tugas praktikum SysAdmin

Tugas SysAdmin DHCP server ubuntu

DHCP server befungsi untuk mengontrol ip address pada clint server.
dibawah ini adalah cara untuk setting DHCP server pada ubuntu.

1. ketikkan " sudo apt-get install dhcp3-server" untuk menginstall dhcp nya.



2.yangkedua ketikkan " gedit etc/default/dhcp3-server" untuk melihat interface yang akan dijadikan dhcp.



3.nanti akan muncul page seperti ini. disini interface nya "eth2"


4.selanjutnya ketikkan "gedi /etc/dhcp3/dhcpd.conf" untuk mengatur range ip nya yang akan dibagikan.



5.setelah itu kita restart dengan ketikkan"/etc/init.d/dhcp3-server restart " untuk mengecek dhcp yang kita install tadi jalan atau tidaknya!

tugas praktikum SysAdmin update kernel pada linux

Pertama kita log in dulu dan masuk sebagai root. Lalu kita cek kernel yang ada sebelun update dengan cara ketik : uname-r pada hasil print screen dibawah ini kernel yang ada adalah 2.6.29.4.167.fc11.i586



Selanjutnya kita ketikkan : yum update kernel untuk melihat fersi kernel yang ada. System akan melakukan proses download.


Dibawah ini print screen proses update kernel tadi dengan kernel yang baru. Update selesai setelah ada
Tulisan “complete” dengan kernel yang baru 2.6.30.10.105.2.23.fc11.


Setelah itu, dari kernel yang ada pada server, kita diberi pilihan untuk mengupdate kernel kita dengan kernel yang ada pada server. Jika iya ketik : Y , jika tidak ketik : N. pada pilihan ini pilih Y maka system akan melakukan update kernel kita dengan kernel yang baru.



Setelah ini kita restart dengan cara ketik : halt . setelah itu cek lagi kernel kita yang baru.



Setelah direstart kita masuk lagi sebagai root.. dan ngecek kernel kita ynag baru. Caranya sama seperti ngecek kernel pertama kali tadi. Hasilnya kernel yang berjalan sekarang adalah :
2.6.30.10.105.2.23.fc11.i586.
Selesai untuk update kernel.

Tugas SysAdmin2

ini tugas subnetting SysAdmin, soalnya mang pake bahasa inggris dan ada 28 soal , tapi jawaban nya saya posting in pake bahasa indonesia, cinta bahasa indonesia .....

yuk...mulai!

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

Dikt.
- Sudah ada 25 subnet
- menggunakan kelas B
- maksimum 300 host per segmen jaringan
- perlu untuk meninggalkan address yang tidak berguna yang paling sedikit tiap subnetnya

Dit. Subnet mask yang paling sesuai

Jawab :

256 – 25 = 231

11111111.11111111.11111111.00000000
2^n =....

2^8 = 254 Host/Komputer

Jawabannya D

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

Dit. IP dan gateway yang paling sesuai ????

Jawab :

11111111.11111111.11111111.11100000

Jawabannya D dan F

Karena jumlah subnet adalah 8 dari
dan jumlah host per subnet nya adalah 30
Blok subnet = 256 – 224 = 32
32, 64, 96, 128, 160, 192

D dan F berada dalam blok dari 65 – 94 yang bisa digunakan, sedangkan 63 tidak bisa digunakarena

merupakan broadcast

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

Answer:

Block subnet = 256 – 248 = 8

8, 16, 24, 32, 40,……160..

Jawabannya E

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

Answer :

Jawabannya D & E

Karena subnetmask default unutk class B adalah 255.255.0.0 dan kombinasi dari 2 oktet

selanjutnya.

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

Answer :

Jawabannya C

6. Which type of address is 223.168.17.167/29?

a. host address

b. multicast address

c. broadcast address

d. subnetwork address

Answer :

IP addressnya termasuk golongan class C

/29 = 11111111.11111111.11111111.11111000

Subnet masknya = 255.255.255.248

Block subnet = 256 – 248 = 8

NET ID RHA 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

……………. …………………………….. ………………

223.168.17.160 223.168.17.161-223.168.17.166 223.168.17.167

………………. ………………………………… ………………

Jawabannya C

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. 30 networks / 6 hosts

d. 62 networks / 2 hosts

Answer :

IP addressnya termasuk golongan class C

/29 = 11111111.11111111.11111111.11111000

Subnet masknya = 255.255.255.248

Block subnet = 256 – 248 = 8

Host = 2³-2 = 6

Subnetwork = 2⁵-2 = 30

Jawabannya C

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

Answer :

255.255.255.224 = / 27

1111111.1111111.1111111.11100000

Dan jumlah host didapay dari jumlah sisa dari bit 0

Maka, host = 2⁵ – 2 = 30

Jawaban C

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

Answer :

