Q: Cisco IOS requirement: To allow the use of all 8 subnets with a Class C mask of 255.255.255.224 (/27), which configuration command must be enabled?

Introduction / Context: Older IOS releases required an explicit command to allow using the first and last subnets (so-called “subnet-zero”). Modern IOS enables it by default, but many exams still test the concept. Given Data / Assumptions: Mask: 255.255.255.224 (/27) on a Class C network. Goal: Use all 8 possible /27 subnets (including subnet-zero and the all-ones subnet). Concept / Approach: Total subnets with 3 borrowed bits is 2^3 = 8. Historically, some implementations avoided using the first and last subnets. The IOS command ip subnet-zero permits their use, enabling the full count. Step-by-Step Solution: Borrowed bits = 3 → possible subnets = 8.Enable first and last subnets by using the command to allow subnet-zero.Configuration line: Router(config)# ip subnet-zero. Verification / Alternative check: On modern IOS, verify with show running-config or confirm addressing acceptance when configuring interfaces. Why Other Options Are Wrong: ip classless: Controls classless routing behavior, not subnet-zero usage. no ip classful: Not a valid IOS command. ip unnumbered: Interface addressing feature, unrelated. ip route-cache: CEF/fast switching, unrelated. Common Pitfalls: Confusing classless routing with subnet-zero; assuming all IOS versions always use subnet-zero without checking. Final Answer: Router(config)# ip subnet-zero

Question 1

Subnet reasoning — interpret 10.16.3.65/23 and choose the correct statements Given host IP 10.16.3.65/23, evaluate the following statements about its subnet address, lowest/last host, and broadcast address, and select the two true statements.

Accepted Answer

Introduction / Context: Understanding CIDR masks lets you compute subnet ranges quickly. With /23, each subnet spans 512 addresses (510 usable). Determining network, broadcast, and host ranges is fundamental for addressing, ACLs, and summarization. Given Data / Assumptions: IP: 10.16.3.65/23. Mask: 255.255.254.0 (since /23 = 8+8+7 bits). We consider standard usable host rules (exclude network and broadcast). Concept / Approach: With /23, the third octet increments in steps of 2. Subnet ranges are 10.16.0.0–10.16.1.255, 10.16.2.0–10.16.3.255, 10.16.4.0–10.16.5.255, etc. Address 10.16.3.65 falls in the 10.16.2.0/23 subnet. Therefore, the subnet (network) address is 10.16.2.0, the broadcast is 10.16.3.255, the first usable host is 10.16.2.1, and the last usable host is 10.16.3.254. Step-by-Step Solution: Compute block size: 256 – 254 = 2 in the third octet.Locate range that contains third octet 3 → the 2–3 block.Network address: 10.16.2.0; Broadcast: 10.16.3.255.Usable hosts: 10.16.2.1 through 10.16.3.254.Validate statements: (1) claims subnet 10.16.3.0 → false; (2) lowest host 10.16.2.1 → true; (3) last host 10.16.2.254 → false; (4) broadcast 10.16.3.255 → true. Verification / Alternative check: Convert the /23 mask to binary and AND with the IP to confirm network bits. A quick sanity check: broadcast equals next subnet start minus one (10.16.4.0 − 1 = 10.16.3.255), which matches. Why Other Options Are Wrong: A/C/D/E: Each includes at least one false statement (wrong network or wrong last host). Common Pitfalls: Mixing /23 with /24 rules and assuming third-octet boundaries always align on 1-octet blocks. Always compute the block size for the variable octet. Final Answer: 2 and 4

Question 2

VLSM practice — /28 subnets in 192.168.10.0/28 with zero subnet disallowed If you require the eighth usable /28 subnet (zero subnet not considered valid) in 192.168.10.0/28 and must assign the last usable IP to interface E0, which exact hos…

