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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
Jawab:
kelas B default subnetmask 255.255.0.0
host: 2n-2 ≥ jumlah host
2n-2 ≥ 300
2n ≥ 302
n ≥ 9 (bit 0)
11111111.11111111.11111110.00000000 = 255.255.255.254

2. Refer to the exhibit. Host A is being manually configured for connectivity to the LAN. Which two addressing scheme combinations are possible configurations that can be applied to the host for connectivity? (Choose two.)

a. Address - 192.168.1.14
Gateway - 192.168.1.33
b. Address - 192.168.1.45
Gateway - 192.168.1.33
c. Address - 192.168.1.32
Gateway - 192.168.1.33
d. Address - 192.168.1.82
Gateway - 192.168.1.65
e. Address - 192.168.1.63
Gateway - 192.168.1.65
f. Address - 192.168.1.70
Gateway - 192.168.1.65
Jawab:
- prefik /27 maka subnet masknya : 11111111.11111111.11111111.11100000
- Blog subnet: 256-224=32
- Range IP: 192.168.1.65 – 192.168.1.94

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
blog subnet 256-248=8

- range IP kelipatan 8 ada di sekitar = 172.31.192.160

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
- Untuk default subnet kelas B adalah 2 oktet pertamanya harus full dan oktet yang lain bebas yaitu 255.255.0.0 dan 255.255.252.0


6. Which type of address is 223.168.17.167/29?

a. host address
b. multicast address
c. broadcast address
d. subnetwork address

- prefik /29 subnetmaskny 11111111.11111111.11111111.11111000 = 255.255.255.248
- blog subnet: 256-248=8
- net ID kelipatan 8 eshingga 223.168.17.167 termasuk broadcast.

7. What is the correct number of usable subnetworks and hosts for the IP network address 192.168.99.0 subnetted with a /29 mask?

a. 6 networks / 32 hosts
b. 14 networks / 14 hosts
c. 30networks /6 hosts
d. 62 networks / 2 hosts

- Prefix /29 subnetmask adalah 11111111.11111111.11111111.11111000 = 255.255.255.248
- Rumus cari network (bit 1) : 25 = 32
- Rumus cari host (bit 0) : 23-2 = 6

8. Company XYZ uses a network address of 192.168.4.0. It uses the mask of 255.255.255.224 to create subnets. What is the maximum number of usable hosts in each subnet?

a. 6
b. 14
c. 30
d. 62

- Subnet 255.255.255.224 = 11111111.11111111.11111111.11100000
- Rumus cari host (bit 0) : 25-2 = 30

9. A company is planning to subnet its network for a maximum of 27 hosts. Which subnet mask would provide the needed hosts and leave the fewest unused addresses in each subnet?

a. 255.255.255.0
b. 255.255.255.192
c. 255.255.255.224
d. 255.255.255.240
e. 255.255.255.248

host (bit 0) :
2n-2 ≥ 27
2n ≥ 29
n ≥ 5 (bit 0)
11111111.11111111.11111111.11100000 = 255.255.255.224

10. An IP network address has been subnetted so that every subnetwork has 14 usable host IP addresses. What is the appropriate subnet mask for the newly created subnetworks?

a. 255.255.255.128
b. 255.255.255.224
c. 255.255.255.240
d. 255.255.255.248
e. 255.255.255.252

- Rumus cari host (bit 0) :
2n-2 ≥ 14
2n ≥ 16
n ≥ 4 (bit 0)
11111111.11111111.11111111.11110000 = 255.255.255.240

11. A company is using a Class B IP addressing scheme and expects to need as many as 100 networks. What is the correct subnet mask to use with the network configuration?

a. 255.255.0.0
b. 255.255.240.0
c. 255.255.254.0
d. 255.255.255.0
e. 255.255.255.128
f. 255.255.255.192

- Rumus cari network (bit 1) :
2n ≥ 100
n ≥ 7(bit 1)
11111111.11111111.11111110.00000000 = 255.255.254.0

12. Given a host with the IP address 172.32.65.13 and a default subnet mask, to which network does the host belong?

a. 172.32.65.0
b. 172.32.65.32
c. 172.32.0.0
d. 172.32.32.0

13. What is the subnetwork number of a host with an IP address of 172.16.210.0/22?

a. 172.16.42.0
b. 172.16.107.0
c. 172.16.208.0
d. 172.16.252.0
e. 172.16.254.0

- 172.16.208.0 merupakan subnetwork

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

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

- prefix /28 sama dengan 11111111.11111111.11111111.11110000 = 255.255.255.240

- blog subnet: 256-240 = 16, jadi kelipatan dari range IP soal diatas adalah 16.

