Refer to the exhibit. A computer that is configured with the IPv6 address as shown in the exhibit is unable to access the internet. What is the problem?
The gateway address is in the wrong subnet.
The DNS address is wrong.
There should not be an alternative DNS address.
The settings were not validated.
2.When subnetting a /64 IPv6 network prefix, which is the preferred new prefix?
3.Which two notations are useable nibble boundaries when subnetting in IPv6? (Choose two.)
Open the PT Activity. Perform the tasks in the activity instructions and then answer the question.
What issue is causing Host A to be unable to communicate with Host B
Host A and host B are on overlapping subnets
The subnet mask of host A is incorrect.
Host A has an incorrect default gateway.
The IP address of host B is not in the same subnet as the default gateway is on.
5.Consider the following range of addresses:
The prefix 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 is /60
6.Fill in the blank.
In dotted decimal notation, the subnet mask_________ will accommodate 500 hosts per subnet.
If the network has to accommodate 500 hosts per subnet, then we need 9 host bits
(2^9 – 2 = 510 hosts). The Class B subnet mask has 16 bits available and if we use 9 bits for hosts, we will have 7 network bits remaining.The subnet mask with 9 host bits is 11111111.11111111.11111110.00000000, which corresponds to 255.255.254.0.
Refer to the exhibit.
Given the network address of 192.168.5.0 and a subnet mask of 255.255.255.224, how many addresses are wasted in total by subnetting each network with a subnet mask of 255.255.255.224?
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
8.In a network that uses IPv4, what prefix would best fit a subnet containing 100 hosts?
Prefix /25 means that 7 bits are reserved for the host address range, which is 2^7-2=126. This is the best option to accommodate 100 hosts with the least waste of IP addresses. /23 gives 2^9-2=510 host addresses, a waste of 410 addresses. /24 gives 2^8-2=254 hosts, wasting 154 host addresses. /26 gives 2^6-2=62 host addresses, not enough to accommodate 100 hosts.
9.Which two reasons generally make DHCP the preferred method of assigning IP addresses to hosts on large networks? (Choose two.)
It eliminates most address configuration errors
It reduces the burden on network support staff
It ensures that addresses are only applied to devices that require a permanent address.
It guarantees that every device that needs an address will get one.
It provides an address only to devices that are authorized to be connected to the network.
10.Fill in the blank.
In dotted decimal notation, the IP address_______ is the last host address for the network
The binary representation of the network address 172.25.0.64 is 10101100.00011001.00000000.01000000, where the last six zeros represent the host part of the address. The last address on that subnet would have the host part equal to 111111, and the last host address would end in 111110. This results in a binary representation of the last host of the IP address as 10101100.00011001.00000000.01111110, which translates in decimal to 172.25.0.126.
11.When developing an IP addressing scheme for an enterprise network, which devices are recommended to be grouped into their own subnet or logical addressing group?
hosts accessible from the internet
mobile and laptop hosts
Hosts that are accessible from the Internet require public IP addresses. End-user clients, mobile and laptop hosts, and workstation clients are internal network devices that would be assigned private IP addresses.
Refer to the exhibit. How many broadcast domains are there?
13.How many usable host addresses are there in the subnet 192.168.1.32/27?
14.A nibble consists of _______ bits
15.How many host addresses are available on the network 172.16.128.0 with a subnet mask of 255.255.252.0?
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 (210 – 2 = 1022).
16.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?
random static addresses to improve security
addresses from different subnets for redundancy
predictable static IP addresses for easier identification
dynamic addresses to reduce the probability of duplicate addresses
17.How many bits must be borrowed from the host portion of an address to accommodate a router with five connected networks?
Refer to the exhibit. The network administrator has assigned the LAN of LBMISS an address range of 192.168.10.0. This address range has been subnetted using a /29 prefix. In order to accommodate a new building, the technician has decided to use the fifth subnet for configuring the new network (subnet zero is the first subnet). By company policies, the router interface is always assigned the first usable host address and the workgroup server is given the last usable host address.
Which configuration should be entered into the properties of the workgroup server to allow connectivity to the Internet?
IP address: 192.168.10.65 subnet mask: 255.255.255.240, default gateway: 192.168.10.76
IP address: 192.168.10.38 subnet mask: 255.255.255.240, default gateway: 192.168.10.33
IP address: 192.168.10.38 subnet mask: 255.255.255.248, default gateway: 192.168.10.33
IP address: 192.168.10.41 subnet mask: 255.255.255.248, default gateway: 192.168.10.46
IP address: 192.168.10.254 subnet mask: 255.255.255.0, default gateway: 192.168.10.1
Using a /29 prefix to subnet 192.168.10.0 results in subnets that increment by 8:
19.What is the subnet address for the address 2001:DB8:BC15:A:12AB::1/64?
The network and subnet fields cover 64 bits. This means that the first 4 groups of hexadecimal digits represent the network and subnet fields. The first address within that range is 2001:DB8:BC15:A::0
Refer to the exhibit. A company uses the address block of 188.8.131.52/16 for its network. What subnet mask would provide the maximum number of equal size subnets while providing enough host addresses for each subnet in the exhibit?
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.
21.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? (Choose two.)
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
22.A network administrator is variably subnetting a network. The smallest subnet has a mask of 255.255.255.248. How many 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 (23 = 8 – 2 = 6).
Refer to the exhibit. Match the correct IP address and prefix that will satisfy the usable host addressing requirements.
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.
24.Match the subnetwork to a host address that would be included within the subnetwork
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
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