Accepted Answer

Introduction / Context: Calculating host ranges inside small subnets is a staple skill when allocating addresses on point-to-point and access networks. Here, we must skip the zero subnet and then choose the eighth /28 and its last usable IP for a router interface. Given Data / Assumptions: Base network: 192.168.10.0. Mask: /28 (255.255.255.240), block size 16 addresses per subnet. Zero subnet not valid for this question. We want the last usable host of the eighth /28. Concept / Approach: With /28, each subnet spans 16 addresses: .0–.15, .16–.31, .32–.47, ... The zero subnet (.0/28) is excluded, so counting starts at .16/28 as the first usable. The last usable in any /28 is subnet_end − 1 (because the broadcast is subnet_end). Step-by-Step Solution: Enumerate /28 subnets (excluding .0): .16 (1), .32 (2), .48 (3), .64 (4), .80 (5), .96 (6), .112 (7), .128 (8).Identify the eighth subnet: 192.168.10.128/28 covering .128–.143.Usable hosts are .129–.142; broadcast is .143.Select the last usable: 192.168.10.142. Verification / Alternative check: Compute broadcast as next subnet start minus one: next subnet after .128 is .144, so broadcast is .143; hence last usable is .142, confirming our choice. Why Other Options Are Wrong: B/E: Belong to different /28 ranges (.64–.79 and .112–.127 respectively), not the eighth subnet’s last host. C: Different third octet entirely (192.168.100.x), invalid network. D: .143 is the broadcast of the eighth /28, not a usable host. Common Pitfalls: Forgetting to exclude the zero subnet when instructed or mistakenly using the broadcast address as the last host. Final Answer: 192.168.10.142

Question 3

CIDR capacity — how many subnets and hosts does 172.16.0.0/19 provide? For the private Class B network 172.16.0.0 with a /19 mask, determine the number of subnets created and the number of usable hosts per subnet.

Accepted Answer

Introduction / Context: Mask length determines both the number of subnets and host capacity. Quickly computing these values is essential for scalable design, especially when planning allocations inside private ranges like 172.16.0.0/12 and its sub-blocks. Given Data / Assumptions: Base classful reference: Class B default /16. Given mask: /19 → 3 more bits beyond /16. Usable hosts per subnet exclude network and broadcast addresses. Concept / Approach: Borrowed subnet bits = 19 − 16 = 3 → number of subnets = 2^3 = 8. Host bits per subnet = 32 − 19 = 13 → usable hosts = 2^13 − 2 = 8192 − 2 = 8190. These computations do not depend on classful “zero subnet” restrictions in modern CIDR practice. Step-by-Step Solution: Compute subnet count: 2^(borrowed) = 2^3 = 8.Compute host bits: 32 − 19 = 13.Compute usable hosts: 2^13 − 2 = 8190.Report as “8 subnets, 8,190 hosts each.” Verification / Alternative check: Block size per subnet in the third octet is 256 − 224 = 32; subnets progress 0, 32, 64, … which corroborates 8 distinct subnets across the /16 space (256/32 = 8). Why Other Options Are Wrong: A/D: 7 subnets is invalid; with 3 borrowed bits you get 8. C/E: Host counts correspond to /21 and /22-style calculations, not /19. Common Pitfalls: Mixing classful defaults with CIDR arithmetic or forgetting to subtract 2 for network and broadcast when host addressing is required. Final Answer: 8 subnets, 8,190 hosts each

Question 4

Apply /23 logic — valid hosts on 172.16.2.1/23 LAN A router interface E0 is 172.16.2.1/23 (mask 255.255.254.0). Which of the following are valid host addresses on that same LAN segment? Consider: 1) 172.16.1.100 2) 172.16.1.198 3) 172.16.2.255 4) 17…

Accepted Answer

Introduction / Context: Hosts must share the same subnet to communicate without routing. With a /23 mask, two contiguous /24 blocks form one larger subnet. Determining whether candidate addresses fall inside that range is a common exam and field task. Given Data / Assumptions: Interface E0: 172.16.2.1/23. Mask: 255.255.254.0 (block size 2 in the third octet). We must select all addresses within 172.16.2.0/23 usable host range. Concept / Approach: A /23 covering 172.16.2.0 spans from 172.16.2.0 to 172.16.3.255, with usable hosts 172.16.2.1 through 172.16.3.254. Therefore, any address in the 172.16.2.x or 172.16.3.x ranges—except the network (.2.0) and broadcast (.3.255)—is valid. Step-by-Step Solution: List usable range: 172.16.2.1–172.16.3.254.Evaluate #1 (172.16.1.100): outside the /23, invalid.Evaluate #2 (172.16.1.198): also outside, invalid.Evaluate #3 (172.16.2.255): within usable range (not the broadcast), valid.Evaluate #4 (172.16.3.0): within usable range (not the network of the /23), valid. Verification / Alternative check: Compute network by ANDing 172.16.2.1 with mask 255.255.254.0 → 172.16.2.0; broadcast = next /23 start (172.16.4.0) − 1 = 172.16.3.255. #3 and #4 sit between these values. Why Other Options Are Wrong: A/B/E: Include addresses from the 172.16.0.0/23 below, not this LAN. D: Incorrect because two candidates are valid. Common Pitfalls: Assuming any .255 is always a broadcast; in /23, 172.16.2.255 is a host. Also mistaking 172.16.3.0 for a network boundary; the network is 172.16.2.0. Final Answer: 3 and 4 only

