CCNA Cram Ch.3
What is the ideal mask to use on point-to-point serial links?
/30, or 255.255.255.252. That provides two valid host IPs, one for each end of the link.
The only possible valid values in any given octet of a subnet mask are:
0, 128, 192, 224, 240, 248, 252, 254, and 255
What is the range of a Class A Private IP address?
10.0.0.0 to 10.255.255.255
Write the following subnet mask in binary format: 255.255.255.240.
11111111.11111111.11111111.11110000
What is the range of a class B Private ip address?
172.16.0.0 to 172.31.255.255
What is the best summary for the following range of subnets? 172.20.32.0/24 to 172.20.47.0/24
172.20.32.0/20 or 172.20.32.0 255.255.240.0
What is the range of the first octet of a Class C address?
192-223
How does VLSM make IP addressing more efficient? A. By increasing the total number of IP addresses. B. By decreasing the total number of IP addresses C. By creating subnets D. By allowing a routed system to include subnets of different mask lengths to suit requirements
D. By allowing a routed system to include subnets of different mask lengths to suit requirements
What is the range of private Class A addresses?
0.0.0.0 to 10.255.255.255
What is 127.0.0.1 in binary?
01111111.00000000.00000000.00000001
What is the range of the first octet of a Class A address?
1-127
What is the loopback (localhost) ip address?
127.0.0.1
What is the range of the first octet of a Class B address?
128-191
What is the range (in decimal) of Class B addresses?
128.0.0 .0 to 191.255.255.255
What is the range of a class C Private ip address?
192.168.0.0 to 192.168.255.255
1. Write the following binary IP in dotted-decimal format: 11000000.10101000.00000001.11111110.
192.168.1.254
What is the default subnet mask for a class A address?
255.0.0.0
What is the default subnet mask for a class B address?
255.255.0.0
What is the default subnet mask for a class C address?
255.255.255.0
Types of IPv6 Addresses, Global unicast:
A global unicast IPv6 address is the equivalent of a public, registered IP address. They are Internet routable, globally registered IPs that must be leased from an ISP.
Which of the following are valid IPv6 addresses? Choose all that apply. A. 2001:0db8:0000:0000:0000:ff00:0042:8329 B. 2001:db8:0:0:0:ff00:42:8329 C. 2001:db8::ff00:42:8329 D. 0000:0000:0000:0000:0000:0000:0000:0001 E. ::1 F. ::192:168:1:1
All answers are correct! Even F, which looks bogus, is sometimes used
Types of IPv6 Addresses, Unicast:
An IPv6 unicast address is the same as an IPv4 unicast address; it is an IP that is assigned to an interface on a host. It can be the source of an IP packet or the destination for one. A packet sent to a unicast address goes to the one host with that address.
Types of IPv6 Addresses, Anycast:
An anycast address is a single address that is assigned to multiple hosts. This is similar to a multicast, except that a packet for the anycast address will be delivered to the one host that is nearest according to the routing protocol's idea of distance. There is no special prefix for anycast addresses.
Which of the following is a valid IPv6 unicast address format? A. 2001:8888:EEEE:1010:0000:0000:0000:0001 B. FF00:0000:0000:0002:00CO:00A8:0001:0042 C. 2001:8888::2FFE::00A8 D. FFFF:FFFF:FFFF:FFFF:FFFF:FFFF:FFFF
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.
If the router's MAC address is 0012.7feb.6b40, which of the following is the correct EUI-64 format for the IPv6 link-local interface address? A. ::0012:7FEB:6B40 B. 2001:DB8::212:7FFF:FEEB:6B40 C. 2001:DB8::212:7FFF:FE80:6B40 D. 2001:DB8::2012:7FEB:6B40 E. 2001:DB8::212:7FFF:0000:6B40
Answer B is correct. Answer A is incorrect because it omits the network number. Answer C is wrong because it uses the wrong value for the EUI expansion. (It uses FE80 instead of the correct FFFE.) Answer D is incorrect because it uses the wrong link-local network number, adding 0x2 to the end of the link-local identifier.
Which of the following is a valid command to apply an IPv6 address to a router interface? A. interface fastethernet 1/0 ip address 2001:AB00:00FF:1::/64 eui-64 B. interface fastethernet 1/0 ipv6 address 2001:AB00:00FF:1::/64 eui-64 C. line con 0 ipv6 address 2001:AB00:00FF:1::/64 eui-64 D. interface fastethernet 1/0 ipv6 address 2001:AB00:000FF:1/64 eui-64
Answer B is correct. Answer A is wrong because it uses ip instead of ipv6. Answer C is wrong because the Console port cannot be given an IPv6 (or IP for that matter) address. Answer D is wrong because the IPv6 address is missing the :: to make it the correct 64-bit length.
