---
title: "Prove Q = √3N; optimal N for 15 dB S/I (n = 4)"
subject: "Wireless Communication"
chapter: 2
key: "semester_7/wireless_communication:reuse-ratio-optimal-n-n4"
source: https://www.bctengineer.com/q/semester_7/wireless_communication/reuse-ratio-optimal-n-n4
site: BCT Engineering
---

# Prove Q = √3N; optimal N for 15 dB S/I (n = 4)

Wireless Communication · Chapter 2: Cellular Mobile Communication Concepts

- Asked in: 2071 Bhadra, 4+4 marks
- Video: [Numerical on Frequency Reuse - Fundamentals of Mobile Communication - Mobile Communication System](https://www.youtube.com/watch?v=-fBJpiGrks4) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/wireless_communication/reuse-ratio-optimal-n-n4
- Whole chapter: https://www.bctengineer.com/md/semester_7/wireless_communication/chapter-2

## Question

Prove that for a hexagonal geometry the co-channel reuse ratio is given by Q = √(3N); where N = i² + j² + ij. A cellular service provider decides to use a digital TDMA scheme which can tolerate a Signal-to-Interference Ratio of 15 dB in the worst case. Find the optimal value of N for a) Omni directional antennas b) 120° Sectoring c) 60° Sectoring. [Use path loss exponent of 4 and consider trunking efficiency]

## Short answer

D² = (i√3R)² + (j√3R)² – 2(i√3R)(j√3R)cos120° = 3R²(i²+ij+j²)=3R²N, so Q = √(3N). For S/I = 15 dB (31.62) and n=4: omni (i₀=6) gives N=7; 120° sectoring (i₀=2) gives N=3; 60° (i₀=1) also N=3. Considering trunking efficiency, 120° with N=3 is optimal.

## Answer

#### Proof that Q = √(3N)
In a hexagonal layout, to reach the nearest co-channel cell you move *i* cells along one chain of hexagons, turn 60°, and move *j* cells more (this is how N = i² + ij + j² is defined).

- Distance between centres of two adjacent hexagons = √3·R (R = centre-to-vertex radius).
- Using axes 60° apart and the cosine rule on the triangle with sides i·√3R and j·√3R and the angle between them 120°:

```text
D² = (i√3R)² + (j√3R)² − 2(i√3R)(j√3R)cos120°
   = 3R²(i² + j² + ij)
   = 3R²·N
D  = R√(3N)
Q  = D/R = √(3N)
```

```text
start --(i cells)--> turn 60° --(j cells)--> co-ch cell
        \___________________ D ______________________/
```

#### Optimal N for S/I = 15 dB, n = 4
Required S/I = 15 dB = 10^1.5 = 31.62. Using S/I = Qⁿ / i₀ = (√(3N))⁴ / i₀ = (3N)²/i₀. Valid N: 1, 3, 4, 7, 9, 12, …

**a) Omnidirectional antennas (i₀ = 6)**

```text
(3N)²/6 ≥ 31.62  →  3N ≥ 13.77  →  N ≥ 4.59
N = 4: Q = 3.464, S/I = 144/6 = 24
       = 13.80 dB < 15 (fails)
N = 7: Q = 4.583, S/I = 441/6 = 73.5
       = 18.66 dB > 15 (ok)
```
**N = 7**

**b) 120° sectoring (i₀ = 2)**

```text
(3N)²/2 ≥ 31.62  →  3N ≥ 7.95  →  N ≥ 2.65
Try N = 3: Q = 3, S/I = 81/2 = 40.5 = 16.07 dB > 15 (ok)
```
(The exact worst-case geometry, S/I = 1/[Q⁻⁴ + (Q+0.7)⁻⁴], gives 17.52 dB for N = 3, also acceptable.) **N = 3**

**c) 60° sectoring (i₀ = 1)**

```text
(3N)² ≥ 31.62  →  3N ≥ 5.62  →  N ≥ 1.87
Try N = 3: S/I = 81 = 19.08 dB > 15 (ok)
(N = 1 gives 9 = 9.54 dB, fails)
```
**N = 3**

#### Effect of trunking efficiency
Sectoring splits each cell's channels among the sectors, so each sector is a smaller trunk and trunking efficiency drops. 60° sectoring gives the same N = 3 as 120° sectoring, but splits channels into 6 groups instead of 3, losing more trunking efficiency for no gain in cluster size.

| Antenna | i₀ | Optimal N | S/I (dB) |
|---|---|---|---|
| Omni | 6 | 7 | 18.66 |
| 120° sector | 2 | 3 | 16.07 |
| 60° sector | 1 | 3 | 19.08 |

**Answer: N = 7 (omni), N = 3 (120°), N = 3 (60°). Considering trunking efficiency, 120° sectoring with N = 3 is the best choice.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In the proof of Q = sqrt(3N), which rule is applied to the triangle with sides i*sqrt(3)R and j*sqrt(3)R?
   A. Cosine rule with an angle of 120 degrees
   B. Sine rule with 60 degrees
   C. Pythagoras theorem with 90 degrees
   D. Cosine rule with an angle of 30 degrees
   Answer: A. The angle between the two moves is 120 degrees, giving D^2 = 3R^2(i^2+j^2+ij) = 3R^2 N. (Key point: Proof of Q)

2. What is 15 dB as a linear S/I ratio?
   A. 15
   B. 3.16
   C. 31.6
   D. 150
   Answer: C. 10^(15/10) = 10^1.5 = 31.62. (Key point: Required S/I)

3. For omnidirectional antennas (i0 = 6, n = 4), which cluster size is the smallest valid one meeting 15 dB?
   A. N = 7
   B. N = 4
   C. N = 3
   D. N = 9
   Answer: A. N = 4 gives S/I = 144/6 = 24 (13.8 dB), which fails; N = 7 gives 73.5 (18.66 dB), which passes. (Key point: Omni antenna N)

4. Why does 120 degree sectoring allow N = 3?
   A. It increases transmit power
   B. It increases the cell radius
   C. It lowers the number of first-tier interferers from 6 to 2, raising S/I
   D. It increases the path loss exponent
   Answer: C. With i0 = 2, S/I = 81/2 = 40.5 (16.07 dB) for N = 3, which exceeds 15 dB. (Key point: Sectoring effect)

5. Why is 120 degree sectoring preferred over 60 degree sectoring when trunking efficiency is considered?
   A. 60 degree sectoring gives the same N = 3 but splits channels into 6 groups, losing more trunking efficiency
   B. 60 degree sectoring needs a larger N
   C. 120 degree sectoring has a higher S/I
   D. 60 degree sectoring has more interferers
   Answer: A. Both give N = 3, but more sectors mean smaller trunks and lower trunking efficiency for no gain. (Key point: Trunking efficiency)

## Related in Wireless Communication

- [GoS and optimal N for 15 dB S/I (n = 3)](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Agos-optimal-cluster-size-n3) — Wireless Communication, chapter 2
- [CCI vs ACI and co-channel reuse ratio proof](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Acci-aci-prove-reuse-ratio) — Wireless Communication, chapter 2
- [Derivation of co-channel reuse ratio](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Aderive-cochannel-reuse-ratio) — Wireless Communication, chapter 2
- [Frequency reuse and co-channel reuse ratio](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Afrequency-reuse-cochannel-reuse-ratio) — Wireless Communication, chapter 2
- [CCI/ACI, reuse ratio proof and channels per cell](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Acci-aci-reuse-duplex-channels) — Wireless Communication, chapter 2
- [Numerical: co-channel distance for 7-cell reuse](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Acochannel-distance-seven-cell) — Wireless Communication, chapter 2