TUGAS NETADMIN PART 2

jawabann soal LPKIA....Check thiss out....^___^'

1.max host = 300
2^N-2>300
2^N>302
N=9

subnet mask = 111111111 11111111 11111110 00000000
= 255.255.254.0

jawaban C

2.Gateway = 196.168.1.65
subnet mask = 11111111 11111111 11111111 11100000
=255.255.255.224
block subnet = 256-224 = 32

net id range ip broadcast
192.168.1.0 192.168.1.1-192.168.1.30 192.168.1.31
1.32 192.168.1.33-192.168.1.62 1.63
1.64 1.65- 1.94 1.95

jawaban D dan F

3.net = (x div block sub).8
= (166 div 8).8
= 20.8
= 160
jadi net id = 172.31.192.160

jawaban E

4.jawaban D,F

5.subnet id 172.16.160.0 dan sebelumnya 172.16.128.0
jd block subnet = 160-128 =32
block subnet ke 3 = 256-32=224
subnet mask = 255.255.255.224

jawaban D

6.subnet mask 11111111.11111111.11111111.11111000
255.255.255.248
block subnet = 2^3 = 8
subnet id 223.168.17.160
broadcast 223.168.17.167
jadiii berupa broadcast address

jawaban C

7.Net = 2^5 - 2 = 30
host = 2^3 - 2 = 6

jawaban C

8.224 = 128+64+32
n = 3
N = 5
maximum host = 2^5-2 = 30 host

jawaban C

9.2^N = 32
block subnet mask= 256-32
=224

jawaban C

10.2^n - 2 = 14
2^n = 16
n = 4
block subnet =256-16
=240

jawaban C

11.block subnet = 256-128 = 128

jawaban E

12.jawaban C

13.n=22-16
=6
N=2
block subnet 2^N = 4
subnet work =210 div 4 . 4
=208
jawaban C

14.N= 24-2
=2
block subnet = 2^N = 4

ip CIDR harus berada pada range 115.64.4.0 - 115.64.8.0

jawaban B,C,E

15.n = 32-28 = 4
block subnet 2^4= 16
subnetwork address = 68 div 16.16
=4.16 =64
jawaban C

16.n = 19 - 16 = 3
N=32-19 =13
net = 2^3 = 8
host = 2^13 - 2 = 8192 - 2 = 8190

jawaban F

17.2^N = 128
N = 7
Subnet mask 255.255.255.128

jawban B

18.N = 24-21 = 3
block subnet 2^N = 8
subnet work =66 div 8.8
=8.8 = 64

jawaban C

19.block subnet 2^9 - 8 = 2
subnet mask = 256 - 2^2 =254

jawaban B

20.N = 32 - 29 = 3 block subnet = 2^N = 8
net mask = 256 - 2^N
= 248

jawaban C

21.N=6 CIDR = /26
Net mask = 11111111 11111111 11111111 1100000
255.255.255.192
jawaban E

22.N = 32-25 =7
block subnet = 2^N = 2^7 =128
Subnet adress = 1 div 128 . 128 = 0 ---> 172.16.112.0

jawaban A

23.N = 10 CIDR =/22
net mask 1111111 1111111 11111100 0000000
255.255.252.0
jawaban D

24.subnet mask 255.255.252.0
block subnet 256-252 = 4
jawaban E

25.N = 32-20
= 12
host = 2^N -2
=2^12 -2 = 4094
jawaban C

26.N = 32-27 = 5
block subnet = 2^5 = 32
valid host tidak boleh kelipatan 0 atau 32

jawaban B C D

27.N = 9 ICDR =/23
Net mask 1111111 1111111 11111110 0000000
255.255.254.0
jawaban C

28.CIDR =/27
net mask = 1111111 11111111 1111111 11100000
255.255.255.224
jawaban A



hhhhuuuuffffttt.....
take such a long time to finished this task....thanks to my friend who is lend me her modem...LOL...^^**poor englishh.....just practice to be better....kekkeke..