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

- prefiks /19 sama dengan 11111111.11111111.11100000.00000000
- network (bit 1) :
23 = 8
- host (bit 0) :
213 – 2 = 8190

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

- network (bit 1) :
2n ≥ 500
n ≥ 9 (bit 1)
- 11111111.11111111.11111111.1000000 = 255.255.255.128

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

- Prefiks /21 = 11111111.11111111.11111000.00000000
- blog subnet : 256 – 248 = 8
- jadi 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

- network (bit 1) :
2n ≥ 100
n ≥ 7 (bit 1)
11111111.11111111.11111110.00000000 = 255.255.254.0

20. You need to configure a server that is on the subnet 192.168.19.24/29. The router has the first available host address. Which of the following should you assign to the server?

a. 192.168.19.0 255.255.255.0
b. 192.168.19.33 255.255.255.240
c. 192.168.19.26 255.255.255.248
d. 192.168.19.31 255.255.255.248
e. 192.168.19.34 255.255.255.240

- Prefiks /29 sama dengan 11111111.11111111.11111111.11111000 = 255.255.255.248
- blog subnet: 256 – 248 = 8
- kelipatan 8 pada range 192.168.19.24 – 192.168.19.31

21. You need a minimum of 300 subnets with a maximum of 50 hosts per subnet. Which of the following masks will support the business requirements? (Choose two.)

a. 255.255.255.0
b. 255.255.255.128
c. 255.255.252.0
d. 255.25.255.224
e. 255.255.255.192
f. 255.255.248.0

- network (bit 1) :
2n ≥ 300
n ≥ 9 (bit 1)
11111111.11111111.11111111.10000000 = 255.255.255.128

- host (bit 0) :
2n – 2 ≥ 50
2n ≥ 52
n ≥ 6 (bit 0)
11111111.11111111.11111111.11000000 = 255.255.255.192

22. If an Ethernet port on a router were assigned an IP address of 172.16.112.1/25, what would be the valid subnet address of this host?

a. 172.16.112.0
b. 172.16.0.0
c. 172.16.96.0
d. 172.16.255.0
e. 172.16.128.0

23. Refer to the exhibit. The internetwork in the exhibit has been assigned the IP address 172.20.0.0. What would be the appropriate subnet mask to maximize the number of networks available for future growth?

a. 255.255.224.0
b. 255.255.240.0
c. 255.255.248.0
d. 255.255.252.0
e. 255.255.254.0

- host (bit 0) :
2n – 2 ≥ 850
2n ≥ 852
n ≥ 10 = 1024 (bit 0)
11111111.11111111.11111100.00000000 = 255.255.255.252

24. You have a network with a subnet of 172.16.17.0/22. Which are valid host addresses?

a. 172.16.17.1 255.255.255.252
b. 172.16.0.1 255.255.240.0
c. 172.16.20.1 255.255.254.0
d. 172.16.16.1 255.255.255.240
e. 172.16.18.255 255.255.252.0
f. 172.16.0.1 255.255.255.0

- Prefiks /22 sama dengan 11111111.11111111.11111100.00000000 = 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

- Prefiks /20 sama dengan 11111111.11111111.11110000.00000000 = 255.255.240.0
- Rumus cari host (bit 0) :
2n – 2 ≥ jumlah host
212 – 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

27. You have a Class B network ID and need about 450 IP addresses per subnet. What is the best mask for this network?

a. 255.255.240.0
b. 255.255.248.0
c. 255.255.254.0
d. 255.255.255.0

- Rumus cari host (bit 0) :
2n – 2 ≥ 450
2n ≥ 452
n ≥ 9 = 512 (bit 0)
11111111.11111111.11111110.00000000 = 255.255.254.0

28. Host A is connected to the LAN, but it cannot connect to the Internet. The host configuration is shown in the exhibit. What are the two problems with this configuration? (Choose two.)

a. The host subnet mask is incorrect.
b. The host is not configured for subnetting.
c. The default gateway is a network address.
d. The default gateway is on a different network than the host.
e. The host IP address is on a different network from the Serial interface of the router.

- Prefik /27 sama dengan 11111111.11111111.11111111.11100000 = 255.255.255.224

- Blog subnet : 256 – 224 = 32

Karena alamat IP pada ruang alamat pribadi tidak akan pernah diberikan oleh Internet Network Information Center (InterNIC) sebagai alamat umum, maka route di dalam Internet router untuk alamat pribadi takkan pernah ada. Alamat pribadi tidak dapat dijangkau di dalam Internet. Oleh karena itu, saat memakai alamat IP pribadi membutuhkan beberapa tipe proxy atau server untuk mengkonversi sejumlah alamat IP pribadi pada jaringan lokal menjadi alamat IP umum yang dapat di-routed. Pilihan lain adalah menerjemahkan alamat pribadi menjadi alamat umum yang valid dengan Network Address Translator (NAT) sebelum dikirimkan di Internet. Dukungan bagi NAT untuk menerjemahkan alamat umum dan alamat pribadi memungkinkan terjadinya koneksi jaringan-jaringan kantor-rumah atau kantor yang kecil ke Internet.

Sebuah NAT menyembunyikan alamat-alamat IP yang dikelola secara internal dari jaringan-jaringan eksternal dengan menerjemahkan alamat internal pribadi menjadi alamat eksternal umum. Hal ini mengurangi biaya registrasi alamat IP dengan cara membiarkan para pelanggan memakai alamat IP yang tidak terdaftar secara internal melalui suatu terjemahan ke sejumlah kecil alamat IP yang terdaftar secara eksternal. Hal ini juga menyembunyikan struktur jaringan internal, mengurangi resiko penolakan serangan layanan terhadap sistem internal.