Sum-Power Iterative Watefilling Algorithm Daniel P. Palomar Hong Kong University of Science and Technolgy (HKUST) ELEC547 - Convex Optimization Fall 2009-10, HKUST, Hong Kong November 11, 2009
Outline of Lecture Basic MIMO channel and waterfilling solution MIMO MAC and the IWFA MIMO BC Sum-power IWFA via primal decomposition Sum-power IWFA via dual decomposition Summary Daniel P. Palomar 1
Basic MIMO Channel Consider a Multiple-Input Multiple-Output (MIMO) channel: y = Hx + n H is the channel matrix x is the transmitted signal with covariance matrix Q = E [ xx ] n is the Gaussian noise with covariance matrix R n = E [ nn ] y is the received signal. The mutual information of such a channel is maximized with a Gaussian signaling and it is then given by log det ( R n + HQH ) log det(r n ). Average power constraint: Tr(Q) P. Daniel P. Palomar 2
Problem Formulation The maximization of the mutual information over the transmit covariance matrix is maximize Q subject to log det ( R n + HQH ) Tr(Q) P Q 0. This problem is convex: the logdet function is concave, the trace constraint is just a linear constraint, and the positive semidefiniteness constraint is an LMI. Hence, we can use a general-purpose method such as an interiorpoint method to solve it in polynomial time. Daniel P. Palomar 3
Closed-Form Solution: MIMO Waterfilling However, this problem admits a closed-form solution as can be derived from the KKT conditions. The Lagrangian is L(Q;µ,Ψ) = log det ( R n + HQH ) +µ(tr(q) P) Tr(ΨQ). The gradient of the Lagrangian is Q L = H ( R n + HQH ) 1 H + µi Ψ. Daniel P. Palomar 4
KKT Conditions The KKT conditions are Tr(Q) P, Q 0 µ 0, Ψ 0 H ( R n + HQH ) 1 H + Ψ = µi µ(tr(q) P) = 0, ΨQ = 0. Can we find a Q that satisfies the KKT conditions (together with some dual variables)? Daniel P. Palomar 5
First, let s simplify the KKT conditions by defining the so-called whitened channel: H = R 1/2 n H. Then, the third KKT condition becomes: H ( I + HQ H ) 1 H + Ψ = µi. To simplify even further, let s write the SVD of the channel matrix as H = UΣV (denote the eigenvalues σ i ), obtaining: Σ ( I + Σ QΣ ) 1 Σ + Ψ = µi. where Q = V QV and Ψ = V ΨV. Daniel P. Palomar 6
The KKT conditions are: Tr( Q) P, Q 0 µ 0, Ψ 0 ( Σ I + Σ QΣ ) 1 Σ + Ψ = µi ( ) µ Tr( Q) P = 0, Ψ Q = 0. At this point, we can make a guess: perhaps the optimal Q and Ψ are diagonal? Let s try... Daniel P. Palomar 7
Define Q = diag (p) (p is the power allocation) and Ψ = diag (ψ). The KKT conditions become: i p i P, p i 0 µ 0, ψ i 0 µ ( i σ 2 i 1 + σi 2p i ) + ψ i = µ p i P = 0,ψ i p i = 0. Let s now look into detail at the KKT conditions. Daniel P. Palomar 8
First of all, observe that µ > 0, otherwise we would have ψ i = 0 which cannot be satisfied. σ 2 i 1+σ 2 i p i + Let s differentiate two cases in the power allocation: if p i > 0, then ψ i = 0 = σ2 i 1+σ 2 i p i = µ = p i = µ 1 1/σ 2 i (also note that µ = σ2 i < 1+σ i 2p σ2 i ) i if p i = 0, then σi 2 + ψ i = µ (note that µ = σi 2 + ψ i σi 2. Equivalently, if σ 2 i > µ, then p i = µ 1 1/σ 2 i if σ 2 i µ, then p i = 0. Daniel P. Palomar 9
Waterfilling Solution More compactly, we can write the well-known waterfilling solution: p i = ( µ 1 1/σ 2 i ) + where µ 1 is called water-level and is chosen to satisfy i p i = P (so that all the KKT conditions are satisfied). Therefore, the optimal solution is given by where Q = Vdiag (p)v the optimal transmit directions are matched to the channel matrix the optimal power allocation is the waterfilling. Daniel P. Palomar 10
