Tuesday, June 19, 2012

Tugas Komjar (Subnetting)

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 :
Diketahui terdapat 25 Subnet dengan maksimum computer yang ada yaitu 300 computer. Jaringan ini menggunakan Class B. Langkah untuk menyelesaikan permasalahan ini adalah 
256 – 25 = 23111111111.11111111.11111111.0000000 
2 pangkat n – 2 = (<300 Computer)
2 pangkat 8 – 2 = 254 Host / Komputer
Jawaban 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 appliedto the host for connectivity? (Choose two.)
a. Address - 192.168.1.14Gateway - 192.168.1.33
b. Address - 192.168.1.45Gateway - 192.168.1.33
c. Address - 192.168.1.32Gateway - 192.168.1.33
d. Address - 192.168.1.82Gateway - 192.168.1.65
e. Address - 192.168.1.63Gateway - 192.168.1.65
f. Address - 192.168.1.70Gateway - 192.168.1.65
Jawaban Eth0 = 192.168.1.65/27
Subnetmask /27 = 11111111.11111111.11111111.1110000 = 255.255.255.224
Host per blok : 256-224 = 32Range 32, 64, 96, 128, 160, 192
Jawaban “ F (Address - 192.168.1.70 Gateway - 192.168.1.65 )”
“D (Address - 192.168.1.82 Gateway - 192.168.1.65)” F dan D berada dalam range 65 –  94 yang bisa digunakan, nuntuk 192.168.1.63 tidak digunakan karena ini merupakan alamat 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
Jawaban :Host per range = 256  –  248 = 8

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
Jawabannya D dan E

5. Which combination of network id and subnet mask correctly identifies all IP addressesfrom 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 255.255.255.192172.16.128.0/  
jawabannya D

6. Which type of address is 223.168.17.167/29?
a. host address
b. multicast address
c. broadcast address
d. subnetwork address
Jawaban: 11111111 11111111 11111111 11111000256 – 248 = 8
Block Range IP :8,16,24,32,40,48,56,64,72,80,88,96,104,112,120,128,136,144,152,160 ,168
Jawaban = C Broadcast Address 168-1 = 167 (Broadcast IP)

7. What is the correct number of usable subnetworks and hosts for the IP network address192.168.99.0 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 : Subnetmask : 11111111.11111111.11111111.11111000 = 255.255.255.248
Untuk mengetahui berapa jumlah networknya, hitung  bit “1” dari blok yang keempat :
2 pangkat5 -2 = 30 network 
Untuk mengetahui jumlah hostnya, hitung bit “0” dari blok yang keempat :
2 pangkat3 -2 = 6 host
Jawaban : C. 30 networks / 6 hosts

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 eachsubnet?
a. 6
b. 14
c. 30
d. 62
Jawaban IP address : 192.168.4.0 (kelas C) subnetmask : 255.255.255.224Host per blok : 256-224 = 32 Jumlah host max : 32-2 = 30 Dilakukan pengurangan sebanyak 2 host dikarenakan akan dialokasikan untuk netid dan IP broadcat Jawaban : C. 30

9. A company is planning to subnet its network for a maximum of 27 hosts. Which subnetmask would provide the needed hosts and leave the fewest unused addresses in eachsubnet?
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
Jawaban NM : 11111111.11111111.11111111.11100000 (255.255.255.224)
Jawaban : C 255.255.255.224

10. An IP network address has been subnetted so that every subnetwork has 14 usable hostIPaddresses. 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
Jawaban Range Jawaban 
A : 256-128 = 128 (dalam 1 blok bisa menampung 128 host)
B : 256-224 = 32 (dalam 1 blok bisa menampung 32 host)
C : 256-240 = 16 (dalam 1 blok hanya bisa menampung 16 host)
D : 256-248 = 8 (dalam 1 blok hanya bisa menampung 8 host)
E : 256-252 = 4 (dalam 1 blok hanya bisa menampung 4 host)
Jawaban Paling Tepat adalah C 255.255.255.240 dikarenakan dapat menampung 14 Host 24 –  2 = 14

11. A company is using a Class B IP addressing scheme and expects to need as many as 100networks. 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
JawabanPada soal dibutuhkan 100 Network pada class B
2pangkat7 = 128
Menentukan Subnetnya dengan cara menambahkan bit 1 pada otket ke tiga 11111111.11111111.11111110.0000000 (255.255.254.0)
Jawaban C = 255.255.254.0

