Calculating links of a fully meshed network

Dist · 12 May 2008 at 15:35

Whats the easiest way to calculate the number of links required to make a fully meshed network? I have a multiple choice question i have to answer, and the only way i know to calculate the number of links is by adding all the numbers between x-1 and 1 (inclusive). This isnt a problem for small ammounts of devices, but i have to work out the answer for hundreds of devices linked together.

Does anyone know a better way to solve this?

FrenchTart · 12 May 2008 at 15:37

From the sounds of how you work it out..can't you just use the iteration function on a calculator?

Visage · 12 May 2008 at 15:39

Dist said:
Whats the easiest way to calculate the number of links required to make a fully meshed network? I have a multiple choice question i have to answer, and the only way i know to calculate the number of links is by adding all the numbers between x-1 and 1 (inclusive). This isnt a problem for small ammounts of devices, but i have to work out the answer for hundreds of devices linked together.

Does anyone know a better way to solve this?

To add up 1 + 2 + 3 + 4 + ... + x is simply, 1/2 * (x)(x+1).

Dist · 12 May 2008 at 15:48

Visage said:
To add up 1 + 2 + 3 + 4 + ... + x is simply, 1/2 * (x)(x+1).

Unless im typing it wrong, then for 150 devices would have 11325 links, but thats not one of the multiple choice options. :confused:

FrenchTart · 12 May 2008 at 15:50

Didn't you say it was one less add than that? How about if you take away the last number (or count x as 149 rather than 150)

Dist · 12 May 2008 at 15:51

Still doesnt give a result that matches my multiple choice options. The closest answer is 11125, but the question doesnt say find the closest ammount of links required or anything like that.

oddjob62 · 12 May 2008 at 15:53

Visage said:
To add up 1 + 2 + 3 + 4 + ... + x is simply, 1/2 * (x)(x+1).

isn't it 1/2 * (x)(x-1)

Visage · 12 May 2008 at 15:55

oddjob62 said:
isn't it 1/2 * (x)(x-1)

If you're counting up to (x-1) as in the OP then yes. I gave the generic answer for 1..x

Visage · 12 May 2008 at 16:00

Dist said:
Unless im typing it wrong, then for 150 devices would have 11325 links, but thats not one of the multiple choice options.

150 devices fully meshed would take 150 * 149 * 0.5 = 11,175 links.

FrenchTart · 12 May 2008 at 16:01

Dist said:
Still doesnt give a result that matches my multiple choice options. The closest answer is 11125, but the question doesnt say find the closest ammount of links required or anything like that.

You need to do 1/2 * 150 x 150 (you need x-1, the normal way to work them out is x +1...so you get x+0).

That totals 11250

...edit: I suck @ this. Ignore me

Dist · 12 May 2008 at 16:14

Thanks for the help everyone

FrenchTart · 12 May 2008 at 16:17

What was the final outcome?

Dist · 13 May 2008 at 01:07

Well the possible answers for the question of 150 elements in a fully meshed network were: 100, 300, 4500, 11125, 22250, 45000. So 11250 must be the correct answer, even if its not the answer i got myself when i tried working it out, its the only sensible one as the other choices would have way to many or way to few links.

FrenchTart · 13 May 2008 at 11:02

Ah right..so I wasn't so wrong as I thought. Woo