NetAdmin 2

(1) terdapat 25 subnet, menggunakan IP class B, maksimum komputer 300 tiap network segment
subnet mask yang cocok untuk digunakan adalah . . .
a. 2^n - 2 >= 25
n = 5 (menambahkan bit 1) = IIIIIIII.IIIIIIII.IIIII000.00000000
255.255.248.0
b. 2^n - 2 >= 300
n = 9 (menambahkan bit 0) = IIIIIIII.IIIIIIII.IIIIIII0.00000000
255.255.248.0
(jawaban : B dan C)

(2) Eth0=168.1.65/27
Subnetmasknya /27 = 11111111.11111111.11111111.1110000 = 255.255.255.224
Host per blok : 256-224 = 32
192.168.1.0 192.168.1.1 – .30 192.168.1.31
192.168.1.32 192.168.1.33 – .62 192.168.1.63
192.168.1.64 192.168.1.65 – .96 192.168.1.95
192.168.1.96 ……. ……
(Jawaban : F). Address - 192.168.1.70 Gateway -192.168.1.65
D. Address - 192.168.1.82 Gateway -192.168.1.65

(3) Diketahui :IP Address: 172.31.192.166
SubnetMask: 255.255.255.248=11111111.11111111.11111111.11111000.
172.31.192.0 172.31.192.1- .6 172.31.192.7
172.31.192.8 172.31.192.9- .14 172.31.192.15
…………………………
172.31.192.160 172.31.192.161- .166 172.31.192.167
(Jawaban : E ). 172.31.192.160.

(4) subnet mask yang valid untuk IP kelas B (255.255.252.0 dan 255.255.255.192)
karena pada IP kelas B oktet 1 dan 2 tetap tidak berubah
(jawaban E dan F)

(5) setelah broadcast id 172.16.159.255 adalah 172.16.160.0 yaitu subnet baru. jadi kombinasi net id dan subnet mask nya adalah . . .

• 160 - 128 (net id yang diketahui) = 32 (blok subnet)
• block subnet (256 - x = 32, x = 224)
• jadi net id (172.16.128.0) dan subnet mask 255.255.224.0 (yang berubah oktet ke 3)
  

(jawaban D)

(6) subnet mask = 255.255.255.248, maka block subnet (256 - 248 = 8)
jadi setelah IP 223.168.17.167 adalah net id (223.168.17.168). berarti jawabannya adalah broadcast id.
(jawaban C)

(7) /29 = 255.255.255.248, IP 192.168.99.0.

• 255.255.255.248 = IIIIIIII.IIIIIIII.IIIIIIII.IIIII000
• bit 1 terakhir = 2^n - 2 >= 5 > networks 30
• bit 0 terakhir = 2^n - 2 >= 3 > hosts 6
  

(jawaban C)

(8) IP company 192.168.4.0, subnet mask 255.255.255.224.
255.255.255.224 = IIIIIIII.IIIIIIII.IIIIIIII.III00000

• banyak hosts 2^n - 2 >= ... (n = sisa bit 0)
• 30 hosts
  

(jawaban C)

(9) subnet mask . . .

• 2^n - 2 >= 27
• n = 5 (sisa bit 0)
• 255.255.255.224

(jawaban C)

(10) subnet mask . . .

• 2^n - 2 >= 14
• n = 4 (sisa bit 0)
• 255.255.255.240

(jawaban C)

(11) subnet mask . . .

• subnet mask default class B = 255.255.0.0
  
• 2^n - 2 >= 100
• n = 7 (bit 1)
  
• 255.255.254.0

(jawaban C)

(12) Diket :Diberikan Ip Addres 172.32.65.13 Class B.
(Jawaban : C) 172.32.0.0

(13) Diket : IP Addrees 172.16.210.0/22 .
Subnetmasknya : 11111111.11111111.11.0000000
172.16.0.0 172.16.1.0 – 172.16.2.0 172.16.3.0
172.16.4.0 172.16.5.0 – 172.16.6.0 172.16.7.0
……
172.16.208.0 172.16.209.0 - 210.0 172.16.211.0
(Jawaban : C) 172.16.208.0

