I have a basic question about the ip/subnetting thing. The assignment sheet says we have 193.188.34.128/26 available. This is a bit confusing for me...
The notes say that /26 includes network+subnet, mela I'm assuming the first 24 bits are for the network and the first 2 bits in the last part (128) are for the subnet.
Issa... look at the last part (tal-128). Kif qed nifhimha jien, since it's 128, mela it's going to be of the form 1Xxxxxxx. This means that we only have 10 and 11 for subnetting, the latter of which is unusable... so the subnetting part is totally useless. This means we are left with 2^6 = 32 IP addresses to use.
I imagine I'm misunderstanding the notation... so could you be patient and clarify things a bit?
> I have a basic question about the ip/subnetting thing. The assignment > sheet says we have 193.188.34.128/26 available. This is a bit > confusing for me...
> The notes say that /26 includes network+subnet, mela I'm assuming the > first 24 bits are for the network and the first 2 bits in the last > part (128) are for the subnet.
> Issa... look at the last part (tal-128). Kif qed nifhimha jien, since > it's 128, mela it's going to be of the form 1Xxxxxxx. This means that > we only have 10 and 11 for subnetting, the latter of which is > unusable... so the subnetting part is totally useless. This means we > are left with 2^6 = 32 IP addresses to use.
> I imagine I'm misunderstanding the notation... so could you be patient > and clarify things a bit?
I didn't say there's anything wrong... I just wanted to know whether my reasoning was correct. However it is odd that you have 2 bits for subnetting which are useless.
Mela unless you state otherwise, I'm going to assume we have 32 IP addresses available, and that my reasoning is correct.
> I didn't say there's anything wrong... I just wanted to know whether > my reasoning was correct. However it is odd that you have 2 bits for > subnetting which are useless.
> Mela unless you state otherwise, I'm going to assume we have 32 IP > addresses available, and that my reasoning is correct.
Was following this thread 'cause I had a similar prob.
Note that 2^6 = 64 :). That is 64 addresses... less 2 which are reserved or whatever. Regarding the subnet bit, I think your reasoning is correct.
If you were refering to to pg7/slide2.pdf in the notes, I think the example there is from an administrator's point of view who decided to subnet the IP further, thus loosing addresses. The ip in the ass. sheet refers to the IPs routed to the company from the ISP. Or not?
You're right about the 64 IPs... that was a silly miscalculation on my part.
Actually I was referring to page 6 of the second set of slides, where it says:
"Subnet Addresses are also represented as /x where x is the number of bits burrowed [supposed to be borrowed :P] for network + subnet."
That site helps... thank you very much. :)
Mr. Cordina... would I be gaining any unfair advantage over my peers if you told me whether my assumption is correct? This is a public discussion area and everyone would benefit from it. I don't understand why you are being so mysterious.
> You're right about the 64 IPs... that was a silly miscalculation on my > part.
> Actually I was referring to page 6 of the second set of slides, where > it says:
> "Subnet Addresses are also represented as /x where x is the number of > bits burrowed [supposed to be borrowed :P] for network + subnet."
> That site helps... thank you very much. :)
> Mr. Cordina... would I be gaining any unfair advantage over my peers > if you told me whether my assumption is correct? This is a public > discussion area and everyone would benefit from it. I don't understand > why you are being so mysterious.
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