- 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 : B dan C
Penyelesaian :
Mencari jumlah subnet :
2^n – 2 ≥ 25
2^n ≥ 27
n ≥ 5
Subnet mask 11111111.11111111.11111000.00000000
255.255.248.0
Mencari jumlah host valid :
2^N – 2 ≥ 300
2^N ≥ 302
N ≥ 9
Subnet mask 11111111.11111111.111111110.00000000
255.255.254.0
Jadi, subnet mask yang memenuhi untuk menampung 25 subnet dan maksimum 300 host adalah 255.255.248.0 dan 255.255.254.0.
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. 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 : D dan F
Penyelesaian :
Mencari gateway host A :
IP router yang menjadi acuan adalah IP router terdekat, yaitu 192.168.1.65/27. IP ini akan menjadi gateway host A.
Mencari IP host A :
Subnet mask 11111111.11111111.11111111.11100000
255.255.255.224
Interval subnet = 256 – 224 = 32
Block subnet :
| Subnet | Range IP |
| 192.168.1.0 192.168.1.32 192.168.1.64 192.168.1.96 192.168.1.128 192.168.1.160 192.168.1.192 | 192.168.1.1 - 192.168.1.30 192.168.1.33 - 192.168.1.62 192.168.1.65 - 192.168.1.94 192.168.1.97 - 192.168.1.126 192.168.1.129 - 192.168.1.148 192.168.1.161 - 192.168.1.190 192.168.1.193 - 192.168.1.222 |
IP router berada dalam subnet 192.168.1.64, sehingga range IP host yang bisa digunakan adalah 192.168.1.65 - 192.168.1.94.
Jadi, gateway host A adalah 192.168.1.65/27 dan yang termasuk kedalam host IP satu jaringan adalah 192.168.1.82 dan 192.168.1.70.
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 : E
Penyelesaian :
Subnet mask 11111111.11111111.11111111.11111000
255.255.255.248
Interval subnet = 256 – 248 = 8
| Subnet | Range IP |
| 172.31.192.0 172.31.192.8 172.31.192.16 .... .... .... 172.31.192.160 .... | 172.31.192.1 - 172.31.192.6 172.31.192.9 - 172.31.192.14 172.31.192.17 - 172.31.192.22 .... .... .... 172.31.192.161 - 172.31.192.164 .... |
Jadi, IP 172.31.192.166 terletak pada subnet 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
Jawaban : D dan E
Penyelesaian :
Misalkan : Net ID = x dan Host ID y, maka format default subnet mask untuk kelas B adalah
x . x . y . y
ketentuan : Net ID (x) memiliki default subnet mask 255 untuk setiap oktetnya, sedangkan untuk host ID (y) adalah relatif bisa diubah-ubah sesuai dengan kebutuhan menjadi 255.255.y.y.
Jadi, subnet mask yang valid untuk kelas B adalah 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
Jawaban : D
Penyelesaian :
Mencari range subnet pada oktet ketiga = 159 – 128 – 1 = 30 host. (1 merupakan IP broadcast)
Interval subnet = banyaknya IP host + 2 = 30 + 2 = 32
Oktet ketiga pada subnet mask = 256 – 32 = 224
Jadi, networknya adalah 172.16.128.0 dan subnet mask 255.255.224.0.
6. Which type of address is 223.168.17.167/29?
a. host address
b. multicast address
c. broadcast address
d. subnetwork address
Jawaban : C
Penyelesaian :
Subnet mask 11111111.11111111.11111111.11111000
255.255.255.248
Interval subnet = 256 – 248 = 8
Block subnet
| Subnet | Range IP | Broadcast |
| 223.168.17.0 223.168.17.8 223.168.17.16 .... 223.168.17.160 .... | 223.168.17.1 - 223.168.17.6 223.168.17.9 - 223.168.17.14 223.168.17.17 - 223.168.17.22 .... 223.168.17.161 - 223.168.17.166 .... | 223.168.17.7 223.168.17.15 223.168.17.23 .... 223.168.17.167 .... |
Jadi, 223.168.17.167/29 termasuk broadcast address.