Which of the following is a valid IPv6 address format? A. G412:AFFA:2001:0000:0000:0000:0000:0001 B. 2001:8888:EEEE:1010:0000:0000:0000:0001 C. 2001::8888::1 D. 2010:2112:5440:1812:1867
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.
What must a routing protocol be able to do to support VLSM? A. Multicast B. Automatically summarize networks to a common mask C. Advertise the mask for each subnet in the routing update
Answer C is correct.
Which of the following is the best summary statement for the following range of networks? 192.168.24.0 /24-192.168.31.0 /24 A. 192.168.24.0 255.255.240.0 B. 192.168.24.0 /28 C. 192.168.24.0 /21 D. 192.168.0.0 /27
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.
Your boss, Duncan, does not seem to be able to grasp subnetting. He comes out of a management meeting and quietly asks you to help him with a subnetting issue. He needs to divide the Class B address space the company uses into six subnets for the various buildings in the plant, while keeping the subnets as large as possible to allow for future growth. What is the best subnet mask to use in this scenario? A. 255.255.0.0 B. 255.255.248.0 C. 255.255.224.0 D. 255.255.240.0 E. 255.255.255.224
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.
Julie's IP address is 192.168.1.21 255.255.255.240. Joost's IP is 192.168.1.14/28. Their computers are connected together using a crossover Ethernet cable. Why can't they ping each other? A. The subnet masks are different B. They can. This is another trick question. C. Because they are in different subnets. D. Because the router does not support subnetting. E. Because it should be a straight-through cable.
Answer C is correct..
Why is VLSM important to modern IP networks? A. Because networks that use the same mask cannot route. B. Because every subnet must use a different mask to avoid conflicts. C. Because it allows us to uniquely identify each subnet by its mask number. D. Because it allows each subnet in a routed system to be correctly sized for the requirement.
Answer D is correct.
How many hosts are on the network 172.16.41.0/27? A. 65,534 B. 32 C. 254 D. 30 E. 27 F. 14
Answer D is correct. -2 for subnet & broadcast
You have been asked to create a subnet that supports 16 hosts. What subnet mask should you use? A. 255.255.255.252 B. 255.255.255.248 C. 255.255.255.240 D. 255.255.255.224
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.
You are a senior network engineer at True North Technologies. Your boss, Mr. Martin, asks you to create a subnet with room for 12 IPs for some new managers. Mr. Martin promises that there will never be more than 12 managers, and he asks you to make sure that you conserve IP address space by providing the minimum number of possible host IPs on the subnet. What subnet mask will best meet these requirements? A. 255.255.255.12 B. 255.255.255.0 C. 255.255.240.0 D. 255.255.255.240 E. 255.255.255.224
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.
Which of the following is a valid IPv6 unicast address? A. FF00:2112:1812:5440::1 B. 1812:2112:5440:1 C. 255:255:255:255:255:255:255:255:255 D. None of the above
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.
Given the mask 255.255.254.0, how many hosts per subnet does this create? A. 254 B. 256 C. 512 D. 510 E. 2
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
Indy and Greg have configured their own Windows 8 PCs and connected them with crossover cables. They can't seem to share their illegally downloaded MP3 files, however. Given their configurations, what could be the problem? Indy's configuration: IP:192.168.0.65 Mask:255.255.255.192 Greg's configuration: IP:192.168.0.62 Mask:255.255.255.192 A. Indy is using a broadcast ID for his IP. B. Greg is using an invalid mask. C. Indy's IP is in one of the zero subnets. D. Greg and Indy are using IPs in different subnets.
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.
Which of the following is the best summary statement for the following range of networks? 192.168.1.0 /24-192.168.15.0 /24 A. 192.168.1.0 B. 192.168.1.0 255.255.240.0 C. 192.168.1.0 0.0.15.0 D. 192.168.1.0 255.255.248.0 E. 192.168.0.0 255.255.240.0
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!
You have purchased several brand-new Cisco routers for your company. Your current address space is 172.16.0.0 /22. Because these new routers support the ip subnet zero command, you realize you are about to gain back two subnets that you could not use with the old gear. How many subnets total will be available to you once the upgrades are complete? A. 4 B. 2 C. 32 D. 62 E. 64
Answer E is correct. With ip subnet zero enabled, all 64 subnets created by the mask in use become available.
