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soal subnetting
by mencoba berkarya on Nov.22, 2009, under netadmin
1. A company has the following addressing scheme requirements:
-currently has 25 subnets
-uses a Class B IP address
-has a maximum of 300 computers on any network segment
-needs to leave the fewest unused addresses in each subnet
What subnet mask is appropriate to use in this company?
a. 255.255.240.0
b. 255.255.248.0
c. 255.255.254.0
d. 255.255.255.0
e. 255.255.255.128
f. 255.255.255.248
jawaban : C
jadi submask yang mencukupi host 300 adalah 2^9-2 = 510, maka submasknya adalah
11111111.11111111.11111110.00000000
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 : D. Address - 192.168.1.82
Gateway -192.168.1.65
F. Address - 192.168.1.70
Gateway -192.168.1.65
3. A NIC of a computer has been assigned an IP address of 172.31.192.166 with a mask of
255.255.255.248. To which subnet does the IP address belong?
a. 172.31.0.0
b. 172.31.160.0
c. 172.31.192.0
d. 172.31.248.0
e. 172.31.192.160
f. 172.31.192.248
jwb. E
172.31.192.166 -> 10101100.00011111.11000000.10100110
255.255.255.248->11111111.11111111.11111111.11111000
172.31.192.160 -> 10101100.00011111.11000000.10100000
4. Which subnet masks would be valid for a subnetted Class B address? (Choose two.)
a. 255.0.0.0
b. 255.254.0.0
c. 255.224.0.0
d. 255.255.0.0
e. 255.255.252.0
f. 255.255.255.192
jwb. D dan E
Karena batas dari submask untuk kelas B adalah 255.255.0.0 – 255.255.254.0
5. Which combination of network id and subnet mask correctly identifies all IP addresses
from 172.16.128.0 through 172.16.159.255?
a. 172.16.128.0 and 255.255.255.224
b. 172.16.128.0 and 255.255.0.0
c. 172.16.128.0 and 255.255.192.0
d. 172.16.128.0 and 255.255.224.0
e. 172.16.128.0 and 255.255.255.192
jwb.
6. Which type of address is 223.168.17.167/29?
a. host address
b. multicast address
c. broadcast address
d. subnetwork address
jwb. D
karna address 223.168.17.167/29 menempati pada submask 255.255.255.248
maka memiliki blok ip 8. Sedangkan alamat 223.168.17.167 merupaka broadcast dari jaringan 223.168.17.160.
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 Which of the following IP addresses fall into the CIDR block of 115.64.4.0/22? (Choose
three.)
a. 115.64.8.32
b. 115.64.7.64
c. 115.64.6.255
d. 115.64.3.255
e. 115.64.5.128
f. 115.64.12.128
jwb. A dan F
Submask 11111111.11111111.11111100.00000000, maka subnetnya 2^6 -> 64 dan blocknya adalah 4. Sehinga subnetwork yang tepat adalah 115.64.8.32 dan 115.64.12.128
15. What is the subnetwork address for a host with the IP address 200.10.5.68/28?
a. 200.10.5.56
b. 200.10.5.32
c. 200.10.5.64
d. 200.10.5.0
jwb. C
Submask 11111111.11111111.11111111.11110000 (255.255.255.240), sehinga mempunyai block =16 dimana IP address200.10.5.68 merupakan subnet dari 200.10.5.64
16. The network address of 172.16.0.0/19 provides how many subnets and hosts?
a. 7 subnets, 30 hosts each
b. 7 subnets, 2046 hosts each
c. 7 subnets, 8190 hosts each
d. 8 subnets, 30 hosts each
e. 8 subnets, 2046 hosts each
f. 8 subnets, 8190 hosts each
jwb. F
17. You need 500 subnets, each with about 100 usable host addresses per subnet. What mask
will you assign using a Class B network address?
a. 255.255.255.252
b. 255.255.255.128
c. 255.255.255.0
d. 255.255.254.0
jwb. C
submask 11111111.11111111.11111111.00000000 (255.255.255.0) dapat memiliki 100 host, dimana host max 2^8-2 = 126.
