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. 255.255.240.0b. 255.255.248.0c. 255.255.254.0d. 255.255.255.0e. 255.255.255.128f. 255.255.255.248Jawaban : B dan CPenyelesaian :Mencari jumlah subnet :2^n – 2 ≥ 252^n       ≥ 27n           ≥ 5Subnet mask 11111111.11111111.11111000.00000000                      255.255.248.0Mencari jumlah host valid :2^N – 2  ≥ 3002^N        ≥ 302N            ≥ 9Subnet mask 11111111.11111111.111111110.00000000                      255.255.254.0Jadi, 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 - 192.168.1.33b. Address - 192.168.1.45  Gateway - 192.168.1.33

c. Address - 192.168.1.32  Gateway - 192.168.1.33d. Address - 192.168.1.82  Gateway - 192.168.1.65e. Address - 192.168.1.63  Gateway - 192.168.1.65f. Address - 192.168.1.70  Gateway - 192.168.1.65Jawaban : D dan FPenyelesaian :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           255.255.255.224Interval subnet = 256 – 224 = 32Block subnet :

Subnet Range IP192.168.1.0192.168.1.32192.168.1.64                                                     192.168.1.96192.168.1.128192.168.1.160192.168.1.192

192.168.1.1 - 192.168.1.30192.168.1.33 - 192.168.1.62192.168.1.65 - 192.168.1.94192.168.1.97 - 192.168.1.126192.168.1.129 - 192.168.1.148192.168.1.161 - 192.168.1.190192.168.1.193 - 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. 172.31.0.0b. 172.31.160.0c. 172.31.192.0d. 172.31.248.0e. 172.31.192.160f. 172.31.192.248Jawaban : EPenyelesaian :Subnet mask  11111111.11111111.11111111.11111000                         255.255.255.248

Interval subnet = 256 – 248 = 8Subnet Range IP172.31.192.0172.31.192.8172.31.192.16                                                     ............172.31.192.160....

172.31.192.1 - 172.31.192.6172.31.192.9 - 172.31.192.14172.31.192.17 - 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. 255.0.0.0b. 255.254.0.0c. 255.224.0.0d. 255.255.0.0e. 255.255.252.0f. 255.255.255.192Jawaban : D dan EPenyelesaian :Misalkan : Net ID = x dan Host ID y, maka format default subnet mask untuk kelas B adalahx . x . y . yketentuan : 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 dan255.255.252.0.

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 255.255.255.224b. 172.16.128.0 and 255.255.0.0c. 172.16.128.0 and 255.255.192.0d. 172.16.128.0 and 255.255.224.0e. 172.16.128.0 and 255.255.255.192Jawaban : DPenyelesaian :Mencari range subnet pada oktet ketiga = 159 – 128 – 1 = 30 host. (1 merupakan IP broadcast)

Interval subnet = banyaknya IP host + 2 = 30 + 2 = 32Oktet ketiga pada subnet mask = 256 – 32 = 224Jadi, 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 addressb. multicast addressc. broadcast addressd. subnetwork addressJawaban : CPenyelesaian :Subnet mask  11111111.11111111.11111111.11111000                        255.255.255.248Interval subnet = 256 – 248 = 8Block subnet

Subnet Range IP Broadcast223.168.17.0223.168.17.8223.168.17.16....223.168.17.160....

223.168.17.1 - 223.168.17.6223.168.17.9 - 223.168.17.14223.168.17.17 - 223.168.17.22....223.168.17.161 - 223.168.17.166....

