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    1K views5 pages

    tập chia mạng con

    Uploaded by

    thaovuthi66

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    Attribution Non-Commercial (BY-NC)
    Download as DOC, PDF, TXT or read online from Scribd
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    of 5

    tp chia mng con (Phn 2)

    Posted by Hong V on May 20, 2011

    1/ Chia mng con lm g?


    - d qun l v qun l mt cch khoa hc
    - tit kim a ch IP
    - p ng nhu cu ca con ngi
    - .
    2/ Cch chia mng con.
    - a ch IP c 2 phn NET ID v HOST ID. Chng ta s mn vi bit trong HOST ID chia mng con theo
    yu cu.
    3/ Mt s lu khi chia mng con

    Gi s bn gp mng 192.168.1.1/24 iu ny tng ng vi IP 192.168.1.1 v subnet mask


    255.255.255.0
    /23 /24 /25 . Tng ng vi subnet mask. Nhn vo n ta c th bit c bao nhiu bit lm Net ID.
    Nh v d trn l mng dung 24 bit lm Net ID.
    Mt mng thng thng lun c 3 loi a ch
    a ch mng (a ch u mi mng, i din cho mng)
    a ch broadcast (a ch cui mi mng, l a ch m s pht tt tt c cc thng tin n mng ).
    Vng a ch IP xi c (nm gia a ch Mng v a ch Broadcast).
    4/ Mn bit chia mng con.(Nn nm k phn ny)
    Gi s mun chia mt mng bt k thuc lp C 194.23.12.0/24 (Net ID 24 bit, Host ID 8bit)

    S bit
    mn

    S bit cn S mng con chia c


    li

    S hosts trong mt mng Khong cch gia 2


    mng lin tip

    2^1 =2

    2^7 -2 = 126

    128 (2^7)

    2^2 =4

    2^6 -2 = 62

    64 (2^6)

    2^5 -2 = 30

    32 (2^5)

    16

    2^4 -2 = 14

    16 (2^4)

    32

    2^3 -2 = 6

    8 (2^3)

    64

    2^2 -2 = 2

    4 (2^2)

    Tng t nh vy i vi cc lp A,B
    Vd mun chia mng 194.23.12.0/24 ra lm 4 mng con
    - Ta cn phi mn 2 bit.
    - Khi Sub net l 255.255.255.192.
    - S host trong mt mng l 64-2 =62 (Tr a ch mng v a ch Broadcast)
    - Mi mng cch nhau 64 n v
    (cc mng s l 194.23.12.0, 192.23.12.64, 192.23.12.128, 192.23.12.192)
    5/ Bi tp Trc Nghim
    1. Mt mng con lp C mn 2 bit chia Subnet th Subnet Mask s l:
    a. 255.255.224.0 b. 255.255.255.192 c. 255.255.255.240 d. 255.255.255.128
    Lp C th Subnet Mask mc nh l 255.255.255.0
    Mn thm 2 bit chia sub net Mask th ta phi bt 2 bit 1 tip theo my tnh bit ti khc l Net ID.
    iu c ngha l
    Subnet Mask : 11111111.11111111.11111111.11000000 => 255.255.255.192
    Cc bn nn thuc lng cch xc nh Subnet Mask (i vi Mng con lp C)

    S bit mn chia
    mng

    Subnet Mask

    255.255.255.128

    255.255.255.192

    255.255.255.224

    255.255.255.240

    255.255.255.248

    2. Mt mng con lp C mn 5 bit chia Subnet th Subnet Mask s l:


    a. 255.255.224.0 b. 255.255.255.1 c. 255.255.255.248 d. 255.255.255.128
    Mn 4 bit th Subnet Mask s l 240 -> Mn 5 bit s l 248. Vic nh gip cc bn lm bi nhanh hn ^^
    3. Mt mng con lp A mn 21 bit chia Subnet th Subnet Mask s l:
    a. 255.255.224.0 b. 255..255.192.0 c. 255.255.248.0 d. 255.255.255.248
    Ch : Lp A c 8 bit u lm Net ID mn 21 bit => c 29 bit lm Host ID
    29> 24 nn 3 octet u ta bt bit 1 ln ht -> n l 255.255.255.0
    Cn 5 bit cn li ta cng bt ln nt. Vy cui cng ta c Subnet Mask l 255.255.255.248
    4. Trong mng my tnh dng giao thc TCP/IP v Subnet Mask l 255.255.255.224, hy xc nh a ch
    broadcast ca mng nu bit rng mt my tnh trong mng c a ch 192.168.1.1:
    a. 192.168.1.31 b. 192.168.1.255 c. 192.168.1.15 d. 192.168.1.96
    Hng dn gii:
    - Khi c Subnet Mask Ta d dng suy ra c Phn Net ID v Host ID ( trong bt ny th Phn Net l 27 bit
    v host l 5 bit).
    - Da vo a ch ca my tnh, ta bit c a ch lp no (trong bt ny l lp C)
    T suy ra Net ny mn 3 bit chia Subnet -> Do ta cn li 5 bit host.
    - 3 bit chia c 8 Subnet, mi Subnet cch nhau 32 n v. v s host trong mi Net l 32-2 =30

    - Ta tm c Mng u tin 192.168.1.0


    - Mng th 2 192.168.1.32..
    - T suy ra Broadcast ca Mng u tin cha a ch 192.168.1.1 l 192.168.1.31
    6 Trong mng my tnh dng Subnet Mask l 255.255.255.0 th cp my tnh no sau y lin thng:
    a. 192.168.1.3 v 192.168.100.1 b. 192.168.15.1 v 192.168.15.254
    c. 192.168.100.15 v 192.186.100.16 d. 172.25.11.1 v 172.26.11.2
    Ci ny Subnet Mc nh lp C. Nn NET ID c 24 bit u, v Host ID l 8 bit 6.
    Ta i chiu cc NET ID vi nhau th s bit c cp no cng mng. p n B
    y l mt vi bi tp c bn. Mnh s post cc bi tp c Lin quan v p n ca mnh trong mt part
    khc. Cc bn nu c thc mc g c li Comment cho mnh nh
    Posted under Network
    Comments (5)

    5 Responses to Bi tp chia mng con (Phn 2)

    1.

    May 24 2011

    nhan
    Cu 6 bn b sai ri phi l B mi ng.

    2.

    May 30 2011

    hung
    em co bai tap nay mong anh huong dan chi tiet gium em voi.
    cho: ip 192.168.1.0,subnetmak 255255.92.0
    cau hoi? mang tren co chia mang con hay khong neu ro tung truong hop?,
    cho biet dia chi mang

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