1:

Cu 1: (3 im)

i cc s IP c biu din bng dng nh phn sang thp phn, hy gch chn phn NetID trong cc a ch IP di y:

a. 11011100.01100101.11011000.00010101=

b. 10111000.00101011.11010100.10101111=

c. 01011101.11011101.11000001.00100001=

d. 10101010.01010101.11001100.11100010=

Cu 2: (3 im)

Hy trnh by khi nim subnet mask? ngha v cch s dng subnet mask xc nh a ch mng.?Cu 3: (4 im)

a ch IP 129.56.7.8 c subnet mask l 255.255.128.0. Hi c bao nhiu subnet? bao nhiu a ch IP trong mi subnet? bao nhiu a ch IP trong mng ?p n:

Cu 1: (3 im) a. 11011100.01100101.11011000.00010101= 220.101.216.21

b. 10111000.00101011.11010100.10101111= 184.43.212.175c. 01011101.11011101.11000001.00100001= 93.221.193.33d. 10101010.01010101.11001100.11100010= 170.85.204.226Cu 2: (3 im)

* Subnet mask:- Khi nim:+ Khi truyn thng tin, mt my cn phi bit a ch IP ca my nhn c trong cng mng vi mnh hay khng, thc hin iu ny ngoi a ch IP, mt thng s khc gi l subnet mask cn c xc nh cho my.+ Subnet mask cng gm 4 s thp phn khng du, mi s gm 8 bits; gi trca subnet mask gm 32 bits c chia lm 2 phn: Bn tri gm nhng bit 1 Bn phi gm nhng bit 0+ Cc bit 0 xc nh nhng a ch IP no cng nm trn cng mt mng con vi n. Thng thng, cc client c cung cp a ch IP i km vi subnet mask khi ni vo mt ISP.* ngha: Kha li mt phn a ch IP phn bit Network ID v Host ID Xc nh mt a ch IP ch thuc mng ni b hay thuc mng khc* Cch s dng:- Dng php AND xc nh a ch mng ca 1 host theo cng thc sau:Network Address = IP Address AND Net mask- V d:Cho a ch host l 198.53.147.45, lp C c subnet mask 255.255.255.0Dng php AND : 198.53.147.45 AND 255.255.255.0 l 198.53.147.0 l network ID.- xc nh xem 2 a ch IP ca my ch v my ngun c cng a ch 1mng hay khng ta lm theo cc bc sau: Tnh kt qu php AND gia a ch IP ch v mask ca mng Tnh kt qu php AND gia a ch IP ngun v mask ca mng Nu 2 kt qu trng nhau th 2 a ch IP cng mt mng l khng phi thit lp ng truyn V d: So snh 2 a ch IP 192.168.1.10 v 192.168.1.102 c cng trn mt mng hay khng? Cho subnet mask 255.255.255.0Gii: 192.168.1.10 = 11000000.10101000.00000001.00001010255.255.255.0 = 11111111.11111111.11111111.00000000192.168.1.0 = 11000000.10101000.00000001.00000000=> network ID = 192.168.1.0192.168.1.102 = 11000000.10101000.00000001.01100110255.255.255.0 = 11111111.11111111.11111111.00000000192.168.1.0 = 11000000.10101000.00000001.00000000=> network ID = 192.168.1.0Vy 2 a ch trn l cng 1 mngCu 3: (4 im)

a ch IP 129.56.7.8 l a ch thuc lp B, c subnetmask mc nh l 255.255.0.0

Theo bi ta c subnetmask 255.255.128.0 nh vy mn 1 bit chia subnet

Da vo cng thc Y S host ti a c trong mi subnet: 214 2 = 16382 Bc 2: Lit k ID ca cc Subnet mi:Trong a ch IP mi (139.12.0.0/18) 18 bits u tin ca 32 bits IP address dng biu din a ch IP ca mt subnet. Nh vy Network ID ca 4 subnets mi c l:

SubnetSubnet ID trong dng nh phnSubnet ID

110001011.00001100.00000000.00000000139.12.0.0/18

210001011.00001100.01000000.00000000139.12.64.0/18

310001011.00001100.10000000.00000000139.12.128.0/18

410001011.00001100.11000000.00000000139.12.192.0/18

Bc 3: Cho bit IP address range ca cc Host ID trong mi Subnet:Ta c 14 bits dng cho Host ID. Da trn lut: dng cho Host ID tt c mi bits khng th u l 0 hoc 1 do ta c bng IP Address cacc Host ID trong mi subnet:

Cu 3: (4 im)

Da vo cng thc : Network Address = IP Address AND Net mask15.100.10.70 = 00001111.01100100.00001010.01000110255.255.255.192 = 11111111.11111111.11111111.1100000015.100.10.64 = 00001111.01100100.00001010.01000000=> network ID = 15.100.10.64

