Variable Length or Nonclassful Subnet Masking 1
3. Your corporate network uses variable length subnetting to make more efficient use of IP addresses. One of the IP addresses for a host is 131.39.161.17 with a subnet mask of 255.255.248.0.What is the proper notation for the network to which this host is connected?
0 A. Using the subnet mask to identify the number of network bits used, you can add bit values, starting from the left, until you reach 248.Thus, you add 128 + 64 + 32 + 16 + 8, or the left-most 5 bits, to generate 248 in the third octet.You know that you're using 21 bits for the network address space because the total number of bits set to 1 in the subnet mask is 21 (counting from left to right).Thus, the notation must end with /21. Next, you need to determine the underlying network address. Using bitwise ANDing to compare the IP address with the subnet mask yields this operation:
131.39.161.17 10000011.00100111.10100001.00010001
255.255.248.0 11111111.11111111.11111000.00000000
Result 10000011.00100111.10100000.00000000
Dotted decimal 131.39.160.0
Therefore, Answer A is correct: 131.36.160.0/21.
0 In this example, the subnet mask's third octet determines the total number of network bits used.To equal 248, you must use the left-most five bits of the third octet.Therefore, the total number of bits set to 1 in the subnet mask is 21.The notation must end with /21, which eliminates Answer B as a possible answer. Performing bitwise ANDing would also help you determine that the underlying network address in the third octet is 160, not 161.To yield 161, the third octet of the subnet mask would have to be 249, not 248. Answer C provides the correct number of network bits (21) but uses the host's IP address. The question specified the underlying network address, not the host address; therefore Answer C is incorrect. Answer D is incorrect because the underlying network address is correct but the notation should be 21, not 20. If you selected this answer, it's possible you made an error in your math.
4. You need to create several subnets for your corporate network. Each subnet should have no more than two host addresses available per subnet.You have a subnet with the address of
136.42.255.0/24.What are the first two subnet addresses that would be created in this configuration?
0 D.To end up with two host IP addresses per subnet, you need two bits for the host address space. Using one bit gives you two numbers (0 and 1) but a host address cannot be all 0s or all 1s.Thus, you need two bits so you can use two of the four possible configurations (01, and 10, discarding 00 and 11).Taking two bits for the host address space leaves 30 bits for the network address space. The notation must end with /30 to meet the requirements. Next, you need to determine the starting network addresses in this configuration. Begin with the given address of 136.42.255.0/24.This uses the first three octets for the network address space.You are extending this by six bits. Using the binary representation, you can see what happens in the fourth octet, showing the first three subnets created.
136.42.255.0/30 10001000.00101010.11111111.00000000
(network bits underlined)
136.42.255.4/30 10001000.00101010.11111111.00000100
136.42.255.8/30 10001000.00101010.11111111.00001000
The lowest bit value associated with the network address space is the bit with the weighted binary value of 4.You now know that your network address will increment by four each time, so each subnet will end with some multiple of 4 (4, 8, 12,16, etc.).The first two addresses are 136.42.255.0/30 and 136.42.255.4/30.Therefore, Answer D is correct.
0 The requirement is for two IP addresses per subnet. Using 31 bits would leave one bit for host addresses, but neither address (0, 1) is legal for a host address. Therefore, you need at least two bits to meet this requirement, reducing the number of network bits to 30, so Answer A is incorrect.. Answer B begins with the correct number of network bits used (as explained in the correct answer), but is incorrect. However, the network addresses increment by two, which is the 31st bit. For this answer to be correct, the network address would have to be 31 bits, making the notation 136.42.255.2/31, 136.42.255.4/31. In addition, these are the second and third addresses in this range (the first is 136.42.255.0/31), the question asked for the first two. Answer C is incorrect because if you used only 29 bits for your network space, you would have three bits left for the host address space. Three bits yields eight addresses, six of which can be used (000, 001, 010, 011, 100,101,110, 111).This does not meet the requirement of two IP addresses per subnet.
5. You've just accepted a position in the IT department at a small, growing company.You've been asked to devise a subnetting scheme for their network that will allow for a maximum of 30 hosts per subnet.The company's assigned network ID is 197.228.69.0. What is the subnet mask for the configuration you must develop?
A. 255.255.255.248
B. 255.255.255.240
C. 255.255.255.224
0 C.Your assigned network is a Class C network, which uses the subnet mask of 255.255.255.0 by default.To subnet this network to yield a maximum of 30 host addresses per subnet, you need to use five bits for the host address space and 27 (32 - 5) for the network address space.Three bits in the fourth octet equals 128 + 64 + 32 or 224.The correct subnet mask for a maximum of 30 hosts per subnet is 255.255.255.224; therefore Answer C is correct.
0 Answer A is incorrect because it uses five bits from the fourth octet for the network address space. That leaves only three bits (8 - 5) for the host address space, which allows for only eight host addresses (six useable) per subnet.This does not meet the requirement. If you selected this answer, you may have gotten your network and host address bits mixed up. Answer B is incorrect because it uses four network bits and four host bits for the fourth octet. In this case, you have 16 host addresses (14 useable) per subnet, which does not meet the requirement. Answer D allows for 11 host address bits because the third octet is now set to 248, which uses the left-most five bits for the network
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