lp mng C. Bn cn 30 my, ta thy 25 = 32, tr i

2 a ch u v cui l a ch network v a ch broadcast ca nhnh mng, bn s cn li 30 a ch. Nh vy, 1 octec c 8 bit, ta s ly 28 - 23 = 25, ngha l bn s mn thm 3 bit chia subnet cho mng ca mnh. Cch chia theo mt th thut nh sau: Lp C c subnet mask 255.255.255.0 hay cn c vit l /24. Mt octet c 8 bit, bn mn i 3 bit th s bit c mn s bt ln gi tr l 1, s bit cn li s vn nm gi tr l 0. 128 64 32 16 8 4 2 1 1 1 1 0 0 0 0 0 Bn ly 3 bit c gi tr 1 cng li: 128 + 64 + 32 = 224. Tng t, nu bn mn 4 bit th subnet mask s l 240. T , cho d nh, dn c trn mng t ra mt bng tra cu. 1 128 -128 2 192 -64 3 224 -32 4 240 -16 5 248 +8 6 252 +4 7 254 +2 8 255 +1 Bn ch cn nh gi tr bit th 4 cn mn s c gi tr subnet mask tng ng l 240, t bn cng, tr theo nh bng trn l tm ra c a ch subnet mask ca cc bit khc. Tr li bi ton trn, s dng lp mng C v mn thm 3 bit (24 + 3 = 27), bn s c subnet mask 255.255.255.224, v mi nhnh mng con s c chia ra tng ng nh sau: 0. 192.168.1.0/27 1: 192.168.1.32/27 2: 192.168.1.64/27 3: 192.168.1.96/27 4: 192.168.1.128/27 5: 192.168.1.160/27 6: 192.168.1.192/27 7: 192.168.1.224/27 8: 192.168.1.254/27 y, bn lu s th t th 8, gi tr 224 + 32 = 256, nhng do lp C ch c 254 a ch IP, a ch 255 l a ch broadcast, do vy network y l 192.168.1.254, v phng s 8 bn c th t a ch IP t 192.168.1.225/27 192.168.1.253/27. Tng t, bn t cho phng s 1 di a ch

IP: 192.168.1.1/27 192.168.1.30/27, trong c th s dng a ch 192.168.1.1 lm a ch default getway cho phng s 1 ny. 2. ng dng xy dng h thng mng Bn hon thnh vic chia subnet, by gi s ng dng vo trong mng ni b. Lc ny cc phng u l mt nhnh mng con, hon ton tch bit. Bn khng th ngi phng s 1 truyn d liu, truy cp hoc dng giao thc ICMP nh lnh ping n 1 my khc phng s 2. cc phng c th kt ni ra internet, bn cn phi c mt my tnh lm chc nng router. My router ny s gip cc my trong tng mng cc b kt ni n modem ADSL v truy cp ra ngoi internet. Nu bn s dng thit b ca Cisco nh Router 2800 th vic chia subnet trong router c gi l k thut Inter Vlan. Ty theo chnh sch bo mt ca cng ty hay c quan m bn c th xy dng thm tng la (firewall), access list v c ch Nat inside hay outside cc my trm truy cp ra ngoi internet, bn ngoi internet remote vo trong mng ni b.

PHNG PHP CHIA SUBNET BNG CCH M LNG NGN TAY 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 bit mn 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 li dng 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 bng cch 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:

m theo s mu en nh! 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 id c c sau khi chia, ta m s bit mn lm subnet id trong octet 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 cng chnh 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 ring octet . 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. Dng phng php m ta c 2 4 8, 3 bit mn ri, vy tng s subnet id l 8. Ta bit trong octet th 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 bit ri, gi tr l 32, v cng chnh l bc nhy ca subnet id., th xem no: -172.35.0.0/19 -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 id ta ly subnet id k tip gim 1. V d, tnh broadcast 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 tr bit sau cng -2, Lu l khng phn bit octet. Nhc li, ta ly gi tr ca bit c m sau cng - 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. Ta m 2 4 8 16 32 64 128 256 512

1024 2048 4096 8192, 13 bit ri, ok, s host trong mng s l 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 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 subnet mask ca mng.

