Subnetting Competition

1. Berapakah jumlah ip yang dapat digunakan dari ip 192.168.88.1/30

Tidak ada

2

1

3

The given IP address 192.168.88.1/30 has a subnet mask of /30, which means it has a subnet mask of 255.255.255.252. This subnet mask allows for only 2 usable IP addresses. The first IP address is the network address (192.168.88.0) and the last IP address is the broadcast address (192.168.88.3). Therefore, there are only 2 usable IP addresses that can be used from the given IP address.

Explanation

The given IP address 192.168.88.1/30 has a subnet mask of /30, which means it has a subnet mask of 255.255.255.252. This subnet mask allows for only 2 usable IP addresses. The first IP address is the network address (192.168.88.0) and the last IP address is the broadcast address (192.168.88.3). Therefore, there are only 2 usable IP addresses that can be used from the given IP address.

2. Berapa subnetmask dari 189.89.20.34/19

255.255.255.224

255.255.192.0

255.255.224.0

255.255.252.0

The subnet mask for the given IP address and CIDR notation (/19) is 255.255.224.0. This can be determined by converting the CIDR notation to binary, which gives us 11111111.11111111.11100000.00000000. When converted back to decimal, this corresponds to 255.255.224.0. This subnet mask allows for 8192 host addresses within the subnet.

Explanation

The subnet mask for the given IP address and CIDR notation (/19) is 255.255.224.0. This can be determined by converting the CIDR notation to binary, which gives us 11111111.11111111.11100000.00000000. When converted back to decimal, this corresponds to 255.255.224.0. This subnet mask allows for 8192 host addresses within the subnet.

3. Berapa subnetmask dari 189.89.20.34/19

255.255.255.224

255.255.192.0

255.255.224.0

255.255.252.0

The correct subnet mask for the IP address 189.89.20.34/19 is 255.255.224.0. This is because the /19 in the IP address indicates that the first 19 bits are the network address and the remaining bits are for host addresses. In binary, the first 19 bits are 11111111.11111111.11100000.00000000, which translates to 255.255.224.0 in decimal notation. This subnet mask allows for 8190 host addresses within the network.

Explanation

The correct subnet mask for the IP address 189.89.20.34/19 is 255.255.224.0. This is because the /19 in the IP address indicates that the first 19 bits are the network address and the remaining bits are for host addresses. In binary, the first 19 bits are 11111111.11111111.11100000.00000000, which translates to 255.255.224.0 in decimal notation. This subnet mask allows for 8190 host addresses within the network.

4. Berapakah jumlah maksimal host untuk 12.13.14.15/25

89

124

126

123

The given IP address is 12.13.14.15 with a subnet mask of /25. A /25 subnet mask allows for 126 usable host addresses. The formula to calculate the number of usable hosts is 2^(32 - subnet mask) - 2. In this case, 32 - 25 = 7, so 2^7 - 2 = 128 - 2 = 126. Therefore, the maximum number of hosts for this network is 126.

Explanation

The given IP address is 12.13.14.15 with a subnet mask of /25. A /25 subnet mask allows for 126 usable host addresses. The formula to calculate the number of usable hosts is 2^(32 - subnet mask) - 2. In this case, 32 - 25 = 7, so 2^7 - 2 = 128 - 2 = 126. Therefore, the maximum number of hosts for this network is 126.

5. Berapa jumlah ip yang dapat digunakan pada 202.159.10.102/25

126

128

64

62

The given IP address 202.159.10.102/25 indicates that it is a Class C IP address with a subnet mask of 25 bits. In a Class C network, the first three octets (202.159.10) are fixed, and the last octet (102) is variable. The subnet mask of /25 means that the first 25 bits are fixed, and the remaining 7 bits are variable. Since there are 7 variable bits, there are 2^7 = 128 possible combinations for the last octet. However, the first and last combinations (all zeros and all ones) are reserved, leaving 126 usable IP addresses. Therefore, the correct answer is 126.

Explanation

The given IP address 202.159.10.102/25 indicates that it is a Class C IP address with a subnet mask of 25 bits. In a Class C network, the first three octets (202.159.10) are fixed, and the last octet (102) is variable. The subnet mask of /25 means that the first 25 bits are fixed, and the remaining 7 bits are variable. Since there are 7 variable bits, there are 2^7 = 128 possible combinations for the last octet. However, the first and last combinations (all zeros and all ones) are reserved, leaving 126 usable IP addresses. Therefore, the correct answer is 126.

