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Given the 172.16.0.0 255.255.255.0 IP segment, how many IP addresses can be assigned to host and network devices?
Suggested Answer: A
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32 comments · newest first
RA
Raylene
11 months ago
I thought it was 126 too, but I guess I was wrong!
LE
Lettie
11 months ago
128 is way off, it's definitely more than that.
DO
Domonique
11 months ago
Wait, are we sure about that?
JO
Josephine
12 months ago
Totally agree, 254 is the right number.
VA
Valentin
12 months ago
It's 254 usable IPs, not 126!
LO
Louvenia
12 months ago
I thought it was 128 usable addresses, but now I'm confused about the network and broadcast addresses. Can someone clarify that?
GA
Galen
1 year ago
I practiced a question like this, and I think the answer is 126 usable IPs. The formula is 2^n - 2, where n is the number of host bits.
RA
Raylene
1 year ago
I'm not entirely sure, but I remember something about needing to calculate usable addresses. Is it 128 or 126? I feel like I've seen similar questions before.
BI
Billi
1 year ago
I think the subnet mask 255.255.255.0 gives us a /24, which means we have 256 total addresses, but we need to subtract 2 for the network and broadcast addresses, right?
KI
Kindra
1 year ago
I've got this one! The subnet mask 255.255.255.0 means we have 8 bits for the host portion. 2^8 = 256 possible host addresses, but we need to subtract the network address (172.16.0.0) and the broadcast address (172.16.0.255), so the answer is 254 available IP addresses.
OW
Owen
1 year ago
This looks straightforward. The subnet mask is 255.255.255.0, so the network has 256 possible IP addresses (2^8). Subtracting the network and broadcast addresses, we get 254 available IP addresses.
AL
Aliza
1 year ago
I'm a bit confused on how to calculate the number of available IP addresses. Can someone walk me through the process?
AN
Annalee
1 year ago
Okay, let's think this through step-by-step. The IP segment is 172.16.0.0 with a subnet mask of 255.255.255.0, so the network portion is 172.16.0.0 and the host portion is 0.0.0.0. The number of host bits is 8, so the number of possible host addresses is 2^8 - 2 = 256 - 2 = 254.
JO
Jodi
1 year ago
I'm just hoping the answer isn't E) 42. That would be way too cosmic for a networking exam!
LO
Louis
1 year ago
I agree, C) 128 sounds right for that IP segment.
DE
Dean
1 year ago
I hope it's C) 128. That seems like a reasonable number.
TA
Tamie
1 year ago
C) 128
RE
Rebeca
1 year ago
A) 126
CA
Carlota
1 year ago
I'm with Jeannetta on this one. A) 126 has got to be the correct answer. Who would waste two IP addresses for the network and broadcast addresses? Silly question.
JA
Jaime
1 year ago
Woah, this is a tricky one. Let me think... B) 6 seems way too low, and D) 4 just doesn't make sense. I'm going with C) 128, it feels right.
MI
Micah
1 year ago
Yes, C) 128 is the correct answer for the number of IP addresses that can be assigned.
JO
Jolanda
1 year ago
I agree, C) 128 sounds right to me too.
AN
Anastacia
1 year ago
I think C) 128 is the correct answer.
JE
Jeannetta
2 years ago
I'm not so sure about that. Doesn't the network address and broadcast address take up two of those addresses? I'm leaning towards A) 126.
LI
Ligia
1 year ago
Great, glad we figured that out together!
MO
Mona
1 year ago
I agree, the correct answer is A) 126.
EA
Earlean
1 year ago
So that leaves us with 126 usable IP addresses for host and network devices.
AS
Ashton
1 year ago
I think you're right, the network and broadcast addresses do take up two of the addresses.
RE
Remona
2 years ago
Hmm, I think the answer is C) 128. That's the number of available host addresses in a /24 subnet, right?
JA
Janessa
2 years ago
So that leaves us with 254 usable IP addresses for host devices. The answer should be C) 128.
IS
Isabella
2 years ago
I agree with Tuyet. With a /24 subnet mask, there are 256 total IP addresses, but 2 are reserved for network and broadcast addresses.
TU
Tuyet
2 years ago
I think the answer is C) 128.