# nerworking ccna question

scheme best defines the address range and subnet mask that meet the requirement and
the fewest subnet and host addresses?

A. 10.10.0.0/18 subnetted with mask 255.255.255.252
B. 10.10.0.0/25 subnetted with mask 255.255.255.252
C. 10.10.0.0/24 subnetted with mask 255.255.255.252
D. 10.10.0.0/23 subnetted with mask 255.255.255.252
E. 10.10.0.0/16 subnetted with mask 255.255.255.252

Please explain me why D is correct
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Commented:

- 2 hosts

113*4 = 452

First network that support that many hosts is /23 = 512 addresses

A. 10.10.0.0/18 - too big - 16382 addresses
B. 10.10.0.0/25 - not enough addresses - 128
C. 10.10.0.0/24 - not enough addresses - 256
D. 10.10.0.0/23 - best fit - 512 addresses
E. 10.10.0.0/16 - too big - 65536 addresses
Author Commented:
hi, you wrote that for ny point-to-point link you will waste 4 IP addresses (mask 255.255.255.252), where did you get 4?
Commented:
- 2 hosts

First subnet
10.10.0.1 - host 1
10.10.0.2 - host 2
Next subnet 10.10.0.5 - network address .....  etc ....
:)
Commented:
Easy to know number of IP addresses in subnet if you have decimal notation, quick and dirty:

last digit is 252
max number of ip addresses is 256 (255 (1 to 155) + 1 (for 0) = 256)
256 - 252 = 4 IP addresses
so that leaves you with 2 hosts per subnet :)

You can apply the same logic to 255.255.255.240
:)
14 hosts per subnet

Logic for 255.255.240.0 would be
256 - 240 = 16 addresses
0 on the end 256 addresses
16 * 256 = 4096 addresses
4094 hosts per subnet
Author Commented:
Predrag, thanks i got first part.
But i still can't understand how you get the number of addresses on each variant (16382,128,256....)?
Commented:
/16 - subnet mask 255.255.0.0         = binary 11111111.11111111.00000000.0000000
/18 - subnet mask 255.255.192.0     = binary 11111111.11111111.11000000.0000000
/24 - subnet mask 255.255.255.0     = binary 11111111.11111111.11111111.0000000
/25 - subnet mask 255.255.255.128 = binary 11111111.11111111.11111111.1000000
/23 - subnet mask 255.255.254.0     = binary 11111111.11111111.11111110.0000000
InstructorCommented:
The (correct) provided address space has 9 bits available.  Total address is 32-bits and subtract the network portion of 23-bits (32-23=9).  When you decide to use two of those bits for hosts, that leaves 7-bits for the subnets.  With 7-bits, you can have 128 subnets which is sufficent for the 113 in the question.

BTW, when you use a /30 mask and have a 2-bit host field, you don't "waste" four addresses.  Two of the addresses are actually used while the other two define the network and broadcast.  So IMHO, only two addresses are "wasted". The other two are used.
Author Commented:
Predrag, i meant these nembers 16382, 128,256, 512, 65536.
Commented:
That is what you need to learn, how to get those.
Example

/25 - subnet mask 255.255.255.128 = binary 11111111.11111111.11111111.1000000
Leading 1s are network part of address, host portion is part with 0.
So you have 7 bits for host part.
7 bits is 2**7 = 128

The same way
/16 - subnet mask 255.255.0.0         = binary 11111111.11111111.00000000.0000000
Leading 1s are network part of address (16 in this case), host part is 16 bits (part with zeroes)
16 bits  is 2**16 = 65536

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Author Commented:
Thanks Buddy, now i get it, i was confused by question itself, Thanks !!!
Commented:
You're welcome.
:)
They will try to confuse you on the exam the same way, read carefully EVERY word.
Author Commented:
Ok, promise ;)
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