Question 5

Host capacity on a /29 LAN — include the router interface in the count A router interface is configured 192.168.192.10/29. Counting all usable IPs on that LAN, including the router interface itself, how many hosts can be addressed on this /29 segmen…

Accepted Answer

Introduction / Context: Small subnets such as /30 and /29 are common on WAN links and access segments. You must know exactly how many usable host addresses each mask provides to avoid over-subscribing a subnet or wasting address space. Given Data / Assumptions: Prefix length /29 means mask 255.255.255.248. Usable host count follows 2^h − 2, where h is the number of host bits. The question explicitly includes the router interface in the total. Concept / Approach: For /29, host bits h = 32 − 29 = 3. Therefore, usable hosts = 2^3 − 2 = 8 − 2 = 6. This total includes any router interface addresses and all attached host devices. The network and broadcast consume the two excluded addresses in each /29 block of 8 addresses. Step-by-Step Solution: Compute host bits: h = 3.Compute usable hosts: 2^3 − 2 = 6.Confirm that 192.168.192.8/29 spans .8–.15 with usable .9–.14, which contains .10.Therefore, total usable addresses on that LAN (including the router) = 6. Verification / Alternative check: Block size for /29 is 256 − 248 = 8 addresses per subnet. Usable = 8 − 2 = 6 corroborates the formula above. Why Other Options Are Wrong: B/D: 8 or 32 count total addresses in unrelated masks or fail to exclude network/broadcast. C: 30 is the /27 usable count. E: 4 corresponds to /30 (2 usable hosts). Common Pitfalls: Confusing “hosts total” with “usable hosts,” and forgetting that the question includes the router interface in the same count. Final Answer: 6

Question 6

Local IP stack test — choose the correct loopback address to ping To verify the TCP/IP stack on your own machine (without hitting the network), which IP address should you ping as the standard IPv4 loopback target?

Accepted Answer

Introduction / Context: Pinging the loopback address tests the local host's TCP/IP stack without requiring any network interface or external connectivity. This is a first diagnostic step if applications cannot reach the network or if you need to confirm that IP is bound and operational on the machine. Given Data / Assumptions: IPv4 loopback range is 127.0.0.0/8. Most tools and documentation use 127.0.0.1 as the canonical host address. Ping reaches the local protocol stack, not the NIC or wire. Concept / Approach: The entire 127/8 range is reserved for loopback, but by convention and in practice the single address 127.0.0.1 is the standard target for diagnostics. Using this address validates local protocol handling, ICMP processing, and the OS networking subsystem. Failure indicates a local stack issue, not a cabling or upstream problem. Step-by-Step Solution: Recognize 127.0.0.0/8 as loopback reserved range.Select the well-known host address 127.0.0.1 as the test endpoint.Execute ping 127.0.0.1; successful replies confirm local stack operation. Verification / Alternative check: Other addresses in 127/8 may also loop back, but most operating systems document and prefer 127.0.0.1. Traceroute to 127.0.0.1 never leaves the host, further confirming the behavior. Why Other Options Are Wrong: A: 127.0.0.0 is the network address of the loopback block, not the standard host target. B: 1.0.0.127 is a public Internet address (Cloudflare uses 1.1.1.1/1.0.0.1 for DNS), not loopback. D: 127.0.0.255 is within 127/8 but not the conventional host address used for tests. E: 0.0.0.0 means “this host” or default route placeholder; it is not pingable. Common Pitfalls: Misinterpreting a successful 127.0.0.1 ping as proof that the NIC or cable works; it validates only the local stack, not external connectivity. Final Answer: 127.0.0.1

Question 7

IPv4 subnetting practice: What is the subnetwork (network) address for a host with IP address 200.10.5.68/28?

