CS4220: Computer Networks - Quiz 9 IP Addressing

The IP address range of 172.16.0.0 - 172.31.255.255 is what address class?

A: Private Class B The IP address range of 172.16.0.0 - 172.31.255.255 belongs to address class B. In classful addressing, the first octet of an IPv4 address determines its address class. In this range, the first octet starts with binary 1010, which corresponds to a class B address. Class A addresses have a first octet range of 1 to 126, class B addresses have a first octet range of 128 to 191, and class C addresses have a first octet range of 192 to 223. 172.16.0.0 - 172.31.255.255 represents the entire range of Class B private IP addresses.

How many subnets do you get with a subnet of 192.168.1.0/28?

CIDR block: 192.168.1.0/28 IP range (network - broadcast): 192.168.1.0 - 192.168.1.15 Subnet Mask: 255.255.255.240 IP Quantity: 16 The binary representation of /28 prefix is 11111111.11111111.11111111.11110000. The formula to determine the number of subnets is 2 ^ X, where X equals to the number of borrowed bits beyond the default mask of 24, for this Class C address. In this case, there are 4 borrowed bits ( 2 ^ 4 = 16); therefore, there are 16 possible subnets. https://www.dan.me.uk/ipsubnets?ip=192.168.1.0 - IP Subnet Calculator A: 16

How many hosts can you have with a subnet of 192.168.1.0/28?

In a subnet with the address range 192.168.1.0/28, the subnet mask /28 means there are 4 bits available for host addresses. With 4 bits, you can have 2^4 = 16 different combinations. However, 2 of these combinations are reserved for the network and broadcast addresses. Therefore, in this subnet, there are14 usable host addresses. 14 The binary representation of /28 prefix means that 28 are network bits. That leaves four bits for hosts in last octet (11110000). The formula to determine the number of hosts is 2 ^ X - 2, where X is equal to the number host bits. Substituting 4 for x = (2 ^ 4 - 2 = 14). therefore, there are 14 possible hosts. Remember that we subtract the network number and the broadcast address.

How many assignable IP addresses exist in the 172.16.1.10/28 network?

In the 172.16.1.10/28 network, there are 16 possible IP addresses because there are 4 bits available for hosts. However, 2 of those addresses are reserved for the network and broadcast addresses. So, there are 16 - 2 = 14 usable IP addresses that can be assigned to devices in this network. 14 There are 4 bits left here for host address assignment, and 2 raised to the fourth power is 16. Subtract 2 for the network ID and broadcast address.

NAT was originally designed to allow what kind of addresses to be translated into Internet-routable IP addresses?

NAT was originally designed to allow RFC 1918 private IP addresses to be translated into Internet-routable IP addresses. RFC 1918 defines a range of IP addresses that are reserved for private use within a private network. These addresses are: 10.0.0.0 to 10.255.255.255 172.16.0.0 to 172.31.255.255 192.168.0.0 to 192.168.255.255 NAT enables these private IP addresses to be translated into a single public IP address or a pool of public IP addresses when accessing resources on the internet. A: RFC 1918 private IP addresses Network Address Translation (NAT) allows private IP addresses (as defined in RFC 1918) to be translated into Internet-routable IP addresses (public IP addresses).

What do sections A and B refer to in this IP address?

Subnet masks are used to separate the network and host portions of an IPv4 address. In a subnet mask, 1s are used to indicate the network portion of the address, and 0s are used to indicate the host portion of the address. In the preceding example, Section A is the network portion of the IPv4 address, and Section B is the host portion of the IPv4 address, due to the mask that is in use. A: Private Class B

What is the network address for APIPA?

The IP address range for APIPA is (169.254. 0.1 to 169.254. 255.254) having 65, 534 usable IP addresses, with the subnet mask of 255.255. 0.0. The APIPA feature allows a networked device to self-assign an IP address from the 169.254.0.0/16 network. However, since it is not 'routable', it is a just a number to represent that private network. A: 169.254.0.0/16

Your company has been assigned the 192.168.48.0/24 network for use at one of its sites. You need to use a subnet mask that will accommodate at least 3 subnets while simultaneously accommodating the maximum number of hosts per subnet. What subnet mask should you use?

To accommodate at least 3 subnets while maximizing the number of hosts per subnet within the 192.168.48.0/24 network, you need to determine the number of bits required for the subnet and host portions. To have at least 3 subnets, you need 2 bits for subnetting because 2^2 = 4, which allows for 4 possible subnets (0, 1, 2, 3). This leaves 6 bits for host addresses within each subnet (since 8 bits - 2 bits for subnetting = 6 bits for hosts). To represent the 6 bits for hosts, you can have 2^6 = 64 host addresses per subnet (from 0 to 63). However, subtracting the network address (all zeros) and the broadcast address (all ones) leaves 64 - 2 = 62 usable host addresses per subnet. Therefore, the subnet mask you should use to accommodate at least 3 subnets and maximize the number of hosts per subnet is 255.255.255.192, which is equivalent to the /26 subnet mask. A: /26 Using a /26 mask gives you 2 bits for subnetting. This creates four potential subnets, satisfying the minimum requirement

What is the result of the binary conversion in the example shown here?

To convert the binary number 10010110 to decimal using the given place values (128, 64, 32, 16, 8, 4, 2, 1), you can sum the values corresponding to the '1' digits: Explanation: To convert a binary number to its decimal equivalent, you need to multiply each digit in the binary number by the corresponding place value in powers of 2 (from right to left) and then sum up these products. In this case, we have the binary number 10010110, and we are given the place values as 128, 64, 32, 16, 8, 4, 2, and 1. Step 2/2 Here's how the conversion works: The leftmost digit is '1', and it corresponds to the place value of 128. The next digit is '0', which does not contribute to the sum (0 * 64). The next digit is '0', which also does not contribute to the sum (0 * 32). The next digit is '1', and it corresponds to the place value of 16. The next digit is '0', which does not contribute to the sum (0 * 8). The next digit is '1', and it corresponds to the place value of 4. The next digit is '1', and it

How many subnets do you get with a subnet of 192.168.1.0/28?

16 The binary representation of /28 prefix is 11111111.11111111.11111111.11110000. The formula to determine the number of subnets is 2 ^ X, where X equals to the number of borrowed bits beyond the default mask of 24, for this Class C address. In this case, there are 4 borrowed bits ( 2 ^ 4 = 16); therefore, there are 16 possible subnets.

The IP address range of 172.16.0.0 - 172.31.255.255 is what address class?

A: Private Class B 172.16.0.0 - 172.31.255.255 represents the entire range of Class B private IP addresses.

In IPv6, what technology allows for the automatic assignment of the host portion of an address? You may need to look this one up if we have not reached it in class.

EUI64 in IPv6 permits the automatic generation of host portions of addresses. A: EUI-64