223.168.17.7223.168.17.15223.168.17.23....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 hostsb. 14 networks / 14 hostsc. 30 networks / 6 hostsd. 62 networks / 2 hostsJawaban : CPenyelesaian :Subnet mask   11111111.11111111.11111111.11111000                        255.255.255.248Jumlah network = 2^n – 2                           = 2^5 – 2                           = 30 networkJumlah host = 2^N – 2                     = 2^3 – 2                     = 6 hostsJadi, 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. 6b. 14c. 30d. 62Jawaban : CPenyelesaian :Jumlah IP persubnet = 256 – 224 = 32Jumlah IP host persubnet = 32 – 2 = 30 hostsJadi, 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. 255.255.255.0b. 255.255.255.192c. 255.255.255.224d. 255.255.255.240e. 255.255.255.248Jawaban : CPenyelesaian :2^N – 2 ≥ jumlah host2^N – 2 ≥ 27N            ≥ 5Subnet mask : 11111111.11111111.11111111.11100000                           255.255.255.224Jadi , 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. 255.255.255.128b. 255.255.255.224c. 255.255.255.240d. 255.255.255.248e. 255.255.255.252Jawaban : CPenyelesaian :2^N – 2 ≥ jumlah host2^N – 2 ≥ 14N           ≥ 4Subnet mask 11111111.11111111.11111111.11110000                       255.255.255.240Jadi, 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. 255.255.0.0b. 255.255.240.0c. 255.255.254.0d. 255.255.255.0e. 255.255.255.128f. 255.255.255.192Jawaban : CPenyelesaian :2^n – 2 ≥ jumlah subnet2^n – 2 ≥ 100n            ≥ 7Default subnet mask kelas B 255.255.0.0, setelah penambahan network jumlah bit 1 subnet mask menjadi :11111111.11111111.11111110.00000000255.255.254.0Jadi, 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. 172.32.65.0b. 172.32.65.32c. 172.32.0.0d. 172.32.32.0Jawaban : CPenyelasaian :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.

10011000.00100000.01000001.0000110111111111.11111111.00000000.00000000___________________________________ AND10011000.00100000.00000000.00000000

Jadi, dari hasil diatas dapat diketahui network IP 172.32.65.13 adalah172.32.0.0.

13. What is the subnetwork number of a host with an IP address of 172.16.210.0/22?a. 172.16.42.0b. 172.16.107.0c. 172.16.208.0

d. 172.16.252.0e. 172.16.254.0Jawaban : CPenyelesaian :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.

10011000.00010000.11010010.0000000011111111.11111111.11111100.00000000___________________________________ AND10011000.00010000.11010000.00000000

Jadi, dari hasil diatas dapat diketahui network IP 172.16.210.0 adalah172.16.208.0.

14. Which of the following IP addresses fall into the CIDR block of 115.64.4.0/22? (Choose three.)a. 115.64.8.32b.  115.64.7.64c. 115.64.6.255d. 115.64.3.255e. 115.64.5.128f. 115.64.12.128Jawaban : B, C, dan EPenyelesaian :CIDR blok 115.64.4.0/22Subnet mask  11111111.11111111.11111100.00000000                          255.255.252.0Interval subnet = 256 – 252 = 4Block subnet

Subnet Range IP115.64.0.0115.64.4.0115.64.8.0

115.64.0.1 - 115.64.3.255115.64.4.1 - 115.64.7.255115.64.8.1 - 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, dan115.64.5.128.

15. What is the subnetwork address for a host with the IP address 200.10.5.68/28?a. 200.10.5.56b. 200.10.5.32c. 200.10.5.64d. 200.10.5.0Jawaban : C

Penyelesaian :Subnet mask 11111111.11111111.11111111.11110000                      255.255.255.240Interval subnet = 256 – 240 = 16Blok subnet

Subnet Range IP200.10.5.0200.10.5.16200.10.5.32200.10.5.48200.10.5.64200.10.5.80

200.10.5.1 - 200.10.5.14200.10.5.17 - 200.10.5.30200.10.5.33 - 200.10.5.46200.10.5.49 - 200.10.5.62200.10.5.65 - 200.10.5.78200.10.5.81 - 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 eachb. 7 subnets, 2046 hosts eachc. 7 subnets, 8190 hosts eachd. 8 subnets, 30 hosts eache. 8 subnets, 2046 hosts eachf. 8 subnets, 8190 hosts eachJawaban : FPenyelesaian :Subnet mask 11111111.11111111.11100000.00000000                        255.255.224.0Net ID = 2^3             = 8Jumlah host = 2^N -2                    = 2^13-2                    = 8190Jadi, 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. 255.255.255.252b. 255.255.255.128c. 255.255.255.0d. 255.255.254.0Jawaban : BPenyelesaian :