Accepted Answer

Introduction / Context: Subnetting with Classless Inter-Domain Routing (CIDR) requires identifying the block size and then finding the exact network address that contains a given host IP. Here we are working with a /28 prefix, a common exam scenario. Given Data / Assumptions: Host IP: 200.10.5.68/28 /28 corresponds to mask 255.255.255.240 We seek the network (subnetwork) address for that host Concept / Approach: The block size is determined by 256 minus the last-octet mask value. For /28, the last octet mask is 240, so the block size is 256 - 240 = 16. Valid network boundaries occur at increments of 16 in the fourth octet: 0, 16, 32, 48, 64, 80, etc. Step-by-Step Solution: Mask /28 => 255.255.255.240 (block size 16).List boundaries in the 4th octet: 0, 16, 32, 48, 64, 80, 96...Locate 68 between 64 and 80.Therefore, the network address is 200.10.5.64 and the broadcast is 200.10.5.79. Verification / Alternative check: Compute: 68 - (68 mod 16) = 68 - 4 = 64, confirming the network base. Why Other Options Are Wrong: 200.10.5.56: 56 is not a /28 boundary (16-multiple) for this range. 200.10.5.32 and 200.10.5.0: Different /28 networks; do not contain 68. 200.10.5.80: Next network boundary above 68, not the one containing it. Common Pitfalls: Mixing up /28 with /29 block sizes; forgetting boundaries must be exact multiples of the block size. Final Answer: 200.10.5.64

Question 8

Subnet mask to host capacity: With subnet mask 255.255.255.224, what is the maximum number of IP host addresses available on a single subnet?

Accepted Answer

Introduction / Context: Calculating usable hosts from a subnet mask is foundational for IP addressing design. The /27 mask (255.255.255.224) is a frequent example in certification exams. Given Data / Assumptions: Subnet mask: 255.255.255.224 (/27) We want usable host addresses per subnet (excludes network and broadcast) Concept / Approach: Usable hosts per subnet = 2^(host_bits) - 2. For /27, there are 32 - 27 = 5 host bits. Step-by-Step Solution: host_bits = 32 - 27 = 5total addresses = 2^5 = 32usable hosts = 32 - 2 = 30 Verification / Alternative check: Block size of /27 is 32 addresses. Removing the network and broadcast leaves 30 hosts. Why Other Options Are Wrong: 14/15/16: Correspond to smaller subnets (/28 or miscounting usable hosts). 62: That is typical of a /26 (64 addresses total minus 2). Common Pitfalls: Forgetting to subtract 2 for network and broadcast; mixing up /27 with /28 calculations. Final Answer: 30

Question 9

Design goal: 5 subnets, each needing at least 16 hosts. Which classful subnet mask should be used to satisfy these requirements?

Accepted Answer

Introduction / Context: When choosing a subnet mask, you must meet both the required number of subnets and the minimum hosts per subnet. This is a classic capacity-planning exercise. Given Data / Assumptions: Need at least 5 subnets. Each subnet needs at least 16 usable hosts. Looking at common Class C–style masks. Concept / Approach: Check host capacity first. For a mask to support at least 16 usable hosts, usable hosts = 2^(host_bits) - 2 must be >= 16. /27 (255.255.255.224) yields 30 hosts, /28 yields 14 hosts (insufficient). Then ensure number of subnets is enough. Step-by-Step Solution: /28 (255.255.255.240): hosts = 14 < 16 → not acceptable./27 (255.255.255.224): hosts = 30 ≥ 16 → acceptable./27 subnets in a Class C: 2^(borrowed_bits) = 2^3 = 8 (with subnet-zero), enough for 5.Therefore choose 255.255.255.224. Verification / Alternative check: Even without subnet-zero (legacy), /27 still provides 6 usable subnets—still ≥5. With subnet-zero (modern default), you get 8. Why Other Options Are Wrong: 255.255.255.240 (/28): Only 14 hosts. 255.255.255.248 (/29): Only 6 hosts. 255.255.255.192 (/26): Valid but overprovisions hosts and reduces subnet count flexibility unnecessarily for the stated minimums. 255.255.255.128 (/25): Similar overprovision; not needed. Common Pitfalls: Confusing “at least 16 hosts” with “16 addresses total.” Remember to subtract network and broadcast. Final Answer: 255.255.255.224

Question 10

Cisco IOS requirement: To allow the use of all 8 subnets with a Class C mask of 255.255.255.224 (/27), which configuration command must be enabled?

Accepted Answer

Introduction / Context: Older IOS releases required an explicit command to allow using the first and last subnets (so-called “subnet-zero”). Modern IOS enables it by default, but many exams still test the concept. Given Data / Assumptions: Mask: 255.255.255.224 (/27) on a Class C network. Goal: Use all 8 possible /27 subnets (including subnet-zero and the all-ones subnet). Concept / Approach: Total subnets with 3 borrowed bits is 2^3 = 8. Historically, some implementations avoided using the first and last subnets. The IOS command ip subnet-zero permits their use, enabling the full count. Step-by-Step Solution: Borrowed bits = 3 → possible subnets = 8.Enable first and last subnets by using the command to allow subnet-zero.Configuration line: Router(config)# ip subnet-zero. Verification / Alternative check: On modern IOS, verify with show running-config or confirm addressing acceptance when configuring interfaces. Why Other Options Are Wrong: ip classless: Controls classless routing behavior, not subnet-zero usage. no ip classful: Not a valid IOS command. ip unnumbered: Interface addressing feature, unrelated. ip route-cache: CEF/fast switching, unrelated. Common Pitfalls: Confusing classless routing with subnet-zero; assuming all IOS versions always use subnet-zero without checking. Final Answer: Router(config)# ip subnet-zero

Question 11

Find the network: What is the subnetwork number for host 172.16.66.0/21?

Accepted Answer

Introduction / Context: Summarization-style masks like /21 require spotting the third-octet block size and then choosing the correct boundary that includes the given host. Given Data / Assumptions: IP: 172.16.66.0/21 /21 mask = 255.255.248.0 We need the corresponding network address. Concept / Approach: For /21, the block size in the third octet is 256 - 248 = 8. Valid third-octet boundaries are 0, 8, 16, ..., 64, 72, etc. We select the boundary less than or equal to the host’s third octet (66). Step-by-Step Solution: Third-octet block size = 8.Find boundary ≤ 66: ... 56, 64, 72 ... → 64 is the correct lower boundary.Thus, the network is 172.16.64.0; the broadcast is 172.16.71.255. Verification / Alternative check: Range check: 64 ≤ 66 ≤ 71 confirms 66.0 lies within the 64.0/21 block. Why Other Options Are Wrong: 36.0 and 48.0: Different /21 blocks; do not contain 66.0. 0.0: Too broad; not the /21 block for this host. 72.0: Next block above; does not include 66.0. Common Pitfalls: Forgetting that /21 spans multiple Class B /24s; miscounting the 8-value steps in the third octet. Final Answer: 172.16.64.0

Question 12

Valid host check for /22: Given subnet 172.16.17.0/22, which of the following is a valid host address (with mask shown)?

Accepted Answer

Introduction / Context: Verifying a valid host requires matching both the IP's range within the subnet and the correct subnet mask. /22 networks often confuse because the broadcast address ends with .255, yet some middle .255 values can still be hosts. Given Data / Assumptions: Subnet: 172.16.17.0/22 /22 mask = 255.255.252.0 We must pick an IP/mask pair that is valid within this /22. Concept / Approach: A /22 spans 4 contiguous /24s. Determine the network range, then confirm the candidate address falls within it and isn’t the network or broadcast address. Step-by-Step Solution: /22 block size in third octet = 4. The 17.x subnet belongs to the 16–19 range.Network: 172.16.16.0/22; Broadcast: 172.16.19.255.Candidate D: 172.16.18.255 with 255.255.252.0 is within 16.0–19.255 and is not the broadcast (which is 19.255). Verification / Alternative check: Check each option's mask and range: only option D pairs the correct /22 mask and an IP inside the range without being network/broadcast. Why Other Options Are Wrong: A: Mask is /30 and not the specified /22. B: IP and mask point to a different network entirely. C: Mask /23 and IP 20.x fall outside 16–19 range for the /22 given. E: Mask /24; also 19.255 would be the /24 broadcast if /24 were used, but the mask is wrong. Common Pitfalls: Assuming any .255 is always a broadcast; it only is when it matches the subnet’s actual broadcast boundary. Final Answer: 172.16.18.255 255.255.252.0

Question 13

Broadcast address determination: An interface has IP 192.168.192.10/29. What broadcast address will hosts on this LAN use?

Accepted Answer

Introduction / Context: For any subnet, the broadcast address is the highest address in that block. With /29, determining the 8-address block is straightforward and frequently tested. Given Data / Assumptions: Host: 192.168.192.10/29 /29 mask = 255.255.255.248 (block size 8) Concept / Approach: Identify the /29 block containing .10 by finding the nearest lower multiple of 8 and then adding 7 for the broadcast address. Step-by-Step Solution: Block boundaries (4th octet) every 8: 0, 8, 16, 24, ....10 lies in the 8–15 block.Network = .8, Broadcast = .15, Hosts = .9–.14. Verification / Alternative check: Compute: 10 - (10 mod 8) = 8; broadcast = 8 + 7 = 15. Why Other Options Are Wrong: .31/.63/.127: Broadcasts for larger subnets (/27, /26, /25), not /29. .11: A valid host, not the broadcast. Common Pitfalls: Forgetting that /29 uses 8-address blocks; misidentifying the correct block around the host. Final Answer: 192.168.192.15

Question 14

Choose a valid server IP: Subnet is 192.168.19.24/29. The router uses the first available host address. Which IP/mask should you assign to the server on this subnet?

Accepted Answer

Introduction / Context: Assigning host addresses requires understanding the subnet range and avoiding the network, broadcast, and already-allocated addresses. /29 subnets are small and easy to enumerate. Given Data / Assumptions: Subnet: 192.168.19.24/29 (mask 255.255.255.248) Router has the first usable host in the subnet. Concept / Approach: A /29 gives 8 addresses: network, 6 hosts, broadcast. For the 24/29 block, hosts run from .25 to .30 if .24 is the network and .31 is broadcast. The router, using the first host, takes .25, so pick another host such as .26. Step-by-Step Solution: Network = 192.168.19.24; Broadcast = 192.168.19.31.Usable hosts: .25–.30; router uses .25.Select a remaining host: 192.168.19.26 with mask 255.255.255.248. Verification / Alternative check: Count addresses (8 total), confirm .26 falls between network and broadcast and is not already assigned to the router. Why Other Options Are Wrong: A: Different network and mask. B: .33 is not in the 24/29 block; mask /28 also wrong. D: .31 is the broadcast for this /29. E: .25 is already taken by the router per the scenario. Common Pitfalls: Misreading the block size; forgetting that the first usable address is already allocated. Final Answer: 192.168.19.26 255.255.255.248

Question 15

Identify the subnet: An Ethernet port has IP 172.16.112.1/25. What is the valid subnet (network) address for this host?

Accepted Answer

Introduction / Context: /25 subnets split a /24 into two halves. To find the network address, determine which half the host belongs to by looking at the most significant bit of the last octet under the /25 mask. Given Data / Assumptions: IP: 172.16.112.1/25 Mask: 255.255.255.128 (/25) Concept / Approach: /25 gives two blocks in the fourth octet: 0–127 and 128–255. Since the host's fourth octet is 1, it must be in the 0–127 block, making the network address .0. Step-by-Step Solution: Block size = 128 addresses.Host 172.16.112.1 lies in 172.16.112.0–172.16.112.127.Network address = 172.16.112.0; Broadcast = 172.16.112.127. Verification / Alternative check: Compute: 1 - (1 mod 128) = 0; confirms network base is .0. Why Other Options Are Wrong: 172.16.0.0 / 172.16.255.0: Not the correct /25 network for this host. 172.16.96.0: Different /24 network. 172.16.112.128: That is the next /25 network, not the one containing .1. Common Pitfalls: Confusing /24 with /25; failing to realize /25 splits the 4th octet in half. Final Answer: 172.16.112.0

Question 16

Planning for 29 subnets: To create at least 29 subnets while maximizing host addresses per subnet, how many bits must be borrowed from the host field?

Accepted Answer

Introduction / Context: Subnet-count planning uses powers of two. The number of subnet bits you borrow from the host field must be large enough so that 2^(borrowed_bits) meets or exceeds the number of required subnets. Given Data / Assumptions: Required subnets: at least 29. Goal: maximize hosts per subnet (so borrow the minimum bits that satisfy subnet count). Concept / Approach: Find the smallest n such that 2^n ≥ 29. The fewer bits we borrow, the more host bits remain, maximizing host capacity. Step-by-Step Solution: 2^4 = 16 → not enough.2^5 = 32 → sufficient for 29.Therefore, borrow 5 bits to create at least 29 subnets and preserve as many host bits as possible. Verification / Alternative check: Check next lower value (4) fails; next higher (6) would reduce host space unnecessarily. Why Other Options Are Wrong: 2/3/4: Provide 4, 8, or 16 subnets—insufficient. 6: Provides 64 subnets but reduces host counts more than needed. Common Pitfalls: Forgetting that modern designs allow using all subnets (subnet-zero included); borrowing more bits than needed and needlessly shrinking host capacity. Final Answer: 5

Question 17

VLSM design for point-to-point WAN links On a Variable Length Subnet Mask (VLSM) IPv4 network, which subnet mask should be assigned to point-to-point serial WAN links to minimize wasted host addresses while remaining standards-compliant?

