Which Subnet Mask Fits 320 Users on One Subnet?
Configure and verify IPv4 addressing and subnetting
Describe private IPv4 addressing
Answer
Correct answer: B — configure ip address 192.168.0.0 255.255.254.0 on the SVI, a /23 supplying 510 usable hosts for about 320 users across 8 floors.
An office has 8 floors with approximately 30-40 users per floor. One subnet must be used. Which command must be configured on the router Switched Virtual Interface to use address space efficiently?
Community Insight
You must size one subnet for the worst-case total user count (~320), so the trap is picking a mask that is either too small (a /25 or /27) or wastefully large (a /16) instead of the tight-fitting /23.
This CCNA subnetting question asks which SVI IP address/mask command covers 8 floors of 30-40 users in a single subnet while wasting the least address space. The answer is a /23 (255.255.254.0), because it is the smallest mask that still provides enough usable host addresses for roughly 320 devices.
The most common wrong pick is C, ip address 192.168.0.0 255.255.255.128, chosen by learners who size the subnet for one 40-user floor (/25 gives 126 usable hosts) and forget that ALL 8 floors share one subnet; D is even worse at 30 usable hosts, and A (/16) works but throws away over 65,000 addresses.
Community Discussion (39 comments)
itone333
👍 1
Selected: B
Write them power of 2s down before you take test for questions like this.
5bc68c8
👍 2
Selected: B
First, calculate the maximum number of hosts needed: 8 floors × 40 users ≈ 320 users We need a subnet mask that supports at least 320 hosts. Host capacities: A. 255.255.0.0 = /16 → 65,534 hosts ❌ too large, wastes address space B. 255.255.254.0 = /23 → 510 hosts ✅ C. 255.255.255.128 = /25 → 126 hosts ❌ not enough * D. 255.255.255.224 = /27 → 30 hosts ❌ not enough So the correct answer is: ✅ B. ip address 192.168.0.0 255.255.254.0 For a /23 subnet: 2^{9}-2=510 That gives 510 usable host addresses, which is enough for about 320 users while still being more efficient than a /16 subnet.
kminh9999
👍 1
Selected: B
maximum total users = 40 * 8 = 320 users. Therefore, required host bits = 9 -> /23 subnet -> 255.255.254.0
snowy23
👍 1
Selected: B
The answer is B since 8x40=320 For supporting 320, you'd need to use the remaining host bits formula, 2^9 - 2 = 510 this will give you enough hosts, any less than that is not enough. then you will skip the 4th octet since you borrowed 9 host bits and not any less than 8.. that means you'd go for the 3rd octet, only 1 answer has changes made to 3rd octet, so B is your answer.
snowy23
👍 1
sorry, option A has changes made to 3rd octet too but it goes 0 and only B goes for 254 which is exactly leaving the last bit off for the 3rd octet and that would be borrowing 9 bits :D
liptechBR
👍 1
Selected: C
The correct answer is: C. ip address 192.168.0.0 255.255.255.128 With a subnet mask of 255.255.255.128 (or /25), we have 128 addresses per subnet, of which 126 can be assigned to devices. This mask is enough to accommodate between 30 and 40 users on each of the 8 floors, totaling up to 320 users, and also saves address space.
JRossJ
👍 1
Posting ChatGPT generated stuff in the answers section shouldn't be allowed. Throws people off. The correct answer is B.
c49555c
👍 6
your chatgpt generated answer is wrong, you would have realized it yourself if you just read what you copy pasted
Ineedicecreamdaily
👍 3
40 cannot fit in 32 so it must fit in the 64, therefor you know that you need (64 addresses) * 8 (floors) = 512 addresses. You know that /23 contains 512 addresses and the subnet mask is 255.255.254.0. Thank you, and sorry for my english.
Agribaba
👍 4
Calculating Total Users: Minimum number of users: 8 floors 30 users = 240 users Maximum number of users: 8 floors 40 users = 320 users To ensure there is enough space for growth and network devices (such as printers, IP phones, and other equipment), it is best to choose a subnet that can accommodate more than the maximum number of users. Choosing the Subnet Mask: A /23 subnet provides 512 IP addresses (510 usable). This will accommodate up to 510 devices, which is sufficient for the current and future needs of the office.
KnackerTopf1
👍 1
literally none of these commands would actually work
T0ga
👍 1
Selected: C
Convert IP address and subnet mask to binary: IP address: 10.10.225.48 becomes 00001010.00001010.11111111.00110000 Subnet mask: 255.255.255.224 becomes 11111111.11111111.11111111.11100000 Compare the first 27 bits: Align the two binary strings and check if the first 27 bits (from left to right) are identical: 00001010.00001010.11111111.00110000 (IP address) 11111111.11111111.11111111.11100000 (Subnet mask) The first 27 bits (highlighted) match perfectly.
