- 2072 Magh · 12 marks
Numerical: 12-cell system capacity at 2% GOS
Short answer
Cell area = 2.598×1.387² ≈ 5 km², so 1500/5 = 300 cells. Total channels = 28.5 MHz/25 kHz = 1140, per cell = 95. Carried traffic = 300×84 = 25,200 E. Users = 25,200/0.03 = 840,000. Mobiles per channel = 840,000/1140 ≈ 737. Max simultaneous = 95×300 = 28,500.
A city with a coverage area of 1500 sq km is covered with a 12-cell system each with a radius of 1.387 km. If the total spectrum allocated is 28.5 MHz with a full duplex channel bandwidth of 25 MHz. Assume a GOS of 0.02 for a blocked calls cleared system, is specified and the offered traffic per user is 0.03 Erlangs and traffic intensity of each cell is 84 Erlang, compute: (a) the number of cells in the service area (b) the number of channels per cell (c) the maximum carrier traffic (d) the total number of users that can be served for 2% GOS (e) the number of mobiles per unique channel (f) theoretical maximum number of users that could be served at one time by the system.
Given: city area = 1500 km²; 12-cell reuse pattern (N = 12); R = 1.387 km; total spectrum = 28.5 MHz; full duplex channel bandwidth = 25 kHz (the "25 MHz" in the paper must be 25 kHz); GOS = 2 % (Erlang B); Au = 0.03 Erlang per user; traffic per cell = 84 Erlangs.
(a) Number of cells in the service area
Area of one cell = 2.598 R² = 2.598 × 1.387²
= 2.598 × 1.9238 = 4.998 ≈ 5 km²
Number of cells = 1500 / 4.998 = 300.1 ≈ 300 cells
(b) Number of channels per cell
Total channels = 28.5 MHz / 25 kHz = 1140 channels
Channels/cell = 1140 / 12 = 95 channels
(c) Maximum carried traffic
For 95 channels at 2 % GOS, the Erlang B chart gives about 84 Erlangs per cell (as given).
Max carried traffic = 300 × 84 = 25 200 Erlangs
(d) Total number of users for 2 % GOS
Users = Total traffic / Au = 25 200 / 0.03 = 840 000 users
(e) Number of mobiles per unique channel
= 840 000 / 1140 = 736.8 ≈ 737 mobiles per channel
(Each unique channel is reused in 300/12 = 25 cells.)
(f) Theoretical maximum users served at one time
All channels in all cells busy at once:
= 95 × 300 = 28 500 users
This is about 28 500 / 840 000 = 3.4 % of the subscribers.
| Quantity | Value |
|---|---|
| Cells | 300 |
| Channels per cell | 95 |
| Carried traffic | 25 200 E |
| Subscribers (2 % GOS) | 840 000 |
| Mobiles per channel | ≈ 737 |
| Max simultaneous users | 28 500 |
✓ Answer checked by Claude Opus 5.5 (high effort)
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