Tugas 2 (Network Administrator)

1.Diketahui : subnet :25,Class:B,Max 300 komputer persegmen
Ditanya : subnet mask.
Jawab :
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
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
Jawab : E. 172.31.192.160.

4.A.255.0.0.0 untuk kelas A
B.255.254.0.0 untuk kelas A dengan prefix 15
C.255.224.0.0 untuk kelas A dengan prefix 11
F.255.0.0.0 untuk kelas C dengan prefix 26
Jawab : D. 255.255.0.0 & E. 255.255.252.0

5. Ditanya : combination of network id and subnet mask correctly identifies all IP addresses from 172.16.128.0 through 172.16.159.255?
Maka kita bisa dapatkan blog subnetnya yaitu 160-128 = 32, sehingga untuk mecari subnet mask yang baru adalah misal subnetmask baru = x, 256-x = 32,maka x adalah 256-32=224
Sehingga subnetmask yang baru adalah 255.255.255.224
Jawab : A. 172.16.128.0 and 255.255.255.224

6. Diket : 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. Diket : IP address : 192.168.99.0/29 kelas C
Ditanya : 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.Diketahui : IP address : 192.168.4.0 kelas C subnetmask : 255.255.255.224
Host/blok : 256-224 = 32 host max : 32-2 = 30
Jawab : C. 30

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

10.Diketahui : setiap subnetwork mempunyai 14 host yang digunakan.
Ditanya : berapakah subnetmask yang baru?
2n-2≥14, n=4 subnetmask : 111111111.1111111.11111111.11110000 = 255.255.255.248
Jawab : C. 255.255.255.248

11.Diketahui :Class B membutuhkan 100 Networks
2n -2≥100, n = 7 Subnetmasknya : 11111111.11111111.11111111.11000000 = 255.255.255.192
Jawab: F. 255.255.255.192.

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.Diket : IP Address dengan CIDR 115.64.4.0/22
Subnetmask : 11111111.11111111.11111100.00000000 = 255.255.252.0
Jawaban: b. 115.64.7.64 c. 115.64.6.25515. e. 115.64.5.128


15.Diket : IP address 200.10.5.68/28
Subnetmask : 11111111.11111111.11111111.11110000 = 255.255.255.240
Host per blok : 256-240 = 16
200.10.5.0 200.10.5.1 - .14 200.10.5.15
……
200.10.5.64 200.10.5.65 - .78 200.10.5.79
Jawaban : C. 200.10.5.64

16. The network address of 172.16.0.0/19
Subnetmask : 11111111.11111111.11100000.00000000 = 255.255.224.0
Subnets bit "1" = 23 = 8
Host bit "0" = 213 = 8190
Jawaban : E. 8 subnets, 8190 hosts each


17. Diket:Subnet =500 subnet(Class B), setiap subnet digunakan 100 host.
Subnetmask = 11111111.11111111.11111111.10000000 = 255.255.255.128
Subnet = 2n > 500, N = 9 bit "1"

Jawaban : B. 255.255.255.128

18.Diket : IP address 172.16.66.0/21
Subnetmask : 11111111.11111111.11111000.00000000 = 255.255.248. 0
Host: 256-248 = 8
172.16.0.0 172.16.1.0 - .6.0 172.16.7.0
….
172.16.64.0 172.16.65.0 - .70.0 172.16.71.0
Jawaban : C. 172.16.64.0

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

20. Diket : IP address 192.168.19.24/29
Subnetmask : 11111111.11111111.11111111.11111000 = 255.255.255.248
Host Id = 23-3 =6host /network
Net id range broadcast
192.168.19.0 192.168.19.1 - 192.168.19.6 192.168.19.7
….
192.168.19.24 192.168.19.25 - .30 192.168.19.31
Jawaban : C. 192.168.19.26 255.255.255.248

21. Diketahui :minimum dari 300 subnets with a maximum of 50 hosts per subnet.
Jawab : 2n-2≥300, n=9 subnetmasknya 11111111.11111111.11111111.10000000 = 255.255.255.128
2n-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
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 2n = 2n-2 > 3350, N = 12

Subnetmask : 11111111.11111111.11111000.00000000 = 255.255.248.0
Jawaban : C. 255.255.248.0

24. Diket : network with a subnet of 172.16.17.0/22.
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. Diket :Sebuah Router dengan IP Ethernet0: 172.16.112.1/20.
Subnetmask : 11111111.11111111.11110000.00000000 = 255.255.240.0
Host n = 12 (bit “0”), 212 – 2 = 4094
Jawaban : C. 4094

26.Diket : You have a /27 subnet mask.
Ditanya : IP yang valid adalah ?
Prefixnya 27 maka subnetmasknya 11111111.11111111.11111111.11100000 = 255.255.255.224, sehingga 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
Jawab : B. 90.10.170.93 , C. 143.187.16.56, dan D. 192.168.15.87

27. Diket : You have a Class B network ID and need about 450 IP addresses per subnet. What is the best mask for this network?
Kelas B, 450 host/subnet
Host 2n - 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 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.