Mencari jumlah subnet :2^n – 2 ≥ jumlah subnet2^n – 2 ≥ 500n             ≥ 9Mencari jumlah host :2^N – 2 ≥ jumlah host2^N – 2 ≥ 100N           ≥ 7Subnet mask 11111111.11111111.11111111.10000000                      255.255.255.128Jadi, subnet mask yang dibutuhkan untuk 500 subnet dan 100 host adalah255.255.255.128.

18. What is the subnetwork number of a host with an IP address of 172.16.66.0/21?a. 172.16.36.0b. 172.16.48.0c. 172.16.64.0d. 172.16.0.0Jawaban : CPenyelesaian :Subnet mask 11111111.11111111.11111000.00000000                      255.255.248.0

Dengan metode eliminasi metode AND untuk mencari subnetwork10101100.00010000.01000010.0000000011111111.11111111.11111000.00000000__________________________________ AND10101100.00010000.01000000.00000000         =         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. 255.255.255.0b. 255.255.254.0c. 255.255.252.0d. 255.255.0.0Jawaban : BPenyelesaian :Mencari jumlah subnet :2^n – 2 ≥ Jumlah subnet2^n – 2 ≥ 100n           = 7Mencari jumlah host :2^N – 2 ≥ Jumlah host2^N – 2 ≥ 500

N           = 9Subnet mask 11111111.11111111.11111110.00000000                      255.255.254.0Jadi, 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. 192.168.19.0 255.255.255.0b. 192.168.19.33 255.255.255.240c. 192.168.19.26 255.255.255.248d. 192.168.19.31 255.255.255.248e. 192.168.19.34 255.255.255.240Jawaban : CPenyelesaian :Subnet mask 11111111.11111111.11111111.11111000                      255.255.255.248Interval subnet = 256 – 248 = 8Block subnet

Subnet Range IP Broadcast192.168.19.0192.168.19.8192.168.19.16192.168.19.24192.168.19.32

192.168.19.1 - 192.168.19.6192.168.19.9 - 192.168.19.14192.168.19.17 - 192.168.19.22192.168.19.25 - 192.168.19.30192.168.19.33 - 192.168.19.38

192.168.19.7192.168.19.15192.168.19.0.23192.168.19.0.31192.168.19.0.39

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. 255.255.255.0b. 255.255.255.128c. 255.255.252.0d. 255.25.255.224e. 255.255.255.192f. 255.255.248.0Jawaban : B dan EPenyelesaian :Mencari jumlah host :2^N – 2 ≥ Jumlah host2^N – 2 ≥ 50

N           = 6  Subnet mask 11111111.11111111.11111111.11000000                      255.255.255.192Mencari jumlah subnet :2^n – 2 ≥ Jumlah subnet2^n – 2 ≥ 300n           = 9  Subnet mask 11111111.11111111.11111111.10000000                      255.255.255.128Jadi, subnet mask yang memenuhi 300 subnets dan 50 hosts adalah255.255.255.128 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. 172.16.112.0b. 172.16.0.0c. 172.16.96.0d. 172.16.255.0e. 172.16.128.0Jawaban : APenyelesaian :Subnet mask 11111111.11111111.11111111.10000000                      255.255.255.128Interval subnet = 256 – 128 = 128Block subnet

Subnet Range IP Broadcast172.16.112.0 172.16.112.1 - 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. 255.255.224.0b. 255.255.240.0c. 255.255.248.0d. 255.255.252.0

e. 255.255.254.0 Jawaban : DPenyelesaian :Jumlah host maksimal = 8502^N – 2 ≥ 8502^N       ≥ 852N           = 10Subnet mask 11111111.11111111.11111100.00000000                      255.255.252.0

Jadi, subnet mask yang memenuhi rancangan jaringan sesuai di gambar adalah255.255.252.0.