MIMO MAC Consider a Multiple-Access Channel (MAC), e.g., the uplink of a cellular wireless system where K different users transmit to the same base station: K y = H k x k + n where k=1 x k is the signal transmitted by the kth user with cov. matrix Q k H k is the channel matrix between kth user and the base station R n is the covariance matrix of the noise. Daniel P. Palomar 11
MIMO MAC Problem Formulation Such system can be partially characterized with one parameter: the sum rate R 1 + R 2 + + R K. The sum rate for a MIMO MAC is ( ) K log det R n + H k Q k H k k=1 log det (R n ) The constraints in the system are transmit power constraints for each of the users: Tr(Q k ) P k. Daniel P. Palomar 12
MIMO MAC Problem Formulation The formulation of the problem is maximize {Q k } log det ( R n + ) K k=1 H kq k H k subject to Tr(Q k ) P k k Q k 0. This problem is convex: the logdet function is concave, the trace constraints are just a linear constraints, and the positive semidefiniteness constraints are LMIs. Hence, we can use a general-purpose method such as an interiorpoint method to solve it in polynomial time. The question is: Can we derive a closed-form solution like in the single-user case with the waterfilling? Daniel P. Palomar 13
KKT Conditions The Lagrangian is L({Q k };{µ k }, {Ψ k }) = log det ( R n + K k=1 H k Q k H k ) + k µ k (Tr(Q k ) P k ) k Tr(Ψ k Q k ). The gradients of the Lagrangian are ( 1 K Qk L = H k R n + H l Q l H l) H k + µ k I Ψ k. l=1 Daniel P. Palomar 14
The KKT conditions are, for all k, Tr(Q k ) P k, Q k 0 H k ( R n + K µ k 0, Ψ k 0 1 H l Q l H l) H k + Ψ k = µ k I l=1 µ k (Tr(Q k ) P k ) = 0, Ψ k Q k = 0. It follows from the KKT conditions that at an optimal point, each user k, must satisfy a set of KKT conditions identical to those for the single-user case, but with the equivalent noise covariance matrix: R k = R n + l k H l Q l H l. Daniel P. Palomar 15
Closed-Form Solution from the KKT Conditions? This implies that at an optimal point, each user must be waterfilling (treating the other users as noise). Unfortunately, this observation does not mean that we can derive a closed-form solution. However, this suggests that perhaps we could optimize one user at a time, each optimization being solved in closed-form by a waterfilling. But, is this going to converge to a stationary point where all users waterfill jointly? The answer is positive: It follows from a classic result in optimization of the so-called Nonlinear Gauss-Seidel algorithm. Daniel P. Palomar 16
Gauss-Seidel Algorithm The Nonlinear Gauss-Seidel algorithm (AKA block-coordinate descent algorithm) optimizes f (x 1,,x N ) sequentially: x (t+1) k = arg max x k where t is the index for a global iteration. Converges if: f ( ) x (t+1) 1,,x (t+1) k 1,x k,x (t) k+1,,x(t) N 1. the function f (x 1,,x N ) is jointly convex 2. each optimization has a unique solution 3. the feasible set is a Cartesian product: X = X 1 X N. Daniel P. Palomar 17
Iterative Waterfilling Algorithm Particularization of the Gauss-Seidel algorithm to the maximization of the sum rate leads to the popular iterative waterfilling algorithm. repeat for t = 1,2, for each user k = 1,,K obtain the update as where Q k (t + 1) = V k diag (p k )V k V k is the eigenvector matrix of the equivalent whitened squared channel: H k R 1 k H k (note: R k depends on the other users) ( +. p k is the waterfilling power allocation p k,i = µ 1 k k,i) 1/σ2 Daniel P. Palomar 18
MIMO BC Consider a Broadcast Channel (BC), e.g., the downlink of a cellular wireless system where the base station transmits to K different users: y k = H k x + n k where x is the signal transmitted by the base station H k is the channel matrix between base station and the kth user R n,k is the covariance matrix of the noise. It turns out that, via uplink-downlink duality, the previous BC formulation can be reformulated as an equivalent MAC like similar to the one before. Daniel P. Palomar 19
MIMO BC Problem Formulation Such system can be partially characterized with one parameter: the sum rate R 1 + R 2 + + R K. The sum rate for a MIMO BC is ( ) K log det R n + H k Q k H k k=1 log det (R n ) The constraint in the system is a total transmit power constraint for the base station: k Tr(Q k) P T. Daniel P. Palomar 20
MIMO BC Problem Formulation The formulation of the problem is maximize {Q k } subject to ( log det R n + ) K k=1 H kq k H k k Tr(Q k) P T Q k 0 k. This problem is very similar to the MAC formulation with the difference that instead of having individual power constraints for each of the users Tr(Q k ) P k we have a total power constraint coupling the feasible sets of all the users. Can we develop here a similar iterative waterfilling algorithm based on the Gauss-Seidel algorithm? Daniel P. Palomar 21
Iterative Waterfilling for the MIMO BC? The answer is negative: the feasible set is not a Cartesian product and we cannot use the Gauss-Seidel algorithm. The problem is the sum-power constraint that couples all the users. We know from the basic theory on decomposition techniques that coupling constraints can be dealt with via dual decomposition. In fact, a primal decomposition can also be effectively employed here. Daniel P. Palomar 22
Using Decomposition Methods For convenience, let s introduce additional variables denoting the power allocation over the users P 1,,P K : ( maximize log det R n + ) K k=1 H kq k H k {P k },{Q k } subject to Tr(Q k ) P k k Q k 0 k P k P T. Primal decomposition: fix the power alocation {P k } and solve each subproblem by the classical iterative waterfilling algorithm. Dual decomposition: relax the coupling constraint k P k P T and solve each subproblem by a modified iterative waterfilling algorithm. Daniel P. Palomar 23
Primal Decomposition For fixed {P k }, we have the subproblem (only one, the objective does not decouple): maximize {Q k } subject to log det Tr(Q k ) P k Q k 0 with optimal value f (P 1,,P K ). ( R n + ) K k=1 H kq k H k It can be solved with the iterative waterfilling algorithm. A subgradient of f (P 1,,P K ) is µ (P 1,,P K ). k Daniel P. Palomar 24
Master primal problem: Primal Decomposition maximize {P k } subject to f (P 1,,P K ) k P k P T (P k 0). It can be solved with a projected subgradient method: P 1 P 1. (t + 1) =. (t) + α k µ(t) P K P K where µ (t) is defined as µ (P 1 (t),,p K (t)) and S is the simplex S = {x i x i P T, x 0}. S Daniel P. Palomar 25
It is not difficult to show that a projection on a simplex is given by [x] S = (x θ) +. The subgradient method is then P 1 P 1. (t + 1) =. P K P K (t) + α k µ(t) θ +. For each update of the power allocation {P k } we need to solve the subproblem by means of the iterative waterfilling algorithm (because the objective is still coupled in the objective). Alternative: we could employ a different decomposition to decouple the objective as well. Daniel P. Palomar 26
Recall the original problem maximize {P k },{Q k } Dual Decomposition log det ( R n + ) K k=1 H kq k H k subject to Tr(Q k ) P k k Q k 0 k P k P T. Form a partial Lagrangian by relaxing the coupling constraint k P k P T : ( ) K L({Q k },{P k } ;µ) = log det R n + H k Q k H k k=1 ( ) +µ P k P T. k Daniel P. Palomar 27
Dual Decomposition For fixed µ, the subproblem is (only one, the objective does not decouple) ( maximize log det R n + ) K k=1 H kq k H k µ( k P k P T ) {P k },{Q k } subject to Tr(Q k ) P k k Q k 0 with optimal value g (µ) and subgradient P T k P k (µ). The master dual problem is minimize µ g (µ) subject to µ 0. Daniel P. Palomar 28
Dual Decomposition: Solving the Subproblem To solve the subproblem, we can use again a Nonlinear Gauss-Seidel method to optimize one user at a time. Each optimization is maximize P k,q k subject to log det Tr(Q k ) P k Q k 0. ( ) R k + H k Q k H k µp k This problem is slightly different from the one that led to the waterfilling solution: here P k is an optimization variable, and we have the additional term µp k in the objective. Daniel P. Palomar 29
Let s derive a closed-form solution from the KKT conditions. The Lagrangian is L(P k,q k ;λ k,ψ k ) = log det ( ) R k + H k Q k H k + µp k +λ k (Tr(Q k ) P k ) Tr(Ψ k Q k ). The gradients of the Lagrangian are Qk L = H k ( R k + H k Q k H k) 1Hk + λ k I Ψ k Pk L = µ λ k. Daniel P. Palomar 30
The KKT conditions are H k Tr(Q k ) P k, Q k 0 λ k 0, Ψ k 0 ( 1Hk R k + H k Q k H k) + Ψ k = λ k I µ = λ k λ k (Tr(Q k ) P k ) = 0, Ψ k Q k = 0. These conditions are the same we encountered when we derived the waterfilling solution with the only different that now λ k = µ, i.e., the waterlevel λ 1 k is now fixed to µ 1. Daniel P. Palomar 31
Thus, the optimal solution is Q k = V k diag (p k )V k P k = Tr(Q k) where p k is the waterfilling power allocation with fixed waterlevel µ 1. p k,i = ( µ 1 1/σk,i 2 ) + Daniel P. Palomar 32
Dual Decomposition: Solving the Master Problem Now we need to solve the dual master problem minimize µ g (µ) subject to µ 0. We can use a simple subgradient method: ( µ(t + 1) = [µ(t) α k P T k P k (µ(t)))] +. Observe that if k P k (µ(t)) > P T, then µ(t) will be increased if k P k (µ(t)) < P T, then µ(t) will be decreased. Daniel P. Palomar 33
Dual Decomposition: Solving the Master Problem Alternatively, since the dual variable is a scalar, we can use the way more effective bisection method: 1. Initialize µ min and µ max. 2. Let µ = (µ min + µ max ) /2. 3. Solve iteratively for all users (using the iterative waterfilling with fixed waterlevel µ 1 ). 4. If k P k > P T, then set µ min = µ; otherwise set µ max = µ. 5. Stopping criterion: If µ max µ min ε, then stop; otherwise, go to step 2. Daniel P. Palomar 34
Summary We have recalled the MIMO waterfilling solution for the single-user case. We have derived the iterative waterfilling algorithm for the MIMO MAC with individual power constraints. For the MIMO BC with a sum-power constraint, we have considered: primal decomposition dual decomposition. Each decomposition leads to a different variation of the basic iterative waterfilling algorithm. Daniel P. Palomar 35
References W. Yu, W. Rhee, S. Boyd, and J. Cioffi, Iterative Water-Filling for Gaussian Vector Multiple-Access Channels, IEEE Trans. on Information Theory, vol. 50, no. 1, Jan. 2004. W. Yu, A Dual Decomposition Approach to the Sum Power Gaussian Vector Multiple Access Channel Sum Capacity Problem, in Proc. Conf. on Information and Systems (CISS), The Johns Hopkins Univ., March 12-14, 2003. N. Jindal, W. Rhee, S. Vishwanath, S. A. Jafar, and A. Goldsmith, Sum Power Iterative Water-Filling for Multi-Antenna Gaussian Broadcast Channels, IEEE Trans. on Information Theory, vol. 51, no. 4, April 2005. Daniel P. Palomar, Convex Primal Decomposition for Multicarrier Linear MIMO Transceivers, IEEE Trans. on Signal Processing, vol. 53, no. 12, Dec. 2005. Daniel P. Palomar 36