--- title: "Numerical: trunking capacity loss with sectoring" subject: "Wireless Communication" chapter: 2 key: "semester_7/wireless_communication:trunking-loss-sectoring" source: https://www.bctengineer.com/q/semester_7/wireless_communication/trunking-loss-sectoring site: BCT Engineering --- # Numerical: trunking capacity loss with sectoring Wireless Communication · Chapter 2: Cellular Mobile Communication Concepts - Asked in: 2081 Baisakh, 2+2+2 marks - Video: [5.6 Trunking Examples](https://www.youtube.com/watch?v=H5295gOK3Mg) — Ali Muqaibel - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/wireless_communication/trunking-loss-sectoring - Whole chapter: https://www.bctengineer.com/md/semester_7/wireless_communication/chapter-2 ## Question 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 | ## 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. ## Answer **Given:** N = 7, B = 1 %, H = 2 min, λ = 1 call/hour, 57 channels per cell, blocked calls cleared (Erlang B). **Traffic per user:** ```text Au = λH = 1 × 2/60 = 0.0333 Erlang ``` #### Omnidirectional antennas All 57 channels form one trunk group. From the table (57 channels, 1 %): ```text 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 %): ```text 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 ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With 1 call/hour and 2 minute calls, the traffic per user is: A. 0.5 Erlang B. 0.0333 Erlang C. 2 Erlang D. 0.333 Erlang Answer: B. Au = lambda x H = 1 x 2/60 = 0.0333 E. (Key point: Traffic per user) 2. From the table, an omnidirectional cell with 57 channels at 1% GOS carries: A. 3.783 Erlangs B. 22.7 Erlangs C. 46.8 Erlangs D. 44.2 Erlangs (about 1326 users) Answer: D. Read 57 channels at 1% from the table: 44.2 E; 44.2/0.0333 = 1326 users. (Key point: Omni capacity) 3. With 60 degree sectoring, how many channels does each sector get and what traffic per sector is carried? A. 9 channels, 3.783 E B. 57 channels, 44.2 E C. 6 channels, 1 E D. 19 channels, 11.23 E Answer: A. 57/6 = 9.5, taken as 9 channels; the table gives 3.783 E at 1%. (Key point: Channels per sector) 4. Why does sectoring reduce traffic capacity (about 48.6% loss here)? A. Each sector has fewer users who call less B. Sectoring raises the blocking probability target C. One large trunk group is broken into six small ones, which are less efficient D. Sectoring increases the cluster size N Answer: C. Small trunk groups have lower trunking efficiency; the gain of sectoring comes from the smaller cluster size it permits. 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