Tugas 2 NETADMIN (jawaban dan pembahasanya)

1. Dik : 300 host
peny : 2^N-2 >= 300
2^N>=302
N=9
subnet : 11111111.11111111.11111110.00000000
maka,255.255.255.254.0

jawaban C

2. gateway : 192.168.1.65 dimana 192.168.1.?/27
peny : sm = 11111111.11111111.11100000.00000000
255.255.224.0
bloksubnet : 256 - 224 = 32
maka ,terletak di ip range 192.168.1.65 - 192.168.1.94

jawaban D dan F

3. peny : blok subnet = 256 - 248 = 8
blok ip terakhir = (x div blok subnet) blok subnet
= (166 div 8)8 = 160

jawaban E

4. Alasanya karena termasuk kelas B

jawaban D dan E

5. berarti blok subnet = 160 - 128
= 32
nah,baru blok subnet berikutnya
= 256 - 32 = 224 ,maka 172.16.224.0

jawaban D

6. sm : 11111111.11111111.11111111.11111000
255.255.255.248 ,blok subnet = 8
dimana pada broadcast 223.168.17.167

jawaban C

7. net id = 2^n-2 =2^5-2 = 30
host id = 2^N-2 = 2^3-2 = 6

jawaban C

8. maka = 11111111.11111111.11111111.11111000
host id = 2^N-2 = 2^5-2 = 30

jawaban C

9. host id 2^N-2>= 27
2^N>=29
N= 5 ,maka 2^N = 2^5 = 32
blok subnet = 256 - 32 = 224
maka 255.255.255.224

jawaban C

10. host id 2^N-2 = 14
2^N = 16
N= 4
blok subnet = 256 - 16 = 240
maka 255.255.255.240

jawaban C

11. host id 2^N-2 >= 100
2^N >=102
N = 7 ,maka 2^N = 2^7 = 128
blok subnetnya = 256 - 128 = 128
maka 255.255.255.128

jawaban E

12. Karena termasuk kedalam kelas B

jawaban C

13. net = 6,host = 2 ,dan blok subnetnya = 2^N = 2^2 =4
subnetmasknya = (210 div 4)4 = 208

jawaban C

14. host = 2 ,blok subnet = 2^N = 2^2 = 4
range ipnya 115.64.4.0 - 115.64.8.0

jawaban B,C, dan E

15. net = 4 ,blok subnetnya 2^4 = 16
subnetmasknya = (68 div 16)16 = 64

jawaban C

16. net = 19 - 16 = 3,host = 32 -16 = 13 dan networknya = 2^n = 2^3 = 8
maka host id = 2^N-2 = 2^13-2 = 8190

jawaban F

17. host id 2^N-2 >= 100
2^N >= 102
N = 7 ,dimana 2^N = 2^7 = 128 ,maka 255.255.255.128

jawaban B

18. host = 24-21 = 3, blok subnet = 2^N = 2^3 = 8
subnetnya (66 div 8) 8 = 64

jawaban C

19. host id 2^N-2 >= 500
2^N >= 502
N = 9
blok subnet = 2^9-8 = 2
blok subnetnya = 256 - 2 = 254

jawaban B

20. N = 32 - 29 = 3 ,dimana netmask = 256 - 8 = 248 dan blok subnet = 2^N =2^3 = 9

jawaban C

21. host id 2^N-2>= 50
2^N>=52
N = 6 , maka netmask 11111111.11111111.11111111.11000000
255.255.255.192

jawaban E

22. host = 32-25 = 7 ,dimana blok subnet = 2^N=2^7 = 128
sehingga subnet addressnya = (1 div 128)128 = 0

jawaban A

23. host id 2^N-2>= 850
2^N>= 852
N = 10
maka netmasknya 11111111.11111111.11111100.00000000
255.255.252.0

jawaban D

24. sm = 255.255.252.0
maka blok subnetnya = 256 - 252 = 4

jawaban E

25. host = 32-20 = 12, host valid = 2^N-2 =2^12-2 = 4094

jawaban C

26. CIDR /27
host = 32-27 = 5 ,blok subnetnya = 2^N = 2^5 = 32

jawaban B ,C dan D

27. CIDR /23
host id 2^N-2 >=450
2^N>=452
N=9
maka netmasknya = 11111111.11111111.11111110.00000000
255.255.254.0

jawaban C

28. CIDR /27
maka netmasknya = 11111111.11111111.11111111.11100000
255.255.255.224

jawaban A