Versions Compared

Key

  • This line was added.
  • This line was removed.
  • Formatting was changed.

...

    • Calculate the pathloss from each UE to all cells and find the smallest pathloss
    • Link the UE randomly to a cell to which the pathloss is within the smallest pathloss plus the HO margin of 3 dB
    • Select K UEs randomly from all the UEs linked to one cell as active UEs. These K active UEs will be scheduled during this snapshot.
    • Note: a full load system is assumed, namely, all available resource blocks (RBs) will be allocated to active UEs. And each UE is scheduled with the same number N of RBs. Thus, the BS transmit power per UE is fixed.
      •  Let Image Removed  denote
        Unit
        body$P_{BS}^{Max}$
         denote the maximum transmit power of BS
      •  Image Removedis
        Unit
        body$M=N\times K$
         is the number of all available RBs in each cellImage Removed
      • Unit
        body$P_{BS}^{UE}$
         is the transmit power from BS to the active UE, andImage Removed
      • Unit
        body$P_{BS}^{UE}=P_{BS}^{Max}\frac{N}{M}$
        .

2. Calculate DL C/I for all active UEs in all cells.

...


    • Loop over all cells from j=1 to Image Removed
      Unit
      body${{N}_{cell}}$
       (the number of cells in the system area e.g. 57 for 19 sites with tri-sector antennas)
    • Loop over all active UEs from k=1 to Kfrom 
      Unit
      body$k=1$
       to 
      Unit
      body$K$
    • For the k-th active UE in the j-th cell (i.e.UE_ 
      Unit
      body$U{{E}_{j,k}}$
      ) its C/I is denoted by Image Removed
      Unit
      body$\frac{C(j,k)}{I(j,k)}$
      ,

3. Determine the throughput for each UE with its C/I according to the link-to-system level mapping.

4. Collect statistics.

...