Numerical: cluster size and users in service area

The paper as printed does not give the data for this question, so a typical set of values is assumed below to show the full method:

• Service area = 1765 km², cell radius R = 1.5 km (hexagonal cells)
• Required S/I = 18 dB, path loss exponent n = 4, omnidirectional antennas (i₀ = 6)
• Total available duplex channels S = 660 (e.g. 33 MHz with 2 × 25 kHz per duplex channel)

a) Cell cluster size

Find the smallest valid N (N = i² + ij + j²) with S/I = (√(3N))ⁿ / i₀ ≥ 18 dB:

N = 3 : (√9)⁴ / 6  = 13.5  → 11.30 dB ✗
N = 4 : (√12)⁴ / 6 = 24.0  → 13.80 dB ✗
N = 7 : (√21)⁴ / 6 = 73.5  → 18.66 dB ✓

N = 7.

b) Number of cell clusters in the service area

Area of one cell = 2.598 R² = 2.598 × 1.5² = 5.846 km²
Number of cells  = 1765 / 5.846 = 301.9 ≈ 302 cells
Number of clusters = 302 / 7 = 43.1 ≈ 43 clusters

c) Maximum number of users at any instant

Channels per cell = 660 / 7 = 94.3 ≈ 94
Max users = 94 × 302 = 28 388
(or by clusters: 43 × 660 = 28 380)

About 28 400 users can be served simultaneously (one user per channel).

General method: (1) choose N from the S/I requirement, (2) number of cells = area / (2.598 R²), clusters = cells / N, (3) maximum simultaneous users = (S/N) × number of cells = M × S.