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

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?

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



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thasiba

thasiba

please explain how is possible!

mr_tienvu

mr_tienvu

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.

iCalvyn.com

iCalvyn.com

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

ismat

ismat

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.

dhina

dhina

please explain how is possible! i’m not understand..?

Daniel

Daniel

@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)

iCalvyn.com

iCalvyn.com

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)

DC

DC

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.