- 2081 Baisakh · 2+2+2 marks
Numerical: trunking capacity loss with sectoring
Short answer
Traffic per user = 0.0333 Erlang. Omnidirectional: 57 channels → 44.2 Erlangs, 1326 users. 60° sectored: 6×9 channels → 22.70 Erlangs, 681 users. Capacity loss =21.5 Erlangs per cell (48.% reduction), about 645 fewer users due to trunking inefficiency.
For a cluster N = 7 system with a blocking probability of 1% and an average call length of two minutes, find the traffic capacity loss for a blocked calls cleared system due to trunking for 57 channels when going from omnidirectional antennas to 60° sectored antennas. The average request rate per user is 1 call per hour.
| GOS/ Channel | 0.5% | 1% | 2% |
|---|---|---|---|
| 1 | .005 | .0101 | .0204 |
| 9 | 3.333 | 3.783 | 4.345 |
| 19 | 10.33 | 11.23 | 12.33 |
| 54 | 39.47 | 41.51 | 44.00 |
| 57 | 42.1 | 44.2 | 46.8 |
Given: N = 7, B = 1 %, H = 2 min, λ = 1 call/hour, 57 channels per cell, blocked calls cleared (Erlang B).
Traffic per user:
Au = λH = 1 × 2/60 = 0.0333 Erlang
Omnidirectional antennas
All 57 channels form one trunk group. From the table (57 channels, 1 %):
A = 44.2 Erlangs per cell
Users per cell = 44.2 / 0.0333 = 1326 users
60° sectoring
Each cell has 6 sectors, channels per sector = 57/6 = 9.5 ≈ 9 channels. From the table (9 channels, 1 %):
A per sector = 3.783 Erlangs
A per cell = 6 × 3.783 = 22.70 Erlangs
Users per cell = 22.70 / 0.0333 ≈ 681 users
Trunking capacity loss
Loss = 44.2 − 22.70 = 21.50 Erlangs per cell
= 21.50 / 44.2 × 100 = 48.6 %
| Antenna | Channels per trunk | Traffic/cell | Users/cell |
|---|---|---|---|
| Omni | 57 | 44.2 E | 1326 |
| 60° sectors | 9 (×6) | 22.70 E | 681 |
Answer: Trunking efficiency loss ≈ 21.5 Erlangs per cell (≈ 48.6 %), about 645 fewer users per cell. Sectoring breaks a large trunk into small ones, which are less efficient; the gain from sectoring must come from the smaller cluster size it permits.
✓ Answer checked by Claude Opus 5.5 (high effort)
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