2488fcf
👍 2
I would reread your book, mate. This is the weirdest gibberish I have ever read. You need 8 floors times 40 people that is 320 in total and the question is to use one subnet for the whole building so 320 hosts are needed in the subnet. /23 is 510 hosts /24 is 254 hosts so answer is /23 or B.
ciscopue
👍 2
The option is B. ip address 192.168.0.0 255.255.254.0. Explanation: An office has 8 floors with approximately 30 to 40 users per floor. They make a total of 320 users, a mask of 255.55.255.0 only fits 254 users, which is not enough, so the next largest mask would be: 255.255.254.0.
duongccna
👍 1
B correct, ~40 x 8 = 320 user 2^9 -2 =510 >320 (ok) 2^8-2=254 < 320 therefore borrow 9 bit, choose B
Freitas512
👍 3
Selected: B
To get a subnet to support 40 users we need at least 6 bits. 32-6 = /26 but we want to be able to get 8 different subnets so we need 3 more bits. /26-3 = /23 . We just need to find the subnet mask of /23
blackcisco
👍 1
B is correct because i quote "One subnet must be used" one subnet is needed for all 8 floors and only A and B provide this with B uses must less address space
Naetan0809
👍 1
8 FLOORS = 8 SUBNET "APPROX" 30-40 USERS = HOST NEEDED 192.168.0.0 = Old Subnet Mask =/24 8 SUBNET= 2,4,8 = 3 BITS BORROWED NewSubnetMask= OldSubnetMask + BITS BORROWED = 24+3 = 27 USABLE HOST = 2^(32-NSM)- 2 =2^(32-27) - 2 = 30 USABLE HOST With “192.168.0.0 255.255.255.224” we have 8 subnets: • + First subnet: 192.168.0.0 to 192.168.0.31 • + Second subnet: 192.168.0.32 to 192.168.0.63 • + Third subnet: 192.168.0.64 to 192.168.0.95 • + Fourth subnet: 192.168.0.96 to 192.168.0.127 • + Fifth subnet: 192.168.0.128 to 192.168.0.159 • + Sixth subnet: 192.168.0.160 to 192.168.0.191 • + Seventh subnet: 192.168.0.192 to 192.168.0.223 • + Eighth subnet: 192.168.0.224 to 192.168.0.255
timtgh
👍 1
They most likely meant one subnet per floor, not just one big subnet for the whole office. It's a judgement call. They didn't say "per floor," but they often word things poorly, and that is probably what they meant. If the whole office is one subnet, then it's a flat network and no subnetting is needed at all, and the mask doesn't matter.
oatmealturkey
👍 2
But subnetting is still needed if they want only one subnet, because with their allocated Class C address that we can see in the answer choices, they do not have enough space for that many hosts on a single subnet.
Maycao
👍 1
Correct
Vile_Yogabear
👍 3
I managed to solve it by I didn't like the way I did it. 8 x 40 = 320 hosts I manually tried to find which subnet can support this many hosts. 512, 254, 128, 64, 32, 16,8 , 4, 2, 1 If a /24 can support 254 - 2 hosts then /23 can support 512 - 2 hosts so its must be a /23 network. I just counted the subnet my memorizing the numbers 128, 192, 224,240,248, 252,254, 255 /17, /18, /19, /20, /21, /22, /23, /24 So the subnet mask for /23 is 255.255.254.0
keokkeo_123
👍 1
Selected: B
BBBBBBB
👍 85
8 floors and 40 user per floor means 320 users (approx.). How many bits do you need to have 320 IP addresses? 8 bits = (2^8 – 2) =254 IP addresses, and it’s not enough. 9 bits = (2^9 – 2) = 510 IP addresses, and this is enough. You have a class C subnet (192.168.0.0). This means a subnet mask like this: 255.255.255.0 But you need 9 bits for the hosts, so you’ve got left with a subnet mask like this: 255.255.1111111x.xxxxxxxx =255.255.254.0 This means you will use VLSM subnetting. Answer B is correct.
bikila123
👍 6
very good explanation!!!
HMaw
👍 13
Reading your explain make me hearing Jeremy voice saying "Save the hosts". Nicely done.
re__roy
👍 3
Well explained brother
Vishalb86
👍 2
For a class C network, the default subnet mask is 255.255.255.0. In CIDR the lowest is a /25 which supports 126 hosts. To answer this question there is supposed to be a class B address with a subnet mask of 255.255.254.0 or /23 which will support 9 subnets and 510 host.
