Figure: Find the Subnet ID: 220.127.116.11, 255.255.240.0
Q1. When you’re thinking about an IP address using classful addressing rules, the address can have three parts: network, subnet, and host. If you examined all the addresses in one subnet, in binary, which of the following answers correctly states which of the three parts of the addresses will be equal among all addresses? (Pick the best answer.) a. Network part only b. Subnet part only c. Host part only d. Network and subnet parts e. Subnet and host parts
Answer: D. When using classful IP addressing concepts as described in Chapter 13, “Analyzing Subnet Masks,” addresses have three parts: network, subnet, and host. For addresses in a single classful network, the network parts must be identical for the numbers to be in the same network. For addresses in the same subnet, both the network and subnet parts must have identical values. The host part differs when comparing different addresses in the same subnet.
Q2. Which of the following statements are true regarding the binary subnet ID, subnet broadcast address, and host IP address values in any single subnet? (Choose two.) a. The host part of the broadcast address is all binary 0s. b. The host part of the subnet ID is all binary 0s. c. The host part of a usable IP address can have all binary 1s. d. The host part of any usable IP address must not be all binary 0s.
Answer: B and D. In any subnet, the subnet ID is the smallest number in the range, the subnet broadcast address is the largest number, and the usable IP addresses sit between those. All numbers in a subnet have identical binary values in the prefix part (classless view) and network + subnet part (classful view). To be the lowest number, the subnet ID must have the lowest possible binary value (all 0s) in the host part. To be the largest number, the broadcast address must have the highest possible binary value (all binary 1s) in the host part. The usable addresses do not include the subnet ID and subnet broadcast address, so the addresses in the range of usable IP addresses never have a value of all 0s or 1s in their host parts.
Q3. Which of the following is the resident subnet ID for IP address 10.7.99.133/24? a. 10.0.0.0 b. 10.7.0.0 c. 10.7.99.0 d. 10.7.99.128
Answer: C. The mask converts to 255.255.255.0. To find the subnet ID, for each octet of the mask that is 255, you can copy the IP address’s corresponding values. For mask octets of decimal 0, you can record a 0 in that octet of the subnet ID. As such, copy the 10.7.99 and write a 0 for the fourth octet, for a subnet ID of 10.7.99.0.
Q4. Which of the following is the resident subnet for IP address 192.168.44.97/30? a. 192.168.44.0 b. 192.168.44.64 c. 192.168.44.96 d. 192.168.44.128
Answer: C. First, the resident subnet (the subnet ID of the subnet in which the address resides) must be numerically smaller than the IP address, which rules out one of the answers. The mask converts to 255.255.255.252. As such, you can copy the first three octets of the IP address because of their value of 255. For the fourth octet, the subnet ID value must be a multiple of 4, because 256 – 252 (mask) = 4. Those multiples include 96 and 100, and the right choice is the multiple closest to the IP address value in that octet (97) without going over. So, the correct subnet ID is 192.168.44.96.
Q5. Which of the following is the subnet broadcast address for the subnet in which IP address 172.31.77.201/27 resides? a. 172.31.201.255 b. 172.31.255.255 c. 172.31.77.223 d. 172.31.77.207
Answer: C. The resident subnet ID in this case is 172.31.77.192. You can find the subnet broadcast address based on the subnet ID and mask using several methods. Following the decimal process in the book, the mask converts to 255.255.255.224, making the interesting octet 4, with magic number 256 – 224 = 32. For the three octets where the mask = 255, copy the subnet ID (172.31.77). For the interesting octet, take the subnet ID value (192), add magic (32), and subtract 1, for 223. That makes the subnet broadcast address 172.31.77.223.
Q6. A fellow engineer tells you to configure the DHCP server to lease the last 100 usable IP addresses in subnet 10.1.4.0/23. Which of the following IP addresses could be leased as a result of your new configuration? a. 10.1.4.156 b. 10.1.4.254 c. 10.1.5.200 d. 10.1.7.200 e. 10.1.255.200
Answer: C. To answer this question, you need to find the range of addresses in the subnet, which typically then means you need to calculate the subnet ID and subnet broadcast address. With subnet ID/mask of 10.1.4.0/23, the mask converts to 255.255.254.0. To find the subnet broadcast address, following the decimal process described in this chapter, you can copy the subnet ID’s first two octets because the mask’s value is 255 in each octet. You write a 255 in the fourth octet because the mask has a 0 on the fourth octet. In octet 3, the interesting octet, add the magic number (2) to the subnet ID’s value (4), minus 1, for a value of 2 + 4 – 1 = 5. (The magic number in this case is calculated as 256 – 254 = 2.) That makes the broadcast address 10.1.5.255. The last usable address is 1 less: 10.1.5.254. The range that includes the last 100 addresses is 10.1.5.155–10.1.5.254.
Q7. A fellow engineer tells you to configure the DHCP server to lease the first 20 usable IP addresses in subnet 192.168.9.96/27. Which of the following IP addresses could be leased as a result of your new configuration? a. 192.168.9.126 b. 192.168.9.110 c. 192.168.9.1 d. 192.168.9.119
Answer: B. To answer this question, you do not actually need to calculate the subnet broadcast address, because you only need to know the low end of the range of addresses in the subnet. The first IP address in the subnet is 1 more than the subnet ID, or 192.168.9.97. The first 20 addresses then include 192.168.9.97–192.168.9.116.