Accepted Answer

Introduction / Context: When you design an IPv4 network using Variable Length Subnet Masking (VLSM), you want to allocate just enough addresses for each segment. Point-to-point WAN links connect exactly two endpoints, so they need the smallest subnet that still supports two usable host addresses and standard routing behavior. Choosing the right mask reduces address waste and simplifies documentation. Given Data / Assumptions: IPv4 addressing using VLSM is permitted. The link is a classic point-to-point WAN segment that must support exactly two IP nodes. Conventional routing and legacy devices are in scope (that is, we assume two usable host addresses plus network and broadcast). Concept / Approach: Each IPv4 subnet historically reserves two addresses: the network ID and the directed broadcast. A /30 network (mask 255.255.255.252) has 4 total addresses: 1 network, 2 hosts, 1 broadcast. This perfectly fits a two-endpoint link. While /31 (255.255.255.254) can be used on many modern routers for point-to-point links, traditional interoperability and some exam objectives assume /30 for two usable hosts plus broadcast semantics. Step-by-Step Solution: Determine minimum usable host count for a P2P link: 2 hosts.Find the smallest prefix with at least 2 usable hosts: /30 (4 total addresses → 2 usable).Validate routing compatibility across devices and protocols: /30 is universally supported.Select /30 for the WAN segment to minimize waste. Verification / Alternative check: Calculate usable hosts by formula: usable_hosts = 2^(32 - prefix) - 2. For /30, usable_hosts = 2^(2) - 2 = 2. This matches the requirement for a two-node link. Field deployments confirm /30s are standard on serial links across vendors. Why Other Options Are Wrong: /27, /28, /29: These provide 30, 14, and 6 usable hosts respectively, which wastes addresses on a two-host link. /31: Although valid on many modern platforms for point-to-point, many curricula and mixed-vendor networks still prefer /30. The question emphasizes minimizing waste while staying broadly compatible. Common Pitfalls: Forgetting that legacy gear might not support /31 semantics. Accidentally assigning a /29 and later discovering unused hosts across many links. Final Answer: /30

Question 18

Choosing the last usable IP in the first valid /28 subnet Given network 192.168.10.0/28, the zero subnet is not considered valid. Using the first valid /28 subnet, what should be the IP address of interface S0 if you must use the last available host…

Accepted Answer

Introduction / Context: Subnetting questions often test your ability to enumerate ranges and pick boundary addresses accurately. Here you are given a Class C–like network with a /28 mask and a constraint that the zero subnet (the first numerically) is not valid. You must identify the first valid subnet's range and then choose the last usable host address within that range for the S0 interface. Given Data / Assumptions: Base network: 192.168.10.0/28. Zero subnet is not considered valid for this question's context. We need the last usable host address in the first valid /28 subnet. Concept / Approach: /28 corresponds to a mask of 255.255.255.240. The block size is 16 in the last octet (since 256 - 240 = 16). Subnets increment by 16: .0, .16, .32, .48, … Each /28 provides 16 addresses: 1 network, 14 usable hosts, 1 broadcast. If the zero subnet .0/28 is disallowed, the first valid subnet is .16/28 whose usable range is .17 through .30, with .31 as broadcast. Step-by-Step Solution: Calculate mask: /28 → 255.255.255.240.Compute increment: 256 - 240 = 16.List ranges: .0–.15, .16–.31, .32–.47, …Skip .0/28 (zero subnet invalid by rule).First valid subnet: 192.168.10.16/28 with usable .17–.30.Pick last usable: 192.168.10.30. Verification / Alternative check: Confirm broadcast of the .16/28 subnet is .31 by formula: last address in the block (16 + 15 = 31). Therefore the last usable host is one less than broadcast, .30. A quick ping sweep or lab test will validate reachability assignments within the .17–.30 span. Why Other Options Are Wrong: .24 sits inside the first valid /28, but it is not the last usable address. .62 and .127 are in different, larger subnets (/26 and /25 boundaries respectively), not in the first /28 block after .0. .14 is in the zero subnet, which the problem statement excludes. Common Pitfalls: Mistaking the block size or mixing broadcast and last-usable addresses. Forgetting the constraint that the zero subnet is considered invalid here. Final Answer: 192.168.10.30

Question 19

Identify the /30 subnetwork A host has IPv4 address 172.16.45.14/30. Which subnetwork does this host belong to when using a /30 mask on that class B private range?

