Tugas NetAdmin08072

1. 300 Host berarti 2n - 2 ≥ 300 n=9
Subnet masknya jadi=11111111.11111111.11111110.0000000
Jawaban : C. 255.255.254.0

2. Eth0=168.1.65/27
Subnetmask   /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
Jawab : E. 172.31.192.160.

4. 255.0.0.0 untuk kelas A .255.254.0.0 untuk kelas A, 255.224.0.0 untuk kelas A
   F.255.0.0.0 untuk kelas C
Jawab : D. 255.255.0.0 & E. 255.255.252.0

5. Kombinasi dari network id and subnet mask yang benar?

blok subnetnya yaitu 160-128 = 32, jadi subnet mask yang baru

dimisalkan subnetmask baru = x, 256-x = 32,maka x adalah 256-32=224

di dapat subnetmask yang baru  255.255.255.224

Jawab : A. 172.16.128.0 and 255.255.255.224

6.Address: 223.168.17.167/29
Subnetmask : 11111111.11111111.11111111.11111000 = 255.255.255.248
Host per blok : 256-248 = 8
223.168.17.0 223.168.17.1 - .6 223.168.17.7
223.168.17.8 223.168.17.9 - .14 223.168.17.15
……..
223.168.17.160 223.168.17.161 - .166 223.168.17.167
Jawaban : C. broadcast address

7. Dik IP address : 192.168.99.0/29.

kombinasi net ID dan host yang benar ?

Subnetmasknya 11111111.11111111.11111111.11111000 = 255.255.255.248, sehingga hostnya adalah 2^3-2 ≥ 6 host, dan jumlah networknya adalah 2^5-2 ≥ 30 network.

Jawab : c. 30 networks / 6 hosts

8. IP address : 192.168.4.0 kelas C subnetmask : 255.255.255.224
Host per blok : 256-224 = 32 host max : 32-2 = 30
Jawab : C. 30

9.Diket : Host Max=27 , 2^N-2>27, N = 5
NetMasknya : 11111111.11111111.11111111.11100000 = 255.255.255.224
Jawaban : C. 255.255.255.224.

10.Setiap subnetwork memiliki 14 host yang digunakan.berapakah subnetmask yang baru?

2^N-2≥14, N=4 subnetmask : 111111111.1111111.11111111.11110000 = 255.255.255.248

Jawab : C. 255.255.255.248

11.Sebuah Class B membutuhkan 100 Networks.
2^n >100, n = 7 Subnetmasknya : 11111111.11111111.11111110.00000000 = 255.255.254.0
Jawab: C. 255.255.254.0

12. Diberikan Ip Addres 172.32.65.13 Class B.dengan default subnetmask
Jawaban :C. 172.32.0.0


13.IP Addrees 172.16.210.0/22 .
Subnetmask : 11111111.11111111.11111100.0000000 : 255.255.252.0 , 256-252=4.
subnet?

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 – 172.16.210.0 172.16.211.0
Jawaban : C. 172.16.208.0

14.Sebuah IP Address dengan CIDR 115.64.4.0/22
Subnetmask : 11111111.11111111.11111100.00000000 = 255.255.252.0

Host/blok: 256-252= 4, 115.64.4.0-115.64.7.255
Jawaban: b. 115.64.7.64 c. 115.64.6.255 e. 115.64.5.128


15.Sebuah IP address 200.10.5.68/28
Subnetmask : 11111111.11111111.11111111.11110000 = 255.255.255.240
 blok subnet : 256-240 = 16, alamat subnetwork pada host tersebut :
200.10.5.0   200.10.5.1 - 200.10.5.14  200.10.5.15
200.10.5.16 200.10.5.17- 200.10.5.30  200.10.5.31
200.10.5.32 200.10.5.33- 200.10.5.46  200.10.5.47                                                            …….           ………………………………………………                                                                         200.10.5.64 200.10.5.65 – 200.10.5.78 200.10.5.79
Jawaban : C. 200.10.5.64