Hodicek
👍 2
Sorry B is the correct answer as 8 X40 = 320 so it should to be B not C
Hodicek
👍 2
Answer is C
AKA1987
👍 10
40*8<=2^H-2, will give H=9 which is a /23 OR 255.255.254.0 = Answer B
HakamCnna
👍 2
how you give H=9 .?
timtgh
👍 1
As stated above, the question probably did mean per floor, and they just worded it badly. There's no way to know for sure.
Adaya
👍 2
Thanks for your explanation
Shamwedge
👍 10
No. 255.255.254.0 is correct. 256 = /24 but you need to use at least 512 to cover the 320 users. 512 hosts is /23 and 255.255.254.0 is the subnet mask for /23
Bne_Pradhan
👍 2
40*8<=2^H-2, will give H=8 hence in between A and B, but As A will waste lot of addresses, correct will be B
Bne_Pradhan
👍 1
i suppose B was meant to be 255.255.255.0,, but in anyways thts the most favoured ans
Doad
👍 4
Answer must be 255.255.255.128 as at last octet bits will 10000000-- only then it can occupy 40 hosts otherwise not.
Heymannicerouter
👍 2
That would be correct if the question said one subnet per floor, however since it's only one in total, you need a subnet that covers at least 320 users, therefore B is correct.
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Expert Analysis
Why the Answer Is Correct
Option B, ip address 192.168.0.0 255.255.254.0, is a /23 prefix, which borrows 7 subnet bits from a Class C boundary and leaves 9 host bits. Nine host bits yield 2^9 = 512 total addresses, or 510 usable hosts after reserving the network and broadcast addresses, comfortably exceeding the 8 x 40 = 320 device worst case. It is also the tightest fit available among the choices, which is exactly what the question's phrase "use address space efficiently" is testing. Option A (a /16 with 65,534 usable hosts) would technically connect everyone but wastes more than 65,000 addresses, so it fails the efficiency requirement.Why the Other Options Are Wrong
Option C is a /25 with only 126 usable hosts — enough for a single 40-user floor but not for all 8 floors sharing one subnet, so hosts on the upper floors would get no address. Option D is a /27 with a mere 30 usable hosts, which does not even cover one floor, let alone the whole office. Option A is not "wrong" in connectivity terms but is grossly oversized; on an exam that asks for efficient use of address space, the smallest sufficient prefix is the graded answer. Note also that on a real Cisco multilayer switch SVI you would normally apply the first usable host address (192.168.0.1/23) rather than the network ID itself, but the mask is the deciding factor in every option here.Community Comment Notes
Most commenters converged on B and showed the math cleanly: one wrote "40*8<=2^H-2, will give H=9 which is a /23 OR 255.255.254.0 = Answer B", and another framed it as "40 cannot fit in 32 so it must fit in the 64" and multiplied 64 x 8 = 512 to land on /23. Freitas512 reached the same mask a different way, going from a /26 per floor and borrowing 3 subnet bits to get /23. Agribaba emphasized covering printers, IP phones and future growth rather than the bare 320. Only one comment pushed for C, arguing "255.255.255.128 as at last octet bits will 10000000", which misreads the requirement by sizing for one floor; another comment bluntly warned that a previously posted "chatgpt generated answer is wrong", which the community math confirms.Official Reference
Exam Strategy
On subnetting questions, first multiply out the worst-case host count (8 x 40 = 320), then solve 2^H - 2 >= 320 to get H = 9 host bits, which means /23 = 255.255.254.0. Always read whether the question asks for the SMALLEST sufficient subnet — phrases like "use address space efficiently" are the exam's signal that a /16 answer is a trap.
Frequently Asked Questions
Why isn't a /25 (255.255.255.128) enough if each floor only has 40 users?
Because all 8 floors share ONE subnet, so the subnet must hold up to 320 addresses at once. A /25 gives only 126 usable hosts, so it would exhaust after roughly three floors.
Why is 255.255.254.0 considered more efficient than 255.255.0.0 if both work?
Both would connect the office, but 255.255.0.0 (/16) reserves 65,534 usable hosts when only about 320 are needed. The /23 wastes nothing unnecessary, which is what "use address space efficiently" asks for.
Shouldn't the SVI use 192.168.0.1 instead of the 192.168.0.0 network address?
On a real Cisco device you would assign the first usable address, 192.168.0.1/23. Every option here lists the network ID, so the intended discriminator is the subnet mask, and only B has the correct one.
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