Achieving Maximum Energy-Efficiency in Multi-Relay OFDMA Cellular Networks: A Fractional Programming Approach
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Citations
Fractional Programming for Communication Systems—Part I: Power Control and Beamforming
Nonlinear Fractional Programming
A Survey of Energy-Efficient Techniques for 5G Networks and Challenges Ahead
Energy Efficiency in Wireless Networks via Fractional Programming Theory
A Survey of Multi-Objective Optimization in Wireless Sensor Networks: Metrics, Algorithms, and Open Problems
References
Cooperative diversity in wireless networks: Efficient protocols and outage behavior
Wireless communications
Multiuser OFDM with adaptive subcarrier, bit, and power allocation
Related Papers (5)
Frequently Asked Questions (17)
Q2. What is the concave form of the fractional program?
The concave form of the fractional program (P) is formed by denoting the OF value as q so that a subtractive form of the OF may be written as F (q) = RT (P,S) − qPT (P,S), which is concave.
Q3. What is the SNR of the AF relaying protocol?
Using the direct transmission protocol, the receiver’s SNR at UE k on subcarrier n may be expressed as ΓD,nk (P), whereas when using the AF relaying protocol, the receiver’s SNR at UE k on subcarrier n may be expressed as [6]
Q4. What is the optimal solution to the problem?
Since the problem is now in a standard concave form, the Karush–Kuhn–Tucker (KKT) conditions [32], which are firstorder necessary and sufficient conditions for optimality, may be used in order to find the optimal solution.
Q5. What is the optimal subcarrier for a given EE?
since each subcarrier may only be used for transmission to a single user, each subcarrier n is allocated to the specific user k having the highest value of max (Ank , D n k ) in order to achieve the highest increase in L (P,S, λ).
Q6. What is the algorithmic complexity of the dual decomposition method?
The algorithmic complexity of this method is dominated by the comparison operations given by (35) and (36), which leads to a total complexity of O (Idual × 2NK) when NK is large, where Idual is the total number of inner iterations required for reaching convergence in Dinkelbach’s method.8 λ(i+ 1) = [ λ(i)− αλ(i) ( Pmax −K∑ k=1 N∑ n=1 P̃D,n∗0,k + P̃ A,n∗ 0,M(k) + P̃ A,n∗ M(k),k)]+ (38)
Q7. What is the optimum SNR of a single subcarrier?
(4)The subcarrier indicator variable sX,nk ∈ {0, 1} is now introduced, which denotes the allocation of subcarrier n for transmission to user k using protocol X for sX,nk = 1, and sX,nk = 0 otherwise.
Q8. What is the simplest way to find increasing q values?
P,S {RT (P,S)− q∗PT (P,S)} = 0. (19)Explicitly, the solution of F (q∗) is equivalent to the solution of the fractional problem (P). Dinkelbach [23] proposed an iterative method to find increasing q values, which are feasible, by solving the parameterized problem of maxP,S {F (qi−1)} at each iteration.
Q9. What is the constraint for a subcarrier?
The constraint (11) guarantees that each subcarrier is only allocated to at most one user, thus intra-cell interference is avoided.
Q10. What is the way to show that the algorithm produces an increasing sequence of q values?
it can be shown that the method produces an increasing sequence of q values, which converges to the optimal value at a superlinear convergence rate.
Q11. What is the optimal power allocation for a given channel?
Observe that the optimal power allocations given by (25) and (29) are indeed customized water-filling solutions [33], where the effective channel gains are given by αD,nk and α A,n k , respectively, and where the water levels are determined both by the cost of allocating power, λ, as well as the current cost of power to the EE given by qi−1.2)
Q12. What is the eigenvalue of the Hessian?
The Hessian has the eigenvalues e1 = 0 ande2 = − 2 ( Gn0,M(k)G n M(k),k )2 ( P̃A,n0,M(k) + P̃ A,n M(k),k ) ∆γN0W ( P̃A,n0,M(k)G n 0,M(k) + P̃ A,n M(k),kG n M(k),k)3 , (16)which are non-positive, indicating that the Hessian is negativesemidefinite.
Q13. What is the effect of the pathloss on the EE?
the pathloss is a major factor in determining the receiver’s signal-tonoise ratios (SNRs) at the UEs, and thus has a substantial effect on the EE.
Q14. How many times does the AF link to UE k need to be asynchronous?
The SE of an AF link to UE k on subcarrier n is then given byRA,nk (P) = 12 log2( 1 + ΓA,nk ) [bits/s/Hz], (3)where the factor of 12 accounts for the fact that two time slots are required for the two-hop AF transmission.
Q15. What is the total transmit power assigned for AF transmission to user k?
The total transmit power assigned for AF transmission to user k over subcarrier n is now denoted by P̃A,nk = P̃ A,n 0,M(k) + P̃ A,n M(k),k.
Q16. What is the optimal power for a given subcarrier?
The process of computing the optimal power as well as subcarrier allocations and subsequently updating λ is repeated until convergence is attained, indicating that the dual optimal point has been reached.
Q17. What is the optimal allocation for subcarrier n?
The optimal allocation for subcarrier n is as follows7sD,n∗k ={ 1, if Dnk = maxj [ max ( Anj , D n j )] and Dnk ≥ 0,0, otherwise, (35)andsA,n∗k ={ 1, if Ank = maxj [ max ( Anj , D n j )] and Ank ≥ 0,0, otherwise.