16. Network address 172.16.0.0/19
Subnetmask : 11111111.11111111.11100000.00000000 = 255.255.224.0
Subnets   (n: 19-16= 3) 2^n = 2^3=8
Host bit (N:32-19=13) 2^N-2 = 2^13-2=8192-2= 8190
Jawaban : E. 8 subnets, 8190 hosts each

17. Subnet : 500 subnet, setiap subnet digunakan 100 host,pada kelas B.

Host per subnet 2^N-2 ≥100, 2^N ≥ 102, n=7 ,jumlah ‘0’
Subnetmask = 11111111.11111111.11111111.10000000 = 255.255.255.128

Jawaban : B. 255.255.255.128

18. Sebuah IP address 172.16.66.0/21
Subnetmask : 11111111.11111111.11111000.00000000 = 255.255.248. 0
blok subnet: 256-248 = 8. Nomor subnetwork?
172.16.0.0   172.16.1.0 – 172.16.6.0   172.16.7.0
172.16.8.0   172.16.9.0 – 172.16.14.0 172.16.15.0
…………..                                                                                                                               172.16.64.0 172.16.65.0 -172.16.70.0 172.16.71.0
Jawaban : C. 172.16.64.0

19. Terdapat 100 subnet dan 500 host/subnet Class B.
Subnet = 2^n > 100, n = 7 bit
Subnetmask = 11111111.11111111.11111110.00000000
Jawaban : B. 255.255.254.0

20. Terdapat IP address 192.168.19.24/29
Subnetmask : 11111111.11111111.11111111.11111000 = 255.255.255.248
blok subnet: 256-248= 8
Net id   range    broadcast
192.168.19.0 192.168.19.1 - 192.168.19.6    192.168.19.7
192.168.19.8 192.168.19.9 – 192.168.19.14  192.168.19.15
………..                                                                                                                               192.168.19.24 192.168.19.25 – 192.168.19.30 192.168.19.31
Jawaban : C. 192.168.19.26 255.255.255.248

21. Terdapat minimum dari 300 subnets dengan max  50 hosts per subnet.

Jawab : 2^n≥300, n=9 subnetmasknya 11111111.11111111.11111111.10000000 = 255.255.255.128

2^N-2≥50, N=6 subnetmasknya 11111111.11111111.11111111.11000000 = 255.255.255.192

Jawab : B. 255.255.255.128 dan E. 255.255.255.192  

22. IP address 172.16.112.1/25
Subnetmask : 11111111.11111111.11111111.10000000 = 255.255.255.128
Host per blok : 256-128 = 128
172.16.112.0  172.16.112.1- 172.16.112.126  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 : 11111111.11111111.11110000.00000000 = 255.255.240.0
Jawaban : C. 255.255.240.0


24. network with a subnet of 172.16.17.0/22.
Subnetmask : 11111111.11111111.11111100.00000000 = 255.255.252.0
Host per blok : 256-252 = 4

172.16.17.0 172.16.18….
Jawaban : E. 172.16.18.255  255.255.252.0


25.Sebuah Router dengan IP Ethernet0: 172.16.112.1/20.
N:32-20= 12, host :2^N-2= 2^12-2= 4096-2= 4094
Jawaban : C. 4094

26.dimiliki prefix /27 subnet mask.Host yang valid adalah ?

subnetmasknya 11111111.11111111.11111111.11100000 = 255.255.255.224,

blok subnetnya adalah 256-224 = 32, di dapat IP valid adalah yang tidak dalam kelipatan 32

Jawab : B. 90.10.170.93 , C. 143.187.16.56, dan D. 192.168.15.87

27.  Class B  network ID dan membutuhkan  skitar 450 IP addresses per subnet. Subnetmask?
Kelas B, 450 host/subnet
Host 2^N - 2 ≥ 450, N = 9 (bit "0")
Subnetmask = 11111111.11111111.11111110.00000000 = 255.255.254.0
Jawab : C. 255.255.254.0
.

28.Eth0 = 198.18.166.33/27
Subnetmask = 11111111.11111111.11111111.11100000 = 255.255.255.224
Address host 198.18.166.65  gateway 198.18.166.33,
Blok Subnet = 256 – 224 = 32.
Jawaban : A. The host subnet mask is incorrect