6. Berapakah ip network yang tepat untuk ip 12.13.13.15/20

12.13.13.0

12.13.12.0

12.13.0.0

12.12.13.0

The correct answer is 12.13.0.0. This is the correct IP network for the given IP address 12.13.13.15/20. The subnet mask /20 indicates that the first 20 bits of the IP address are the network portion, and the remaining 12 bits are the host portion. Therefore, the network address is obtained by setting the host portion to all zeros, resulting in 12.13.0.0.

Explanation

The correct answer is 12.13.0.0. This is the correct IP network for the given IP address 12.13.13.15/20. The subnet mask /20 indicates that the first 20 bits of the IP address are the network portion, and the remaining 12 bits are the host portion. Therefore, the network address is obtained by setting the host portion to all zeros, resulting in 12.13.0.0.

7. Berapakah ip broadcast dari ip 17.8.19.45/20

17.8.31.255

17.8.32.248

17.8.30.255

17.8.29.248

The given IP address is 17.8.19.45/20, which means it has a subnet mask of 20 bits. The subnet mask determines the network and host portions of the IP address. In this case, the first 20 bits represent the network portion, and the remaining 12 bits represent the host portion.

To find the IP broadcast address, we need to set all the host bits to 1 within the network portion. In this case, the network portion is 17.8.16.0, and the host bits are 19.45.

Converting the host bits to binary, we get 00010011.00101101. Flipping all the bits to 1, we get 11111111.11111111.

Therefore, the IP broadcast address is 17.8.31.255.

Explanation

The given IP address is 17.8.19.45/20, which means it has a subnet mask of 20 bits. The subnet mask determines the network and host portions of the IP address. In this case, the first 20 bits represent the network portion, and the remaining 12 bits represent the host portion.

To find the IP broadcast address, we need to set all the host bits to 1 within the network portion. In this case, the network portion is 17.8.16.0, and the host bits are 19.45.

Converting the host bits to binary, we get 00010011.00101101. Flipping all the bits to 1, we get 11111111.11111111.

Therefore, the IP broadcast address is 17.8.31.255.

8. Berapa ip pertama dan terahir yang dapat kita gunakan pada 130.12.12.195/26

130.12.12.193 dan 130.12.12.254

130.12.12.192 dan 130.12.12.254

130.12.12.191 dan 130.12.12.255

130.12.12.190 dan 130.12.12.252

The given IP address is 130.12.12.195/26, which means it has a subnet mask of 26 bits. This subnet mask allows for a total of 64 IP addresses in the subnet. The first IP address in the subnet is obtained by setting all the host bits to 0, which gives us 130.12.12.192. The last IP address in the subnet is obtained by setting all the host bits to 1, which gives us 130.12.12.255. Therefore, the first and last usable IP addresses in the subnet are 130.12.12.193 and 130.12.12.254, respectively.

Explanation

The given IP address is 130.12.12.195/26, which means it has a subnet mask of 26 bits. This subnet mask allows for a total of 64 IP addresses in the subnet. The first IP address in the subnet is obtained by setting all the host bits to 0, which gives us 130.12.12.192. The last IP address in the subnet is obtained by setting all the host bits to 1, which gives us 130.12.12.255. Therefore, the first and last usable IP addresses in the subnet are 130.12.12.193 and 130.12.12.254, respectively.

9. Saat kita membuat firewall untuk blokir 100.100.100.100/30, ip mana saja yang terblokir??

100.100.100.0

100,100,100,104

100,100,100,190

100,100,100,102

The given answer is 100,100,100,102. When creating a firewall to block the IP range 100.100.100.100/30, it means that all IP addresses from 100.100.100.100 to 100.100.100.103 will be blocked. Therefore, the IP address 100,100,100,102 falls within this range and will be blocked by the firewall.

Explanation

The given answer is 100,100,100,102. When creating a firewall to block the IP range 100.100.100.100/30, it means that all IP addresses from 100.100.100.100 to 100.100.100.103 will be blocked. Therefore, the IP address 100,100,100,102 falls within this range and will be blocked by the firewall.

10. Berapa jumlah ip yang bisa digunakan pada 178.56.54.34/22??

1022

1024

2048

2046

The given IP address 178.56.54.34/22 has a subnet mask of 22 bits. In a subnet with a subnet mask of 22 bits, the first 22 bits are used to identify the network, and the remaining 10 bits can be used to identify individual hosts. With 10 bits available for hosts, we can have 2^10 - 2 = 1022 usable IP addresses. The 2 is subtracted because the network address and the broadcast address cannot be assigned to individual hosts. Therefore, the correct answer is 1022.

Explanation

The given IP address 178.56.54.34/22 has a subnet mask of 22 bits. In a subnet with a subnet mask of 22 bits, the first 22 bits are used to identify the network, and the remaining 10 bits can be used to identify individual hosts. With 10 bits available for hosts, we can have 2^10 - 2 = 1022 usable IP addresses. The 2 is subtracted because the network address and the broadcast address cannot be assigned to individual hosts. Therefore, the correct answer is 1022.