2⁴<27<2⁵, karena hosat dihitung dari kanan kekiri…

11111111.11111111.11111111.11100000

Subnetnya 255.255.255.224

Jawabannya C

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

Answer :

2³<14<2^{4 ,} karena hosat dihitung dari kanan kekiri…

11111111.11111111.11111111.11110000

Subnetnya 255.255.255.240

Jawabannya C

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

Answer :

2⁶<100<⁷ karena subnet dihitung dari kiri kekanan…

11111111.11111111.11111110.00000000

Subnetnya 255.255.254.0

Jawabannya C

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

Answer :

172.32.65.13 = 10101100.00100000.01000001.00001101

255.255.0.0 = 11111111.11111111.00000000.00000000

10101100.00100000.00000000.00000000

172 . 32 . 0 . 0

Jawabannya C

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

Answer :

/22 = 11111111.11111111.11111100.00000000 = 255.255.252.0

172.16.210.0 = 10101100.00010000.11010010.00000000

255.255.252.0 = 11111111.11111111.11111100.00000000

10101100.00010000.11010000.00000000

172 . 16 . 208 . 0

Jawabannya C

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

Answer :

/22 = 255.255.252.0

Block subnet 256 – 252 = 4

115.64.4.1 – 115.64.7.255

Jawabannya C, D, & E

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

Answer :

/28 = 11111111.11111111.11111111.11110000 = 255.255.255.240

200.10.5.68 = 11001000.00001010.00000101.01000100

255.255.252.0 = 11111111.11111111.11111111.11110000

11001000.00001010.00000101.01000000

200 . 10 . 5 . 64

Jawabannya C

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

Answer :

/ 19 = 11111111.11111111.11100000.00000000 = 255.255.224.0

Host = 2¹³ – 2 = 8190

Subnetwork = 2³ = 8

Jawabannya 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

Answer :

2⁸<500<2⁹ karena subnet dihitung dari kiri kekanan…

11111111.11111111.11111111.10000000

Subnetnya 255.255.255.128

Jawabannya B

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

Answer :

/21 = 11111111.11111111.11111000.00000000 = 255.255.248.0

172.16.66.0 = 10101100.00010000.01000010.00000000

255.255.248.0 = 11111111.11111111.11111000.00000000

10101100.00010000.01000000.00000000

172 . 16 . 64 . 0

Jawabannya C

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

Answer :

2⁶<100<2⁷ karena subnet dihitung dari kiri kekanan…

11111111.11111111.11111110.00000000

Subnetnya 255.255.254.0

Jawaban B

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

Answer :

/ 29 = 11111111.11111111.11111111.11111000 = 255.255.255.248

Block subnet = 256 – 248 = 8

NET ID RHA BROADCAST

192.168.19.0 192.168.19.1 – 192.168.19.6 192.168.19.7

192.168.19.8 192.168.19.9 - 192.168.19.14 192.168.19.15

192.168.19.16 192.168.19.17 – 192.168.19.22 192.168.19.23

192.168.19.24 192.168.19.25 – 192.168.19.30 192.168.19.31

Jawaban C

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

Answer :

2⁸<300<2⁹ karena subnet dihitung dari kiri kekanan…

11111111.11111111.11111111.10000000

Subnetnya 255.255.255.128

2⁵<50<2⁶ karena host dihitung dari kanan kekiri…

11111111.11111111.11111111.11000000

Subnetnya 255.255.255.192

Jawabanya B & E

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

Answer :

/25 = 11111111.11111111.11111111.10000000 = 255.255.255.128

172.16.112.1 = 10101100.00010000.01110000.00000001

255.255.255.128 = 11111111.11111111.11111111.10000000

10101100.00010000.01110000.00000000

172 . 16 . 112 . 0

Jawabannya A

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

Answer :

Jawabannya D

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

Answer :

/22 = 11111111.11111111.11111100.00000000 = 255.255.252.0

Block subnet = 256 – 252 = 4

172.16.17.0 172.16.17.1 – 172.168.17.2 172.16.17.3

…………… .……………………………. ……………

Jawaban E

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

Answer :

/20 = 11111111.1111111.11110000.00000000 = 255.255.240.0

Host = 2¹² – 2 = 4094

Jawaban C

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

Answer :

/27 = 255.255.255.224

Block subnet = 256 – 224 = 32

32, 64, 96, 128,…..

Jawabannya B, C, & D

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

Answer :

2⁸<450<2⁹ karena host dihitung dari kanan kekiri…

11111111.11111111.11111110.00000000

Subnetmasknya 255.255.254.0

Jawabannya C

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.

Answer :

/ 27 = subnet mask = 255.255.255.224

Blocksubnetnya = 256 – 224 = 32

32, 64, 96, 128,…

Jawabannya A & D