You have been asked to come up with a subnet mask that will allow all three web servers to be on
the same network while providing the maximum number of subnets. Which network address and
subnet mask meet this requirement?
A.
192.168.252.0 255.255.255.252
B.
192.168.252.8 255.255.255.248
C.
192.168.252.8 255.255.255.252
D.
192.168.252.16 255.255.255.240
E.
192.168.252.16 255.255.255.252
/30 allows for 2 hosts & 64 subnets
/29 allows for 6 hosts & 32 subnets
/28 allows for 14 hosts & 16 subnets
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Ok, but i think in this situation the right answer is D and no B, because it is “/28”, while B is “\29”.
256 – 248 = 8
log 2 (8) = 3
256 – 240 = 16
log 2 (16) = 4
32 – 3 -> “/29”
32 – 4 -> “/28”
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Marco. I’m not sure where you are getting your numbers from, but you are making it too difficult. You know 255.255.255.252 (/30) would only get you 2 hosts (think point-to-point serial link). You have three servers so you need a bigger mask. A 255.255.255.248 (/29) will give you 6 usable hosts which is plenty. That also maximizes the amount of available subnets. Answer D is a /28 which yields 14 hosts which is overkill in this case. You would lose 16 available subnets if you went with D.
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Ok, do you mean, i have 3 web servers, i have to add 2 address, 1 for the network and one for the broadcast, i need 5 ip address and i can get it with 3 bits because 2^3 = 8 > 5.
Do you agree?
The last time i didn’t understand the question.
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Hi,
It simply says that you need a subnet mask that will allow minimum 3 hosts and maximum subnets.
in case of 255.255.255.248 (/29) it gives you 6 hosts which is fulfilling your requirements of hosts and also give you max subnets i.e 32.
/28 give you 14 hosts but which is more than your requirement i.e 3 host and less subnets i.e 16
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256 128|64| 32 16 8 4 2 RED
HOST 2 |4|8 16 32 64 128 256
*255.255.255.252 /30 0 0 |0 0 0 0 0 0
32-30=2
2^2=4-2(red&broadcast)=2host
RESULTADO= con /30 tengo 2host y 64 subred
*255.255.255.248 /29
32-29=3
2^3=8-2=6host
RESULTADO=con /29 tengo 6host y 32 subred
*255.255.255.240 /28
32-28=4
2^4=16-2=14host
RESULTADO=con /28 tengo 14host y 16 subred
La repuesta es la B. 192.168.252.8 255.255.255.248
porque tiene los host necesario para que los 3 servidores
web esten en la misma red y mayor cantidad de subred
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