子网划分专项练习题

子网划分专项练习（附答案）1.What valid host range is the IP address 172.16.10.22 255.255.255.240 a part of?（在172.16.10.22 255.255.255.240这个网络里，有效IP地址的范围是？）EA. 172.16.10.20 through 172.16.10.22B. 172.16.10.1 through 172.16.10.255C. 172.16.10.16 through 172.16.10.23D. 172.16.10.17 through 172.16.10.31E. 172.16.10.17 through 172.16.10.302. What is the broadcast address of the subnet address 172.16.8.159 255.255.255.192?（在172.16.8.159 255.255.255.192这个网络里，广播地址是？） CA. 172.16.255.255B. 172.16.8.127C. 172.16.8.191D. 172.16.8.2553. What is the broadcast address of the subnet address 192.168.10.33 255.255.255.248? （在192.168.10.33 255.255.255.248这个网络里，广播地址是？）DA. 192.168.10.40B. 192.168.10.255C. 192.168.255.255D. 192.168.10.394. If you wanted to have 12 subnets with a Class C network ID, which subnet mask would you use?（对一个C类网络，要划分出12个子网，那么，划分后的子网掩码是？）CA. 255.255.255.252B. 255.255.255.248C. 255.255.255.240D. 255.255.255.2555. If you need to have a Class B network address subnetted into exactly 510 subnets, what subnet mask would you assign?（对一个B类网络进行子网划分，要划分出510个子网，则子网掩码是？）BA. 255.255.255.252B. 255.255.255.128C. 255.255.255.0D. 255.255.255.1926. If you are using a Class C network ID with two subnets and need 31 hosts per network, which of the following masks should you use?（对于一个C类网络，要划分出两个子网，每个子网里有31台计算机，则子网掩码是？）BA. 255.255.255.0B. 255.255.255.192C. 255.255.255.224D. 255.255.255.2487. How many subnets and hosts can you get from the network 192.168.254.0/26?（把默认的C类网络划分为192.168.254.0/26，则能获得多少个子网，每子网里有多少主机）BA. 4 networks with 64 hostsB. 2 networks with 62 hostsC. 254 networks with 254 hostsD. 1 network with 254 hosts8. You have the network 172.16.10.0/24. How many subnets and hosts are available? （对B类地址进行了划分，划分为172.16.10.0/24，则可用的子网数和主机数是多少？）DA. 1 subnet with 10 hostsB. 1 subnet with 254 hostsC. 192 subnets with 10 hostsD. 254 subnets with 254 hosts9. 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?（对于172.16.0.0，如果你需要100个子网，每子网500台主机，则掩码是什么？）BA. 255.255.255.0B. 255.255.254.0C. 255.255.252.0D. 255.255.0.010. You have a Class C 192.168.10.0/28 network. How many usable subnets and hosts do you have?（现有一个网络ID192.168.10.0/28，则有多少可用的子网和主机？）BA. 16 subnets, 16 hostsB. 14 subnets, 14 hostsC. 30 subnets, 6 hostsD. 62 subnets, 2 hosts11. You have the network 192.168.10.0/24. How many subnets and hosts are available? （现在网络ID为192.168.10.0/24，则有多少可用的子网和主机？）BA. 1 subnet with 10 hostsB. 1 subnet with 254 hostsC. 192 subnets with 10 hostsD. 254 subnets with 254 hosts12. You have a 255.255.255.240 mask. Which two of the following are valid host IDs? （有子网掩码255.255.255.240，则下面哪两个是有效的IP地址？）ACA. 192.168.10.210B. 192.168.10.32C. 192.168.10.94D. 192.168.10.112E. 192.168.10.12713. You have a Class B network ID and need about 450 IP addresses per subnet. What is the best mask for this network?（现在一个B类网络，要对其进行划分，要求每个子网里有450台主机，则最合适的子网掩码是？）CA. 255.255.240.0B. 255.255.248.0C. 255.255.254.0D. 255.255.255.014. Which of the following is a valid host for network 192.168.10.32/28?（对于网络192.168.10.32/28，下面哪个IP地址是属于此网络的合法IP地址？）AA. 192.168.10.39B. 192.168.10.47C. 192.168.10.14D. 192.168.10.5415. What is the valid host range that host 192.168.10.22/30 is a part of?（现有主机的IP地址192.168.10.22/30，那么，这个IP地址属于下面哪个网络？）