11. Berapakah range ip host dari 10.20.30.40/27

10.20.30.33 – 10.20.30.61

10.20.30.32 – 10.20.30.62

10.20.30.32 – 10.20.30.61

10.20.30.33 – 10.20.30.62

The given IP range is 10.20.30.33 - 10.20.30.62. This is determined by the subnet mask /27, which means that the first 27 bits of the IP address are the network address and the remaining bits are used for host addresses. In this case, the network address is 10.20.30.32 and the broadcast address is 10.20.30.63. Therefore, the range of host IP addresses is from 10.20.30.33 to 10.20.30.62.

Explanation

The given IP range is 10.20.30.33 - 10.20.30.62. This is determined by the subnet mask /27, which means that the first 27 bits of the IP address are the network address and the remaining bits are used for host addresses. In this case, the network address is 10.20.30.32 and the broadcast address is 10.20.30.63. Therefore, the range of host IP addresses is from 10.20.30.33 to 10.20.30.62.

12. Berapakah Ip network dari IP 17.8.19.45/21

17.8.17.0

17.8.19.0

17.8.18.0

17.8.16.0

The given IP address is 17.8.19.45/21, which means it has a subnet mask of 21 bits. A subnet mask of 21 bits indicates that the first 21 bits of the IP address are used to identify the network, while the remaining 11 bits are used for host addresses. By looking at the given options, the only network address that falls within the range of the subnet mask is 17.8.16.0. Therefore, the correct answer is 17.8.16.0.

Explanation

The given IP address is 17.8.19.45/21, which means it has a subnet mask of 21 bits. A subnet mask of 21 bits indicates that the first 21 bits of the IP address are used to identify the network, while the remaining 11 bits are used for host addresses. By looking at the given options, the only network address that falls within the range of the subnet mask is 17.8.16.0. Therefore, the correct answer is 17.8.16.0.

13. Berapa ip network dan subnetmask dari 195.20.30.1/23

195.20.30.0 dan 255.255.255.252

195.20.30.0 dan 255.255.254.0

195.20.30.1 dan 255.255.252.0

195.20.29.0 dan 255.255.254.0

The given IP address 195.20.30.1/23 belongs to the subnet 195.20.30.0 with a subnet mask of 255.255.254.0. This is because the /23 notation indicates that the first 23 bits of the IP address are the network portion, leaving 9 bits for the host portion. The subnet mask 255.255.254.0 has the first 23 bits set to 1, indicating the network portion, and the remaining 9 bits set to 0, indicating the host portion. Therefore, the correct answer is 195.20.30.0 and 255.255.254.0.

Explanation

The given IP address 195.20.30.1/23 belongs to the subnet 195.20.30.0 with a subnet mask of 255.255.254.0. This is because the /23 notation indicates that the first 23 bits of the IP address are the network portion, leaving 9 bits for the host portion. The subnet mask 255.255.254.0 has the first 23 bits set to 1, indicating the network portion, and the remaining 9 bits set to 0, indicating the host portion. Therefore, the correct answer is 195.20.30.0 and 255.255.254.0.

14. Berapakah ip network dan ip broadcast dari ip 10.10.10.80/18

10.10.10.0 dan 10.10.60.254

10.10.10.0 – 10.10.60.255

10.10.0.0 dan 10.10.62.252

10.10.0.0 – 10.10.63.255

The given IP address 10.10.10.80/18 has a subnet mask of /18, which means the first 18 bits are used for the network portion and the remaining 14 bits are used for the host portion.

To find the network address, we need to set all the host bits to 0. So the network address is 10.10.0.0.

To find the broadcast address, we need to set all the host bits to 1. So the broadcast address is 10.10.63.255.

Therefore, the correct answer is 10.10.0.0 – 10.10.63.255.

Explanation

The given IP address 10.10.10.80/18 has a subnet mask of /18, which means the first 18 bits are used for the network portion and the remaining 14 bits are used for the host portion.

To find the network address, we need to set all the host bits to 0. So the network address is 10.10.0.0.

To find the broadcast address, we need to set all the host bits to 1. So the broadcast address is 10.10.63.255.

Therefore, the correct answer is 10.10.0.0 – 10.10.63.255.

15. Dari ip 223.18.0.0 akan dibagi menjadi 13 subnet, masing-masing subnet tersebut memiliki 14 host, berapakah subnet mask yang cocok digunakan

255.255.255.248 (/29)

255.255.255.224 (/27)

255.255.255.240 (/28)

255.255.255.192 (/26)

The correct answer is 255.255.255.240 (/28). This subnet mask allows for 16 subnets and 14 hosts per subnet. Since there are 13 subnets required, this subnet mask is suitable as it provides enough subnets and hosts for the given scenario.