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 : C
Penyelesaian :
Subnet mask 11111111.11111111.11111111.11111000
255.255.255.248
Jumlah network = 2^n – 2
= 2^5 – 2
= 30 network
Jumlah host = 2^N – 2
= 2^3 – 2
= 6 hosts
Jadi, jumlah subnetwork dan host persubnet adalah 30 network dan 6 host.
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
Jawaban : C
Penyelesaian :
Jumlah IP persubnet = 256 – 224 = 32
Jumlah IP host persubnet = 32 – 2 = 30 hosts
Jadi, banyak host persubnet adalah 30 host.
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
Jawaban : C
Penyelesaian :
2^N – 2 ≥ jumlah host
2^N – 2 ≥ 27
N ≥ 5
Subnet mask : 11111111.11111111.11111111.11100000
255.255.255.224
Jadi , subnet mask yang palig sedikitnya ada 27 host adalah 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
Jawaban : C
Penyelesaian :
2^N – 2 ≥ jumlah host
2^N – 2 ≥ 14
N ≥ 4
Subnet mask 11111111.11111111.11111111.11110000
255.255.255.240
Jadi, subnet mask dari 14 IP host adalah 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
Jawaban : C
Penyelesaian :
2^n – 2 ≥ jumlah subnet
2^n – 2 ≥ 100
n ≥ 7
Default subnet mask kelas B 255.255.0.0, setelah penambahan network jumlah bit 1 subnet mask menjadi :
11111111.11111111.11111110.00000000
255.255.254.0
Jadi, subnet mask yang mencukupi 100 network adalah 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
Jawaban : C
Penyelasaian :
IP address 172.32.65.13 termasuk ke kelas B memiliki default subnet mask 255.255.0.0. Cara pencarian network address bisa dengan menggunakan operator AND yang membandingkan antara IP address dan subnet mask.
10011000.00100000.01000001.00001101
11111111.11111111.00000000.00000000
___________________________________ AND
10011000.00100000.00000000.00000000
Jadi, dari hasil diatas dapat diketahui network IP 172.32.65.13 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
Jawaban : C
Penyelesaian :
IP address 172.16.210.0 memiliki prefiks 22 jadi subnet mask 255.255.252.0. Cara pencarian network address bisa dengan menggunakan operator AND yang membandingkan antara IP address dan subnet mask.
10011000.00010000.11010010.00000000
11111111.11111111.11111100.00000000
___________________________________ AND
10011000.00010000.11010000.00000000
Jadi, dari hasil diatas dapat diketahui network IP 172.16.210.0 adalah 172.16.208.0.
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
Jawaban : B, C, dan E
Penyelesaian :
CIDR blok 115.64.4.0/22
Subnet mask 11111111.11111111.11111100.00000000
255.255.252.0
Interval subnet = 256 – 252 = 4
Block subnet
| Subnet | Range IP |
| 115.64.0.0 115.64.4.0 115.64.8.0 | 115.64.0.1 - 115.64.3.255 115.64.4.1 - 115.64.7.255 115.64.8.1 - 115.64.11.255 |
IP host harus berada pada range IP 115.64.4.1 - 115.64.7.255.
Jadi, host IP yang memenuhi adalah 115.64.7.64, 115.64.6.255, dan 115.64.5.128.
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
Jawaban : C
Penyelesaian :
Subnet mask 11111111.11111111.11111111.11110000
255.255.255.240
Interval subnet = 256 – 240 = 16
Blok subnet
| Subnet | Range IP |
| 200.10.5.0 200.10.5.16 200.10.5.32 200.10.5.48 200.10.5.64 200.10.5.80 | 200.10.5.1 - 200.10.5.14 200.10.5.17 - 200.10.5.30 200.10.5.33 - 200.10.5.46 200.10.5.49 - 200.10.5.62 200.10.5.65 - 200.10.5.78 200.10.5.81 - 200.10.5.94 |
Jadi , IP address 200.10.5.68/28 berada pada subnetwork 200.10.5.64.
