Tugas ke 2 Net admin

1. Dik : host =300
2n- 2 ≥300, maka n =9
29 - 2 =510
subnet mask menjadi :
11111111.11111111.11111110.0000000
C. 255.255.254.0

2. Eth0 = 192.168.1.65/27
Subnetmask :11111111.11111111.11111111.11100000
Host = 25-2 = 30 host/subnet
Net = 23 - 2 =6 subnet
Net id range broadcast
192.168.1.0 192.168.1.1 – 192.168.1.30 192.168.1.31
192.168.1.32 192.168.1.33 – 192.168.1.62 192.168.1.63
192.168.1.64 192.168.1.65 – 192.168.1.96 192.168.1.95
192.168.1.96 ……. ……
Syarat saling terhubung : Berada pada range yang sama
Jawaban : F. Address - 192.168.1.70
Gateway -192.168.1.65
D. Address - 192.168.1.82
Gateway -192.168.1.65

3. IP address = 172.31.192.166
Subnet Mask = 11111111.11111111.11111111.11111000
Host Id = 23-2= 6
Net Id = 25-2 =30
Net id range broadcast
172.31.192.0 172.31.192.1- 172.31.192.6 172.31.192.7
172.31.192.8 172.31.192.9- 172.31.192.14 172.31.192.15
…
172.31.192.160 172.31.192.161- 172.31.192.166 172.31.192.167
Jawaban : E. 172.31.192.160

4. Analisa pada setiap pilihan
A. 255.0.0.0 digunakan untuk kelas A
B. 255.254.0.0 digunakan untuk kelas A dengan CIDR (/15)
C. 255.224.0.0 digunakan untuk kelas A dengan CIDR (/11)
F. 255.0.0.0 digunakan untuk kelas C dengan CIDR (/26)
Jawaban : D. 255.255.0.0 E. 255.255.252.0

5. Net id range broadcast
172.16.128.0 172.16.159.255 172.16.159.255
172.16.160.0 …….
Net id :160-128 = 32 = 25
Subnetmask : 11111111.11111111.11111000.00000000
Jawaban : D. 172.16.128.0 and 255.255.224.0

6. Ip Address : 223.168.17.167/29
Subnetmask : 11111111.11111111.11111111.11111000
Host Id =23-2=6 Host/subnet
Net id range broadcast
223.168.17.0 223.168.17.1 – 223.168.17..6 223.168.17.7
223.168.17.8 223.168.17.9 – 223.168.17.14 223.168.17.15
……..
223.168.17.160 223.168.17.161 - .166 223.168.17.167
Jawaban : C. broadcast address
7. IP address : 192.168.99.0/29 (kelas C)
Subnetmask : 11111111.11111111.11111111.11111000
Host Id : 23-2 = 6 host/subnet
Net Id : 25-2 = 30 subnet
Jawaban : C. 30 networks / 6 hosts

8. IP address : 192.168.4.0 (kelas C)
subnetmask : 255.255.255.224 = 11111111.1111111111.1111111.11100000
Host Id : 25-2 = 30 host/subnet
*Dikurangi 2 untuk pemakaian broadcast dan localhost/loopback
Jawaban : C. 30

9. 27 host /subnet = 2n-2 ≥27 , n = 5
2^5-2=30
jumlah host id = 30 /subnet, maka subnet mask = 11111111.11111111.11111111.11100000
Jawaban : C. 255.255.255.224
10. 14 host/subnet ,maka 2n-2 ≥14, n = 4 ,karena 24-2 = 14
Untuk Jumlah Host Id= 14/subnet ,maka subnetmask :11111111.11111111.1111111.11110000
Jawaban : C. 255.255.255.240

11. Pada kelas B ,membutuhkan 100 networks
2n -2≥100, n = 7
Subnetmask : 11111111.11111111.11111111.10000000 = 255.255.255.128
Jawaban : F. 255.255.255.128

12. IP Address = 172.32.65.13(Kelas B)
Default Mask = 255.255.0.0
Jawaban :C. 172.32.0.0

13. IP address of 172.16.210.0/22
Subnetmask : 11111111.11111111.11.0000000
Host Id : 210-2= 1022
Net id range broadcast
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. IP address 115.64.4.0/22
Subnetmask : 11111111.11111111.11111100.00000000
Host Id = 210-2 = 1022
Net id range broadcast
115.64.4.0 115.64.4.1 – 115.64.4.6 115.64.4.7
115.64.8.0 …………….
Jawaban : B, C, E

15. IP address 200.10.5.68/28
Subnetmask : 11111111.11111111.11111111.11110000
Host Id : 24-2 =14
Net id range broadcast
200.10.5.0 200.10.5.1 – 200.10.5.14 200.10.5.15
……
200.10.5.64 200.10.5.65 – 200.10.5.78 200.10.5.79
Jawaban : C. 200.10.5.64

16. 172.16.0.0/19
Subnetmask : 11111111.11111111.11100000.00000000
Net Id : 23-2 = 6
Host Id: = 213 = 8190 /subnet
Jawaban : E. 8 subnets, 8190 hosts each

17. Kelas B, 500 subnet, setiap subnet digunakan 100 host, mask??
Subnet = 2N > 500, N = 9 (bit “1”)
Subnetmask = 11111111.11111111.11111111.10000000 = 255.255.255.128
Jawaban : B. 255.255.255.128

18 . IP address 172.16.66.0/21
Subnetmask : 11111111.11111111.11111000.00000000 = 255.255.248. 0
Host per blok : 256-248 = 8
Net id range broadcast
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. Kelas B, 100 subnet dan 500 host/subnet
Subnet = 2N > 100, N = 7 (bit “1”)
Subnetmask = 11111111.11111111.11111110.00000000
Jawaban : B. 255.255.254.0

20. 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. subnet = 300 subnet , Host = 50 host/subnet
• 26-2 =62 ≥50 ,11111111.11111111.11111111.11000000 =255.255.255.192
• 27 -2 =126 ≥ 50 11111111.11111111.11111111.10000000 =255.255.255.128
Jawaban : B dan E

22. IP address 172.16.112.1/25
Subnetmask : 11111111.11111111.11111111.10000000 = 255.255.255.128
Host Id = 27- 2 =126
Net id range broadcast
172.16.112.0 172.16.112.1- .126 172.16.112.127
172.16.112.128 ………..
Jawaban : A. 172.16.112.0

23. Jumlah host yang ada = 3350
Host 2n - 2 > 3350, n = 12
Subnetmask : 11111111.11111111.11111000.00000000 = 255.255.248.0
Jawaban : C. 255.255.248.0

24. Subnet 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. Eth0: 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. Prefix /27 Class C
11111111.11111111.11111111.11100000
Subnet 23-2=6
Host Id 25-2 = 30 subnet
Jawaban : D,E,F

27. Kelas B, 450 host/subnet
Host Id 2n - 2 > 450, n = 9
Subnetmask = 11111111.11111111.11111110.00000000 = 255.255.254.0
Jawaban : C. 255.255.254.0

28. Eth0 = 198.18.166.33/27
Subnetmask = 11111111.11111111.11111111.11100000 = 255.255.255.224
IP Address 198.18.166.65 dihubungkan dengan Eth0 gateway 198.18.166.33, maka harus mengikuti ethernetnya.
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.