Explanation

The correct answer is 255.255.255.240 (/28). This subnet mask allows for 16 subnets and 14 hosts per subnet. Since there are 13 subnets required, this subnet mask is suitable as it provides enough subnets and hosts for the given scenario.

16. Berapakah ip broadcast dari ip 15.16.17.18/15

15.16.255.255

15.18.255.255

15.17.255.255

15.19.255.255

The given IP address is 15.16.17.18/15, which means that the first 15 bits of the IP address are the network portion and the remaining 17 bits are the host portion. In the network portion, the first three octets (15.16.17) remain the same, while the last octet (18) becomes all 1s in the broadcast address. Therefore, the IP broadcast for this network is 15.17.255.255.

Explanation

The given IP address is 15.16.17.18/15, which means that the first 15 bits of the IP address are the network portion and the remaining 17 bits are the host portion. In the network portion, the first three octets (15.16.17) remain the same, while the last octet (18) becomes all 1s in the broadcast address. Therefore, the IP broadcast for this network is 15.17.255.255.

17. Berapakah ip valid terakhir pada subnet 192.168.4.32/29

192.168.4.36

192.168.5.38

192.168.4.37

192.168.4.38

The last valid IP address in the subnet 192.168.4.32/29 is 192.168.4.38. In a /29 subnet, there are 8 IP addresses in total. The first IP address is the network address (192.168.4.32) and the last IP address is the broadcast address (192.168.4.39). Therefore, the valid IP addresses range from 192.168.4.33 to 192.168.4.38, with 192.168.4.38 being the last valid IP address.

Explanation

The last valid IP address in the subnet 192.168.4.32/29 is 192.168.4.38. In a /29 subnet, there are 8 IP addresses in total. The first IP address is the network address (192.168.4.32) and the last IP address is the broadcast address (192.168.4.39). Therefore, the valid IP addresses range from 192.168.4.33 to 192.168.4.38, with 192.168.4.38 being the last valid IP address.

18. Dari IP 17.8.19.45 berapakah netmask yang digunakan untuk kebutuhan 60 host

255.255.252

255.255.255.254

255.255.255.248

255.255.255.192

The correct answer is 255.255.255.192. This netmask allows for 64 IP addresses, which is enough to accommodate the 60 hosts needed.

Explanation

The correct answer is 255.255.255.192. This netmask allows for 64 IP addresses, which is enough to accommodate the 60 hosts needed.

19. Berapa ip terahir yang dapat digunakan pada 208.10.12.33/28

208.10.12.48

208.10.12.46

208.10.12.47

208.10.11.46

The given IP address is 208.10.12.33 with a subnet mask of /28. This means that the first 28 bits of the IP address are the network address, and the remaining 4 bits can be used for host addresses. To find the last usable IP address, we need to increment the host bits to their maximum value, which is 15 (1111 in binary). Adding 15 to the host bits of the IP address gives us 208.10.12.48. However, the question asks for the last usable IP address, so we need to subtract 1 from this value. Therefore, the last usable IP address is 208.10.12.47. The answer provided, 208.10.12.46, is incorrect.

Explanation

The given IP address is 208.10.12.33 with a subnet mask of /28. This means that the first 28 bits of the IP address are the network address, and the remaining 4 bits can be used for host addresses. To find the last usable IP address, we need to increment the host bits to their maximum value, which is 15 (1111 in binary). Adding 15 to the host bits of the IP address gives us 208.10.12.48. However, the question asks for the last usable IP address, so we need to subtract 1 from this value. Therefore, the last usable IP address is 208.10.12.47. The answer provided, 208.10.12.46, is incorrect.

20. Jika ip gateway 197.16.12.202/29, mana saja ip address yang mungkin digunakan oleh client??

197.16.12.207

197.16.12.206

197.16.12.208

197.16.12.200

The given IP gateway is 197.16.12.202/29. In a /29 subnet, there are 8 IP addresses available. The usable IP addresses for clients are from the range of the network address plus 1 to the broadcast address minus 1. In this case, the network address is 197.16.12.200 and the broadcast address is 197.16.12.207. Therefore, the possible IP addresses that can be used by clients are 197.16.12.201, 197.16.12.202, 197.16.12.203, 197.16.12.204, 197.16.12.205, 197.16.12.206, and 197.16.12.207. Hence, the correct answer is 197.16.12.206.