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 : F
Penyelesaian :
Subnet mask 11111111.11111111.11100000.00000000
255.255.224.0
Net ID = 2^3
= 8
Jumlah host = 2^N -2
= 2^13-2
= 8190
Jadi, IP tersebut memiliki 8 subnets dan 8190 host persubnet
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 : B
Penyelesaian :
Mencari jumlah subnet :
2^n – 2 ≥ jumlah subnet
2^n – 2 ≥ 500
n ≥ 9
Mencari jumlah host :
2^N – 2 ≥ jumlah host
2^N – 2 ≥ 100
N ≥ 7
Subnet mask 11111111.11111111.11111111.10000000
255.255.255.128
Jadi, subnet mask yang dibutuhkan untuk 500 subnet dan 100 host adalah 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
Jawaban : C
Penyelesaian :
Subnet mask 11111111.11111111.11111000.00000000
255.255.248.0
Dengan metode eliminasi metode AND untuk mencari subnetwork
10101100.00010000.01000010.00000000
11111111.11111111.11111000.00000000
__________________________________ AND
10101100.00010000.01000000.00000000 = 172.16.64.0
Jadi, IP address dari 172.16.66.0/21 berada si subnetwork 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
Jawaban : B
Penyelesaian :
Mencari jumlah subnet :
2^n – 2 ≥ Jumlah subnet
2^n – 2 ≥ 100
n = 7
Mencari jumlah host :
2^N – 2 ≥ Jumlah host
2^N – 2 ≥ 500
N = 9
Subnet mask 11111111.11111111.11111110.00000000
255.255.254.0
Jadi, subnet mask yang menampung 100 subnet dan 500 host dari network ID 172.16.0.0 adalah 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
Jawaban : C
Penyelesaian :
Subnet mask 11111111.11111111.11111111.11111000
255.255.255.248
Interval subnet = 256 – 248 = 8
Block subnet
| Subnet | Range IP | Broadcast |
| 192.168.19.0 192.168.19.8 192.168.19.16 192.168.19.24 192.168.19.32 | 192.168.19.1 - 192.168.19.6 192.168.19.9 - 192.168.19.14 192.168.19.17 - 192.168.19.22 192.168.19.25 - 192.168.19.30 192.168.19.33 - 192.168.19.38 | 192.168.19.7 192.168.19.15 192.168.19.0.23 192.168.19.0.31 192.168.19.0.39 |
Jadi, host address dan subnet mask yang termasuk ke dalam subnetwork 192.169.19.24 adalah 192.168.19.26 dan 255.255.255.248.
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 : B dan E
Penyelesaian :
Mencari jumlah host :
2^N – 2 ≥ Jumlah host
2^N – 2 ≥ 50
N = 6
Subnet mask 11111111.11111111.11111111.11000000
255.255.255.192
Mencari jumlah subnet :
2^n – 2 ≥ Jumlah subnet
2^n – 2 ≥ 300
n = 9
Subnet mask 11111111.11111111.11111111.10000000
255.255.255.128
Jadi, subnet mask yang memenuhi 300 subnets dan 50 hosts adalah 255.255.255.128 dan 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
Jawaban : A
Penyelesaian :
Subnet mask 11111111.11111111.11111111.10000000
255.255.255.128
Interval subnet = 256 – 128 = 128
Block subnet
| Subnet | Range IP | Broadcast |
| 172.16.112.0 | 172.16.112.1 - 172.16.112.126 | 172.16.112.127 |
Jadi, IP 172.16.112.1 berada dalam subnetwork 172.16.112.0.
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
Jawaban : D
Penyelesaian :
Jumlah host maksimal = 850
2^N – 2 ≥ 850
2^N ≥ 852
N = 10
Subnet mask 11111111.11111111.11111100.00000000
255.255.252.0
Jadi, subnet mask yang memenuhi rancangan jaringan sesuai di gambar adalah 255.255.252.0.
