Maths

[craigie]

Just reading through the 70-291 Network Infrastructure Chapter 2 Understanding TCP/IP.

It's awesome to actually be using some of the Maths that I learnt in school back in the 1990's!

Gotta say, so far it's all going in, and makes sense, but is pretty intense.

Anyways, back to the maths err studying err learning about networks

[craigie]

Just finished the exercises in Chapter 2 of the 70-291 book.

I believe MS have made a mistake on Exercise 8, you need to convert Hosts into a Subnet Mask.

The requirement is 63 Hosts, which would convert to 64 (2 power of 6), 32- 6 = 26.

/26 equates to 255.255.255.192

In the MS book they give the answer as 255.255.255.128, I make that /25 which would give 128 Hosts.

Am I right?

[Qs]

I calculate that with a 255.255.255.192 mask and a /26 count you'd get 64 possible hosts, 62 of which are available.

Depends how the question is worded as to whether it's correct or not. What specifically does it say?

EDIT - Forgot to add... Yeah you're semi correct. With a 255.255.255.128 mask and a /25 count you'd get 128 possible hosts, 126 of which are available.

Qs

[craigie]

Originally Posted by Qs

I calculate that with a 255.255.255.192 mask and a /26 count you'd get 64 possible hosts, 62 of which are available.

Depends how the question is worded as to whether it's correct or not. What specifically does it say?

EDIT - Forgot to add... Yeah you're semi correct. With a 255.255.255.128 mask and a /25 count you'd get 128 possible hosts, 126 of which are available.

Qs

Cool, so in future, when I calculate, I will add +2 to the Host value to make sure that it falls within range.

[dmarsh]

You lose two hosts for the broadcast and the network address.

So you have 1+2+4+8+16+32=63-2 = 61 possible hosts, Not enough !

1+2+4+8+16+32+64=127-2 = 125 possible hosts, Enough !

Leaving last bit 128 for network, 8+8+8+1 = 25 So its /25 or 255.255.255.128.

Each octet is 8 bits or 0-255, 128 64 32 16 8 4 2 1.

You can do it with powers of 2 direct but if you get confused just go back to simple math.

For example you notice that first part came out 63, turns out every summation of the max possible representation in the lower columns is (2 power N) -1. This is like our decimal number system which is (10 power N) -1.

Digits column max representation 9, its one less than next power of ten which is 10 and the value of a unit in the next column !

2 power 6 is 64, minus 1 gives 63, which is the maximum number you can represent in 5 bits or the result of the summation. The only problem was you forgot the -2 part.

[craigie]

Originally Posted by dmarsh26

You lose two hosts for the broadcast and the network address.

So you have 1+2+4+8+16+32=63-2 = 61 Not enough !

1+2+4+8+16+32+64=127-2 = 125 Enough !

Leaving last bit 128 for network, 8+8+8+1 = 25 So its /25 or 255.255.255.128.

Yep, when I was doing the calulations, I forgot the -2

[Qs]

Originally Posted by dmarsh26

You lose two hosts for the broadcast and the network address.

So you have 1+2+4+8+16+32=63-2 = 61 possible hosts, Not enough !

1+2+4+8+16+32+64=127-2 = 125 possible, hosts, Enough !

DMarsh is pretty much on the ball... but...

1+2+4+8+16+32= 64

then you minus the two for broadcast and network

64 - 2 = 62 available hosts

It still isn't enough and your concepts are correct, but your calculations are off.

Same with 1+2+4+8+16+32+64 = 128

128 - 2 = 126 available hosts

Unless I'm missing something here?

Qs

[craigie]

Originally Posted by Qs

DMarsh is pretty much on the ball... but...

1+2+4+8+16+32= 64

then you minus the two for broadcast and network

64 - 2 = 62 available hosts

It still isn't enough and your concepts are correct, but your calculations are off.

Same with 1+2+4+8+16+32+64 = 128

128 - 2 = 126 available hosts

Unless I'm missing something here?

Qs

Nope thats exactly right, on the calcualtion I just forgot the -2.

[dmarsh]

You are missing something ! (Check with a calculator if you don't believe me 1+2+4+8+16+32=63) the (base power N) - 1 rule holds for all number systems, binary, octal, decimal, hexidecimal, etc.

Another way to think of it is like this, 32+32 = 64, Well if I could represent 32 in the columns below why do I need a 32 column ? I can only represent 0-31 in the columns below, thats why I need a 32 column ! 31+32=63

There a 32 possible numbers or representations including zero, but zero doesn't count !

Number of combinations of 8 bits is 2 power 8 = 256, but each octet can only contain 255 ! We lose one combination/representation to represent the zero.

[Qs]

*thinks*

[dmarsh]

Okay imagine I've only got three bits (three columns)...

4 2 1

These are my representations :=

000 = 0
001 = 1
010 = 2
011 = 3
100 = 4
101 = 5
110 = 6
111 = 7

Thats eight representations, but the highest representation is 7 because I lost one for the zero.

Hence

(2 power 3)-1 = 8-1
7 is highest number.

(2 power 3) = 8
8 is number of representations.


Same with decimal :-

With one column 0-9

0
1
2
3
4
5
6
7
8
9

10 power 1 representations, or 10.

But maximum number is only 9, (10 power1)-1.

Same holds for other number systems and numbers of columns, just as combinations go up examples get more tedious!

[Qs]

I get what you're saying dmarsh but I still believe you're wrong within the context of calculating available hosts.

Admittedly when you're adding numbers together then it follows your concept (I must have been having an off day when I previously calculated this!), for example:-

1+2+4+8+16+32 = 63

But...

2^6 (or 2*2*2*2*2*2) = 64

You don't add the individual components to calculate the available hosts on a subnet. That's the point.

You calculate the power and then minus 2 (for the broadcast and network address of each subnet.)

So in craig's original example:-

Calculating for 63 available hosts...

2^6 = 64 - 2 = 62 available hosts

2^7 = 128 - 2 = 126 available hosts

As the latter has the required 63 hosts and the former doesn't - then the subnet mask MS are looking for is 255.255.255.128/25.

If you don't believe me then check Cisco's Host and Subnet Quantities table - http://www.cisco.com/en/US/tech/tk36...3.shtml#classc

K?

Qs

[dmarsh]

Ok I see your point you are correct, the zero and network address are the same thing so no need to minus the one in this case.

Take this example :=

000 = 0 All zeros / network.
001 = 1
010 = 2
011 = 3
100 = 4
101 = 5
110 = 6
111 = 7 All ones broadcast.

So in this case its (2 power N) -2 for number of hosts, you are correct. Theres no need to minus the zero/network twice ! My Bad, I knew I shouldn't have got involved with a networks thread !

[Qs]

Originally Posted by dmarsh26

Ok I see your point you are correct, the zero and network address are the same thing so no need to minus the one in this case.

Take this example :=

000 = 0 All zeros / network.
001 = 1
010 = 2
011 = 3
100 = 4
101 = 5
110 = 6
111 = 7 All ones broadcast.

So in this case its (2 power N) -2 for number of hosts, you are correct. Theres no need to minus the zero/network twice ! My Bad !

Np. Thought I was losing it for a second there too, lol. I lost the ability to add! Show's what work can do eh? *passes dmarsh a beer*

Qs