- This leads to the following calculations:
- Frequencies per cell = 24 / 12 = 2
- Traffic channels per cell = 2 x 8 - 2 (control channels) = 14 TCH
- Traffic per cell = 14 TCH with a 2% GOS implies 8.2 Erlangs per cell
- The number of subscribers per cell = 8.2E / 25mE = 328 subscribers per cell
- If there are 10,000 subscribers then the number of cells needed is 10,000 / 328 = 30 cells.
- Therefore, the number of three sector sites needed is 30 /3=10
Part of Erlang’s B Table Example 1 taking into account that the generated traffic per subscriber is 25 mE and the GoS is 0.02 Solution: Generated traffic in the cell: 500 * 25mE = 12.5 Er From Erlange B table n= 20 TCH Number of TRX = 3. Example 2 A cell has 4 TRX ,how many subscribers can be served taking into account the previous assumptions.
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