Explanation

The given IP gateway is 197.16.12.202/29. In a /29 subnet, there are 8 IP addresses available. The usable IP addresses for clients are from the range of the network address plus 1 to the broadcast address minus 1. In this case, the network address is 197.16.12.200 and the broadcast address is 197.16.12.207. Therefore, the possible IP addresses that can be used by clients are 197.16.12.201, 197.16.12.202, 197.16.12.203, 197.16.12.204, 197.16.12.205, 197.16.12.206, and 197.16.12.207. Hence, the correct answer is 197.16.12.206.

21. Berapa jumlah IP yang dapat digunakan pada /27?

30

64

16

32

Pada subnetting dengan /27, terdapat 32 alamat IP yang tersedia. Namun, 2 alamat IP digunakan untuk alamat jaringan dan broadcast, sehingga hanya 30 alamat IP yang dapat digunakan untuk perangkat yang terhubung ke subnet tersebut.

Explanation

Pada subnetting dengan /27, terdapat 32 alamat IP yang tersedia. Namun, 2 alamat IP digunakan untuk alamat jaringan dan broadcast, sehingga hanya 30 alamat IP yang dapat digunakan untuk perangkat yang terhubung ke subnet tersebut.

22. Berapakah sumarisasi yang lebih cocok dari ip tersebut:

192.168.1.0/25

192.168.1.0/23

192.168.1.0/24

192.168.1.0/22

The given IP addresses are in CIDR notation, where the number after the slash represents the network prefix length. The network prefix length determines the number of bits that are used to identify the network portion of the IP address.

To find the most suitable summarization, we need to look for the common network prefix among the given IP addresses.

Among the given options, the IP address 192.168.1.0/24 has the longest common network prefix with the other IP addresses. Therefore, it is the most suitable summarization.

Explanation

The given IP addresses are in CIDR notation, where the number after the slash represents the network prefix length. The network prefix length determines the number of bits that are used to identify the network portion of the IP address.

To find the most suitable summarization, we need to look for the common network prefix among the given IP addresses.

Among the given options, the IP address 192.168.1.0/24 has the longest common network prefix with the other IP addresses. Therefore, it is the most suitable summarization.

23. Dibawah ini adalah ip address satu network, kecuali

192.168.1.17/30

192.168.1.18/30

192.168.1.19/30

192.168.1.20/30

The given IP addresses are all in the same network range of 192.168.1.16/30. However, the IP address 192.168.1.20/30 is not included in the given options, making it the correct answer.

Explanation

The given IP addresses are all in the same network range of 192.168.1.16/30. However, the IP address 192.168.1.20/30 is not included in the given options, making it the correct answer.

24. Berapa ip network dan subnetmask dari 195.20.30.1/27

195.20.30.0 dan 255.255.255.252

195.20.30.1 dan 255.255.252.0

195.20.30.0 dan 255.255.255.224

195.20.29.0 dan 255.255.254.0

The given IP address 195.20.30.1/27 has a subnet mask of 255.255.255.224. This means that the first 27 bits of the IP address are used to identify the network, leaving 5 bits for host addresses. Therefore, the network address would be 195.20.30.0 and the subnet mask would be 255.255.255.224.

Explanation

The given IP address 195.20.30.1/27 has a subnet mask of 255.255.255.224. This means that the first 27 bits of the IP address are used to identify the network, leaving 5 bits for host addresses. Therefore, the network address would be 195.20.30.0 and the subnet mask would be 255.255.255.224.

25. Jika kita ingin memblokir ip 192.168.16.4, 192.168.16.100, 192.168.16.125, 192.168.16.2, dan 192.168.16.1 berapa src-address yang paling tepat untuk kita gunakan??

192.168.16.0/24

192.168.16.3/27

192.168.16.0/25

192.168.16.0/26

The correct answer is 192.168.16.0/25. This is the most appropriate src-address to use because it represents a subnet with a network address of 192.168.16.0 and a subnet mask of 255.255.255.128 (/25). This subnet mask allows for 126 usable IP addresses, which is sufficient to accommodate the given IP addresses that need to be blocked (192.168.16.4, 192.168.16.100, 192.168.16.125, 192.168.16.2, and 192.168.16.1) within the same subnet.

Explanation

The correct answer is 192.168.16.0/25. This is the most appropriate src-address to use because it represents a subnet with a network address of 192.168.16.0 and a subnet mask of 255.255.255.128 (/25). This subnet mask allows for 126 usable IP addresses, which is sufficient to accommodate the given IP addresses that need to be blocked (192.168.16.4, 192.168.16.100, 192.168.16.125, 192.168.16.2, and 192.168.16.1) within the same subnet.

26. Berapa ip terakhir yang dapat digunakan pada 208.10.12.30/27

208.10.12.32

208.10.12.31

208.10.11.31

208.10.12.30

The given IP address is 208.10.12.30 with a subnet mask of /27. This means that the first 27 bits of the IP address represent the network address, and the remaining 5 bits represent the host address. The maximum number of hosts that can be accommodated in this subnet is 2^5 - 2 = 30. Therefore, the last usable IP address in this subnet is 208.10.12.30.

Explanation

The given IP address is 208.10.12.30 with a subnet mask of /27. This means that the first 27 bits of the IP address represent the network address, and the remaining 5 bits represent the host address. The maximum number of hosts that can be accommodated in this subnet is 2^5 - 2 = 30. Therefore, the last usable IP address in this subnet is 208.10.12.30.

27. Berapa range valid ip host dari ip address 192.88.25.148 255.255.255.240

192.88.25.144 – 192.88.25.158

192.88.25.145 – 192.88.25.158

192.88.25.145 – 192.88.25.159

192.88.25.144 – 192.88.25.159

The given IP address is 192.88.25.148 with a subnet mask of 255.255.255.240. This subnet mask indicates that the last 4 bits of the IP address are used for host identification, allowing for a total of 16 possible host addresses. The valid range of host addresses within this subnet would be from 192.88.25.145 to 192.88.25.158, which matches the given answer.

Explanation

The given IP address is 192.88.25.148 with a subnet mask of 255.255.255.240. This subnet mask indicates that the last 4 bits of the IP address are used for host identification, allowing for a total of 16 possible host addresses. The valid range of host addresses within this subnet would be from 192.88.25.145 to 192.88.25.158, which matches the given answer.

28. Berapakah ip network dan ip broadcast dari ip 128.126.125.124 255.255.240.0

128.126.112.0 – 128.127.126.255

128.126.112.0 – 128.126.127.255

128.126.112.0 - 128.126.128.255

128.126.112.0 - 128.127.129.128

The given IP address 128.126.125.124 with subnet mask 255.255.240.0 belongs to the network range 128.126.112.0 - 128.126.127.255. The IP network is the first address in the range, which is 128.126.112.0. The IP broadcast is the last address in the range, which is 128.126.127.255.

Explanation

The given IP address 128.126.125.124 with subnet mask 255.255.240.0 belongs to the network range 128.126.112.0 - 128.126.127.255. The IP network is the first address in the range, which is 128.126.112.0. The IP broadcast is the last address in the range, which is 128.126.127.255.

29. Berapa range ip yang bisa digunakan pada 178.20.21.90/29

178.20.21.0-178.20.21.255

178.20.21.88-178.20.22.95

178.20.20.88-178.20.23.94

178.20.20.89-178.20.23.94

The given IP address range 178.20.21.90/29 indicates that it is a subnet with a subnet mask of /29. In a /29 subnet, there are 6 usable IP addresses. The first IP address is the network address (178.20.21.88), the last IP address is the broadcast address (178.20.21.95), and the remaining 4 IP addresses are usable for hosts. Therefore, the range of usable IP addresses in this subnet is from 178.20.21.89 to 178.20.21.94.

Explanation

The given IP address range 178.20.21.90/29 indicates that it is a subnet with a subnet mask of /29. In a /29 subnet, there are 6 usable IP addresses. The first IP address is the network address (178.20.21.88), the last IP address is the broadcast address (178.20.21.95), and the remaining 4 IP addresses are usable for hosts. Therefore, the range of usable IP addresses in this subnet is from 178.20.21.89 to 178.20.21.94.

30. Berapa ip summarization dari 178.100.1.10/25, 178.100.3.45/24, 178.100.5.73/25, dan 178.100.6.200/25

178.100.0.0/21

178.100.13.0/21

178.100.15.0/21

178.100.12.0/21

The given IP addresses can be summarized as 178.100.0.0/21. This is because the first 21 bits in all the given IP addresses are the same, which is 178.100.0.0. The /21 indicates that the first 21 bits are the network address, and the remaining 11 bits can be used for host addresses. Therefore, the IP addresses 178.100.1.10/25, 178.100.3.45/24, 178.100.5.73/25, and 178.100.6.200/25 can all be summarized as 178.100.0.0/21.

Explanation

The given IP addresses can be summarized as 178.100.0.0/21. This is because the first 21 bits in all the given IP addresses are the same, which is 178.100.0.0. The /21 indicates that the first 21 bits are the network address, and the remaining 11 bits can be used for host addresses. Therefore, the IP addresses 178.100.1.10/25, 178.100.3.45/24, 178.100.5.73/25, and 178.100.6.200/25 can all be summarized as 178.100.0.0/21.

31. Berapa jumlah ip yang dapat digunakan pada 202.159.10.102/26

128

64

62

126

The given IP address is 202.159.10.102/26, which means it has a subnet mask of 26 bits. The subnet mask determines the number of available IP addresses in the network. In this case, the subnet mask of 26 bits leaves 6 bits for host addresses. Since 2^6 equals 64, we subtract 2 (for the network address and broadcast address) from 64 to get the number of usable IP addresses, which is 62. Therefore, the correct answer is 62.

Explanation

The given IP address is 202.159.10.102/26, which means it has a subnet mask of 26 bits. The subnet mask determines the number of available IP addresses in the network. In this case, the subnet mask of 26 bits leaves 6 bits for host addresses. Since 2^6 equals 64, we subtract 2 (for the network address and broadcast address) from 64 to get the number of usable IP addresses, which is 62. Therefore, the correct answer is 62.

32. Berapakah range ip host valid 172.17.8.45 255.255.224.0

172.17.8.1 – 172.17.31.254

172.17.0.1 - 172.17.8.254

172.17.0.1 – 172.17.31.254

172.17.0.1 - 172.17.3.224

The given subnet mask 255.255.224.0 indicates that the first 19 bits are used for the network address and the remaining 13 bits are used for host addresses. Therefore, the valid host IP range would start from the network address 172.17.0.1 and end at the broadcast address 172.17.31.254.

Explanation

The given subnet mask 255.255.224.0 indicates that the first 19 bits are used for the network address and the remaining 13 bits are used for host addresses. Therefore, the valid host IP range would start from the network address 172.17.0.1 and end at the broadcast address 172.17.31.254.

33. Berapa jumlah client maksimal jika ip pertama telah digunakan gateway pada 189.10.10.0/27

30

32

29

28

The given IP address range 189.10.10.0/27 has a subnet mask of 27 bits, which means that there are 32 IP addresses in total. However, the first and last IP addresses in the range are reserved for network and broadcast addresses respectively. Therefore, the maximum number of usable client IP addresses is 32 - 2 = 30. However, since the gateway is already using one of the IP addresses, the actual maximum number of client IP addresses is 30 - 1 = 29.

Explanation

The given IP address range 189.10.10.0/27 has a subnet mask of 27 bits, which means that there are 32 IP addresses in total. However, the first and last IP addresses in the range are reserved for network and broadcast addresses respectively. Therefore, the maximum number of usable client IP addresses is 32 - 2 = 30. However, since the gateway is already using one of the IP addresses, the actual maximum number of client IP addresses is 30 - 1 = 29.

34. Berapa ip gateway yang mungkin digunakan jika ip client adalah 192.168.20.254/30??

192.168.20.252

192.168.20.253

192.168.20.255

192.168.20.254

The given IP address for the client is 192.168.20.254/30, which means it has a subnet mask of /30. A /30 subnet mask allows for only 2 usable IP addresses in the network. The first IP address is reserved for the network address, the second IP address is reserved for the client, and the last IP address is reserved for the broadcast address. Therefore, the only possible IP gateway that can be used in this scenario is 192.168.20.253.

Explanation

The given IP address for the client is 192.168.20.254/30, which means it has a subnet mask of /30. A /30 subnet mask allows for only 2 usable IP addresses in the network. The first IP address is reserved for the network address, the second IP address is reserved for the client, and the last IP address is reserved for the broadcast address. Therefore, the only possible IP gateway that can be used in this scenario is 192.168.20.253.

35. Berapa network id dan brodcast id dari 160.132.34.23/23

160.132.34.0 dan 160.132.35.255

160.132.34.0 dan 160.132.34.255

160.132.34.1 dan 160.132.34.254

160.132.34.1 dan 160.132.35.255

The given IP address is 160.132.34.23/23, which means that the first 23 bits represent the network ID and the remaining 9 bits represent the host ID. To find the network ID, we keep the first 23 bits and set the rest to 0, resulting in 160.132.34.0. To find the broadcast ID, we keep the first 23 bits and set the rest to 1, resulting in 160.132.35.255. Therefore, the correct answer is 160.132.34.0 and 160.132.35.255.

Explanation

The given IP address is 160.132.34.23/23, which means that the first 23 bits represent the network ID and the remaining 9 bits represent the host ID. To find the network ID, we keep the first 23 bits and set the rest to 0, resulting in 160.132.34.0. To find the broadcast ID, we keep the first 23 bits and set the rest to 1, resulting in 160.132.35.255. Therefore, the correct answer is 160.132.34.0 and 160.132.35.255.

