You work as a network administrator at ABC.com. The ABC.com network consists of a single
domain named ABC.com. All servers on the ABC.com network have Windows Server 2012 R2
installed.
The network as a server named ABC_SR05 that provides DHCP services. You need to create the
DHCP scopes to support three subnets. Subnet A has 2,500 hosts, Subnet B has 800 hosts and
Subnet C has 200 hosts. You want to minimize the number of unused IP addresses on each
subnet.
How would you configure DHCP to support these subnets?
A.
You should assign the network 172.16.0.0/16 for Subnet A, the network 172.16.1.0/16 for
Subnet B and the network 172.16.2.0/16 for Subnet C.
B.
You should assign the network 172.16.0.0/19 for Subnet A, the network 172.16.1.0/20 for
Subnet B and the network 172.16.2.0/24 for Subnet C.
C.
You should assign the network 172.16.0.0/20 for Subnet A, the network 172.16.1.0/20 for
Subnet B and the network 172.16.2.0/20 for Subnet C.
D.
You should assign the network 172.16.0.0/18 for Subnet A, the network 172.16.1.0/20 for
Subnet B and the network 172.16.2.0/23 for Subnet C.
E.
You should assign the network 172.16.0.0/20 for Subnet A, the network 172.16.1.0/22 for
Subnet B and the network 172.16.2.0/25 for Subnet C.
F.
You should assign the network 172.16.0.0/21 for Subnet A, the network 172.16.1.0/23 for
Subnet B and the network 172.16.2.0/26 for Subnet C.
Explanation:
Explanation
It looks like none of these are correct. In answer E it says subnet C should be /25 however /25 only allows 128 hosts.
/20 = 4096 hosts
/22 = 1024 hosts
/24 = 256 hosts
/25 = 128 hosts
The Answer E should be:
You should assign the network 172.16.0.0/20 for Subnet A, the network 172.16.1.0/22 for
Subnet B and the network 172.16.2.0/24 for Subnet C
confirm
Liron is right!
No right. Subnet C – only /24. Choice B
Suppose there are NO Typo’s The Correct answer would be C!
B.
172.16.0.0/19;172.16.1.0/20 8190 + 4094 + 254 = 12538 (+6 if you want to)
172.16.2.0/24
C.
172.16.0.0/20;172.16.1.0/20 4094 + 4094 + 4094 = 12282 (+6 if you want to)
172.16.2.0/20
D.
172.16.0.0/18;172.16.1.0/20 16382 + 4096 + 510 = 20988 (+6 if you want to)
172.16.2.0/23
I think the question is if YOU realy understand the doubling(!) issue…
I agree to me it should be C, otherwise how are you going to put 200 clients on a /25 scope as Answer E is contemplating (126 clients)
Absolutely! Did the calculation and you get the least number of IPs with option C while making sure you have enough. Has to be a typo with E though…/24 instead of /25
To allow 2500, 800 and 200 addresses in different subnets, the correct prefix lenghts are /20, /22 and /24 as said in the first comment.
But All answers are wrong!
Didn’t you notice that ALL subnets overlap each other, in ANY answer?
For example:
SubnetA, 172.16.0.0/20= 172.16.0.1 -> 172.16.15.254 (4096 addresses)
SubnetB, 172.16.1.0/22= 172.16.0.1 -> 172.16.3.254 (1024 addresses)
SubnetC, 172.16.2.0/24= 172.16.2.0 -> 172.16.2.254 (256 addresses)
So written may be much clear that subnetA contains SubnetB, than contains subnetC
And this is true for any answer. Even if you take the narrower SubnetA (172.16.0.0/21, answer F), tha also allows 2048 addresses only.
If you consider answerC, all nets with a /20 prefix lenght, it is not a 4096 + 4096 + 4096 =12282 addresses!
Are always THE SAME 4096 addresses. Because they are not 3 different subnets, they are the same subnet.
“SubnetA”, 172.16.0.0/20= 172.16.0.1 -> 172.16.15.254 (4096 addresses)
“SubnetB”, 172.16.1.0/20= 172.16.0.1 -> 172.16.15.254 (4096 addresses)
“SubnetC”, 172.16.2.0/20= 172.16.0.1 -> 172.16.15.254 (4096 addresses)
But leaving back the networks and the routing options and focusing on the DHCP server configuration, that is the core of the question, whichever answer you think is correct, could you please give a working example of the configuration of the three address ranges for the 3 different subnets?
Assuming you have configured a scope for subnetA with a range: 172.16.0.1 -> 172.16.15.254 (/20) to allow the required 2500 addresses, i’m sure Windows (any system) will not allow you to configure another range that overlaps.
The ONLY way I can think this question about is they wanted to check the ability of the candidate to figure out that them are all the same subnet, with the correct prefix lenght of /20, and tryed to confuse him as much as possible… going completely out of sense.
Since it is the same subnet, even a superscope doens’t apply in this case.
With this in mind, dropping the too many senseless elements of the answers, the only solution to this question I can see is the following.
I see that there are not 3 subnets, but only one
2500 + 800 + 200 = 3500 addresses 172.16.15.254
With this in mind, I would answer C.
If you would configure the IP address manually on your clients you would write:
172.16.0.x, 255.255.240.0
172.16.1.x. 255.255.240.0
172.16.2.x. 255.255.240.0
and so on…
Wich is similar to answerC:
”
You should assign the network 172.16.0.0/20 for Subnet A, the network 172.16.1.0/20 for
Subnet B and the network 172.16.2.0/20 for Subnet C.
”
But is a real guess.
All answers are wrong.
agree, all answers have overlap in ranges. some aren’t even correct ranges (172.16.1.0/23 is not a valid subnet, it should start with 172.16.0.0 or 172.16.2.0)
purely based on subnet size it should be
/20 (up to 4k)
/22 (up to 1k)
/24 (up to 256)
not even 1 of the answer fit this. > B would be the answer that’s got the least amount of overhead.
anyway it’s a crap incorrect question
robber is right none of the answers fit. check with a subnet calculator if you are still unsure.
I think this answer should be B.
In E ,there are only 128 IPs for subnet C, so it’s a wrong answer.
In B, /20 and /24 are suit for subnet B and C. Even /19 for subnet A is a waste for IPs, but it’s the only right answer.