The Effect of BSC Reduction on Handover Failures

 

 

1. Introduction

 

Consider the following rectangular grid of square cells, each of which represents a BSC. A handover can occur within a particular BSC (intra-BSC handover) or can cross BSC’s (inter-BSC handover). Also, a handover can fail.

 

           

 

 

We investigate the effect of reducing the number of BSC’s. Consider the following new grid, each cell of which replaces six cells of the old grid. We present a quantitative analysis of the effect of this reduction on the number of handover failures. (Although we consider a 4 x 6 grid, our discussion can be easily generalized to cover grids of different sizes. In section 5 below, we present a generalization.)

 

           

 

 

The probability of inter-BSC handover failure is much higher than the probability of intra-BSC handover failure. Intuitively, a reduction in the number of BSC’s should decrease the number of inter-BSC handovers, hence decrease the overall number of handover failures.

 

2. Quantifying the Reduction in Handover Failures

 

For convenience, call the first grid the “old system” or OS, and the second grid the “new system” or NS. Let

 

x = number of intra-BSC handover failures

y = number of intra-BSC handover attempts

r = number of inter-BSC handover failures

s = number of inter-BSC handover attempts

F = total number of failures in OS

G = total number of failures in NS

f = intra-BSC failure rate in OS

g = inter-BSC failure rate in OS

T = total number of handover attempts (the same for both systems).

 

Clearly, the intra-BSC and inter-BSC failure rates in OS are equal to their corresponding rates in NS. Hence,

 

            x / y = x’ / y’ = f                                                           (1)

 

            r / s = r’ / s’ = g                                                            (2)

 

(Primed quantities denote the corresponding values in the NS.)

 

Also,

 

            y + s = y’ + s’ = T, i.e.,

y – y’ = s’ - s                                                               (3)

 

Now,

 

            E = F – G = (x + r) – (x’ + r’)

                =  (fy + gs) – (fy’ + gs’)                    (by equations (1) and (2))

                =  f(y – y’) – g(s’ – s)

                =  f(y – y’) – g(y – y’)                       (by equation (3))

                =  (f – g)(y – y’)

 

gives the reduction in the number of failures in migrating from OS to NS. We wish to compute E.

 

Let us assume that y’ = C(T) y where C(T) depends on the total number T of handover attempts. (In the next section, we estimate C(T).) Then the formula for E becomes

 

            E = (g – f)(C(T) – 1)y.                                      (4)

 

3. Estimation of C(T)

 

We consider handover attempts as uniformly filling the BSC grids, in the same way as molecules of a gas evenly fill a container in the steady state. For simplicity, we distribute the T handover attempts, represented by points, in a regular lattice pattern, e.g.:

 

           

 

In particular, each square cell in OS will be filled by such a pattern. Since this cannot be done in general, we consider the following approximation for sufficiently large T. Each of the 24 cells in OS will be occupied in this pattern by M2 points, where

 

                                                                     (5)

 

and [x] denotes the result of rounding x to the nearest integer. (Note. The uniform distribution assumption is not completely accurate since handover attempts are actually distributed radially from EBTS’s. However, it provides a reasonable approximation when T is large.)

 

Consider the points adjacent to BSC boundaries; these are the inter-BSC handover “candidates”. Let us assume that ¼ of these represent inter-BSC handovers. (If such a point were to make a random motion in either N, S, E, or W directions, then in most cases, only one direction, out of four, will point to a boundary.) How many of these adjacent points are there in OS?

 

For the cells numbered 8, 9, 10, 11, 14, 15, 16, 17, there are about 4M such points per cell; hence, there are 8 x 4M = 32M such points corresponding to these cells.

 

For the cells numbered 2, 3, 4, 5, 7, 12, 13, 18, 20, 21, 22, 23, there are about 12 x 3M = 36M such points. (We do not consider the outermost rectangle as an inter-BSC boundary.)

 

For the corner cells numbered 1, 6, 19, 24, there are about 4 x 2M = 8M such points.

 

Therefore, there are (32 + 36 + 8)M = 76M such points in OS, one-fourth of which, i.e., 19M, represent inter-BSC handovers.

 

Similary, in NS, there are 20M = 4 x 5M such inter-BSC handover candidates, one-fourth of which, i.e., 5M, represent inter-BSC handovers. (A BSC in NS is a group of 3 x 2 BSC’s in OS; there are 3M + 2M = 5M inter-BSC handover candidates in such a new BSC.)

 

Thus, there are about T – 19M intra-BSC handovers in OS, and T – 5M intra-BSC handovers in NS. Then

 

                                

           

approximately. 

 

                                                                                    (6)

 

4. Graph of C(T)

 

Note that the equation for C(T) is non-linear; the following graph illustrates the dependence of C(T) on T.

 

           

 

It is easy to see from equation (6) that

 

                                                      (7)

 

that is, the line C = 1 is a horizontal asymptote of the graph.

 

4. Sample Calculation of E and C(T)

 

The following numbers are based on actual iDEN statistics.

 

f = 0.09

g = 0.9

y = 8967009

T = 13310841.

 

Substituting these in equation (6), C(13310841) = 1.00078. Using this in equation (4),

E = 5697, approximately. NS can be expected to have about 5697 less handover failures than OS.

 

(We know that OS has at least 807031 handover failures since, using equation (1), x = number of intra-BSC handover failures in OS = fy = 807031.)

 

5. A Generalization

 

Suppose we have the grid pattern of NS as above, and that each BSC of NS is an m x n (m rows, n columns; m, n > 1) sub-grid of BSC’s in OS. Then OS has 4mn BSC’s. The number of inter-BSC handover candidates in OS is

 

            [4(m - 1)(n – 1) + 6(m-1) + 6(n – 1) + 8]M = 2(mn + 2m + 2n)M

 

and the number of inter-BSC handovers is

 

            ½ (mn + 2m + 2n)M.

 

For NS, the number of inter-BSC handover candidates is

 

            4(m + n)M

 

and the number of inter-BSC handovers is

 

            (m + n)M

 

Then the equation for C(T) becomes

 

           

                                                                                   

                                                                                    (8)

 

The reduction E in handover failures is calculated from equation (4).

 

 

J. L. Pe

System Engineering Tools and Statistics