soal latihan subnetting
jawaban soal latihan subnetting:
1) terdapat 25 subnet dan 300 host
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) Gateway = 192.168.1.65
host = 192.168.1.?/27
Subnet mask = 255.255.255.11100000 > 255.255.255.224
Blok subnet = 256-224=32
Range IP 192.168.1.0 – 192.168.1.31
192.168.1.32 – 192.168.1.63
192.168.1.64 – 192.168.1.95 < lokasi gateway
Sehingga IP yang valid adalah 192.168.1.82 dan 192.168.1.70
Jawaban (D,F)
Subnet mask = 255.255.255.11100000 > 255.255.255.224
Blok subnet = 256-224=32
Range IP 192.168.1.0 – 192.168.1.31
192.168.1.32 – 192.168.1.63
192.168.1.64 – 192.168.1.95 < lokasi gateway
Sehingga IP yang valid adalah 192.168.1.82 dan 192.168.1.70
Jawaban (D,F)
3) IP = 172.32.192.166
subnet mask = 255.255.255.248
Blok = 256-248=8
Subnet = 166/8 ± 20
Subnet = 8x20=160 IP = 172.31.192.16
Blok = 256-248=8
Subnet = 166/8 ± 20
Subnet = 8x20=160 IP = 172.31.192.16
Jawaban (E)
4) subnet mask yang valid pada kelas B =255.255.252.0 dan 255.255.255.192
jawaban (E dan F)
5) broadcast id 172.16.159.255 adalah 172.16.160.0
Subnet = 159-128=31 Host = 31x255=7905
2N-2=7905
2N=7905+2
2N=7907
N=13
n = 16-13=3
Subnet mask kelas B 255.255.11100000.0
255.255.224.0
Jawaban (D)
2N-2=7905
2N=7905+2
2N=7907
N=13
n = 16-13=3
Subnet mask kelas B 255.255.11100000.0
255.255.224.0
Jawaban (D)
6) IP 223.168.17.167/29
Subnet mask 255.255.255.248
Blok IP = 256-248=8
Subnet = 167/8 ± 20
Subnet = 8x20=160 > 223.168.17.160
Broadcast = 223.168.17.167
Jawaban (C)
Subnet mask 255.255.255.248
Blok IP = 256-248=8
Subnet = 167/8 ± 20
Subnet = 8x20=160 > 223.168.17.160
Broadcast = 223.168.17.167
Jawaban (C)
7) IP 192.168.99.0/29
Subnet mask 255.255.255.248
n = 5 N = 8-5=3
jumlah subnet = 2n-2 = 25-2 = 32-2 = 30
jumlah host = 2N-2 = 23-2 = 8-2 = 6
Jawaban (C)
8) IP 192.168.4.0
Subnet mask 255.255.255.224
Blok = 256-224 = 32
IP yang bisa dipakai = 32 – Net ID – Broadcast = 30
Jawaban (C)
9) 27 host
2n-2 ≥ 27
2n ≥ 27+2
2n ≥ 29
n = 5
Subnet mask = 255.255.255.248
Jawaban (E)
10) 2^n - 2 >= 14
n = 4 (sisa bit 0)
subnet mask= 255.255.255.240
jawaban (C)11) subnet mask class B = 255.255.0.0
2^n - 2 >= 100
n = 7 (bit 1)
subnet mask= 255.255.254.0
jawaban (C)
12) IP 172.32.65.13
Subnet mask 255.255.0.0
Subnet 172.32.65.0 Broadcast 172.32.65.255
Jawaban (A)
13) IP 172.16.210.0/22
Subnet mask 255.255.252.0
Blok = 256-252 = 4
Subnet = 210/4 ± 52
Subnet = 4x52 = 208
IP subnet 172.16.208.0
Jawaban (C)
14) IP 155.64.4.0/22
Subnet mask 255.255.252.0
Blok = 256-252 = 4
Range IP kelas B
155.64.4.0 - 155.64.7.255
IP yang valid adalah 155.64.5.128, 155.64.6.255, dan 155.64.7.64
Jawaban (B,C,E)
15) IP 200.10.5.68/28
Subnet mask 255.255.255.240
Blok = 256-240 = 16
Subnet = 68/16 ± 4
Subnet = 16x4 = 64
IP 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 id= 2^13 = 8190
jawaban (F)
17) jumlah subnet kelas B= 500
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) IP 172.16.0.0 subnet 100 host 5002^n-2 >= 100
n >= 7 (menambahkan bit 1)
subnet mask = 255.255.254.0
jawaban (B)
20) IP 192.168.19.24/29
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) 300 subnet
2n ≥ 300
n = 9
Subnet mask 255.255.255.128
50 host
2N-2 ≥ 50
2N ≥ 50+2
2N ≥ 52
N = 6
2N-2 = 26-2 = 64-2 = 62
Subnet mask 255.255.255.192
Jawaban (B,E)
22) IP gateway 172.16.112.1/25
Subnet mask 255.255.255.128
Blok = 256-128 = 128
Range IP 172.16.112.0 - 172.16.139.255
Jawaban (A)
23) IP 172.20.0.0 kelas B
Jumlah host = 850+500+550+650+800 = 2850
2N-2 ≥ 2850
2N ≥ 2850+2
2N ≥ 2852
N = 12
n = 16-12 = 4
Subnet mask 255.255.240.0
Jawaban (B)
24) IP 172.16.17.0/22
Subnet mask 255.255.252.0
Blok = 256-252 = 4
IP yang valid adalah 172.16.18.255 dengan subnet mask 255.255.252.0
Jawaban (E)
25) IP 172.16.112.1/20
n = 4
N = 16-4 = 12
Jumlah host = 2N-2 = 212-2 = 4096-2 = 4094
Jawaban (C)
26) /27= 255.255.255.224
Blok = 256-224 = 32
Host yang valid adalah 90.10.170.93, 143.187.16.56, dan 192.168.15.87
Jawaban (B, C, D)
27) 450 IP di kelas B
2N-2 ≥ 450
2N ≥ 450+2
2N ≥ 452
N = 9
n = 16-9 = 7
Subnet mask 255.255.254.0
Jawaban (C)
28) /27 =255.255.255.224
Subnet mask= 255.255.255.240
seharusnya = 255.255.255.224
Blok = 256-224 = 32
range 198.18.166.32 - 198.18.166.63
sementara jaringan router berada pada range 198.18.166.64 - 198.18.166.95 (berbeda jaringan)
Jawaban (A, E)
