Most Efficient Subnet Summary for Four Office Floors
Answer
Correct answer: C — use 192.168.0.0/25 as the summary and 192.168.0.0/27 for each floor, giving 30 usable hosts per floor and 126 usable in the summary.
A corporate office uses four floors in a building. ✑ Floor 1 has 24 users. ✑ Floor 2 has 29 users. Floor 3 has 28 users. [sampleXML/xmlfile-23_3.png] ✑ Floor 4 has 22 users. Which subnet summarizes and gives the most efficient distribution of IP addresses for the router configuration?
Community Insight
The test is efficient IPv4 subnet sizing: you must pick the smallest block that fits the largest floor and a summary that fits the grand total, and the trap is forgetting that usable hosts = 2^(32-prefix) − 2, so /28 (16) and /29 (8) cannot hold 29 users.
This subnetting scenario asks for a summary address plus per-floor subnets that cover 24, 29, 28, and 22 users with minimal address waste. The answer is 192.168.0.0/25 as the summary and 192.168.0.0/27 for each floor, because 103 total users need a /25 and the largest floor (29 users) needs 30 usable addresses, which only a /27 provides.
Learners often choose B or D, assuming /25 or /26 for each floor or /29 blocks, but /29 gives only 8 usable hosts and /28 only 16, neither of which covers a 29-user floor; option A's /28 per floor wastes room in the sense that it still fails, while a /24 summary wastes half the block.
Community Discussion (34 comments)
RudeGrimmy
👍 1
Selected: B
This question is difficult to answer because there is not enough detail in it. Is there a router per floor or are all floors running L2 access switches cable to a single router. The question also does not state if each floor is using its own subnet or if all floors are using the same subnet. Based on the little info in the question my answer would gravitate to B not C since the question states "the router" and the floor descriptions just state how many users are on each floor with no subnet info given. That makes me think there is a single router with one subnet. A /25 would cover all of the users listed for all floors connected to "the router"
croftg1
👍 1
Selected: C
Max users for each subnet is the 32 bit delta. 2 power of 5 = 32 max ips. The 32 bit delta = /27. (32 - 5, or 24 + 3). 4 subnets to be summarized, 2 to the power of 2 = 4. 2 bits needed to summarize. 2 - 27 = 25
kminh9999
👍 1
Selected: C
calculate the required bits each floor 2^n - 2 > users -> n = 5 -> subnet each floor /27
CollabGuru
👍 2
Selected: B
According to ChatGPT the answer is B. It explains why like this: B. 192.168.0.0/23 as summary and 192.168.0.0/25 for each floor: A /23 subnet provides 512 IP addresses (510 usable IPs), which can cover multiple /25 subnets. A /25 subnet provides 128 IP addresses (126 usable IPs), which is more than enough for any of the floors. This option allows for efficient use of IP addresses, providing enough space for each floor and a summary network that covers them all.
Orson_TheOne
👍 1
C is the one believe me :)
Andu93
👍 1
Selected: C
2 to the power of 5 is 32 address . 32 -2 ( Network add and broadcast add ) = 30 on subnet -- that`s hosts on subnet. We need 3 bits for netw and 5bits for host - thats /27
ale72949
👍 1
Can we keep tables during the exam?
oranjo
👍 1
before the exam this should be in your head group size 128, 64, 32, 16, 8, 4, 2, 1 subnet 128, 192, 224, 240, 248, 252, 254, 255 CIDR /25 /26 /27 /28 /29 /30 /31 /32 e.g 10.1.1.55/28 from the table above the subnet is 255.255.255.240 the ip add has a group size of 16 usable ip add is the group size you subtract 2 {16-2} correct answer is C
duongccna
👍 1
max = 29 => 2^5-2=30 => borrow 5 bit => 32-5 = 27 therefore /27 is corect
habbey2080
👍 1
The answer is c
hayo
👍 6
Number of addresses per subnet
MSTAHIR
👍 3
total user 103, must be /25 Mask as for all floors, /27 Mask for each floor, refer to 2^5 = 32 - NW ID and Broad cast address, total usable IP 30.
keokkeo_123
👍 2
Selected: C
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Customexit
👍 76
write this down first thing in the exam: /32 1 /31 2 /30 4 /29 8 /28 16 /27 32 /26 64 /25 128 /24 256 /23 512 /22 1024 /21 2048
NICE_ANSWERS
👍 3
please, what's it's significance?
flash93933
👍 7
love you
lock12333
👍 1
Selected: C
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kentsing
👍 6
16 addresses per floor is not enough so 32 per floor is needed simply count from /32=1 /31=2 /30=4....../27=32 /27 per floor is the answer
DaveDaSpade
👍 2
That's how I got the answer quickly :)
Hanagaki_Shinjiro
👍 1
8-7=1 BRO
ScorpionNet
👍 3
N is the Network, U is the usable host, H is the host
Shamwedge
👍 4
Subnet Mask: 128 192 224 240 Hosts: 128 64 32 16 /Cider 25 26 27 28 /27 is the smallest number that will meet the number of hosts required for all the floors
Bne_Pradhan
👍 34
network summary each floor, max user to each floor=30<=2^H-2 H=5, will give N=3 therefore /27 For Network Summary, Total Users,= 103 103<=2^H-2 H=7 will give N=1 Therefore/25,,, i hope u got ans in short, tht is C
Danielki
👍 2
Where did N came from? I’m lost….
