␡
- Binary
- Hexadecimal
- Conversion Between Binary, Hex, and Decimal
- IP Address Components
- Subnetting
- VLSM
- Route Summarization
- IPv6
- Review Questions
- Answers to Review Questions
- What Next?
This chapter is from the book
This chapter is from the book
This chapter is from the book
This chapter is from the book
Answers to Review Questions
- Answers B and D are correct. Answer A in decimal would be 551. Answer C in decimal would be 323. Answer E in decimal is 230.
- Answers C and D are correct. Answer A in hex is 0xB4. Answer B in hex is 0xBF. Answer E is simply an attempt to trick you—the correct decimal answer is incorrectly expressed as a hex value.
- Answer D is correct. A will only support 2 hosts; B only 6, and C only 14. Watch out for the minus 2 in the host calculation! Answer C creates 16 IP addresses on the subnet, but we lose 2—one for the network ID and one for the broadcast ID.
- Answer D is correct. The mask 255.255.254.0 gives us nine 0s at the end of the mask; 29 – 2 = 510. Answer A is checking to see if you missed the 254 in the third octet because you are used to seeing 255. Answer B does the same thing plus tries to catch you on not subtracting 2 from the host calculation. Answer C tries to catch you on not subtracting 2, and Answer E is the increment of the given mask that you might pick if you were really off track.
- Answer D is correct. Disregarding for the moment the possibility that Mr. Martin might be wrong, let’s look at the requirements. He says make room for 12 managers, and make the subnets as small as possible while doing so. You need to find the mask that has sufficient host IP space without making it bigger than necessary. Answer A is invalid; 12 is not a valid mask value. Remember, a mask is a continuous string of 1s followed by a continuous string of 0s. In Answer B, the mask is valid, but it is not correct. This mask has eight 0s at the end, which, when we apply the formula 28 – 2 gives us 254 hosts. That makes more than enough room for the 12 managers, but does not meet the “as small as possible” requirement. Answer C has the correct mask value in the wrong octet. That mask gives us eight 0s in the fourth octet, plus another four in the third octet; that would give us 4094 hosts on the subnet. Answer E gives us 30 hosts per subnet, but that only meets half the requirement. This mask does not provide the minimum number of hosts.
- Answer C is correct. The default mask for a Class B is 255.255.0.0. Answer C extends that mask by three bits, creating eight subnets (23 = 8). Although we only need six of them, we have to use the mask that creates eight because the next smaller mask would only create four, and that isn’t enough. Answer A is incorrect because it is the default mask for a Class B and not subnetted at all. Answers B and D are incorrect because although they create sufficient subnets, they do not maximize the number of hosts per subnet and so are not the best answer. Answer E uses the correct mask in the wrong octet.
- Answer E is correct. With ip subnet zero enabled, all 64 subnets created by the mask in use become available. Answers A, B, and C are not even close and are simply distracters. Answer D wants to catch you by subtracting the zero subnets.
- Answers B, D, F, G, H, and J are correct. Answers A and C are incorrect because this is a Class A address. Answer E is incorrect because only 16 bits were stolen. Answer I is incorrect because it does not subtract the two IPs for the network ID and broadcast ID.
- Answer D is correct. With that mask, the increment is 64. Greg is in the first subnet, and Indy is in the second. Without a router between them, their PCs will not be able to communicate above Layer 2. Answer A is incorrect; the broadcast ID for Indy would be .63. Answer B is incorrect; nothing is wrong with the mask. Answer C is incorrect; the zero subnets are the first and last created, and Indy is in the second subnet. The question does mention the zero subnets, so we can use them, and in any case Windows 8 fully supports them.
Answers C and F are correct. This is an increment question. The increment here is 8, so you should start by jotting down the multiples of 8 (those are all the network IDs), and then noting what 1 less than each of the network IDs is (those are the broadcast IDs). From there, it is easy to find what the first and last IPs in each subnet are. (Remember that Dave says we can use the zero subnets.)
The eighth subnet network ID is 192.168.1.56; the first valid IP is 192.168.1.57. The twelfth subnet network ID is 192.168.1.88; the last valid IP is 192.168.1.94. Answers A and D are incorrect because they do not use the subnetted address space Dave requested. Answer B is incorrect because it is a network ID. Answer E is incorrect because it is a broadcast ID.
SuperChallenge answers:
- Incorrect (same subnet as Fa1/0)
- Incorrect (network ID)
- Incorrect (not enough hosts)
- Incorrect (Fa1/0 IP already assigned)
- Correct
- Incorrect (wrong subnet—not on the same network as the connected interface on Main)
- Correct
- Incorrect (not enough hosts)
- Correct
- Incorrect (overlaps with Fa1/1)
- Correct
- Incorrect (overlaps with Fa2/0)
- Incorrect (wrong mask)
- Incorrect (no Fa0/2 interface on Genoa)
- Correct
- Incorrect (network ID)
- Incorrect (overlaps with Genoa)
- Incorrect (network ID)
- Correct
- Answer E is correct. Answers A and C use incorrect syntax; Answer D uses the wrong mask. Answer B looks correct, but it does not use the correct network ID; the range should always start at a binary increment (in this case 0, not 1). (In other words, this is a case of “best answer.”) Note that the correct summary does include the 192.168.0.0/24 network as well (not just 192.168.1–15.0/24). This is intended to confuse and distract you!
- Answer C is correct. Answer A uses the wrong mask and summarizes more than the specified networks. Answer B subnets instead of summarizes. Answer D uses the wrong address and mask.
- Answers B, D, and E are correct. The networks in Answers A and C are out of the range, which is 172.16.0.0 through 172.23.0.0.
- Answers B, D, and F are correct. Answer A might have an element of truth, but Cisco does not have much of a sense of humor. Answer C is incorrect because discontiguous subnets are a real problem if you intend to summarize. Answer E is incorrect; route summarization does not identify but rather hides the effects of flapping interfaces.
- Answer B is correct. IPv6 addresses must have (or at least indicate, perhaps with ::) eight sets of four valid hex characters. Answer A is wrong because G is not a valid hex character. Answer C is wrong because it uses the :: notation twice, which is invalid. Answer D is wrong because it uses only five sets.
- Answer A is correct. IPv6 addresses must have eight sets of four valid hex characters. Answer B is wrong because it starts with FF, which indicates a multicast address, not a unicast. Answer C is wrong because it uses :: twice, which is invalid. Answer D is wrong because there are only seven sets.
- Answers A, C, and E are correct. These are the same address, represented in three valid notations: A is not truncated, C has dropped leading 0s, and E has compressed the contiguous all-zero groups with the ::. Answer B is wrong because it drops trailing 0s, not leading 1s. D is wrong because it uses the :: twice, which is invalid.
- Answer D is correct. None of these is a valid unicast format. Answer A is an IPv6 multicast (starts with FF00/8). Answer B has only four sets instead of eight. Note that if there had been a double colon before the last 1, it could have been correct. Answer C uses the decimal 255 to confuse you; it could have been correct except that there are nine sets.