18. What is the subnetwork number of a host with an IP address of 172.16.66.0/21?
a. 172.16.36.0
b. 172.16.48.0
c. 172.16.64.0
d. 172.16.0.0
jwb. C
submask 11111111.11111111.11111000.00000000 (255.255.248.0), maka blocknya adalah 256-248 = 8. Sedangkan IP address 172.16.66.0 terdapat pada subnet 172.16.64.0
19. What mask would you assign to the network ID of 172.16.0.0 if you needed about 100
subnets with about 500 hosts each?
a. 255.255.255.0
b. 255.255.254.0
c. 255.255.252.0
d. 255.255.0.0
jwb. B
Untuk host 500, maka2^9-2 = 510 sehinga submasknya adalah 11111111.11111111.111111110.00000000 (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 Your router has the following IP address on Ethernet0: 172.16.112.1/20. How many hosts
can be accommodated on the Ethernet segment?
a. 1024
b. 2046
c. 4094
d. 4096
e. 8190
jwb. C
submask 11111111.11111111.11110000.00000000 (255.255.240.0), maka banyakny ahost yang tersedia adalah 2^12-2 = 4094
26. You have a /27 subnet mask. Which of the following are valid hosts? (Choose three.)
a. 11.244.18.63
b. 90.10.170.93
c. 143.187.16.56
d. 192.168.15.87
e. 200.45.115.159
f. 216.66.11.192
jwb. B, C, dan D
Submask 11111111.11111111.11111111.11100000 (255.255.255.224) dengan blok 256-224 = 32, maka host yang memenuhi adalah 90.10.170.93, 143.187.16.56, dan 192.168.15.87
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.
-currently has 25 subnets
-uses a Class B IP address
-has a maximum of 300 computers on any network segment
-needs to leave the fewest unused addresses in each subnet
What subnet mask is appropriate to use in this company?
a. 255.255.240.0
b. 255.255.248.0
c. 255.255.254.0
d. 255.255.255.0
e. 255.255.255.128
f. 255.255.255.248
jawaban : C
jadi submask yang mencukupi host 300 adalah 2^9-2 = 510, maka submasknya adalah
11111111.11111111.11111110.00000000
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 : D. Address - 192.168.1.82
Gateway -192.168.1.65
F. Address - 192.168.1.70
Gateway -192.168.1.65
3. A NIC of a computer has been assigned an IP address of 172.31.192.166 with a mask of
255.255.255.248. To which subnet does the IP address belong?
a. 172.31.0.0
b. 172.31.160.0
c. 172.31.192.0
d. 172.31.248.0
e. 172.31.192.160
f. 172.31.192.248
jwb. E
172.31.192.166 -> 10101100.00011111.11000000.10100110
255.255.255.248->11111111.11111111.11111111.11111000
172.31.192.160 -> 10101100.00011111.11000000.10100000
4. Which subnet masks would be valid for a subnetted Class B address? (Choose two.)
a. 255.0.0.0
b. 255.254.0.0
c. 255.224.0.0
d. 255.255.0.0
e. 255.255.252.0
f. 255.255.255.192
jwb. D dan E
Karena batas dari submask untuk kelas B adalah 255.255.0.0 – 255.255.254.0
5. Which combination of network id and subnet mask correctly identifies all IP addresses
from 172.16.128.0 through 172.16.159.255?
a. 172.16.128.0 and 255.255.255.224
b. 172.16.128.0 and 255.255.0.0
c. 172.16.128.0 and 255.255.192.0
d. 172.16.128.0 and 255.255.224.0
e. 172.16.128.0 and 255.255.255.192
jwb.
6. Which type of address is 223.168.17.167/29?
a. host address
b. multicast address
c. broadcast address
d. subnetwork address
jwb. D
karna address 223.168.17.167/29 menempati pada submask 255.255.255.248
maka memiliki blok ip 8. Sedangkan alamat 223.168.17.167 merupaka broadcast dari jaringan 223.168.17.160.