12. Given a host with the IP address 172.32.65.13 and a default subnet mask, to whichnetwork 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 Mempunyai Net ID 172.32.65.0 (Class B) dengan default 255.255.0.0
JAWABAN : A (172.32.65.0)

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

14. Which of the following IP addresses fall into the CIDR block of 115.64.4.0/22? (Choosethree.)
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

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 NM: 11111111.11111111.11100000.00000000 = 255.255.224.0
Subnets (bit “1”) = 2pangkat3 = 8
Host (bit “0”) = 2pangkat13 = 8190 
Jawaban : F. 8 subnets, 8190 hosts each

17. You need 500 subnets, each with about 100 usable host addresses per subnet. WhatMask 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 Dibutuhkan 500 Subnet dengan tersedia 100 Host masing-masing Subnet
Subnet = 2pangkatN > 500, N = 9 (bit “1”)
NM = 11111111.11111111.11111111.10000000 = 255.255.255.128
Jawaban : B. 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

19. What mask would you assign to the network ID of 172.16.0.0 if you needed about 100subnets 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 : Dibutuhkan 100 Subnet dengan 500 host per subnetSubnet = 2pangkatN > 100, N = 7 (bit “1”) Subnetmask = 11111111.11111111.11111110.00000000 = 255.255.254.0
Jawaban : B. 255.255.254.0

20. You need to configure a server that is on the subnet 192.168.19.24/29. The router hasthe 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

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
100 subnet, tiap subnet 50 hostSubnet = 2pangkatN > 100, N = 7 (bit “1”)
Host = 2pangkatn - 2 > 50, n = 6 (bit “0”)
Mask berdasarkan subnet = 11111111.11111111.11111111.10000000 = 255.255.255.128
Mask berdsarkan host = 11111111.11111111.11111111.11000000 = 255.255.255.192
Jawaban : E. 255.255.255.192 B. 255.255.255.128

22. If an Ethernet port on a router were assigned an IP address of 172.16.112.1/25, whatwould 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 address172.20.0.0. 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 Jumlah host maximum 800+650 = 1450Host 2pangkatn - 2 > 1450, n = 11 (bit “0”)
NM: 11111111.11111111.11111000.00000000 = 255.255.248.0
Jawaban : C. 255.255.248.0

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

25 Your router has the following IP address on Ethernet0: 172.16.112.1/20. How manyhosts can be accommodated on the Ethernet segment?
a. 1024
b. 2046
c. 4094
d. 4096
e. 8190
JAWABAN Subnetmask : 11111111.11111111.11110000.00000000 = 255.255.240.0
Host n = 12 (bit “0”), 2pangkat12 –  2 = 4094
Jawaban : C. 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
JAWABAN Prefix /27 biasanya digunakan untuk kelas C.
Jawaban : 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 isthe 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 Pada class B dibutuhkan 450 IP untuk masing-masing SubnetHost 2pangkatn - 2 > 450, n = 9 (bit “0”) Subnetmask = 11111111.11111111.11111110.00000000 = 255.255.254.0
Jawaban : C. 255.255.254.0

28. Host A is connected to the LAN, but it cannot connect to the Internet. The hostconfiguration 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
Subnetmask = 11111111.11111111.11111111.11100000 = 255.255.255.224 Untuk bisa terhubung, host haru memiliki subnet mask yang sama denganEthernet.IP addres host (198.18.166.65) dihubungkan dengan eth0 Gateway(198.18.166.33) harus mengikuti ethernetnya.
Jawabannya A : The Host subnet Mask is incorrect

12 comments:

  1. just share, ntong komen nu macem2 nya :D

    ReplyDelete
    Replies
    1. waduh, gmn tuh 7:( blaasssssssss nggak tau sedikitpunnnn :p

      Delete
    2. wkwkw, itu soal materi subnetting di jaringan komputer. :D

      Delete
  2. Mumet itu, saya memang Newbie tingkat RT kagak ngerti blesss
    semoga saja berguna bagi yang membutuhkan.......

    ReplyDelete
  3. Perlu pemahaman seruius nieh sop,maklum ane dah aki-aki hehe.
    Ijin menyimak.
    Nice share thanks ya,happy blogging.

    ReplyDelete
  4. :O itu soal udah pake bahasa ingris byk angka bikin pusing http://reader-download.googlecode.com/svn/trunk/images/emo/drummer.gif

    ReplyDelete
    Replies
    1. emg tujuan yang bikin soal emg pengen bikin pusing kali :rolled:

      Delete