The network administrator is asked to configure 113 point-to-point links.Which IP addressing scheme best defines the address range and subnet mask that meet the requirement and waste 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 how is possible!
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Each network will have four address’s networks so if you just do 4 x 113 = 452 and then think about where that will fit.
+ B & C are not enough adresses
+ A & E are good but waste more subnet and host addresses.
+ D is the best choice and wastes the fewest subnet and host addresses.
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Base on your explanation, total address require is 452.
We have to use /23 since it allow 512 address.
The rest is either a lot more wasted or less to fit 452. Correct me if I am wrong
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Dear tienvu,
first you look to /23 bit we have 2 network one 10.10.0.0/30 and second is 10.10.1.0/30 on every network we have 64 subnets.
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please explain how is possible! i’m not understand..?
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@dhina:
If you need to have 113 networks you must borough 7 bits in order to fit the request (128). Now you’ll have 128 networks with /30 (255.255.255.252 – 4 addresses, 2 for host)
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We need 113 point-to-point links which equal to 113 sub-networks < 128 so we need to borrow 7 bits (because 2^7 = 128).
The network used for point-to-point connection should be /30.
So our initial network should be 30 – 7 = 23.
So 10.10.0.0/23 is the correct answer.
You can understand it more clearly when writing it in binary form:
/23 = 1111 1111.1111 1110.0000 0000
/30 = 1111 1111.1111 1111.1111 1100 (borrow 7 bits)
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113 poin-to-point = 113 X 4 <- (network address 2 free address and broadcast)
then I need 452 address. I have to use 512. and the network 10.10.0.0 /23 is a 512 network.
I will use before a mask 255.255.255.252 to generate networks whit 2 free address.
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