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14. D. Learning In this state, the bridge learns information about MAC addresses but does not yet forward frames. 15. C. VLAN stacking VLAN stacking allows a provider to stack its own VLAN information in front of the customer s VLAN information to support customers with overlapping VLANs.
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1. C. Ethernet switches cannot build forwarding tables. Ethernet switches, in fact, do build FDB tables of MAC addresses. The other answers are all factors that prevent Ethernet from creating global networks. 2. B. It provides for a logical hierarchy. This is one of the main reasons that IP can support global networks the use of hierarchical addressing. 3. D. The current version is IPv5. The current version is IPv4, and the next version is IPv6. 4. D. 10.254.1.1 10.254.1.1 is the only valid host address. The other addresses are either out of supported IP address ranges, are in the broadcast range (255), or are multicast (224). 5. B. 193 Calculate this as 28 + 27 + 20 = 128 + 64 + 1 = 193. 6. D. None of the above This is a Class D or multicast address. You can tell this by examining the first octet in binary: 224 = 128 + 64 + 32 = 11100000. The first 3 bits are 1, so this indicates that it is a Class D address. 7. C. 200.1.1.254 The private address ranges are 10.0.0.0/8, 172.16.0.0/12, and 192.168.0.0/16. 8. B. You can identify the host portion of the address without the need for a mask. This is true for classful addressing, but not for classless addressing. Using classless addressing, you need the subnet mask to determine how many bits are used for the subnet and how many for the host.
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9. A. 16 If you need seven hosts per subnet, this means that you need to leave at least 4 bits for each subnet because 24 2 = 14 available hosts. This leaves 4 bits (8 4 = 4) for the subnet, which means you can have 24 = 16 subnets. 3 bits for each subnet is not sufficient because two addresses (all zeros and all ones) cannot be used. 10. D. 128 The trick here is recognizing that if you need 300 hosts per subnet, then you need to use more than 8 bits of the IP address for hosts because you can only have 255 max hosts for an 8-bit octet. If you borrow one of the bits from the third octet, this gives you a maximum of 510 hosts, and leaves 7 bits of the third octet for subnets. 27 = 128, so you can have 128 subnets. 11. C. /10 The key here is to remember that 255 means all 1s, so you know that the first octet has all the bits used and therefore the mask is at least /8. 192 translates into the highest 2 bits of the next octet: 27 + 26 = 128 + 64 + 192, so this means that 10 bits are used so the subnet mask is /10. 12. B. 2 The key here is usable host addresses. A /30 leaves 2 bits for hosts, which would translate into 22 = 4. However, the 0 host is the subnet address and the all 1s host is the broadcast, so those addresses cannot be used. This leaves only two usable host addresses. 13. D. subnet 10.1.1.0 B is not as correct because technically 10.1.1.0 is a subnet of network 10.0.0.0. 14. B. Route aggregation CIDR is a technology that implements route aggregation, but the question asks what the concept is called, and the concept is route aggregation. 15. B. 218 The key is to determine how many bits are available for subnetting. We know we need 31 hosts, so we know we need at least 5 bits because 25 = 32. However, we lose 2 bits for each subnet for the subnet and broadcast address, so we actually need 6 bits for the hosts because using 5 bits would only give us 30 hosts per subnet. Since there are 24 total available bits for subnets on network 10.0.0.0/8, 24 6 host bits = 18 subnet bits, and 218 = 262,144 available subnets.
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