MANDRA SUPRABA PCE 0807 / 30208002
Jawaban Soal 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
· Diketahui dibutuhkan kelas B maka default subnetmask nya 255.255.0.0
· Untuk rumus mencari host kita gunakan 2n-2 ≥ jumlah host, maka:
o 2n-2 ≥ 300
o 2n ≥ 302
o n ≥ 9 (untuk jumlah bit 0) jadi 11111111.11111111.11111110.00000000
o sehingga didapat subnet mask 255.255.254.0
· Untuk rumus mencari subnet kita gunakan 2n ≥ jumlah network, maka:
o 2n ≥ 25
o n ≥ 5 (untuk jumlah bit 1) jadi 11111111.11111111.11111000.00000000
o sehingga didapat subnet mask 255.255.248.0
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. 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
· Untuk tipe soal diatas kata kuncinya adalah pada prefik yang menyertainya, dari soal diatas mempunyai prefik /27 dimana subnetmask nya 255.255.255.224 didapat dari perhitungan jumlah bit 1 yaitu 11111111.11111111.11111111.11100000
· Setelah tahu subnetmask nya kita tentukan Blog subnetnya untuk mengetahui range ip address yang bisa dipakai, rumusnya : 256 – 224 (subnet terakhir) = 32.
· Didapat range IP
Net ID | Range IP | Broadcast |
192.168.1.32 | 192.168.1.33 – 192.168.1.62 | 192.168.1.63 |
192.168.1.64 | 192.168.1.65 – 192.168.1.94 | 192.168.1.95 |
192.168.1.96 | 192.168.1.97 – 192.168.1.126 | 192.168.1.127 |
· Dari table diatas kita dapat mengetahui range IP tiap subnet dan memakainya sesuai dengan topologi soal diatas (berada pada subnet urutan kedua).
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
· Untuk tipe soal diatas kita perhatikan subnet terakhir memiliki nilai 248 sama dengan prefix /29 dengan perhitungan 11111111.11111111.11111111.11111000
· Kemudian cari blog subnet 256-248=8
· Jadi range IP nya kelipatan 8, dan didapat IP pada soal ini berada pada kisaran 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
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
6. Which type of address is 223.168.17.167/29?
a. host address
b. multicast address
c. broadcast address
d. subnetwork address
· Perhatikan prefiknya /29 berarti punya nilai mask 248.
· Cari blog subnetnya : 256-248 = 8, jadi range IP nya kelipatan 8 didapat bahwa IP diatas adalah sebagai broadcast address dari range IP 223.168.17.160 – 223.168.17.168
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. 32 networks / 6 hosts
d. 62 networks / 2 hosts
· Prefix /29 sama dengan 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
· Rumus cari host (bit 0) :
o 2n-2 ≥ 27
o 2n ≥ 29
o n ≥ 5 (bit 0) sama dengan 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) :
o 2n-2 ≥ 14
o 2n ≥ 16
o n ≥ 4 (bit 0) sama dengan 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) :
o 2n ≥ 100
o n ≥ 7(bit 1) sama dengan 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
· Untuk default subnet kelas B adalah 2 oktet pertama, pada soal diatas default subnet mask nya adalah 172.32.0.0
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
· Prefik yang diketahui dari soal diatas adalah / 22 sama dengan 11111111.11111111.11111100.00000000 = 255.255.252.0
· Jadi subnetwork nya adalah 172.16.252.0
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
· Prefik yang diketahui dari soal diatas adalah /22 sama dengan 11111111.11111111.11111100.00000000 = 255.255.252.0
· Cari blog subnetnya 256-252 = 4, jadi selisi range untuk soal ini adalah kelipatan 4
· Berikut adalah tabelnya : (Perlu diperhatikan perhitungan untuk kelas B)
Net ID | Range IP | Broadcast |
115.64.4.0 | 115.64.4.1 – 115.64.7.254 | 115.64.7.255 |
125.64.8.0 | 115.64.8.1 – 115.64.11.254 | 115.64.11.255 |
115.64.12.0 | 115.64.12.1 – ……….. | …… |
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
· Soal diatas punya Prefix /28 sama dengan 11111111.11111111.11111111.11110000 = 255.255.255.240
· Cari Blog subnetnya : 256-240 = 16, jadi kelipatan dari range IP soal diatas adalah 16.