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Introduction / Context: A /30 subnet provides exactly two usable host addresses, making it ideal for point-to-point links. To determine the correct network address for a given host, you must compute the block size and align the host's last octet to the network boundary for that block. This exercise reinforces fast mental math for subnetting in the field. Given Data / Assumptions: IP address: 172.16.45.14. Prefix length: /30 (mask 255.255.255.252). We are interested in the precise subnetwork (network ID). Concept / Approach: The /30 mask means the block size in the last octet is 4 (256 - 252 = 4). Valid network boundaries occur at multiples of 4: .0, .4, .8, .12, .16, etc. Each block of 4 addresses consists of network, two usable hosts, and broadcast. We must find the multiple of 4 that is less than or equal to 14 and whose next boundary is 16; that network is .12 and its range is .12–.15. Step-by-Step Solution: Compute block size: 256 - 252 = 4.List subnets: x.x.x.0, .4, .8, .12, .16, …Locate .14: it lies between .12 and .16.Therefore network ID is 172.16.45.12, usable hosts are .13 and .14, broadcast is .15. Verification / Alternative check: Binary check: AND 14 with mask 252 (11111100) yields 12 (00001100), confirming the network is .12. A quick lab ping between .13 and .14 would succeed with mask /30 and the broadcast observed at .15. Why Other Options Are Wrong: .0, .4, .8, .16 are valid /30 boundaries, but .14 does not fall inside those networks except .12–.15. Common Pitfalls: Using /29 math (block size 8) by mistake. Confusing broadcast with network address when the host is near the top of the block. Final Answer: 172.16.45.12

Question 20

Directly connecting two Ethernet hosts Two PCs are cabled back-to-back via their Ethernet NICs using a straight-through patch cord. Pings fail. Which actions will restore connectivity in this simple host-to-host scenario?

Accepted Answer

Introduction / Context: When two end hosts are connected directly without a hub or switch, the physical and Layer 3 parameters must be appropriate for link negotiation and IP communication. Many classic NICs require a crossover cable for transmit/receive pair alignment. Additionally, both hosts must share a common IP subnet mask so that ARP resolution and local delivery occur without a gateway. Given Data / Assumptions: Items listed: (1) Use a crossover cable, (2) Use a rollover cable, (3) Set mask 255.255.255.192, (4) Configure a default gateway, (5) Set mask 255.255.255.0. Auto-MDI/MDIX is not assumed (older hardware). The goal is direct L2/L3 communication between two hosts only. Concept / Approach: Back-to-back host links traditionally need a crossover cable so that one host’s transmit pair hits the other’s receive pair. A rollover cable is for console connections, not Ethernet data. At Layer 3, hosts must share an IP network; a common /24 like 255.255.255.0 is typical for lab setups. A default gateway is unnecessary for local traffic; ARP resolves peer MACs directly on the same subnet. Step-by-Step Solution: Replace straight-through with a crossover cable (item 1) if auto-MDI/MDIX is absent.Configure both hosts with IPs in the same /24 and mask 255.255.255.0 (item 5).Verify link LEDs and test ping again.Confirm ARP tables populate with the peer’s MAC address. Verification / Alternative check: Run ipconfig /all (Windows) or ip a (Linux) to verify mask and ARP table entries. Modern NICs with auto-MDI/MDIX may work with straight-through cabling; the core requirement remains that both hosts be in the same IP subnet. Why Other Options Are Wrong: (2) Rollover is console wiring; it will not pass Ethernet frames. (3) 255.255.255.192 (/26) could work only if both hosts share matching addresses within the same /26; it is not intrinsically required by the scenario. The question pairs it with default gateway (4) which is unnecessary for local pings. (4) Default gateway is only for off-subnet traffic. Common Pitfalls: Assuming a default gateway is needed for every ping; if peers are local, it is not. Overlooking NIC auto-crossover capabilities on newer equipment. Final Answer: 1 and 5 only

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