CA. 192.168.10.0B. 192.168.10.16C. 192.168.10.20D. 192.168.0.0＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝Answers to Review Questions1. E. This is a Class B network address with 12 bits of subnetting—8 in the third octet and 4 in the fourth octet. The subnet in the third octet is 10, and the subnets in the fourth octet are 256 – 240 = 16, 32, 48, etc. Since the fourth octet is using 22, the host is in the 16 subnet, and since the next subnet is 32, the broadcast address for the 16 subnet is 31. The valid host range is the numbers in between, or 17–30.2. C. This is a Class B network address with 10 bits of subnetting—8 in the third octet and 2 in the fourth octet. The subnet in the third octet is 8, and the subnets in the fourth octet are 256 – 192 = 64, and 128. However, as long as all the subnet bits in the third octet are not all on at once, the subnets in the fourth octet really can be 0, 64, 128 and 192. This means that the host is in the 128 subnet and since the next subnet is 192, our broadcast address is 8.191.3. D. This is a Class C network address with 5 bits of subnetting. The valid subnets are 256 – 248 = 8, 16, 24, 32, 40, etc. Since the host ID is 33, we are in the 32 subnet. The next subnet is 40, so our broadcast address is 39.4. C. Take a look at the answers and see which subnet mask will give you what you need for subnetting. 252 gives you 62 subnets, 248 gives you 30 subnets, 240 gives you 14 subnets, and 255 is invalid. Only the third option (240) gives you what you need.5. B. If you use the mask 255.255.255.0, that only gives you 8 subnet bits, or 254 subnets. You are going to have to use 1 subnet bit from the fourth octet, or 255.255.255.128. This is 9 subnet bits (29 – 2 = 510).6. B. To answer this, you must be able to determine which Class C mask provides how many hosts and subnets. The 255.255.255.0 mask provides one network with 254 hosts. The 255.255.255.192 provides two subnets each with 62 hosts. The 255.255.255.224 provides 6 subnets, each with 30 hosts, and the 255.255.255.248 mask provides 30 subnets, each with 6 hosts.7. B. The Class C mask of 255.255.255.192 provides two subnets (four if you are using subnetzero— which Cisco is not!), each with 62 hosts.8. D. The third octet is used for all subnets, and the fourth octet is used only for hosts. 8 bits for subnetting, 8 bits for hosts. However, a subnet is already listed, so you have one subnet with 254 hosts. If the question stated 172.16.0.0/24, then the answer would be 254 subnets each with 254 hosts.9. B. This one takes some thought. 255.255.255.0 would give you 254 hosts each with 254 subnets. Doesn’t work for this question. 255.255.254.0 would provide 126 subnets, eachwith 510 hosts; the second option looks good. 255.255.252.0 is 62 subnets, each with 1022 hosts. So 255.255.254.0 is the best answer.10．B. To answer this, you must know that /28 is 255.255.255.240. 256 – 240 = 16. You subtract 2 from this number for all subnet bits and host bits on/off, so the answer is 14 subnets with 14 hosts each.11. B. This is a Class C network using a default mask. This provides a simple single network with 254 hosts.12. A, C. To answer this, just do 256 – 240 = 16. Keep adding 16 together until you reach the subnet mask value. 16 + 16 = 32. We’ll keep adding 16 until we have all our subnets: 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, 192, 208, 224. The broadcast addresses for each subnet are the numbers right before the next subnet and are not valid hosts. The answers are 210 and 94.13. C. Start with 255.255.255.0. This provides 254 subnets, each with 254 hosts. Move the subnet bits right if you need more subnets; move them left if you need more hosts. Since we need more hosts, we’re going to take away subnet bits. The next mask then is 255.255.254.0, which provides 7 subnet bits and 9 host bits, or 126 subnets, each with 510 hosts.14. A. You have to know that /28 is 255.255.255.240. 256 – 240 = 16, 32, 48, etc. The host ID of 34 is in the 32 subnet; the next subnet is 48, so the broadcast address is 47. The valid host range is 33–46, so answer A is correct.15. C. This is as easy as they get. A /30 is a 255.255.255.252 mask. 256 – 252 = 4, 8, 12, 16, 20, 24. This host is in the 20 subnet, the broadcast address is 23 and the valid host range is 21 and 22.。