(14) subnet mask 255.255.252.0, IP 115.64.4.0 (class A)

• block subnet 256 - 252 = 4
• hosts 2^9 - 2 = 254

Net ID Range IP Broadcast ID
115.64.4.0 115.64.4.1 - 115.64.7.254 115.64.7.255
115.64.8.0 115.64.8.1 - 115.64.12.254 115.64.12.255

jadi yang tidak termasuk dalam subnet IP 115.64.4.0 adalah jawaban (A, D, F)

(15) /28 = 255.255.255.240, block subnet 256 - 240 = 16
Net ID range IP
200.10.5.0
200.10.5.16
200.10.5.32
200.10.5.64 (200.10.5.68)
200.10.5.80
jadi IP 200.10.5.68 terdapat pada subnet 200.10.5.64 (jawaban C)

(16) /19, sisa bit 0 = 13, bit 1 pada oktet ke 3 = 3
subnet = 2^3 = 8
hosts = 2^13 = 8190
(jawaban F)

(17) 500 subnet class B, maka subnet mask nya . . .

• 2^n-2 >= 500
• n = 9 (bit 1) subnet mask 255.255.255.128

(jawaban B)

(18) /21 = 255.255.255.248
block subnet 256-248 = 8
jadi IP 172.16.66.0 berada pada net ID 172.16.64.0
(jawaban C)

(19) 2^n-2 >= 100
n >= 7 (menambahkan bit 1)
subnet mask = 255.255.254.0
(jawaban B)

(20) /29 = subnet mask 255.255.255.248
block subnet 256 - 248 = 8
jadi IP 192.168.19.26 255.255.255.248 berada pada subnet 192.168.19.24
(jawaban C)

(21) 2 subnet mask . . .

• subnet 300
• 2^n-2 >= 300
• n >= 9 (bit)
• 255.255.255.128
• hosts 50
• 2^n-2 >= 50
• n >= 6 (bit 0)
• 255.255.255.224

(jawaban B dan D)

(22) Subnetmask : 11111111.11111111.11111111.10000000 = 255.255.255.128
Host per blok : 256-128 = 128
72.16.112.0 172.16.112.1- .126 172.16.112.127
172.16.112.128 ………..
(Jawaban : A) 172.16.112.0

(23) jumlah host yang ada = 3350
Host 2^n-2 > 3350, n = 12
Subnetmask : 11111111.11111111.11111000.00000000 = 255.255.248.0
(Jawaban : C) 255.255.248.0

(24) Subnetmask : 11111111.11111111.11111100.00000000 = 255.255.255.252
Host per blok : 256-252 = 4
(Jawaban : E) 172.16.18.255 255.255.252.0

(25) Subnetmask : 11111111.11111111.11110000.00000000 = 255.255.240.0
Host n = 12 (bit 0),
2^n-2=
212 – 2 = 4094
(Jawaban : C)

(26) /27 = 11111111.11111111.11111111.11100000 = 255.255.255.224,
blog subnetnya adalah 256-224 = 32,
sehingga IP yang valid adalah IP yang tidak digunakan untuk Net ID dan Broadcast, maka jawaban yang benar adalah 90.10.170.93 , C. 143.187.16.56, dan D. 192.168.15.87
(Jawaban : B)

(27) Kelas B, 450 host/subnet
Host 2^n - 2 >= 450, n = 9 (bit 0)
Subnetmask = 11111111.11111111.11111110.00000000 = 255.255.254.0
(Jawaban : C)

(28) IP 98.18.166.33/27
Subnetmask = 11111111.11111111.11111111.11100000 = 255.255.255.224
Address host 198.18.166.65 dengan Eth0 gateway 198.18.166.33,
Blok Subnet = 256 – 224 = 32
32, 64, 96, 128, 160, 192
Jawaban :
A. The host subnet mask is incorrect
B. The host IP address is on a different network from the Serial interface of the router.