Help me to understand this mysterious question !!!!!!!!!!

[Amr Ali]

hello dears,

i am looking for a help as i am a little junior

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Q - 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

I know answer is D as we need 113 subnet , hereby we need 7 bits to do it so 30-7 = 23 but i can't understand this answer , my point is we need only 7 bits why we chose /23 in addition i can't get what is the meaning of /23 subnetted with /30

[Bilal Nawaz]

This means, you have a /23 but you calve it up in to /30's.

E.g.
10.10.0.0 /23. Out of this /23 how many /30's can I make

Network 1: 10.10.0.0/30
Network 2: 10.10.0.4/30
Network 3: 10.10.0.8/30
And so on...

But all these networks have to fit within the 10.10.0.0/23 scope

So should include the range 10.10.0.0 - 10.10.1.255

So the question is asking which one can you calve up most efficiently without any wastage. With other subnets you will have IPs left over, therefore get wasted.

Hope this helps

Sent from Cisco Technical Support iPhone App

[Amr Ali]

i can't get if he meant 113 subnets or hosts , as if he meant subnets i think any number higher than /7 will be true

[chamira dil]

First understand what a Point-to-Point link is (just google and c images for a better understanding)

In a Single Point-to-Point link there’re 2 Points (as the name implies), one point  each sides 

And those Points can be considered as Hosts.

So in a Single Point-to-Point link there are 2 Hosts.

So for a single point-to-point link we must assign 2 valid IPs from a same subnet.

So to have 2 valid IPs the chosen subnet should have at least 4 IPs (2+2=4) (Remember we have to reserve 1 IP for the network ID and 1 IP for Broadcast ID)

I think up to this u have understood what I’m telling

So to have 1 Point-to-Point link we should have

1 Subnet and

4 IPs 

So to have 113 Point-to-Point links we should have (at least)

1 x 113 = 113 Subnets

4 x 113 = 452 IPs (Hosts, u may call in subletting world)

So to have 452 IPs we should use a subnet of /27 

Because /23 has 9 host bits so it can offer 512 (2^9) IPs and 510 (512-2) valid IPs we can have our required 452 IPs wasting a minimum of IPs

P.S.

/24 subnet can only have 254 ((2^8)-2) valid IPs so not sufficient for our requirement of 452 IPs

and anything less than /23 like /22 /21 /20 /18 can be used for accommodate 452 IPs but we would have to waste more IPs 

Another way to approach (not adviced to use)

And the networks with subnet mask /31 (borrowed 7 hosts bits from class c network) , /23 (borrowed 7 hosts bits from class b network) and /15 (borrowed 7 hosts bits from class a network), we can have 113 subnetworks 
So by looking at the answers there's only one subnetwork that maches thoes cidr notation which is /23

__

Regards,

Dilshan

[Lassek]

Your subnetting sequence is incorrect.

I know this is old but just in case it causes confusion to other readers.

The correct subnetting is :

Network 1: 10.10.0.0/30
Network 2: 10.10.128.0/30
Network 3: 10.10.0.4/30

Network 4: 10.10.128.4/30 and so on

[cadet alain]

Hi,

you need 113 /30 networks, these are /30 because point-to-point.

So you need  113 subnets so what power of 2 is greater than or equal to 113: this is 7 because 2 to the power 7= 128

So 30-7= 23 which is the initial subnet you're going to split in 113 point-to-point networks.

Regards

Alain

Don't forget to rate helpful posts.

[Amr Ali]

i know your answer but i can't get why , as i understand that /30 will offer 2^30 subnets , so why we need to decrease it by 7

[Joseph W. Doherty · Accepted solution]

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Posting

A /30 consumes 4 IP addresses.  So, if we need 113, we're going to need 452 IP addresses.

So looking at the different sized address blocks offered . . .

The /25 and /24 are too small at 128 and 256 IP addresses each.

The /23, at 512 IP addresses, can provide the needed amount of IP addresses.

The /18 and /16, can also provide the needed IP addresses, but as they provide many more, for example 32K for the /16, they are larger than needed.

[Amr Ali]

thanks very much Mr. JosephDoherty  now i got it , really i appreciate your effort

[paulstone80]

Hi Amr,

The question asks for 113 point-to-point links. You can interpret this as requiring 113 /30 subnets.

A /30 network uses 4 IP addresses;
1 x network address
1x broadcast address
2 x usable IP addresses

The total number of IP address required for 113 /30 subnets is 113 x 4 = 452

How many bits are needed to represent a value of 452?

7 bits = 128
8 bits = 256
9 bits = 512

You need took subtract the 9 bits from the mask.

32 - 9 = 23

This gives you a network of 10.10.0.0/23.

HTH

Sent from Cisco Technical Support Android App

[Amr Ali]

thanks very much Mr. paulstone80  now i got it , really i appreciate your effort