In Multiple Access Protocol for cellular networks, there is this Frequency Reuse Patterns. And I am given a formula to one of the patterns that states: \$\text{Reuse factor }\mathbf{K} = i^2+j^2+ij\$

I know that the reuse factor \$K\$ meant the number of cells that are reusing the frequencies. But what do the \$i\$ and \$j\$ represent? And how can I use this formula?


"K" in your formula is the number of cell sites that can form a valid "cluster: which meets certain rules - more on that below.

i and j are simply terms that make a formula that returns values of K that are valid in practice.

They may have "real" meaning but there is no need to know it. You can deal with the "abstraction" one level above that.

Here is pages 11-12 from here that shows how cells may be packed in a multi cell frequency resue system.

Here is how cells might be packed in a system with frequency reuse. Not that a hexagon is NOT the only shape that allows syymmetric repacking - it just works better than most in building shapes that fit together well.

enter image description here

Real meaning:

Here is the actual answer to your question BUT to make use of it you are probably going to have to pore over the better drawings in the references below.

  • Cell frequencies in a cellular system are reused in other cells a certain distance away.

  • On an open level surface true cell radiation amplitude patterns are circular but when packing sites together diagrammatically the assumption is made that the cell sites are hexagonal see diagrams.

  • To pack in an infinitely even manner cells are packed in clusters of cells with a certain number of cells in the cluster. These clusters then pack into a larger overall system.

  • When hexagons are packed and there are symmetrically distributed sites around the current one which use the same frequency, they can be reached by travelling along two straight lines through the centres of hexagons. The geometry of hexagons is such that only certain combinations of hexagons can be packed together to achieve symmetric reuse patters (see diagrams).

  • i & j are simply the lengths in number of heaxagons traversed to reach the nearest cells using the same frequency. The formula relating the length of these two "arms" and the number of cells in a "cluster" thus formed is not immediately intuitive but is obvious enough from working through the formulae in the diagram below.

enter image description here

Empirical approach:

If desired you can ignore the above reason and treat it as a set of empirical rules.

  • K is the number of cells in a "cluster" that meets certain packing rules.
    eg Playing with coloured hexagonal blocks will demonstrate this.

  • Values of K which match what can be achieved in reality are given by your formula

  • i and j are +ve integers starting at 0 (although 1,0 or 0,1 is the smallest sensible combination. As can be seen, i and j are interchangeable in the formula so 1,2 is the same as 2,1 Using that formula the number of cells in a valid cluster are

K = i^2 + j^2 + i x j

i j K
0 1 1 ........ = 1^1 + 0 x 1 + 0^2
1 1 3
0 2 4
1 2 7 ......... = 1^2 + 2 x 1 + 2^2 = 1 + 2 + 4 = 7
0 3 9
2 2 12

Making it clearer:

Starting with

  • "i & j are the arm lengths between two cells of the same frequency in a system of reusable frequency hexagons"

there are many references on the web which deal with this, many of which can be found by eg gargoyling - Reuse factor K=i2+j2+ij
But some are overly simplistic and some are horrendously complex.

One which may be 'just right' is found here as a slide show. Not enough words but enough to get a feel.
Pages 8-12 lead up to this, 11 & 12 directly address the question and 14-15 discuss a reuse proof which has pictures but which would benefit from more words.

This paper - also a slideshow again has good pictures but too few words but the real world meaning of i + j can be seen in the diagrams (about 8 pages starting at about page 12 - pages not numbered)


  • \$\begingroup\$ Thank you so much for the informative answer. It is really very helpful. :) \$\endgroup\$ – xenon Nov 2 '11 at 10:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.