--- a/doc/manual/source/lte-testing.rst Fri May 13 14:42:45 2011 +0200
+++ b/doc/manual/source/lte-testing.rst Fri May 13 15:50:47 2011 +0200
@@ -49,17 +49,17 @@
---------------------------------
This system test program creates different test cases with a single eNB and several UEs, all having the same Radio Bearer specification. In each test case, the UEs see the same SINR from the eNB; different test cases are implemented by using different distance among UEs and the eNB (i.e., therefore having different SINR values) and different numbers of UEs. The test consists on checking that the obtained throughput performance is equal among users and matches a reference throughput value obtained according to the SINR perceived within a given tolerance.
-The test vector is obtained according to the values of transport block size reported in table 7.1.7.2.1-1 of [TS36.213]_, considering an equal distribution of the physical resource block among the users using Resource Allocation Type 0 as defined in Section 7.1.6.1 of [TS36.213]_. Let :math:`\tau` be the TTI duration, :math:`N` be the number of UEs, :math:`B` the transmission bandwidth configuration in number of RBs, :math:`G` the RBG size, :math:`M` the modulation and coding scheme in use at the given SINR and :math:`S(M, B)` be the transport block size in bytes as defined by 3GPP TS 36.213. We first calculate the number :math:`L` of RBGs allocated to each user as
+The test vector is obtained according to the values of transport block size reported in table 7.1.7.2.1-1 of [TS36.213]_, considering an equal distribution of the physical resource block among the users using Resource Allocation Type 0 as defined in Section 7.1.6.1 of [TS36.213]_. Let :math:`\tau` be the TTI duration, :math:`N` be the number of UEs, :math:`B` the transmission bandwidth configuration in number of RBs, :math:`G` the RBG size, :math:`M` the modulation and coding scheme in use at the given SINR and :math:`S(M, B)` be the transport block size in bits as defined by 3GPP TS 36.213. We first calculate the number :math:`L` of RBGs allocated to each user as
.. math::
L = \left\lfloor \frac{B}{NG} \right\rfloor
-The reference throughput :math:`T` in bps achieved by each UE is then calculated as
+The reference throughput (in bytes per second) :math:`T` in bps achieved by each UE is then calculated as
.. math::
- T = \frac{8 \; S(M, L G)}{\tau}
+ T = \frac{S(M, L G)}{8 \; \tau}
@@ -75,10 +75,12 @@
T = \frac{S(M,B)}{\tau N}
-
+The second category of tests is aimed at verify the fairness behavior in presence of UEs with different SINRs (and therefore different estimated throughput from PFS). The test consists of checking whether the throughput obtained by each UE over the whole simulation matches with the steady-state throughput expected by the PF scheduler according to the theory (see for instance [Kushner2004cpfsa]_).
+Let :math:`Ri` the estimation done by PFS of the throughput of the :math:`i` UE for the next TTI according to the CQIs received and :math:`Ti` the throughput preceived by the :math:`i` UE . The test verifies that the ratio of the obtained throughput value respect to the global one (i.e. the sum of the ones of all users) is equal to the steady-state throughput of the PFS, that is
-The second category of tests is aimed at verify the fairness behavior in presence of UEs with different SINRs (and therefore different estimated throughput from PFS). The test consists of checking whether the throughput obtained by each UE over the whole simulation matches with the steady-state throughput expected by the PF scheduler according to the theory (see for instance [Kushner2004cpfsa]_).
-In detail, each TTI the PFS estimates the throughput of each UE for the next TTI according to the CQIs received; the ratio of this value respect to the global of all users represent the long term trend for the bandwidht allocated by the scheduler. In this specific case, this results in allocating proportionally the maximum bandwidht achievable (i.e., all PRB allocated to a single users) among the users in a time division fashion.
+.. math::
+
+ K = \frac{Ri}{\sum_{k=1}^N Ri} = \frac{Ti}{\sum_{k=1}^N Ti}