How many subnets are created by the address 192.168.1.0 255.255.255.248? A. 1 B. 2 C. 4 D. 8 E. 16 F. 32 G. 64
Answer F is correct.
What is the broadcast ID of the seventh subnet created using 172.16.0.0/28? A. 172.16.111.0 B. 172.16.0.0 C. 172.16.0.7 D. 172.16.0.96 E. 172.16.0.110 F. 172.16.0.111 G. 172.16.0.112
Answer F is correct. The seventh subnet ranges from the network ID of 172.16.0.96 to the broadcast ID of 172.16.0.111.
Which is a valid alternate expression of FE80:0000:0000:0000:0202:B3FF:0E1E:8329? Choose all that apply. A. FE80::0202:B3FF:0E1E:8329 B. FE80::0202:B3FF::E1E:8329 C. FE80::202:B3FF:E1E:8329 D. FE80::0202:B8FF:0E1E:8329
Answers A and C are correct. Answer B is incorrect because it uses the "::" twice. Answer D is incorrect because of the B8FF, which should be B3FF:.
Which of the following are valid types of IPv6 address? Choose all that apply. A. Global unicast B. Unique local C. Link local D. Multicast E. Anycast F. Broadcast G. Directed broadcast
Answers A, B, C, D, and E are correct. IPv6 does not broadcast
Which of the following are valid IPv6 address formats for the same address? Choose three. A. 2001:0000:0000:0200:0222:0000:0000:0001 B. 2001:0000:0000:02:0222:0000:0000:0001 C. 2001:0:0:200:222:0:0:1 D. 2001::200:222::1 E. 2001:0:0:200:222::1
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.
Which of the following are valid transition strategies when moving from IPv4 to IPv6? Choose all that apply. A. NAT-PT B. SNARD encapsulation C. 4-in-4-out tunneling D. Short stacking E. 6-to-4 tunneling F. Dual stacking
Answers A, E, and F are correct.
Which of the following are alternate representations of the decimal number 227? Choose two. A. 0x227 B. 11100011 C. 0x143 D. 0xE3 E. 11100110
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.
Which of the following are private IP addresses that can be assigned to a host? A. 12.17.1.45 B. 10.255.255.254 C. 172.15.255.248 D. 192.168.1.5 E. 239.0.0.1
Answers B and D are correct. The other addresses are not in the private unicast ranges.
Which of the following are true about the following address and mask pair: 10.8.8.0 /24? Assume that all subnets are available. Choose all that apply. A. This is a Class B address. B. This is a Class A address. C. This is a Class C address. D. 16 bits were stolen from the host field. E. 24 bits were stolen from the host field. F. The default mask for this address is 255.0.0.0. G. The mask can also be written as 255.255.255.0. H. The mask creates 65,536 subnets total from the default address space. I. Each subnet supports 256 valid host IPs. J. Each subnet supports 254 valid host IPs.
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.
Which of the following networks are included in the summary 172.16.0.0 /13? Choose all that apply. A. 172.0.0.0 /16 B. 172.16.0.0 /16 C. 172.24.0.0 /16 D. 172.21.0.0 /16 E. 172.18.0.0 /16
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.
What are the advantages of route summarization? Choose three. A. Ensures job security for network admins because of difficulty of configuration B. Reduces routing update traffic overhead C. Reduces the impact of discontiguous subnets D. Reduces CPU and memory load on routers E. Identifies flapping interfaces F. Hides network instability
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.
Your boss complains that manual route summarization is difficult and complex, and wonders if maybe you should not bother with it. What are the most compelling arguments in favor of route summarization? Choose all that apply. A. Utilizes the full RAM and CPU performance capacity of the routers B. Can suppress the effects of an unstable or "flapping" interface C. Advertises complete and detailed route tables D. Increases security by advertising "fake" networks E. Reduces the size of the route tables F. Reduces the load on RAM, CPU, and bandwidth of routers
Answers B, E, and F are correct. Answer A is incorrect because summarization actually reduces the load on routers, and maxing out your router is not a good idea to begin with. Answer C is incorrect; summarization sends out summary routes that represent the detailed routes. Answer D is incorrect, but tricky: Summarization does in fact send out fake routes, but this does nothing to increase security.
Which of the following are alternate representations of 0xB8? Choose two. A. 10110100 B. 10111111 C. 10111000 D. 184 E. 0x184
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.
