5.5
Allocation of Resource Blocks by the BS
5.5.1
Open Questions and Preliminaries
In this section, the open questions and preliminaries are given for the allocation of resource blocks solved by the BS. The BS transmits in the first subframe of a frame. For the BS t = 0, the questions must be answered which resource blocks are allocated to the link (0, r), where r ∈ {R0∪ RRS}. The BS is aware of:
• the subframe size Sn for each subframe n, where n = 1, 2, ...NSF.
• SINR value for each resource block k and for each link (0, r), where k = 1, 2, ...K0 and r ∈ {R0∪ RRS}.
• the SINR threshold γ,r for each number of bits, where ∈ E, and for each receiving station r ∈ {R0∪ RRS} in order to guarantee the bit error probability BEP0,r for each link (0, r).
• the minimum data rate Rmin,t,r for each link (t, r) in the cell if the objective is to maximize the sum rate as claimed in problem (P2).
• the initial data rate values ˆRt,r for each link (t, r), where t ∈ RRS and r ∈ Rt.
Using the SINR values and the SINR thresholds in (4.10), the BS determines like an RS the bits which can be transmitted per resource block if a resource block is allocated. These bits are denoted by t,kS1. Then, the allocation of resource blocks to the links is performed. Since the values of the subframe sizes and of the initial data rates are known, the BS can allocate the resource blocks with respect to the maximum number of bits which can be forwarded by an RS during the frame.
5.5.2
Non-Adaptive Allocation of Resource Blocks by the BS
In this section, the non-adaptive algorithm applied by the BS is presented. Actually, the weighted round robin algorithm as introduced for the BS in Section 4.4.2 is applied, but it is extended. Based on the subframe sizes and initial data rate values, the BS detects if resource blocks are allocated to an RS although it cannot forward received bits.
Assume that the resource blocks are allocated from k = 1 to k = K0 sequentially. If the resource block k0 is allocated to the link (0, r), where r ∈ R
RS, the bits carried by the resource blocks already allocated are summed up and represented by
R0,r = k0 X
k=1
ur,kr,kS0. (5.13)
Assume that the RS r forwads in subframe n, where the subframe size is Sn. The sum of the initial data rate values of the links served by the RS r is given by
ˆ Rr= X r0∈Rr ˆ Rr,r0. (5.14)
If so many resource blocks are allocated to RS r that the RS cannot forward all received bits, i.e.,
R0,r > ˆRr· Sn, (5.15)
the RS t is not considered in the weighted round robin any longer.
5.5.3
Adaptive Allocation of Resource Blocks by the BS Aim-
ing at Maximizing the Minimum User Rate
In this section, the allocation of resource blocks is presented if the objective is the maximization of the minimum user rate according to (P1). The subproblem addressed in this section is rather similar to subproblem of the initial allocation of the resource blocks, which is solved by the RS and described by (5.10). As in the previous Section 5.5.2, differences occur since the BS is aware of the data rate values which can be forwarded by the RSs. This knowledge is included in the formulation of the subproblem derived as follows.
The objective is the maximization of the minimum data rate. This is equivalent to the maximization of the data rate on the link (0, r0), where r0 ∈ R
0 and r0 is chosen arbitrarily, if two constraints are fulfilled. At first, the data rate of each BS-to-UE link (0, r), where r ∈ R0 and r 6= r0, is larger or equal to the maximized data rate. Secondly, the data rate on the BS-to-RS link (0, r), where r ∈ RRS, shall be a multiple of the maximized one according to the inverse of the weight wr defined in (4.24) and representing the number of UEs served by the RS. However, the data rate which is forwarded by the RS r in subframe n shall be upper bounded by ˆRr·Sn, which represents the bits which can be forwaded by the RS r in a frame, where ˆRr is given by (5.14). Hence, the data rate R0,r shall be larger than or equal to the minimum of the weighted maximized data rate and the upper bound. This is expressed in
min R0,r0 wr
; ˆRr
5.5 Allocation of Resource Blocks by the BS 101
Additionally, it must be fullfilled that at most one resource block of the same time- frequency is allocated to a link and that each resource block is allocated exclusively. Taking into account that the BS is transmitting in subframe n = 1, the subproblem is described by max ur,k K0 X k=1 ur0,kr0,kS1 (5.17a) subject to: K0 X k=1 ur,kr,kS1 ≥ R0,r0, r ∈ R0, (5.17b) min R0,r0 wr ; ˆRr ≤ R0,r, r ∈ RRS, (5.17c) f G0 X k=(f −1)G0+1 ur,k ≤ 1, 1 ≤ f ≤ F, r ∈ {R0∪ RRS}, (5.17d) X r∈{R0∪RRS} ur,k = 1, 1 < k ≤ K0, (5.17e) ur,k ∈ {0, 1}, 1 < k ≤ Kt, r ∈ {R0∪ RRS}, (5.17f) where the data rate R0,r is a function of ur,k as defined in (2.22). The integer program is rather similar to (5.10) describing the subproblem of the initial allocation of resource blocks by the RS if the objective is the maximization of the minimum user rate. The different objective and constraint (5.17c) occur only since the number of bits, which can be forwarded by an RS, is predefined. Thus, the greedy algorithm depicted in Fig. 4.3 and proposed to solve (5.10) is modified slightly in order to solve the integer program. Following modifications are made:
• The set Rtis replaced by the union {R0∪ RRS} in order to represent the stations receiving from the BS.
• The link (0, r), where r ∈ RRS, is not considered any longer in the third step if R0,r > ˆRr in order to prevent that more bits are transmitted to an RS than it can forward.
5.5.4
Adaptive Allocation of Resource Blocks by the BS Aim-
ing at Maximizing the Sum Rate
In this section, the allocation of resource blocks is considered with respect to the objective of maximizing the sum rate given by (P2). It is assumed that each BS-to-UE
link and each BS-to-RS link achieve its minimum data rate. Otherwise, a UE must be dropped or the minimum data rate must be reduced according to Section 4.5.3. The subproblem is described by modifying integer program (5.11) representing the initial allocation of resource blocks by the RS if the objective is the maximization of the sum rate. Since the data-rate on the RS-to-UE links is known, the constraint is introduced that so many resource blocks are not allocated to an RS that it receives more bits than it can forward. Then, integer program (5.11) is modified to
max ur,k X r∈{R0∪RRS} K0 X k=1 ur,kr,kS0 (5.18a) subject to: f G0 X k=(f −1)G0+1 ur,k ≤ 1, 1 ≤ f ≤ F, r ∈ {R0∪ RRS}, (5.18b) X r∈{R0∪RRS} ur,k = 1, 1 ≤ k ≤ K0, (5.18c) K0 X k=1 ur,kr,kS0 ≥ Rmin,0,r, r ∈ {R0∪ RRS}, (5.18d) K0 X k=1 ur,kr,kS0 ≤ ˆRr, r ∈ RRS, (5.18e) ur,k ∈ {0, 1}, 1 < k ≤ K0, r ∈ {R0∪ RRS}. (5.18f) The objective function representing the maximization of the bits transmitted by the BS includes the subframe size. Constrains (5.18b) and (5.18c) represent that at most one resource block of the same time-frequency unit is allocated to a link and that each resource block is allocated exclusively. Constraint (5.18d) claims that each link achieves its minimum data rate. Constraint (5.18e) ensures that an RS does not receive more than it can forward. Since the integer program is rather similar to (5.11), the subproblem is solved by nearly the same greedy algorithm as depicted in Fig. 5.2. The only difference is that the BS-to-RS link is only considered in the last step of the algorithm as long as the RS keeps able to forward all received bits.