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LATIHAN 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 Jawab : Analisa : max 300 Host >> 29-2=510 Host/Subnet Subnet mask yang tebentuk 11111111. 11111111. 11111110.00000000 255.255.254.0 (C) 2. Pilih 2!! 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 Dik : IP Router 192.168.1.65/27 Dit : IP Host & Gateway?? Jawab : Analisa /27 = 11111111.1111111.1111111.11100000 Jumlah Bit 0 =5 25=32 (Range)>>>>0-32-64-96-… Jadi IP Router 192.168.1.65 >>>> Masuk Range 64-95 192.168.1.65 (Sebagai Gateway Host) Address - 192.168.1.82 (Host) (D) Address - 192.168.1.70 (Host) (F)
Karena Kedua IP diatas Termasuk dalam Range Subnet yang sama 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 Jawab : Analisa : Subnet Mask 255.255.255.248 Biner 11111111.11111111.11111111.11111000 Range 23 = 8 (0-8-16-24-32…….160-168) IP 172.168.192.166 Masuk Range 160-167 NetID dari Rangenya 172.31.192.160 (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 Jawab : Diketahui IP kelas B Kelas B = 2 NetID dan 2 HostID Jadi yang termasuk kelas B hanya (D dan 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 Jawab : (B) Memakai default subnet mask untuk mencapai 172.16.159.255
6. Which type of address is 223.168.17.167/29? a. host address b. multicast address
c. broadcast address d. subnetwork address Dik : 223.168.17.167/29 Dit : Masuk IP jenis apa?? Jawab : Analisa : /29 = 11111111.11111111.11111111.11111000 Range = 23 =8 (0-8-16-24-32…..160-168) 223.168.17.167 Masuk Range 160-167 Jadi Broadcastnya 223.168.17.167 (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 Dik : IP 192.168.99.0/99 Dit : Berapa Jumlah Subnet dan Jumlah Host?? Jawab: Analisa /29 = 11111111.11111111.11111111.11111000 Subnet = 25 = 32 Subnet Host = 23 -2= 8-2 = 6 Host Jadi Subnet yang terbentuk 32 Subnet dan 6 Host/Subnet
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 Dik
; IP : 192.168.4.0 Netmask = 255.255.255.224 Dit : Max Host/Subnet?? Jawab : Analisa 255.255.255.224 = 11111111.11111111.11111111.11100000 Host = 25-2 = 32-2 =30 Host/Subnet (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 Dit : Subnet Mask Untuk Max Host 27/subnetnya?? Jawab : Analisa 27 = 2 5 25-2 = 32-2= 30 Host Jadi Subnetmasknya = 11111111.11111111.11111111.11100000 Desimalnya = 255.255.255.224 (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 Dit : Subnet Mask Baru Lebih dari 14 Host ?? Jawab : 24-2 = 16-2 = 14 (Yang sudah ada) Untuk Penambahan Host 25-2 = 32 -2 = 30Host/Subnet Jadi Subnetmask Untuk 30 Host = 11111111.11111111.11111111.11100000 Desimalnya = 255.255.255.224 (B)
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 Dit : Subnet Mask yang valid untuk jumlah Netword 100 buah?? Jawab : Analisa 100 = 2 7 = 128 Subnetmasknya = 11111111.11111111.11111110.00000000 Desimalnya = 255.255.254.0 (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 Dik
: IP : 172.32.65.13 Subnetmask : Default Dit : NetID IP diatas ?? Jawab : Analisa Subnetmask : 255.255.0.0 (Kelas B) 216 -2 = 65536-2 = 65534 Host/Subnet Jadi Jawabanya (C) karena semua IP akan masuk dalam satu range jika menggunakan subnet mask default. 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 Dik : IP address of 172.16.210.0/22 Dit : Net ID pada range? Jawab : Analisa : IP diatas Kelas B 16+x=22 x=6 26=64 (range: 0-64-128-192-256) Range IP diatas 192-255, jadi Net ID-nya 172.16.192.0 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
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
Dik : IP 200.10.5.68/28 Dit : subnet untuk host IP diatas? Jawab : Analisa : IP diatas termasuk Kelas C Subnet mask : 11111111.11111111.11111111.11110000 Range = 24= 32 (0-32-64-96) Jadi IP diatas masuk pada range 64-95, berarti Net ID 200.10.5.64 (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 Dik : Net ID 172.16.0.0/19 Dit : Jumlah Subnet dan host yang terbentuk? Jawab: Analisa :Net ID diatas Kelas B Subnet Mask : 11111111.11111111.11100000.00000000 23= 8 subnet yang terbentuk 213-2= 8192-2= 8190 Host (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 Dit : Dibutuhkan 500 subnet dan 100 host per subnet, memakai kelas B? Jawab: Host= 100= 27 128-2= 126, subnet mask : 11111111.11111111.11111110.00000000 Decimal : 255.255.254.0 (D)
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 Dik : IP address of 172.16.66.0/21 Dit : Subnet pada IP address 172.16.66.0?? Jawab : IP : 172.16.66.0 Subnetmask : 1111111.1111111.11111000.00000000
Range Host : 211=2048 (0-2048-4096…) Jawabannya D Karena semua Host dapat tertampung 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 Jawab : 29= 512 -2 =510 Host Subnetnya 11111111.11111111.11111110.00000000 Desimalnya : 255.255.254.0 (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 Dik : IP subnet 192.168.19.24 Dit : Subnetmask?? Jawab : Analisa Subnetmask 11111111.11111111.11111111.11111000 Range 23=8 (Range 0-8-16-24-32….) Masuk Pada Range 24-31 Jadi decimal dari subnet diatas 255.255.255.248 (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 Jawab : Analisa max 50 Host 50 = 2 5= 32 Host/Subnet Subnetmask :11111111.11111111.1111111.11100000 Desimalnya : 255.255.255.224 (D) Subnetnya : 211 = 2048 Subnet
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 Jawab : Analisa : 172.16.112.1 Subnetmask : 11111111.11111111.11111111.10000000 Range : 27= 128 (Range 0-128-256…) Jadi masuk ke Range 0-128 Dan NetID nya 172.16.0.0 (B) 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 Jawab : Host Terbanyak Perjaringan 1450 → 211 Jadi Subnetmasknya : 11111111.11111111..11111000.00000000 255.255.248.0 (B)
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 Jawab : Analisa /22 Subnetmasknya : 11111111.11111111.11111100.00000000 Desimalnya 255.255.252.0 (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 Jawab : Analisa IP address 172.16.112.1/20 Subnetmask : 11111111.11111111.11110000.00000000 212 – 2 = 4096 – 2 = 4094 (C) 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 Jawab : Jumlah Host 450 jadi 2 9 -2=512-2=510 Subnetmasknya 11111111.11111111.11111110.00000000 Desimalnya : 255.255.254.0 28.
Sebutkan kesalahan dari jaringan diatas: 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. Jawab A dan D