You are given an old router to practice for your CCNA. Your boss, Dave, has spent a lot of time teaching you subnetting. Now he challenges you to apply your knowledge. He hands you a note that says the following: "Given the subnetted address space of 192.168.1.0 /29, give the E0 interface the first valid IP in the eighth subnet. Give the S0 interface the last valid IP in the 12th subnet. The zero subnets are available. You have 10 minutes. Go." Which two of the following are the correct IP and mask configurations? Choose two. A. E0: 192.168.1.1 255.255.255.0 B. E0: 192.168.1.56 255.255.255.248 C. E0: 192.168.1.57 255.255.255.248 D. S0: 192.168.1.254 255.255.255.0 E. S0: 192.168.1.95 255.255.255.248 F. S0: 192.168.1.94 255.255.255.248
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.
What is the decimal value of binary 00001110 ? A. 13 B. 14 C. 1,110 D. 16 E. 15
B. 14
What is the binary value of decimal 256? A. 11111111 B. 1111111111 C. 100000000 D. 10000000
C. 100000000
What are the eight binary values found in a single octet of an IP address? A. 256 128 64 32 16 8 4 2 B. 254 62 30 14 6 4 0 C. 128 64 32 16 8 4 2 1 D. 0 2 4 6 8 10 12 14
C. 128 64 32 16 8 4 2 1
Why is summarization so important to an efficient routed system? A. It adds detail to the route tables of routers. B. Summarization sends all subnets as classful networks, eliminating the overhead of transmitting the mask in routing updates. C. Summarization reduces the size of route tables, prevents route table instability due to flapping routes, and reduces the size of routing updates. D. Summarization enforces router authentication, preventing spurious updates from excessively loading the router.
C. Summarization reduces the size of route tables, prevents route table instability due to flapping routes, and reduces the size of routing updates.
What does Dual Stacking mean?
Dual stacking means that the host (router, PC, printer, and so on) runs both the IPv4 and IPv6 protocol stacks and can send and receive both types of packets, probably (but not necessarily) on the same interface.
Types of IPv6 Addresses, Unique local:
Equivalent to a private IPv4 address; not registered with an ISP and not Internet routable.
Types of IPv6 Addresses, Link local:
Every IPv6 interface gives itself a link-local address. The address range is FE80::/10, and usually combines this prefix with the last 64 bits in EUI-64 format.
True or false: It is impossible to subnet a subnet.
False
What is the format of an IPv6 broadcast address?
IPv6 cannot broadcast, ever!
What is a quick way to identify if a network has been subnetted?
If a mask is longer than it is supposed to be, that network has been subnetted
VLSM has two main advantages:
It makes network addressing more efficient & It provides the capability to perform route summarization
What route summarization (VLSM)?
It reduces the number of entries in the route table, which reduces load on the router and network overhead, and hides instability in the system behind the summary, which remains valid even if summarized networks are unavailable.
Types of IPv6 Addresses, Multicast:
Just like in IPv4, a single IPv6 multicast address is assigned to multiple hosts so that a packet sent to the address may be delivered to multiple hosts more or less at the same time. IPv6 multicast addresses always start with the prefix FF00::/8.
If the mask assigned to a private Class C address is 24 bits, is the address subnetted?
No. /24 is the default mask, so the address is not subnetted.
The following questions are part of a Subnetting SuperChallenge. This monster question will stretch your subnetting skills, especially if you give yourself a time limit. Start with 10 minutes and see if you can get down to 5. The Vancouver Sailing Company has four locations: a head office and three branch offices. Each of the branches is connected to the head office by a point-to-point T1 circuit. The branches have one or more LANs connected to their routers. The routers are called Main, Jib, Genoa, and Spinnaker. The company has been assigned the 172.16.0.0/20 address space to work within. Your task is to choose the correct IP address and mask for each interface, based on the information provided. Remember that no IP address may overlap with any address in another subnet, and that the required number of hosts for each subnet will affect your decision as to which address to use. Here are the known IP configurations for the routers: Main: S0/0:172.16.0.1 /30 S0/1:172.16.0.5 /30 S0/2:172.16.0.9 /30 Fa1/0:172.16.4.1 /23 Fa1/1:172.16.6.1 /23 Jib: S0/0: Connects to Main S0/0 Fa1/0: 172.16.8.33/27 Fa1/1: 30 hosts needed Genoa: S0/0: Connects to Main S0/1 Fa1/0: 172.16.8.129/26 Fa1/1: 100 hosts needed Fa2/0: 100 hosts needed Fa2/1: 172.16.13.0/24. Spinnaker: S0/0: Connects to Main S0/2 Fa1/0: 500 hosts needed Choose the correct IP and mask assignments for each router: A. Jib Fa1/1: 172.16.8.62/27 B. Jib Fa1/1: 172. 