CCNA 1 Chapter 8 v5.1 + v6.0 Test Online 2018-2019

Hint

When two or more switches are connected together, the size of the broadcast domain is increased and so is the number of collision domains. The number of broadcast domains is increased only when routers are added.

Hint

A router is used to route traffic between different networks. Broadcast traffic is not permitted to cross the router and therefore will be contained within the respective subnets where it originated.

Hint

Two reasons for creating subnets include reduction of overall network traffic and improvement of network performance. Subnets also allow an administrator to implement subnet-based security policies. The number of routers or switches is not affected. Subnets do not simplify network design.

Hint

The largest subnet in the topology has 100 hosts in it so the subnet mask must have at least 7 host bits in it (27-2=126). 255.255.255.0 has 8 hosts bits, but this does not meet the requirement of providing the maximum number of subnets.

Hint

Using a /29 prefix to subnet 192.168.10.0 results in subnets that increment by 8: 192.168.10.0 (1) 192.168.10.8 (2) 192.168.10.16 (3) 192.168.10.24 (4) 192.168.10.32 (5)

Hint

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.

Hint

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.

Hint

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).

Hint

Each network that is directly connected to an interface on a router requires its own subnet. The formula 2n, where n is the number of bits borrowed, is used to calculate the available number of subnets when borrowing a specific number of bits.

Hint

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.

Hint

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

Hint

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 (23 = 8 – 2 = 6).

Hint

The network IP address 192.168.5.0 with a subnet mask of 255.255.255.224 provides 30 usable IP addresses for each subnet. Subnet A needs 30 host addresses. There are no addresses wasted. Subnet B uses 2 of the 30 available IP addresses, because it is a serial link. Consequently, it wastes 28 addresses. Likewise, subnet C wastes 28 addresses. Subnet D needs 14 addresses, so it wastes 16 addresses. The total wasted addresses are 0+28+28+16=72 addresses.

Hint

Addresses 10.16.10.0 through 10.16.10.63 are taken for the leftmost network. Addresses 10.16.10.192 through 10.16.10.207 are used by the center network.The address space from 208-255 assumes a /28 mask, which does not allow enough host bits to accommodate 25 host addresses.The address ranges that are available include 10.16.10.64/26 and10.16.10.128/26. To accommodate 25 hosts, 5 host bits are needed, so a /27 mask is necessary. Four possible /27 subnets could be created from the available addresses between 10.16.10.64 and 10.16.10.191: 10.16.10.64/27 10.16.10.96/27 10.16.10.128/27 10.16.10.160/27

Hint

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.

Hint

DHCP is generally the preferred method of assigning IP addresses to hosts on large networks because it reduces the burden on network support staff and virtually eliminates entry errors. However, DHCP itself does not discriminate between authorized and unauthorized devices and will assign configuration parameters to all requesting devices. DHCP servers are usually configured to assign addresses from a subnet range, so there is no guarantee that every device that needs an address will get one.

Hint

If the block of addresses allocated to the pool is 192.168.10.0/24, there are 254 IP addresses to be assigned to hosts on the network. As there are 3 printers which need to have their addresses assigned statically, then there are 251 IP addresses left for assignment.

Hint

Because only one hexadecimal character is used to represent the subnet, that one character can represent 16 different values 0 through F.

Hint

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.

Hint

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.

Hint

Subnet 192.168.1.32/27 will have a valid host range from 192.168.1.33 – 192.168.1.62 with the broadcast address as 192.168.1.63 Subnet 192.168.1.64/27 will have a valid host range from 192.168.1.65 – 192.168.1.94 with the broadcast address as 192.168.1.95 Subnet 192.168.1.96/27 will have a valid host range from 192.168.1.97 – 192.168.1.126 with the broadcast address as 192.168.1.127

Hint

Network A needs to use 192.168.0.0 /25 which yields 128 host addresses. Network B needs to use 192.168.0.128 /26 which yields 64 host addresses. Network C needs to use 192.168.0.192 /27 which yields 32 host addresses. Network D needs to use 192.168.0.224 /30 which yields 4 host addresses.