Please Help!
Need an addressing scheme to support.
500 hosts
300 hosts
150 hosts
150 hosts
50 hosts
10 hosts
2 Hosts
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Question
Clarifications
homework assignment?
PurpleSkys
11th Feb
You haven't provided enough information - at the very least it is necessary to know what network space is available to use - and if IPv4/IPv6/both.
Beyond that, if it is a real world question, it would be helpful to explain where the numbers come from, and what is expected in terms of future growth. For example a /23 could accommodate 500 hosts, but there is hardly any free network space left. It is also a lot to have on one LAN segment; it would normally be preferable to split into 3+ /24 subnets. It might also be more worth considering route aggregation over maximum density utilisation.
If it is in fact a homework question, as PurpleSkys suggests, then you should say so, and also say how far you've got, or what specifically you don't understand; there isn't much point in just asking for an answer!
Beyond that, if it is a real world question, it would be helpful to explain where the numbers come from, and what is expected in terms of future growth. For example a /23 could accommodate 500 hosts, but there is hardly any free network space left. It is also a lot to have on one LAN segment; it would normally be preferable to split into 3+ /24 subnets. It might also be more worth considering route aggregation over maximum density utilisation.
If it is in fact a homework question, as PurpleSkys suggests, then you should say so, and also say how far you've got, or what specifically you don't understand; there isn't much point in just asking for an answer!
DomBenson
12th Feb
Answers (1)
-1
Votes
.
Yes homework.
No network space has been given. IPv4 addressing. Seven sites connected over a WAN architecture.
No network space has been given. IPv4 addressing. Seven sites connected over a WAN architecture.
13th Feb
Replies
OK, well let's start at the beginning. What's the smallest subnet that can hold each of those numbers of hosts? (e.g., as I said previously, a /23 can hold 500.) So work it out for 300, 150, 50, 10, 2.
DomBenson
14th Feb