15.100.10.210 = 00001111.01100100.00001010.11010010255.255.255.192 = 11111111.11111111.11111111.1100000015.100.10.192 = 00001111.01100100.00001010.11000000=> network ID = 15.100.10.192 3:Cu 1: (3 im)

v sao ngi qun tr mng chia mng ca mnh thnh nhng mng con? a ch qung b (Broadcast) l gi?Cu 2: (3 im)

Xt a ch IP:

a) 10.67.0.56

b) 192.168.80.129

c) 132.46.7.11

c) 87.56.42.15

Vit li cc a ch IP trn di dng nh phn, chng thuc lp no? phn no l Net ID, phn no l Hot ID, subnet mask?Cu 3: (4 im)

Cho a ch mng: 192.161.121.0/24. Trnh by v chia mng ny thnh 9 mng con. Sau lit k cc a ch mng, Subnetmask, a ch broadcast, vng phm vi a ch c th gn cho my tnh ca cc mng con?p n:Cu 1: (3 im)

- Ngi qun tr mng i khi cn phi phn chia mng, t mt mng ln phn chia thnh nhng mng con v: Gim kch thc, Ph hp vi m hnh mng hin ti ca Cng ty

Phn cp qun l. Gim nghn mng bng cch gii hn phm vi ca cc thng ip qung b. Gii hn trong phm vi tng mng con cc trc trc c th xy ra (khng nh hng ti ton mng LAN) Tng cng bo mt (cc chnh sch bo mt c th p dng cho tng mng con) Cho php p dng cc cu hnh khc nhau trn tng mng con

=>V vy cn phi phn chia thnh nhng mng con ( Khi phn chia mng con th nhng ngI bn ngai mng (outside network) nhn mng chng ta ch l mt single network.- Qung b (Broadcast) l vic m mt host gi d liu n tt c cc host cn li trong cng mt network ID number. C 2 loi a ch qung b: a ch qung b trc tip ( directed broadcast address) l a ch IP m trong tt c cc bit trong trng Host ID u l 1.a ch qung b cc b (local broadcast address) l a ch IP m trong tt c cc bits trong Network ID v Host ID u l 1.Cu 2: (3 im)

a.10.67.0.56Lp aDng nh phn00001010.01000011.00000000.00111000

NetID10

a) HostID67.0.56

subnet Mask255.0.0.0

b.192.168.80.129Lp cDng nh phn11000000.10101000.01010000.10000001

NetID192.168.80

HostID129

subnet Mask255.255.255.0

c.132.46.7.11Lp bDng nh phn10000100.00101110.00000111.00001011

NetID132.46

HostID7.11

b) subnet Mask255.255.0.0

d.87.56.42.15Lp aDng nh phn01010111.00111000.00101010.00001111

NetID87

c) HostID56.42.15

subnet Mask255.0.0.0

Cu 3: (4 im)

Subnet mi:

Theo cng thc : Y 255)

- 252.12.6.12 ( v nm ngoi dy biu din ca 3 lp a, b, c. s 252 nm lp d hin cha s dng)

- 192.168.1.0 ( v a ch thuc lp c, c phn HostID bng 0 -> ch dng cho a ch dng ring, khng dng khai bo cho 1 my tnh trong mng)

- 170.12.255.255 ( v a ch thuc lp b, c phn Host ID bng 1 -> ch dng cho a ch Broadcast, khng dng khai bo cho 1 my tnh trong mng)Cu 2: (3 im)

1. Cu trc ca a ch IP:- Mi thit b trong h thng mng ca chng ta c t nht 2 a ch. Mt a ch l Media Access Control (MAC) v mt a ch Internet Protocol (IP). a chMAC l a ch ca card mng gn vo bn trong thit b, n l duy nht v khng h thay i. a ch IP c th thay i theo ngi s dng ty vo mi trng mng.- a ch IP ang c s dng hin ti (IPv4) c 32 bits chia thnh 4 octet (mi octet c 8 bits, tng ng 1 byte) v c phn cch nhau bi 3 du chm (.). Gi tr ca mi byte c vit li di dng thp phn,vi gi tr hp l nm trong khong t 0 255. a ch IP c chia lm 2 phn: Network ID: phn nhn dng mng, c dng nhn dng 1 mng v phi c gn bi trung tm thng tin mng internet. Host ID: phn nhn dng my tnh: nhn dng 1 my tnh trong mt mng.V d: 1 a ch IP: 192.168.1.108. Mi k t trn i din cho 1 con s dongi s dng my tnh, modem hoc my ch c chc nng ring gn cho chng.2. ngha ca s phn lp a ch IP:- Ban u, mt a ch IP c chia thnh 2 phn: Network ID: xc lp bi octet u tin Host ID: xc nh bi 3 octet cn li- Vi cch chia ny, s lng network b gii hn con s 256, qu t so vinhu cu thc t. vt qua gii hn ny, vic phn lp mng c nh ngha, to nn mt tp hp lp mng y (classful).- to ra cc mng vi s mng hoc s host khc nhau ty vo mc ch s dng ca tng ni m ngi ta chia a ch IP ra thnh cc lp khc nhau: A, B, C, D v E. Trong ch c lp A, B, C c dng cho cc mc ch thng mi. Lp D c dng cho vic pht cc thng tin broadcast/multicast IP. Lp E c dnh ring cho nghin cu.

