Chapter 8
255.255.0.0.
100 subnets. 65532 hosts.=/16
/8
1000 subnets. 16382 hosts.=/
How many host addresses are available on the network 172.16.128.0 with a subnet mask of 255.255.252.0?
1022 A mask of 255.255.252.0 is equal to a prefix of /22. A /22 prefix provides 22 bits for the network portion and leaves 10 bits for the host portion. The 10 bits in the host portion will provide 1022 usable IP addresses (2^10 - 2 = 1022).
If a network device has a mask of /28, how many IP addresses are available for hosts on this network?
14. A /28 mask is the same as 255.255.255.240. This leaves 4 host bits. With 4 host bits, 16 IP addresses are possible, but one address represents the subnet number and one address represents the broadcast address. 14 addresses can then be used to assign to network devices.
A company has a network address of 192.168.1.64 with a subnet mask of 255.255.255.192. The company wants to create two subnetworks that would contain 10 hosts and 18 hosts respectively. Which two networks would achieve that?
192.168.1.96/28. 192.168.1.64/27. Subnet 192.168.1.64 /27 has 5 bits that are allocated for host addresses and therefore will be able to support 32 addresses, but only 30 valid host IP addresses. Subnet 192.168.1.96/28 has 4 bits for host addresses and will be able to support 16 addresses, but only 14 valid host IP addresses.
What is the prefix for the host address 2001:DB8:BC15:A:12AB::1/64?
2001:DB8:BC15:A. The network portion, or prefix, of an IPv6 address is identified through the prefix length. A /64 prefix length indicates that the first 64 bits of the IPv6 address is the network portion. Hence the prefix is 2001:DB8:BC15:A.
A DHCP server is used to assign IP addresses dynamically to the hosts on a network. The address pool is configured with 192.168.10.0/24. There are 3 printers on this network that need to use reserved static IP addresses from the pool. How many IP addresses in the pool are left to be assigned to other hosts?
251
A network administrator wants to have the same network mask for all networks at a particular small site. The site has the following networks and number of devices: IP phones - 22 addresses PCs - 20 addresses needed Printers - 2 addresses needed Scanners - 2 addresses needed The network administrator has deemed that 192.168.10.0/24 is to be the network used at this site. Which single subnet mask would make the most efficient use of the available addresses to use for the four subnetworks?
255.255.255.224. If the same mask is to be used, then the network with the most hosts must be examined for the number of hosts, which in this case is 22 hosts. Thus, 5 host bits are needed. The /27 or 255.255.255.224 subnet mask would be appropriate to use for these networks.
Which subnet mask would be used if 5 host bits are available?
255.255.255.224 (The subnet mask of 255.255.255.0 has 8 host bits. The mask of 255.255.255.128 results in 7 host bits. The mask of 255.255.255.224 has 5 host bits. Finally, 255.255.255.240 represents 4 host bits.)
what is the formula for needed hosts?
2^n - 2
How many bits must be borrowed from the host portion of an address to accommodate a router with five connected networks?
3. Each network that is directly connected to an interface on a router requires its own subnet. The formula 2^n, where n is the number of bits borrowed, is used to calculate the available number of subnets when borrowing a specific number of bits.
how many bits is a subnet mask?
32 bits. first for network. last for host.
2001:0DB8:BC15:0600:: to 2001:0DB8:BC15:0FFF:: The prefix length for the range of addresses is
52
IPv6 Interface ID of 2001:DB8::1000:A9CD:47FF:FE57:FE94/64?
A9CD:47FF:FE57:FE94 . The interface ID of an IPv6 address is the rightmost 64 bits, or last four hextets, of the address if no interface ID bits have been used for subnets.
Consider the following range of addresses: 2001:0DB8:BC15:00A0:0000:: 2001:0DB8:BC15:00A1:0000:: 2001:0DB8:BC15:00A2:0000:: ... 2001:0DB8:BC15:00AF:0000:: The prefix-length for the range of addresses is
All the addresses have the part 2001:0DB8:BC15:00A in common. Each number or letter in the address represents 4 bits, so the prefix-length is /60.
Broadcast domain
Broadcast traffic is not permitted to cross the router and therefore will be contained within the respective subnets where it originated.
Which two reasons generally make DHCP the preferred method of assigning IP addresses to hosts on large networks?
It ELIMINATES most address configuration ERRORS. It reduces the burden on network support staff.
A network administrator needs to monitor network traffic to and from servers in a data center. Which features of an IP addressing scheme should be applied to these devices?
PREDICTABLE STATIC IP addresses for easier identification. When monitoring SERVERS, a network administrator needs to be able to quickly identify them. Using a predictable static addressing scheme for these devices makes them easier to identify. Server security, redundancy, and duplication of addresses are not features of an IP addressing scheme.
In an Ethernet LAN, devices use broadcasts to locate:
Services(using DHCP) and other devices(using ARP)
A network administrator is variably subnetting a network. The smallest subnet has a mask of 255.255.255.248. How many usable host addresses will this subnet provide?
The mask 255.255.255.248 is equivalent to the /29 prefix. This leaves 3 bits for hosts, providing a total of 6 usable IP addresses (2^3 = 8 - 2 = 6).
What is a result of connecting two or more switches together?
The size of the broadcast domain is increased.
Longer prefix lengths ....
decreases the number of hosts per subnet.
What are two reasons a network administrator might want to create subnets?
easier to implement security policies improves network performance
Classless Subnetting
subnets can borrow bits from any host bit position to create other masks.
For each bit borrowed in the fourth octet ...
the number of subnetworks available is doubled while reducing the number of host addresses per subnet.
Variable Length Subnet Masks
traditional subnetting creates subnets of equal size. VLSM allows a network space to be divided into unequal parts. for example, 192.168.20.224/27 can be further subnetted. Because there are 5 host bits in the subnetted 192.168.20.224/27 address space, 3 more bits can be borrowed, leaving 2 bits in the host portion. The calculations at this point are exactly the same as those used for traditional subnetting. The bits are borrowed, and the subnet ranges are determined.