Alibaba
👍 2
here should be add vlan also, in this situation question was a little misunderstand, but its cisco tricky question
ZUMY
👍 3
C is correct /27 mask will give 30 host for each subnet (2^5-2)=30
ZUMY
👍 3
/25 gives us 126 maximum hosts per subnet (Total no. hosts in the building) C. 192.168.0.0/25 as summary and 192.168.0.0/27 for each floor
bigbux
👍 6
We are Keeping in mind not to waste IPs. /27 gives us 30 maximum hosts per subnet (per floor) /25 gives us 126 maximum hosts per subnet (Total no. hosts in the building) C. 192.168.0.0/25 as summary and 192.168.0.0/27 for each floor
hokieman91
👍 2
C. 192.168.0.0/25 as summary and 192.168.0.0/27 for each floor This gives each floor separate networks with 30 hosts (plus network id and plus BC address). /25 allows us to summarize the four /27 networks with 30 hosts
admin1982
👍 2
C is correct. given a /27 mask
Siegfried
👍 1
Not really. Chose C but in real scenario that would be really dumb thing to do as you want to have at least little bit of free IPs for scalability. That's why they completely missed /26 subnet as an option as that one would be ideal - so I clicked C of course. But in reality If I was forced -from some dumb reason - to choose from wasting some IPs and scalability I would go for scalability 100% (and choosing /25 then).
jasten
👍 10
The main goal is always to make efficient use of IPs (Waste as little as possible). "B" & "D" it is not enough to cover all the users for each floor. "A" waste many IPs. In my opinion the correct answer is C, due to a /27 gives 30 usable ip per floor.
Ali526
👍 3
This question does not have sufficient information to reach at a unique conclusion.
Ali526
👍 3
Sorry. It does. C is correct.
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Expert Analysis
Why the Answer Is Correct
The largest floor has 29 users, so each floor subnet must supply at least 29 usable host addresses. Using the block table (/27 = 32 addresses, /26 = 64), a /27 yields 32 − 2 = 30 usable addresses, which is the smallest block that fits 29 hosts with only one address to spare. Four /27 subnets consume 4 × 32 = 128 addresses, which fits exactly inside a single /25 block (128 addresses, 126 usable), and the building total of 24 + 29 + 28 + 22 = 103 users also requires at least a /25 (since /26 offers only 62 usable hosts). Therefore the most efficient pairing is summary 192.168.0.0/25 with 192.168.0.0/27 per floor.Why the Other Options Are Wrong
Option A pairs a /24 summary with /28 floors, but a /28 provides only 16 − 2 = 14 usable hosts, far short of the 29-user floor, and it also wastes an entire /24 when a /25 suffices. Option B uses a /25 summary with /25 per floor, so each floor would need its own /25, and four /25s cannot fit inside one /25 summary block; in practice each floor would require a separate /25, which is not an efficient distribution. Option D's /29 per floor offers only 8 − 2 = 6 usable hosts per floor, the worst mismatch for 22–29 users, and the /26 summary (62 usable) cannot even cover the 103 total users. Only option C satisfies both the per-floor minimum and the aggregate requirement.Community Comment Notes
The community converges on C, and the reasoning is sound: as Bne_Pradhan computed, the max floor of 30 or fewer usable hosts forces H=5 bits (a /27), and the 103-user total forces a /25 summary. jasten reinforced that "'B' & 'D' it is not enough to cover all the users for each floor" while A wastes addresses, and bigbux noted that "/27 gives us 30 maximum hosts per subnet (per floor)" with "/25 gives us 126 maximum hosts" for the building. kentsing's memorization tip of counting up from /32 = 1, /31 = 2, /30 = 4 … /27 = 32 is exactly the shortcut this objective rewards.Official Reference
Exam Strategy
On subnet sizing questions, first compute the usable hosts for each candidate prefix (2^n − 2) and pick the smallest block that fits the largest requirement, then verify the summary can contain all the subnets. Memorize the /24–/32 block table so you can answer without a calculator.
Frequently Asked Questions
Why doesn't a /28 per floor work when Floor 1 only has 24 users?
The design must fit the largest floor: Floor 2 has 29 users, and a /28 provides only 14 usable hosts, so every floor needs at least a /27 (30 usable).
Why is /25 the summary instead of /26 for 103 total users?
A /26 offers just 62 usable hosts, which is below the 103-user total, so the smallest summary that holds all four /27 subnets is 192.168.0.0/25.
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