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 Which of the following IP addresses fall into the CIDR block of 115.64.4.0/22? (Choose
three.)
a. 115.64.8.32
b. 115.64.7.64
c. 115.64.6.255
d. 115.64.3.255
e. 115.64.5.128
f. 115.64.12.128
jwb. A dan F
Submask 11111111.11111111.11111100.00000000, maka subnetnya 2^6 -> 64 dan blocknya adalah 4. Sehinga subnetwork yang tepat adalah 115.64.8.32 dan 115.64.12.128
15. What is the subnetwork address for a host with the IP address 200.10.5.68/28?
a. 200.10.5.56
b. 200.10.5.32
c. 200.10.5.64
d. 200.10.5.0
jwb. C
Submask 11111111.11111111.11111111.11110000 (255.255.255.240), sehinga mempunyai block =16 dimana IP address200.10.5.68 merupakan subnet dari 200.10.5.64
16. The network address of 172.16.0.0/19 provides how many subnets and hosts?
a. 7 subnets, 30 hosts each
b. 7 subnets, 2046 hosts each
c. 7 subnets, 8190 hosts each
d. 8 subnets, 30 hosts each
e. 8 subnets, 2046 hosts each
f. 8 subnets, 8190 hosts each
jwb. F
17. You need 500 subnets, each with about 100 usable host addresses per subnet. What mask
will you assign using a Class B network address?
a. 255.255.255.252
b. 255.255.255.128
c. 255.255.255.0
d. 255.255.254.0
jwb. C
submask 11111111.11111111.11111111.00000000 (255.255.255.0) dapat memiliki 100 host, dimana host max 2^8-2 = 126.
18. What is the subnetwork number of a host with an IP address of 172.16.66.0/21?
a. 172.16.36.0
b. 172.16.48.0
c. 172.16.64.0
d. 172.16.0.0
jwb. C
submask 11111111.11111111.11111000.00000000 (255.255.248.0), maka blocknya adalah 256-248 = 8. Sedangkan IP address 172.16.66.0 terdapat pada subnet 172.16.64.0
19. What mask would you assign to the network ID of 172.16.0.0 if you needed about 100
subnets with about 500 hosts each?
a. 255.255.255.0
b. 255.255.254.0
c. 255.255.252.0
d. 255.255.0.0
jwb. B
Untuk host 500, maka2^9-2 = 510 sehinga submasknya adalah 11111111.11111111.111111110.00000000 (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 Your router has the following IP address on Ethernet0: 172.16.112.1/20. How many hosts
can be accommodated on the Ethernet segment?
a. 1024
b. 2046
c. 4094
d. 4096
e. 8190
jwb. C
submask 11111111.11111111.11110000.00000000 (255.255.240.0), maka banyakny ahost yang tersedia adalah 2^12-2 = 4094
26. You have a /27 subnet mask. Which of the following are valid hosts? (Choose three.)
a. 11.244.18.63
b. 90.10.170.93
c. 143.187.16.56
d. 192.168.15.87
e. 200.45.115.159
f. 216.66.11.192
jwb. B, C, dan D
Submask 11111111.11111111.11111111.11100000 (255.255.255.224) dengan blok 256-224 = 32, maka host yang memenuhi adalah 90.10.170.93, 143.187.16.56, dan 192.168.15.87
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.
pembagian ip v4
by mencoba berkarya on Nov.22, 2009, under netadmin
pembagian ip v4
Kelas A
1 st byte = 0
Net length = 8
Host Range = 24
IP Range = 0-127
IP use = 1-126
∑ net = 2n (28-1 = 27)
∑ host = 2n – 2 (224-2 =16777214 )
Subnet mask = 255.0.0.0
Kelas B
1 st byte = 10
Net length = 16
Host Range = 16
IP Range = 128-191
IP use = 129-190
∑ net = 2n (216-1 = 215)
∑ host = 2n – 2 (216-2 = 65534)
Subnet mask = 255.255.0.0
Kelas C
1 st byte = 110
Net length = 24
Host Range = 8
IP Range = 192-223
IP use = 193-222
∑ net = 2n (224-1 = 223)
∑ host = 2n – 2 (28-2 =254)
Subnet mask = 255.255.255.0
kenapa kita bisa ping ke 127.0.0.1.....?