HYPERLINK "http://4.bp.blogspot.com/-uWWPdUiYGUU/T7jErvfEMLI/AAAAAAAAAoQ/jgb75uE_198/s1600/thi+van+dap+mon+mang+may+tinh2.png" \t "_blank"

- Tuy nhin, cc nh qun tr mng thng phn chia mng ca h ra thnh nhiu mng nh hn gi l mng con subnet.Cu 3: (4 im) Da vo cng thc : Network Address = IP Address AND Net mask190.160.100.164= 10111110.10100000.01100100.10100100255.255.255.248= 11111111.11111111.11111111.11111000190.160.100.160 = 10111110.10100000.01100100.10100000=> network ID = 190.160.100.160 5:Cu 1: (3 im)

Nu cch qun l cc a ch IP? Trnh by chi tit v cc a ch IP dng ring?.Cu 2: (3 im)

i cc s IP c biu din bng dng nh phn sang thp phn, hy gch chn phn NetID trong cc a ch IP di y:

a. 11011100.01100101.11011000.00010101=

b. 10111000.00101011.11010100.10101111=

c. 01011101.11011101.11000001.00100001=

d. 10101010.01010101.11001100.11100010=

Cu 3: (4 im) Cho 2 a ch IP: 100.101.102.70 v IP: 100.101.102.210 c subnet mask l: 255.255.255.192. Hy trnh by v cho bit a ch mng, a ch broadcast ca chng l bao nhiu?p n:Cu 1: (3 im)

1. Cch qun l cc a ch:- a ch IP cn c qun l mt cch hp l nhm trnh xy ra cc xung t khi ng thi c 2 a ch IP ging nhau trn cng mt cp mng my tnh.- cp mng ton cu (internet), mt t chc ng ra qun l cp pht cc di IP cho cc nh cung cp dch v kt ni Internet (IXP, ISP) cc di IP cung cp cho khch hng ca mnh.- cp mng nh hn (WAN), ngi qun tr mng cung cp n cc lp cho cc mng nh hn thng qua my ch DHCP.- cc mng nh hn na (LAN) th vic qun l a ch IP ni b thng do cc modem ADSL (c DHCP) gn a ch IP cho tng my tnh (khi thit t ch t ng trong h iu hnh) hoc do ngi s dng t thit t.2. Cc a ch IP dng ring:- a ch IP cng cng l danh nh c tnh ton cu v c tiu chun ha. Cc a ch IP cng cng l duy nht v c cung cp t 1 nh cung cp dch v hay ng k vi mt chi ph no .- a ch IP ring: Cc mng ring khng kt ni vo internet c th dng bt k1 a ch host no, min l mi host trong mng ring l duy nht.- Cc khi a ch IP ring:A: 10.0.0.0 10.255.255.255B: 172.16.0.0 172.31.255.255C: 192.168.0.0 192.168.255.255- Cc a ch IP dng ring ca cc lp nh sau:Lp A: 10.0.0.1 n 10.255.255.254Lp B: 172.16.0.1 n 172.31.255.254.Cu 2: (3 im)

a. 11011100.01100101.11011000.00010101= 220.101.216.21

b. 10111000.00101011.11010100.10101111= 184.43.212.175c. 01011101.11011101.11000001.00100001= 93.221.193.33d. 10101010.01010101.11001100.11100010= 170.85.204.226Cu 3: (4 im) Da vo cng thc : Network Address = IP Address AND Net mask100.101.102.70 = 01100100.01100101.01100110.01000110255.255.255.192= 11111111.11111111.11111111.11000000100.101.102.64 = 01100100.01100101.01100110.01000000=> network ID = 100.101.102.64

a ch broadcast:

o bit subnetmask ta c: 00000000.00000000.00000000.00111111

OR vi IP 01100100.01100101.01100110.01000110

=> Broadcast: : 01100100.01100101.01100110.01111111 = 100.101.102.117100.101.102.210 = 01100100.01100101.01100110.11010010255.255.255.192= 11111111.11111111.11111111.11000000100.101.102.192 = 01100100.01100101.01100110.11000000=> network ID = 100.101.102.192

a ch broadcast:

o bit subnetmask ta c: 00000000.00000000.00000000.00111111

OR vi IP 01100100.01100101.01100110.11010010

=> Broadcast: : 01100100.01100101.01100110.11111111 = 100.101.102.255