36. Berapa ip summarization dari 178.100.12.4/25, 178.100.13.150/24, 178.100.12.150/25, dan 178.100.14.200/25

178.100.12.0/23

178.100.10.0/22

178.100.10.0/24

178.100.12.0/22

The given IP addresses can be summarized as 178.100.12.0/22. This is because the first three IP addresses have the same network address of 178.100.12.0, and the last IP address has a different network address of 178.100.14.0. By using the smallest possible subnet mask that encompasses all the given IP addresses, we get a network address of 178.100.12.0 and a subnet mask of /22, which gives us the summarized IP address of 178.100.12.0/22.

Explanation

The given IP addresses can be summarized as 178.100.12.0/22. This is because the first three IP addresses have the same network address of 178.100.12.0, and the last IP address has a different network address of 178.100.14.0. By using the smallest possible subnet mask that encompasses all the given IP addresses, we get a network address of 178.100.12.0 and a subnet mask of /22, which gives us the summarized IP address of 178.100.12.0/22.

37. Dalam suatu jaringan kita membutuhkan alokasi ip address untuk 128 host, maka netmask yang dapat digunakan adalah..

255.255.255.0

255.255.255.128

255.255.255.192

255.255.255.224

The correct answer is 255.255.255.0. This netmask can be used to allocate IP addresses for 128 hosts because it provides a subnet mask of 255.255.255.0, which means that the first three octets (255.255.255) are fixed and the last octet (0) can vary for host addresses. This allows for a maximum of 256 possible host addresses (2^8 = 256), but since the network and broadcast addresses are reserved, the usable range is 254 addresses. Therefore, this netmask can accommodate 128 hosts.

Explanation

The correct answer is 255.255.255.0. This netmask can be used to allocate IP addresses for 128 hosts because it provides a subnet mask of 255.255.255.0, which means that the first three octets (255.255.255) are fixed and the last octet (0) can vary for host addresses. This allows for a maximum of 256 possible host addresses (2^8 = 256), but since the network and broadcast addresses are reserved, the usable range is 254 addresses. Therefore, this netmask can accommodate 128 hosts.

38. Mana saja host id yang valid pada 178.167.10.234/29

178.167.10.236

178.167.10.231

178.167.10.233

178.167.10.238

The given IP address range is 178.167.10.234/29, which means it has a subnet mask of 29 bits. In this subnet, the first 29 bits represent the network address, and the remaining 3 bits represent the host address. Therefore, the valid host IDs within this subnet would be 178.167.10.231 and 178.167.10.233. The other two options, 178.167.10.236 and 178.167.10.238, are not valid host IDs within this subnet.

Explanation

The given IP address range is 178.167.10.234/29, which means it has a subnet mask of 29 bits. In this subnet, the first 29 bits represent the network address, and the remaining 3 bits represent the host address. Therefore, the valid host IDs within this subnet would be 178.167.10.231 and 178.167.10.233. The other two options, 178.167.10.236 and 178.167.10.238, are not valid host IDs within this subnet.

39. Berapa range ip yang bisa digunakan pada 178.20.21.13/22

178.20.21.0-178.20.21.255

178.20.21.1-178.20.22.254

178.20.20.1-178.20.23.254

178.20.20.0-178.20.23.255

The given IP address range 178.20.21.1-178.20.22.254 falls within the subnet 178.20.21.13/22. The /22 subnet mask indicates that the first 22 bits of the IP address are used for network identification, leaving 10 bits for host identification. Therefore, the valid IP range within this subnet is 178.20.20.1-178.20.23.254. The given answer, 178.20.21.1-178.20.22.254, falls within this valid IP range and is therefore correct.

Explanation

The given IP address range 178.20.21.1-178.20.22.254 falls within the subnet 178.20.21.13/22. The /22 subnet mask indicates that the first 22 bits of the IP address are used for network identification, leaving 10 bits for host identification. Therefore, the valid IP range within this subnet is 178.20.20.1-178.20.23.254. The given answer, 178.20.21.1-178.20.22.254, falls within this valid IP range and is therefore correct.

40. Saat kita membuat firewall untuk blokir 197.16.12.128/26, ip mana saja yang terblokir??

197.16.12.139

197.16.12.150

197.16.12.190

197.16.12.195

The given IP addresses 197.16.12.139, 197.16.12.150, and 197.16.12.190 will be blocked by the firewall because they fall within the specified range of 197.16.12.128/26. This range includes IP addresses from 197.16.12.128 to 197.16.12.191, with a subnet mask of 255.255.255.192 (/26). Therefore, any IP address within this range will be blocked by the firewall.

Explanation

The given IP addresses 197.16.12.139, 197.16.12.150, and 197.16.12.190 will be blocked by the firewall because they fall within the specified range of 197.16.12.128/26. This range includes IP addresses from 197.16.12.128 to 197.16.12.191, with a subnet mask of 255.255.255.192 (/26). Therefore, any IP address within this range will be blocked by the firewall.

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