24. You have a network with a subnet of 172.16.17.0/22. Which are valid host addresses?a. 172.16.17.1 255.255.255.252b. 172.16.0.1 255.255.240.0c. 172.16.20.1 255.255.254.0d. 172.16.16.1 255.255.255.240e. 172.16.18.255 255.255.252.0f. 172.16.0.1 255.255.255.0Jawaban : EPenyelesaian :Subnet mask 11111111.11111111.11111100.00000000                        255.255.252.0Interval subnet = 256 – 252 = 4Block subnet

Subnet Range IP172.16.0.0172.16.4.0172.16.8.0172.16.12.0172.16.16.0172.16.24.0....

172.16.0.1 - 172.16.3.255172.16.4.1 - 172.16.7.255172.16.8.1 - 172.16.11.255172.16.12.1 - 172.16.15.255172.16.16.1 - 172.16.23.255172.16.24.1 - 172.16.27.255....

Jadi, IP yang termasuk ke dalam subnetwork tersebut adalah 172.16.18.255dengan 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.      1024b.      2046c.       4094d.      4096e.      8190

Jawaban : C

Penyelesaian :Subnet mask 11111111.11111111.11110000.00000000Jumlah host = 2^N – 2                    = 2^12 – 2                    = 4096 – 2                    = 4094Jadi, banyaknya host yang dimuat dalam satu segmen ethernet IP tersebut adalah4094 host.

26. You have a /27 subnet mask. Which of the following are valid hosts? (Choose three.)a. 11.244.18.63b. 90.10.170.93c. 143.187.16.56d. 192.168.15.87e. 200.45.115.159f. 216.66.11.192Jawaban : C, D, dan EPenyelesaian :Subnet mask 11111111.11111111.11111111.11100000                      255.255.255.224Interval subnet = 256 – 224 = 32Block subnet

Subnet Range IP Broadcastx.x.x.0x.x.x.32x.x.x.64x.x.x.96x.x.x.128x.x.x.160x.x.x.192

x.x.x.1 - x.x.x.30x.x.x.33 - x.x.x.62x.x.x.65 - x.x.x.94x.x.x.97 - x.x.x.126x.x.x.129 - x.x.x.158x.x.x.161 - x.x.x.190x.x.x.193 - x.x.x.222

x.x.x.31x.x.x.63x.x.x.95x.x.x.127x.x.x.159x.x.x.191x.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. 255.255.240.0b. 255.255.248.0c. 255.255.254.0d. 255.255.255.0

Jawaban : CPenyelesaian :Mencari jumlah host ;2^N – 2 ≥ jumlah host2^N – 2 ≥ 450N           = 9Subnet mask 11111111.11111111.11111110.00000000                      255.255.254.0Jadi, subnet mask yang memenuhi 450 IP address per subnet adalah255.255.254.0.

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 EPenyelesaian :Subnet mask 11111111.11111111.11111111.11100000                      255.255.255.224 (subnet mask yang benar)Interval subnet = 256 – 224 = 32Block subnet

Subnet Range IP Broadcast192.18.166.0192.18.166.32192.18.166.64192.18.166.96192.18.166.128192.18.166.160

192.18.166.1 - 192.18.166.30192.18.166.33 - 192.18.166.62192.18.166.65 - 192.18.166.94192.18.166.97 - 192.18.166.126192.18.166.129 - 192.18.166.158192.18.166.161 -

192.18.166.31192.18.166.63192.18.166.95192.18.166.127192.18.166.159192.18.166.191192.18.166.223

192.18.166.192

192.18.166.190192.18.166.193 - 192.18.166.222

Jadi, subnet mask yang terlihat di gambar salah dan IP host A juga salah karena sudah berada di jaringan lain.