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
Jawaban : E
Penyelesaian :
Subnet mask 11111111.11111111.11111100.00000000
255.255.252.0
Interval subnet = 256 – 252 = 4
Block subnet
| Subnet | Range IP |
| 172.16.0.0 172.16.4.0 172.16.8.0 172.16.12.0 172.16.16.0 172.16.24.0 .... | 172.16.0.1 - 172.16.3.255 172.16.4.1 - 172.16.7.255 172.16.8.1 - 172.16.11.255 172.16.12.1 - 172.16.15.255 172.16.16.1 - 172.16.23.255 172.16.24.1 - 172.16.27.255 .... |
Jadi, IP yang termasuk ke dalam subnetwork tersebut adalah 172.16.18.255 dengan subnet mask 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
Jawaban : C
Penyelesaian :
Subnet mask 11111111.11111111.11110000.00000000
Jumlah host = 2^N – 2
= 2^12 – 2
= 4096 – 2
= 4094
Jadi, banyaknya host yang dimuat dalam satu segmen ethernet IP tersebut adalah 4094 host.
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 : C, D, dan E
Penyelesaian :
Subnet mask 11111111.11111111.11111111.11100000
255.255.255.224
Interval subnet = 256 – 224 = 32
Block subnet
| Subnet | Range IP | Broadcast |
| x.x.x.0 x.x.x.32 x.x.x.64 x.x.x.96 x.x.x.128 x.x.x.160 x.x.x.192 | x.x.x.1 - x.x.x.30 x.x.x.33 - x.x.x.62 x.x.x.65 - x.x.x.94 x.x.x.97 - x.x.x.126 x.x.x.129 - x.x.x.158 x.x.x.161 - x.x.x.190 x.x.x.193 - x.x.x.222 | x.x.x.31 x.x.x.63 x.x.x.95 x.x.x.127 x.x.x.159 x.x.x.191 x.x.x.223 |
Kemungkinan IP :
- Melihat dari subnet mask, IP host berada di kelas C atau D (option B merupakan IP kelas A).
- Melihat dari block subnet, IP host tidak termasuk dalam subnet (option F ternasuk IP subnet) dan broadcast (option A termasuk IP broadcast).
Jadi, melihat kemuungkinan tersebut, IP host yang valid adalah 143.187.16.56, 192.168.15.87, dan 200.45.115.159.
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 : C
Penyelesaian :
Mencari jumlah host ;
2^N – 2 ≥ jumlah host
2^N – 2 ≥ 450
N = 9
Subnet mask 11111111.11111111.11111110.00000000
255.255.254.0
Jadi, subnet mask yang memenuhi 450 IP address per subnet adalah 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.
Jawaban : A dan E
Penyelesaian :
Subnet mask 11111111.11111111.11111111.11100000
255.255.255.224 (subnet mask yang benar)
Interval subnet = 256 – 224 = 32
Block subnet
| Subnet | Range IP | Broadcast |
| 192.18.166.0 192.18.166.32 192.18.166.64 192.18.166.96 192.18.166.128 192.18.166.160 192.18.166.192 | 192.18.166.1 - 192.18.166.30 192.18.166.33 - 192.18.166.62 192.18.166.65 - 192.18.166.94 192.18.166.97 - 192.18.166.126 192.18.166.129 - 192.18.166.158 192.18.166.161 - 192.18.166.190 192.18.166.193 - 192.18.166.222 | 192.18.166.31 192.18.166.63 192.18.166.95 192.18.166.127 192.18.166.159 192.18.166.191 192.18.166.223 |
Jadi, subnet mask yang terlihat di gambar salah dan IP host A juga salah karena sudah berada di jaringan lain.
Referensi :
- Sumber: Cisco Networking Academy Program, CCNA 1 and 2 Companion Guide, Cisco Press, 2005 Todd Lammle, CCNA Study Guide 5th Edition, Sybex, 2005.
Apabila ada kesalahan dalam pembahasan soal, mohon dimaklumi dan diperbaiki ..
Hihiii ..
Trimaa kasiih .. ^,^
Smoga bermanfaat .. ~.^