16.8.64/27 C. Jib Fa1/1: 172.16.8.65/28 D. Jib Fa1/0: 172.16.8.65/27 E. Jib Fa1/1: 172.16.8.65/27 F. Genoa S0/0: 172.16.0.2/30 G. Genoa S0/0:172.16.0.6/30 H. Genoa Fa1/1:172.16.12.1/26 I. Genoa Fa1/1:172.16.12.1/25 J. Genoa Fa2/0:172.16.12.129/24 K. Genoa Fa2/0:172.16.12.129/25 L. Genoa Fa2/1:172.16.12.193/25 M. Genoa Fa2/1:172.16.13.1/25 N. Genoa Fa0/2:172.16.13.1/24 O. Spinnaker S0/0: 172.16.0.10/30 P. Spinnaker S0/0: 172.16.0.12/30 Q. Spinnaker Fa1/0: 172.16.13.0/23 R. Spinnaker Fa1/0: 172.16.14.0/23 S. Spinnaker Fa1/0: 172.16.14.1/23
SuperChallenge answers: A. Incorrect (same subnet as Fa1/0) B. Incorrect (network ID) C. Incorrect (not enough hosts) D. Incorrect (Fa1/0 IP already assigned) E. Correct F. Incorrect (wrong subnet—not on the same network as the connected interface on Main) G. Correct H. Incorrect (not enough hosts) I. Correct J. Incorrect (overlaps with Fa1/1) K. Correct L. Incorrect (overlaps with Fa2/0) M. Incorrect (wrong mask) N. Incorrect (no Fa0/2 interface on Genoa) O. Correct P. Incorrect (network ID) Q. Incorrect (overlaps with Genoa) R. Incorrect (network ID) S. Correct
Which version of addressing has a fixed header size and what is it?
The IPv6 header is fixed at 40 bytes or 320 bits.
What is Static configuration using EUI-64:
The administrator manually configures the address with the local /64 network prefix followed by the host's MAC in EUI-64 format.
What is the class & default mask for 24.64.208.5?
The first octet is between 1 and 126: Class A Default mask for Class A: 255.0.0.0
What is the class & default mask for 188.21.21.3?
The first octet is between 128 and 191: Class B Default mask for Class B: 255.255.0.0
What is the class & default mask for 192.168.1.66?
The first octet is between 192 and 223: Class C Default mask for Class C: 255.255.255.0
IPv6, what is Stateless Autoconfiguration?
This feature allows a host to choose and configure an address for itself. The host that wants an address learns what the /64 network prefix is on the local link, then appends its MAC address (in a special 64-bit format called EUI-64), thus generating a 128-bit IPv6 address that is unique to that host because it incorporates the unique MAC of the host.
You are given the following ranges of subnets: 192.168.1.0/29 192.168.1.128/29 192.168.1.8/29 192.168.1.136/29 192.168.1.16/29 192.168.1.144/29 192.168.1.24/29 192.168.1.152/29 192.168.1.32/29 192.168.1.160/29 192.168.1.40/29 192.168.1.168/29 192.168.1.48/29 192.168.1.176/29 192.168.1.56/29 192.168.1.184/29 Your task is to summarize these two ranges of subnets. Do not include any subnets not named in the ranges in your summary. (Hint: You may use more than one summary address).
Two summary statements are required. Because these two network ranges are discontiguous, we cannot use a single statement without including the ranges between and after, which is both not allowed in the question and not generally a good idea in practice. The two summary statements are 192.168.1.0/26 and 192.168.1.128/26. The /26 in each statement can also be expressed as 255.255.255.192.
What is the EUI-64 format?
We simply take the 48-bit MAC address and put a special pattern, FFFE, after the first 24 bits (the six OUI characters), followed by the rest of the six hex characters in the host MAC. The only trick is that according to IPv6 rules, the seventh bit in an EUI-64 address must be 1, which identifies that the burned-in MAC address has been modified.
What class of IP address is 191.168.1.0?
What class of IP address is 191.168.1.0?
What is stateful addressing?
Where the DHCP server keeps track of what hosts have been assigned what IPv6/v4 address
Is the address 172.16.1.0/24 subnetted?
Yes. The default mask is /16; /24 is longer so the address is subnetted.
Is 172.16.255.0/18 a valid host IP?
Yes. The subnet's valid host IPs range from 172.16.192.1 to 172.16.255.254 .
What is 6-to-4 Tunnel?
You can picture taking an IPv6 packet from the head office, encapsulating it inside an IPv4 packet to transition across the provider network, then decapsulating it on the other side and forwarding the IPv6 packet into the remote branch office.