C:\Documents and Settings\Tackhin>ping 127.0.0.1
Pinging 127.0.0.1 with 32 bytes of data
Reply from 127.0.0.1: bytes=32time<1ms TTL=128
Reply from 127.0.0.1: bytes=32 time<1ms TTL=128
Reply from 127.0.0.1: bytes=32 time<1ms TTL=128
jadi ip 127.0.0.1 itu adalah alamat komputer kita sendiri, bisa dibilang itu adalah localhost komputer kita
ip private
Kelas A
1 st byte = 0
Net length = 8
Host Range = 24
IP Range = 0-127
IP use = 1-126
∑ net = 2n (28-1 = 27)
∑ host = 2n – 2 (224-2 =16777214 )
Subnet mask = 255.0.0.0
Kelas B
1 st byte = 10
Net length = 16
Host Range = 16
IP Range = 128-191
IP use = 129-190
∑ net = 2n (216-1 = 215)
∑ host = 2n – 2 (216-2 = 65534)
Subnet mask = 255.255.0.0
Kelas C
1 st byte = 110
Net length = 24
Host Range = 8
IP Range = 192-223
IP use = 193-222
∑ net = 2n (224-1 = 223)
∑ host = 2n – 2 (28-2 =254)
Subnet mask = 255.255.255.0
kenapa kita bisa ping ke 127.0.0.1.....?
C:\Documents and Settings\Tackhin>ping 127.0.0.1
Pinging 127.0.0.1 with 32 bytes of data
Reply from 127.0.0.1: bytes=32time<1ms TTL=128
Reply from 127.0.0.1: bytes=32 time<1ms TTL=128
Reply from 127.0.0.1: bytes=32 time<1ms TTL=128
jadi ip 127.0.0.1 itu adalah alamat komputer kita sendiri, bisa dibilang itu adalah localhost komputer kita
ip private
- 10.0.0.0/8
- 172.16.0.0/12
- 192.168.0.0/16
10.0.0.0/8
Jaringan pribadi (private network) 10.0.0.0/8 merupakan sebuah network identifier kelas A yang mengizinkan alamat IP yang valid dari 10.0.0.1 hingga 10.255.255.254. Jaringan pribadi 10.0.0.0/8 memiliki 24 bit host yang dapat digunakan untuk skema subnetting di dalam sebuah organisasi privat.
172.16.0.0/12
Jaringan pribadi 172.16.0.0/12 dapat diinterpretasikan sebagai sebuah block dari 16 network identifier kelas B atau sebagai sebuah ruangan alamat yang memiliki 20 bit yang dapat ditetapkan sebagai host identifier, yang dapat digunakan dengan menggunakan skema subnetting di dalam sebuah organisasi privat. Alamat jaringan privat 17.16.0.0/12 mengizinkan alamat-alamat IP yang valid dari 172.16.0.1 hingga 172.31.255.254.
192.168.0.0/16
Jaringan pribadi 192.168.0.0/16 dapat diinterpretasikan sebagai sebuah block dari 256 network identifier kelas C atau sebagai sebuah ruangan alamat yang memiliki 16 bit yang dapat ditetapkan sebagai host identifier yang dapat digunakan dengan menggunakan skema subnetting apapun di dalam sebuah organisasi privat. Alamat jaringan privat 192.168.0.0/16 dapat mendukung alamat-alamat IP yang valid dari 192.168.0.1 hingga 192.168.255.254.
sumber:http://id.wikipedia.org/wiki/Alamat_IP_versi_4