Net ID | Range IP | Broadcast |
200.10.5.16 | 200.10.5.16 – 200.10.5.16 | 200.10.5.31 |
200.10.5.32 | 200.10.5.16 – 200.10.5.16 | 200.10.5.47 |
200.10.5.48 | 200.10.5.16 – 200.10.5.166 | 200.10.5.65 |
200.10.5.64 | 200.10.5.16 – 200.10.5.16 | 200.10.5.79 |
200.10.5.80 | …………… | ………… |
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 soal diatas /19 sama dengan 11111111.11111111.11100000.00000000
· Rumus cari network (bit 1) :
23 = 8
· Rumus cari 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
· Rumus cari host (bit 0) :
2n-2 ≥ 100
2n ≥ 102
n ≥ 7 (bit 0) sama dengan 11111111.11111111.11111111.1000000 = 255.255.255.128
· Rumus cari network (bit 1) :
2n ≥ 500
n ≥ 9 (bit 1) sama dengan 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 soal diatas /21 sama dengan 11111111.11111111.11111000.00000000
· Cari blog subnet : 256 – 248 = 8, jadi kelipatan range IP nya adalah 8, dan didapat bahwa subnetwork IP soal diatas adalah 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
· Rumus cari network (bit 1) :
2n ≥ 100
n ≥ 7 (bit 1) sama dengan 11111111.11111111.11111110.00000000 = 255.255.254.0
· Rumus cari host (bit 0) :
2n – 2 ≥ 500
2n ≥ 502
n ≥ 9 (bit 0) sama dengan 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 soal diatas /29 sama dengan 11111111.11111111.11111111.11111000 = 255.255.255.248 adalah subnet mask nya.
· Cari blog subnetnya : 256 – 248 = 8, jadi kelipatan range IP nya adalah 8, didapat bahwa IP pada soal diatas berada pada range 192.168.19.24 – 192.168.19.31, jadi address pertama yang bisa dipakai adalah 192.168.19.26
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
· Rumus cari network (bit 1) :
2n ≥ 300
n ≥ 9 (bit 1) sama dengan 11111111.11111111.11111111.10000000 = 255.255.255.128
· Rumus cari host (bit 0) :
2n – 2 ≥ 50
2n ≥ 52
n ≥ 6 (bit 0) sama dengan 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
· Prefiks soal diatas /25 sama dengan 11111111.11111111.11111111.10000000 = 255.255.255.128 adalah subnet mask nya.
· Cari blog subnetnya : 256 – 128 = 128, jadi subnet valid nya 172.16.112.128
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
· Dari studi kasus soal diatas kita harus memilih jumlah max host yg paling banyak karena untuk perkembangan network masa depannya. Maka didapat 850 host/
· Rumus cari host (bit 0) :
2n – 2 ≥ 850
2n ≥ 852
n ≥ 10 = 1024 (bit 0)
sama dengan 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 soal diatas /22 sama dengan 11111111.11111111.11111100.00000000 = 255.255.252.0 adalah subnet masknya.
· Cari blog subnetnya : 256 – 252 = 4, jadi kelipatan dari range IP soal diatas adalah 4, sedangkan subnet nya 172.16.17.0 maka host yg valid dari jawaban soal diatas adalah 172.16.18.255
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 soal diatas /20 sama dengan 11111111.11111111.11110000.00000000 = 255.255.240.0 adalah subnet mask nya.
· 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)
sama dengan 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.
· Prefiks soal diatas /27 sama dengan 11111111.11111111.11111111.11100000 = 255.255.255.224 adalah subnet mask nya.
· Cari Blog subnet : 256 – 224 = 32, jadi kelipatan range IP soal diatas adalah 32.
· Seharusnya untuk address host harus sesuai dengan table berikut :
Net ID | Range IP | Broadcast |
192.18.166.32 | 192.18.166.33 – 192.18.166.62 | 192.18.166.63 |
192.18.166.64 | 192.18.166.65 – 192.18.166.94 | 192.18.166.95 |
192.18.166.96 | 192.18.166.97 – 192.18.166.126 | 192.18.166.127 |