Mn Mn Mn Mn Mn Mn Mn Mn

1 2 3 4 5 6 7 8

bit bit bit bit bit bit bit bit

: : : : : : : :

128 192 224 240 248 252 254 255

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 rt nhanh!!

Subnet Masks

Khi ta chia mt Network ra thnh nhiu Network nh hn, cc Network nh ny c gI l Subnet. Theo quy c, cc a ch IP c chia ra lm ba Class (lp) nh sau: Address Class Class A Class B Class C Subnet mask trong dng nh phn 11111111 00000000 00000000 00000000 11111111 11111111 00000000 00000000 11111111 11111111 11111111 00000000 Subnet mask 255.0.0.0 255.255.0.0 255.255.255.0

Subnet Mask ca Class A bng 255.0.0.0 c ngha rng ta dng 8 bits, tnh t tri qua phi (cc bits c set thnh 1), ca a ch IP phn bit cc NetworkID ca Class A. Trong khi , cc bits cn st li (trong trng hp Class A l 24 bits uc reset thnh 0) c dng biu din computers, gi l HostID. Nu bn cha quen cch dng s nh phn hy c qua bi H thng s 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 a ch th ny c th cha 65,534 nodes hay computers (65,534 = (2^16) 2 ) . y l mt con s qu 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 mask 2) Lit k ID ca cc Subnet mi 3) 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^X m 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 bits a ch IP mi s l 139.12.0.0/18 ( con s 18 thay v 16 nh trc y). Con s hosts ti a c 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 mi Trong a ch IP mi (139.12.0.0/18) con s 18 ni n vic ta dng 18 bits, m t bn tri, ca 32 bit IP address biu din a ch IP ca mt Subnet. Subnet mask trong dng nh phn 11111111 11111111 11000000 00000000 Subnet mask 255.255.192.0

Nh th NetworkID ca bn Subnets mi c l: Subnet 1 2 3 4 Subnet ID trong dng nh phn 10001011.00001100.00000000.00000000 10001011.00001100.01000000.00000000 10001011.00001100.10000000.00000000 10001011.00001100.11000000.00000000 Subnet ID 139.12.0.0/18 139.12.64.0/18 139.12.128.0/18 139.12.192.0/18

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 dng cho Host ID l tt c mi bits khng th u l 0 hay 1. Subnet 1 2 HostID IP address trong dng nh phn

10001011.00001100.00000000.00000001 10001011.00001100.00111111.11111110 10001011.00001100.01000000.0000000110001011.00001100.01111111.111

3 4

10001011.00001100.10000000.00000001 10001011.00001100.10111111.11111110 10001011.00001100.11000000.0000000110001011.00001100.11111111.111

Bn c thy trong mi Subnet, ci range ca HostID t con s nh nht (mu xanh) n con s ln nht (mu cam) 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 li l 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 tnh xem mnh cn bao nhiu bits cho 4000 hosts. Con s hosts ta c 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 (=3212) bits. Qu trnh tnh ton ni trn ny mang tn l Variable Length Subnet Mask(VLSM). Supernetting Gi t ta mng ca ta c 3 Subnets: Accounting: gm 200 hosts Finance : gm 400 hosts Marketing: gm 200 hosts Bn ha mng vi Internet v c Internet Service Provider (ISP) cho 4 Class C IP addresses nh sau: 192.250.9.0/24

192.250.10.0/24 192.250.11.0/24 192.250.12.0/24 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 ti a trong mi Network l [(2^X) 2] = (2^8) 2 = 254. Nh th segments Accounting v Marketing khng b tr 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 192.250.9.0/24 192.250.10.0/24 192.250.11.0/24 192.250.12.0/24 11000000 11111010 00001001 00000000 (1) 11000000 11111010 00001010 00000000 (2) 11000000 11111010 00001011 00000000 (3) 11000000 11111010 00001100 00000000 (4)

Bc 2: Nhn din network prefix notation 23 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 cp 508 hosts. IP address ca mi segment tr thnh: Accounting: gm 200 hosts: 192.250.9.0/24 Finance: gm 400 hosts: 192.250.10.0/23 Marketing: gm 200 hosts: 192.250.12.0/24 By gi IP address 192.250.11.0 tr thnh mt HostID tm thng trong Subnet 192.250.10.0/23. 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 l SUPERNETTING.

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).