Example1: Decrease in trunking efficiency for constant N
Let N = 7, each cell has C = 100 channels, and users who make calls with λ = 0.01 per minute with
average holding time 3 minutes. For blocked-calls-cleared and a GOS of 2%, what is the number
of users which can be supported in this cell? Next, using 120 degree sectoring, and otherwise
identical system parameters, what is the number of users which can be supported in this cell?
What percentage reduction in capacity does sectoring with constant N cause?
Solution:
For C = 100 and GOS= 0.02, from Figure 3.6, I read A ≈ 99. Thus with Au = 0.01(3) =
0.03, we could support U = 99/0.03 = 3300. For the sectoring case, C = 33.3 in each sector, and
from Figure 3.6, A = 24. So we could support U = 24/0.03 ≈ 800 per sector, or 2400 total in the
cell. The number of users has reduced by 28%.
Example 2: Reducing N with sector antennas
For the same system, now assume that with 120 degree sectoring, that N can be reduced from 7
to 4. What number of users can be supported?
Solution:
Now, the number of channels in each cell goes up to 100(7/4) = 175. So each sector has C = 58 channels.
With GOS = 2%, from Figure 3.6, A ≈ 48, so U ≈ 1600, for a total of 4800 users per cell. This is a 45%
increase upon the N = 7 non-sectored cell.
Why does i0 reduce? Consider again a mobile at the edge of a cell. We need to determine which
of the first tier BSes contribute significantly to the interference signal. Refer to Figures 3.10, 3.11,
for N = 7, P3.28(b) for N = 3, and to Figure 7 for N = 4.
Example 3: Increase in capacity by sectoring due to change in S/I
Assume we have S = 533 full-duplex channels. Assume blocked-calls cleared with a GOS of 2%, and per
user offered traffic of 0.015 Erlang. Further assume we’re using modulation with minimum required
SIR(dB) of 19.5 dB and we’ve measured for our deployment area that n = 3.3. Find the total number of
users possible per channel assuming (a) omni-directional antennas and (b) 120o sector antennas.
Solution:
Since the given SIR is a minimum, we need N ≥ 15.0. Since there is no 15-cell reuse, we need to increase
to N = 16, which is possible with i = 4 and j = 0. Thus there are 533/16 = 33 channels per cell available.
With a GOS of 2%, from the Erlang B chart, A ≈ 25. With Au = 0.015, this means U = A/Au = 25/0.015 =
1667 users per cell. (b) For 120o antennas, we need to guess at N since i0 is a function of N. For larger N,
i0 = 2 when using 120o antennas. So let’s plug in i0 = 2 and see what N we get:
So N = 9 would work. (Checking, sure enough, i0 = 2 for N = 9.) Thus there are 533/9 = 59.22 channels
per cell or 533/(9 · 3) = 19.7 channels per sector available. With a GOS of 2%, from the Erlang B chart, A
≈ 14 per sector. With Au = 0.015, this means U = A/Au = 14/0.015 = 933 users per sector, or 2800 per
cell. This is a (2800 − 1667)/1667 = 68% improvement over the omni case.
Example 4: Hand Off Margin
Let the speed of a mobile be v= 35 m/s. For n=4, a cell radius of 500 meters( the distance at which the
power is at the threshold), and a 2 second hand off, what amont of Hand Off margin is required?
Solution:
