Suboptimality of minmax MPC. Seungho Lee. ẋ(t) = f(x(t), u(t)), x(0) = x 0, t 0 (1)

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1 Suboptimality of minmax MPC Seungho Lee In this paper, we consider particular case of Model Predictive Control (MPC) when the problem that needs to be solved in each sample time is the form of min max problem. We find an analytical expression of the suboptimality that is a degree of degradation of the min max MPC solution with respect to the optimal control solution. We transfer the min max problem to the standard Bolza form in a frame of Method of Outer Approximations (MOA). Then we apply Phase I- Phase II method to solve the problem. Our method establishes forward invariant set and hence the feasibility is guaranteed without any additional constraint or modification of the cost function. We show that as the horizon length is increased sufficiently large, MPC solution approaches to the optimal control solution. 1 Introduction The principal of optimality is well studied since Bellman developed one of the most pioneering research about determining the optimal sequence of decisions from the fixed state of the system [1]. The approach that realizes his idea is referred as Dynamic Programming (DP) associated with discrete-time optimization problems. In continuous time optimization problems, the analogous equation is a partial differential equation which is usually referred as Hamilton-Jacobi-Bellman (HJB) equation. Suppose that we have a dynamical system that is modeled by a differential equation of the form ẋ(t) = f(x(t), u(t)), x(0) = x 0, t 0 (1) where x(t) is the state of the system and u(t) is the control. Now suppose that one wants to optimize its behavior by solving an optimal control problem of the form min u(t) U,x(t) X 0 f 0 (x(t), u(t))dt, (2) subject to the differential equation constraint (1). Let û(t) be solution of this optimal control problem and ˆx(t) = x(t, û(t)), t 0 the resulting trajectory. 1

2 Now suppose that the model (1) is not perfect, so the actual trajectory resulting from the control û(t), x (t, û(t)), is quite different from ˆx(t). To remedy this situation, it was proposed that every time units, the actual state be measured and the problem (1) re-solved, to obtain a corrected control, effectively creating a feedback mechanism. Experiments have shown that this idea, called Model Predictive Control (MPC) is an excellent idea. MPC is a form of feedback control in which the current control action is obtained by solving online, at each sample time, an open-loop optimal control problem over a fixed time window with the current system state as the initial condition. The solution of this optimization problem yields a finite sequence of control actions over the optimization window; only the first value is applied to the system, then the window is advanced one sample time and the optimization process repeated [2]. We are particularly interested in the min max MPC which refers the case when the problem that needs to be solved in every sample time is a minmax problem. See [3] for detailed discussion of DP in min max. This type of problem naturally arises in many engineering applications. The pursuit-evasion game applications are presented in [4] and [5]. Network and image processing applications are discussed in [6], [7] and [8], respectively. Biped robot application is given in [9]. However, as MPC is an approximation of the optimal control, the quality of solution obtained by solving MPC problem is inherently degraded from the optimal control problem. Some researches about MPC include finding analytical expression of this degradation, called suboptimality. However, suboptimality in min max MPC has not been explored yet. In this research, we present the analytical expression of it. Perhaps the most closely related works to this research are [10], [11], and [12]. They are in common in a sense that the given MPC problem is a Bolza form and the problem is unconstrained. The fact that the constrained min max problem is considered differentiates this work from the aforementioned researches. Another key feature of this work is a recursive feasibility. The recursive feasibility is one of well known drawback of the MPC. See [13], [14] for the feasibility issue in MPC. This issue is normally resolved by introducing terminal constraint or cost [14], [15]. In this work, we adopt Phase I-Phase II method [16] so that we can explicitly construct forward invariant set and hence the recursive feasibility is guaranteed without any additional constraint or cost. Our approach is to utilize some numerical optimization method. First, we use the method of the Method of Outer Approximations (MOA) [16] to separate continuous min max problem to the sequence of finite min max problems and 2

3 max problems that converges to the initial problem. Next, we adopt Phase I-Phase II method to the separated problems so that the solution of separated problems guarantee the existence of the forward invariant set. Also the sequence of the separated problems are constructed so that the problem can be transformed to the discrete Bolza form. Then we present properties of values of the problem based on the principal of optimality. Finally, analytical expression of the suboptimality that is leveraged by the relaxed dynamic programming [10],[11] is presented. 2 Problem statement In a given sample time t, consider two plays, X and Y whose nonlinear, discrete dynamics are given as x t+k+1;t = f x (x t+k;t, u t+k;t ), y t+k+1;t = f y (y t+k;t, v t+k;t ) (1),where u t+k;t R nu, v t+k;t R nv, x t+k;t R px, y t+k;t R py, x t+k;t χ, with given x t;t = x 0 and y t;t = y 0. Let u t = {u t+0;t, u t+1;t, u t+h 1;t } be a control sequence with a length H and x t = {x t+1;t, x t+2;t, x t+h;t } denote the prediction of the X player trajectory obtained by iterating the dynamics (1). The Y player trajectory is defined similarly. We are interested in solving the following problem in a frame of MPC, in a point of X player s view. min u t U t max v t Y (u t ) c(x k+t;t, y k+t;t ) s.t. f l (u k+t;t, x k+t;t ) 0, l {1, 2,, q x }, k H := {0, 1,, H 1}, x t+k+1;t = f x (x t+k;t, u t+k;t ), y t+k+1;t = f y (y t+k;t, v t+k;t ) (2),where U t R nu H, p : R px R py R, Y (u t ) := {v t R nv H f l c(u k+t;t, x k+t;t, v k+t;t, y k+t;t ) 0, v t+k;t R nv, k H}, l {1, 2,, q y }. Since we are interested in any sample time t, in given x 0 3

4 and y 0, we rewrite (2) in a compact form by omitting t min u U max v Y (u) c(x k, y k ) s.t. f l (u k, x k ) 0, l {1, 2,, q x }, k H x k+1 = f x (x k, u k ), y k+1 = f y (y k, v k ) (3) Problem is to find analytical expression of suboptimality of (3). 3 Method of Outer Approximation Problem in (3) is a generalized minmax problem and it is solved by separating it to the finite minmax problem and the maximimzaion problem. Algorithm 1: Method of Outer Approximation. Step 0: Set P 0 = {ŷ 0 }, I 0 = {0}, i = 0. Step 1: X player solves finite min max problem: û i = arg min u U max j I i c j (x k ) s.t. f l (u k, x k ) 0, l {1, 2,, q}, x k+1 = f x (x k, u k ), y k+1 = f y (y k, v k ), where c j (x k ) := c(x k, y j k ) (4) The trajectory ˆx i that corresponds to û i is obtained through (1). Step 2: X player assumes that Y player solves the following problem: ˆv i = arg max c(ˆx i k, y k ) v Y (û i ) s.t. f l (u k, v k, x k, y k ) 0, l {1, 2,, q}, x k+1 = f x (x k, u k ), y k+1 = f y (y k, v k ). (5) The trajectory ŷ i+1 that corresponds to ˆv i is obtained through (1). Step 3:. Update P i+1 = P i {ŷ i+1 }, I i+1 = I i {i + 1}, i = i + 1 go to Step 1. 4

5 ˆx 1ˆx 2ˆx MOA Horizon 1 2 H t 1 t Sample time Figure 1: An example trajectory generated by MOA at sample time t, when the horizon length is H. ˆx is obtained as lim i ˆx i = ˆx To analyze the convergence of the algorithm 1, let us translate max operation as follow. max c(ˆx i k, y k ) = max µ k c(ˆx i µ Σ H k, y k ), (6) where Σ q = {µ = {µ 1, µ 2,, µ q } µ i 0, Σ q i=1 µi = 1} is a unit simplex. Let v := (v, µ), c (ˆx i k, y k) := µ k c(ˆx i k, y k), then we can rewrite max operation in a compact form: ˆv i = arg max v Y (û i ) c (ˆx i k, y k ) s.t. f l (u k, v k, x k, y k ) 0, l {1, 2,, q} (7) Let us define φ(x) and φ P i(x) functions as max functions in Y and P i Y, respectively. φ(x) := max v Y (u) c (x, y) (8) and φ P i(x) := max v P i c (x, y) (9) Below Lemma 1 to Theorem 6 are from [16], Chapter 2 and 3. We summarize the result for the completeness of the Algorithm 1. Lemma 1. Suppose that the Algorithm 1 has constructed an infinite sequence {ˆx i } i=1, in solving (3). If {ˆxi } i=1 ˆx, then φ P i (ˆx i ) φ(ˆx) Proof. φ(ˆx i ) φ Pi (ˆx i ) c (ˆx i, ŷ i 1 ) (10) 5

6 Note that ŷ i 1 is a maximizer in i 1 th iteration in Algorithm 1. Since φ( ) is a continuous function, Since c ( ) is a continuous function, φ(ˆx i ) φ(ˆx) (11) c (ˆx i, ŷ i 1 ) c (ˆx i 1, ŷ i 1 ) 0 as i. (12) Because c (ˆx i 1, ŷ i 1 ) = φ(ˆx i 1 ), Therefore, c (ˆx i, ŷ i 1 ) φ(ˆx i 1 ) φ(ˆx) as i. (13) φ Pi (ˆx i ) φ(ˆx) (14) Theorem 2. Suppose Algorithm 1 constructed {ˆx i } i=1. If ˆx is an accumulation point, then ˆx is a minimizer of (3). Proof. From Lemma.1, φ Pi (ˆx i ) φ(ˆx) (15) as {ˆx i } i=1 ˆx. Let ˆv = min u U φ(x). Now, suppose ˆx is not a minimizer of φ(x). Then we see that there exists i [1, ] such that φ Pi (x) > φ(x) for some x. This contradict φ Pi (x) φ(x). Therefore, ˆx is a minimizer of φ(x). To solve finite min max problem (4) and max problem (5) in the algorithm 1, we use Phase I-Phase II method. We summarize the result in following Definition 3. to Theorem 6. Interested reader can find detailed discussion in [16], [17], and [18]. Definition 3. Following is called optimality function. { p θ i := min µ 0 ν [ ] } k φ(x i ) c k (x i ) + γψ(x i ) + (16) µ Σ 0 q Lemma 4. Searching direction ( h(x i ) = 1 δ µ 0 x p νx c k i (x i ) + ) q µ j x f j (x i ) j=1 (17) 6

7 , where (µ x, ν x ) is a solution of (16), is the direction that minimizes the cost of (3), while satisfying constraints. Lemma 5. Step size associated with the searching direction (17) is following. λ i = max{β k F (x i, x i + β k h i ) β k αθ i }, (18), where F (z, x) := max{φ(x) φ(z) γψ(z) +, ψ(x) ψ(z) + }, ψ(x) := max j l f l (x), ψ(x) + := max{0, ψ(x)}. Theorem 6. Algorithm solves problem (3) and hence the forward invariant set χ is established. 4 Transformation Remark 7. We have established χ by using P hasei P haseii method. This implies that lim i ŷ i = ŷ and lim i ˆx i = ˆx. Lemma 8. Suppose that the sequence x is given and ŷ i and ˆx i are maximizer and minimizer, respectively and c(x k, y k ) 0. Then following hold. c(x, ŷ i ) 1 max c(ˆx i k, ŷk) i = max v Y (u) c(x, y) 1 c(ˆx i k, y k ) (19) Proof. First, Suppose ˆx is a maximizing sequence of both f(x) and g(x) and maximum of the sequence is obtained at k th element. Then we see that ˆx is also a maximizing sequence of f(x)g(x) because it satisfies f(x)t g(x) + f(x) g(x) = 0 x x when x = ˆx. Now, by definition, ŷ i := arg max v Y (û i ) c(ˆx i k, y k) through f y. We see that ŷ i is a sequence such that c(ˆx i k, ŷi ) is a nondecreasing sequence because any decreasing does not beneficial for maximization. Hence, the maximum element is obtained on the boundary. This implies that every subinterval starting from ŷ 1 to ŷ H is also maximized. In turn, for given x, this implies ŷ i maximizes H c(x, y) = c(x, y) 1 as well. Therefore, (19) holds. Now let us discuss about the transformation from min max to Bolza form. In the Step 2 for some i in Method of Outer Approximation (MOA), ŷ i is obtained. Since ˆx i is obtained in the Step 1, following holds: max c(ˆx i k, ŷk) i = c(ˆx i, ŷ i ) (20) 7

8 Assumption 9. c(ˆx i, ŷ i ) 0. Therefore, max c(ˆx i k, ŷi k ) c(ˆx i, ŷ i ) = 1 (21) Also, for any x = {x 1,, x H }, and for the maximizer (ŷ i ) obtained from Step 2 of MOA, c(x, ŷ i ) 1 = c(x 1, ŷ1) i + c(x 2, ŷ2) i + + c(x H, ŷh) i H = c(x k, ŷk) i (22) Let l i (x k ) := c(x k, ŷk i ), then (22) becomes c(x, ŷ i ) 1 = H l i (x k ) (23) Note that the superscript i in l i (x k ) indicates that the function implicitly depends on ŷ i. Multiplying (21) to the LSH of (23) yields, Since c(x, ŷ i ) 1 max c(ˆx i c(ˆx i, ŷ i ) k, ŷk) i = H l i (x k ) (24) max c(ˆx i k, ŷk) i = max v Y (u) c(ˆx i k, y k ) (25) (24) becomes 1 c(ˆx i, ŷ i ) max v Y (u) c(x, y) 1 c(ˆx i k, y k ) = H l i (x k ) (26) Let c (ˆx i k, x, y) := c(x, y) 1c(ˆx i k, y k) and by taking min u U, 1 min c(ˆx i, ŷ i ) u U max v Y (u) c (ˆx i k, x, y) = min u U H l i (x k ) (27) 8

9 Finally, let l i (x k ) := c(ˆx i, ŷ i ) l i (x k ). Then from (27), min u U max v Y (u) c (ˆx k, x, y) = min u U H l i (x k ) (28) Remark 10. RHS form in (28) is not an implementable form because it requires advanced knowledge ˆx and ŷ. However, it is a favorable form for checking the suboptimality as the suboptimality is defined by the value of the optimization problem which implies that the sequences of ˆx and ŷ is obtained. Note that it is an i th iteration of the MOA. Therefore, a sequence of suboptimality is generated as i. Generated sequence of suboptimality is the case when the objective function is c (ˆx k, x, y). As i, we found relation between c (ˆx k, x, y) and the original objective function from the Proposition 11. Proposition 11. Suppose i so that ˆx i ˆx and ŷ i ŷ. Then c (ˆx k, x, y) = c(ˆx, ŷ) 1 c(ˆx k, ŷ k ). Proof. Because c (,, ) is a continuous function, as ˆx i ˆx and ŷ i ŷ, c (ˆx k, x, y) c (ˆx k, ˆx, ŷ). Therefore, c (ˆx k, x, y) = c(ˆx, ŷ) 1 c(ˆx k, ŷ k ). From now we omit superscript i that implies that we are currently i th iteration of the outer approximations. First, let us define the value of the infinite horizon problem V (x) := l(x n, µ (x n )), (29) n=0, where µ (x n ) is an infinite horizon control solution. The original objective function that is required in V (x) is defined as { l(x k ) := c(ˆx, ŷ ) c(x k, ŷk) i c(ˆx t, ŷ = t) c(x k+t;t, ŷk+t;t i ), if k H c(ˆx t, ŷ t) c(x k+t;t, ŷh+t;t i ), else (30), where ŷ t := {ŷ t+1;t, ŷ t+2;t, ŷ t+h;t, ŷ t+h;t, },i.e., the H th entry is infinitely repeated. Remark 12. The infinite horizon problem (29) describes the situation when the Y player uses finite horizon control law, but X player utilizes infinite horizon control law. Note that Y player control sequence and trajectory are extended by duplicating the last element to well define the cost function. Clearly, this preserves the value of the maximization. The value of the problem (28) is V H (x). The point 9

10 Pursuer obstacle Evader Figure 2: Illustrative example: different predicted trajectories x in V (x) is a starting point of entire problem and the point x in V H (x) is a starting point of current horizon. Fig.2 illustrated the different predicted trajectories resulting from the infinite horizon and finite horizon problem as appears in [19]. Consider the pursue-evasion game between a single pursuer and an evader. The goal of the pursuer is to minimize the distance between two players while avoiding the obstacle. The goal of the evader is to maximize the distance between them while avoiding the obstacle as well. It is intuitively clear that the evader trajectory is opposite direction to the pursuer. Knowing this, the pursuer trajectory is toward the evader. Suppose there is an obstacle between two players. If the pursuer solves finite horizon problem which is not long enough to activate the constraint about the obstacle, resulting trajectory is still straight line to the evader. However, if the pursuer solves infinite horizon problem, he takes the presence the obstacle into the consideration, and hence generates better trajectory. Lemma 13. For two nonempty sets U and U, let U = {u = {u 1, u 2,, u U }}, and U = {u = {u 1, u 2,, u U }} such that U U and hence U U. Then V U (x) V U (x) and hence V (x) V U (x) Proof. Let ψ(x) := max v Y (u) c(x k, y k ). From U U, we immediately see k H min u U ψ(x) min u U ψ(x). Remark 14. In the non-min max problem, for two non-empty sets such that H M, V M (x) V H (x) holds and hence V (x) V H (x). It is because cost function is a sum of the nonpositive running cost. Let l(x n, µ H (x)) be a running cost at a point x n, with a control µ H (x). Definition 15. V µ H(x) (x n ) := l(x n, µ H (x n )), (31) n=0 10

11 where the feedback law µ H (x) is a minimizing sequence of the problem (28). µ H (x) = arg min u U = arg min u U H l i (x k ) max v Y (u) c (ˆx k, x, y) (32) From the Bellman s principle, V H+1 (x n ) = min u {V H (x n+1 ) + l(x n, u)}. It is known that if u = µ H (x), V H+1 (x n ) = V H (x n+1 ) + l(x n, µ H (x)) (33) From the property of min max MPC, V H+1 (x n ) V H (x n ). Therefore, min u {V H (x n+1 ) + l(x n, u)} V H (x n ) (34) and V H (x n+1 ) + l(x n, µ H (x)) V H (x n ) (35) Definition 16. The suboptimality parameter of the min max MPC problem is α 1, such that, for any given x, and non-maximizer y V H (x n ) αv (x n ) αv µ H(x n) (x n ) (36) The parameter α quantifies the effectiveness of the finite horizon control law with respect to the the case when it is applied to the infinite horizon problem. In both cases, Y player uses finite horizon control law. Remark 17. Inequality (36) trivially satisfied by a large α. Suppose α = 1 then due to the property of the value of the minmax problem (Lemma 13), equality should hold: V H (x n ) = V (x n ), and V H (x n ) = V µ H(x) (x n ) (37) This implies that V (x n ) = V µ H(x) (x n ) and hence µ H (x) is the optimal control. Proposition 18. Suppose some α 1 satisfies αl(x n, µ H (x n )) V H (x n ) + V H (x n+1 ) (38) 11

12 Then it also satisfies V H (x n ) αv (x n ) αv µ H(x n) (x n ). Proof. From the hypothesis, αl(x n, µ H (x n )) V H (x n ) + V H (x n+1 ), summing up both side from n to infinity yields αv µ H(x n) (x n ) V H (x k ) + V H (x k+1 ) V H (x n ). (39) k=n Therefore, αv µ H(x n) (x n ) V H (x n ). Clearly αv (x n ) αv µ H(x n) (x n ) and for large α 1, αv (x n ) V H (x n ). Hence, the desired result is obtained. Our definition of the suboptimality is similar to the one in [10],[11],and [12]. However, the fundamental property of the value of the problem from Lemma 13 differentiate the procedure. 5 Sequence of suboptimality Since we are solving (2) in a recursive scheme (outer approximation), the parameter α that is associated with the suboptimality creates a sequence. Step 1. Perform the i th iteration of MOA. Step 2. Check suboptimality using (39). Step 3. Set i = go to Step 1. Step 4. As MOA terminates, compute the suboptimality of the original objective function. 6 Suboptimality Lemma 19. Suppose V H 1 (x n+1 ) + V H (x n+1 ) (α 1)l(x n, µ H (x n )) (40) holds for some α 1. Then, proposition 18 also holds. Proof. Adding l(x n, µ H (x n )) in both side of (40) yields V H 1 (x n+1 ) + V H (x n+1 ) + l(x n, µ H (x)(x n )) αl(x n, µ H (x)(x n )) (41) Since V H (x n ) = V H 1 (x n+1 ) + l(x n, µ H (x n )), V H (x n ) + V H (x n+1 ) αl(x n, µ H (x n )) (42) 12

13 It coincide to (38). Assumption 20. There exists η k 1 and γ 1 such that η k V k (x n ) V k 1 (x n ), V k (x n ) γl(x n ), for k H, V H+1 (x n ) l(x n ) (43) Theorem 21. The value of α that satisfies (40) is α = γ (γ+1) H+1 (γ+1) H+1 γ H+1 Proof. Note that the assumption V k (x n ) γl(x n ) implies V k 1 (x n+1 ) γl(x n ) because V k (x n ) V k 1 (x n+1 ). V k (x n ) = V k 1 (x n+1 ) + l(x n ) V k 1 (x n+1 ) + l(x n ) + ɛ(γl(x n ) V k 1 (x n+1 )) = (1 ɛ)v k 1 (x n+1 ) + (1 + ɛγ)l(x n ) η k (1 ɛ)v k (x n+1 ) + (1 + ɛγ)l(x n ) (44) Set ɛ = η 1 then (44) becomes γ+η V k (x n ) η k γ + 1 γ + η k (V k (x n+1 ) + l(x n )) = η k γ + 1 γ + η k V k+1 (x n ) (45) Suppose Then it satisfies η k = (γ + 1)k (γ + 1) k γ k (46) η k γ + 1 γ + η k = η k+1 (47) From the assumption, V H 1 (x n ) η H V H (x n ), adding V H (x n+1 ) both side V H (x n ) + V H (x n+1 ) η H+1 V H+1 (x n ) + V H (x n+1 ) (48) Since V H+1 (x n ) = V H (x n+1 ) + l(x n ), V H (x n ) + V H (x n+1 ) η H+1 V H+1 (x n ) + V H+1 (x n ) l(x n ). (49) 13

14 Since V H+1 (x n ) l(x n ) 0, (49) still holds without this term. Since V H+1 (x n ) V H (x n ), and V H (x n ) γl(x n ), and hence, V H (x n ) + V H (x n+1 ) γ(η H+1 )l(x n ) (50) From (40) and (51), V H 1 (x n+1 ) + V H (x n+1 ) [γ(η H+1 ) 1]l(x n ) (51) (α 1)l(x n ) = [γ(η H+1 ) 1]l(x n ) (52) Therefore, (γ + 1) H+1 α = γ (53) (γ + 1) H+1 γ H+1 Now we are interested in finding γ without evaluating V k 1 (x n+1 ) γl(x n ). Consider the following inequality. min c(x, u U ŷi 1 ) V H 1 (x n+1 ) γl(x n ). (54) Hote that ŷ i 1 is a maximizer obtained from the previous iteration i 1 and hence in general, it is not a maximizer in i th iteration of MOA. Let ˆx i be a resulting trajectory from min u U c(x, ŷ i 1 ). Then min c(x, u U ŷi 1 ) = min{c(ˆx i 1, ŷ1 i 1 ), c(ˆx i 2, ŷ2 i 1 ),, c(ˆx i H, ŷ i 1 H )} = min{c i 1, c i 2,, c i H} (55),where c i k := c(ˆxi k, ŷi 1 k ). Hote the following holds for any two scalar a and b. min{a, b} = a + b a b 2, (56) 2 2 Let us recursively define minimums as follow d k = d k 1 + c k+1 2 dk 1 + c k (57) 14

15 , with Then d 1 = c 1 + c 2 2 c1 + c (58) min c(x, u U ŷi 1 ) = d H 1 (59) and hence from (41) the approximation of the γ which does not require the knowledge of ˆx and ŷ is found γ = d H 1 l(x H ). (60) Remark 22. The condition γ = 1 means V k (x n ) l(x n ), for k H and its approximation is d H 1 = l(x H ). The latter indicates that the minimum is archived at the last entry of the sequence. 7 Conclusion From (40), if γ = 1, for every ɛ > 0, there exists H such that 1 + ɛ > α. This implies that from (25), optimality is recovered in a frame of MOA as H. References [1] R. Bellman, Dynamic programming and lagrange multipliers, Proceedings of the National Academy of Sciences of the United States of America, vol. 42, no. 10, p. 767, [2] D. Q. Mayne, Model predictive control: Recent developments and future promise, Automatica, [3] B. O Donoghue, Y. Wang, and S. Boyd, Min-max approximate dynamic programming, in Computer-Aided Control System Design (CACSD), 2011 IEEE International Symposium on, pp , IEEE, [4] S. Lee, E. Polak, and J. Walrand, A receding horizon control law for harbor defense, in Proc. 51th annual Allerton Conference on Communication, Control, and Computing, October

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