INSTALASI DIRECTORY SERVER 2003

untuk menginstal active directory pada windows server 2003, hal pertama yang harus di lakukan adalah ketik dcpromo pada windows run.maka akan muncul tampilan seperti d bawah ini.

jika tampilan seperti d atas sudah muncul, klik next sampai muncul tampilan seperti d bawah ini,

gambar di bawah ini menunjukan nama domain yang harus kita isi.

Langkah selanjutnya klik next sampai muncul tampilan seperti dibawah ini, centang pada bagian ke dua seperti pada gambar di bawah ini.
Lalu isi password dengan karakter unik(huruf,angka,dan simbol).
gambar di bawah ini menunjukan proses instalasi active directory.
setelah penginstalan active directory selesai, akan muncul tampilan seperti di bawah ini, klik restart now untuk merestart windows server 2003.
gambar di bawah ini menunjukan ping dari windows server 2003 ke julz.com
gambar di bawah ini menunjukan ping dari julz.com ke windows server 2003
untuk membuat group baru pada windows server 2003 langkah-langkahnya sebagai berikut:

klik start, pilih directory tools, pilih active directory user and computer,klik kanan pada julz.com pilih new, pililh organization unit, lalu isi group name nya
di bawah ini merupakan tampilan untuk memasukan user ke dalam sebuah group, dengan cara klik kanan pada user, lalu pilih add to group

cara mengistal dhcp server

untuk menginstal dhcp server kita masuk terlebih dahulu ke root

untuk menginstal nya kita tuliskan "apt-get instal dhcp3"

setelah selesai menginstal dhcp server, kita mengkonfigurasi ethernet dengan cara menuliskan"gedit/etc/default/dhcp3-server"

selanjutnya mengkonfigurasi IP,broadcast,dan range. dengan cara menuliskan "gedit/etc/dhcp3/dhcpd.conf"

langkah terakhir, setelah menseting konfigurasi,kita restart configurasi dengan mengetikan "/etc/init.d/dhcp3-server restart"

Update kernel Fedora

YULIAN RACHMAN K

30208289

PCE-0801

untuk mengupdate kernel dari fedora kita harus masuk sebagai root, setelah itu ketik "# uname -r" untuk mengetahui versi kernel

langkah kedua yaitu mengupdate kernel dengan perintah "# yum update kernel", seperti pada gambar di bawah ini,
selanjutnya ketik "y" untuk menyetujui proses update kernel
gambar di bawah ini merupakan proses update kernel
kemudian setelah kita menyetujuinya kita tinggal menunggu sampai update complete

setelah proses complete, kita ulangi lagi untuk melihat versi kernel, dengan perintah "# uname -r"

<<==FINISHED==>>

soal subneting

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

Diketahui :

- Uses class B memiliki 25 subnet

- max host 300 computer.

2⁴<>5 jadi,

- default subnet mask class B 11111111.11111111.0000000.00000000

Ditanyakan:

25 subnet

Jawab:

yang mendekati adalah 2⁵ jadi lima buah bit 1menjadi:

11111111.11111111.11111000.00000000 ,subnet mask nya

255 . 255 . 248 . 0

Jawaban: B

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

Jawaban:

/27 = 11111111.11111111.11111111.11100000

- Subnet masknya = 255.255.255.224

- Block subnetnya 256 – 224 = 32

Net ID

192.168.1.31

-----------------

192.169.1.64

RHA

192.168.1.1 – 192.168.1.30

----------------------------------

192.168.1.65-192.168.1.94

Broadcast

192.168.1.0

----------------

192.168.1.95

Jawaban: D & F

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

Jawaban:

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

Jawaban:

Subnetmask default class B adalah 255.255.0.0 dan kombinasi dari 2 oktet selanjutnya.

Jawaban: D & E

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

jawaban :

Jawaban: C

6. Which type of address is 223.168.17.167/29?

a. host address

b. multicast address

c. broadcast address

d. subnetwork address

Jawaban:

IP addressnya class C

/29 = 11111111.11111111.11111111.11111000

Subnet masknya = 255.255.255.248

Block subnet = 256 – 248 = 8

- NET ID

223.168.17.0

223.168.17.8

-------------------

223.168.17.160

- RHA

223.168.17.1- 223.168.17.6

223.168.17.9- 223.168.17.14

----------------------------------------

223.168.17.161-223.168.17.166

- BROADCAST

223.168.17.7

223.168.17.15

--------------------

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

Jawaban:

IP addressnya class C

/29 = 11111111.11111111.11111111.11111000

Subnet masknya = 255.255.255.248

Block subnet = 256 – 248 = 8

Host = 2³-2 = 8-2=6

Subnetwork = 2⁵-2 = 32-2=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

jumlah host didapat dari jumlah sisa bit 0

jadi host = 2⁵ – 2 = 32-2=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, host dihitung dari kanan ke kiri

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 ,}host dihitung dari kanan ke kiri

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 subnet dihitung dari kiri ke kanan

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

jawaban:

172.32.65.13 = 10101100.00100000.01000001.00001101

255.255.0.0 = 11111111.11111111.00000000.00000000

10101100.00100000.00000000.00000000

172 . 32 . 0 . 0

Jawaban: 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

Jawaban:

/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

Jawaban: 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

Jawaban:

/22 = 255.255.252.0

Block subnetnay: 256 – 252 = 4

115.64.4.1 – 115.64.7.255

Jawaban: 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

Jawaban:

/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

Jawaban: 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

Jawaban:

/ 19 = 11111111.11111111.11100000.00000000 = 255.255.224.0

Host = 2¹³ – 2 = 8190

Subnetwork = 2³ = 8

Jawaban: 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

Jawaban:

2⁸<500<2⁹ subnet dihitung dari kiri ke kanan

11111111.11111111.11111111.10000000

Subnetnya 255.255.255.128

Jawaban: 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

Jawaban:

/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

Jawaban: 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

Jawaban:

2⁶<100<2⁷ subnet dihitung dari kiri ke kanan

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

Jawaban:

/ 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

Jawaban:

2⁸<300<2⁹ subnet dihitung dari kiri ke kanan

11111111.11111111.11111111.10000000

Subnetnya : 255.255.255.128

2⁵<50<2⁶ host dihitung dari kanan ke kiri

11111111.11111111.11111111.11000000

Subnetnya: 255.255.255.192

Jawaban: 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

Jawaban:

/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

Jawaban: 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

Jawaban:

Jawaban: 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

Jawaban:

/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

Jawaban:

/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

Jawaban:

/27 = 255.255.255.224

Block subnet = 256 – 224 = 32

32, 64, 96, 128,…..

Jawaban: 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

Jawaban:

2⁸<450<2⁹ host dihitung dari kanan ke kiri

11111111.11111111.11111110.00000000

Subnetmasknya : 255.255.254.0

Jawaban: 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.

Jawaban:

/ 27 = subnet mask = 255.255.255.224

Blocksubnetnya = 256 – 224 = 32

32, 64, 96, 128,…

Jawaban: A & D