ECE Optimization for wireless networks Final. minimize f o (x) s.t. Ax = b,

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1 ECE Optimization for wireless networks Final Please provide clear and complete answers. PART I: Questions - Q.. Discuss an iterative algorithm that converges to the solution of the problem minimize f o () s.t. A = b, where f o () is a strictly conve function in R n. Specify the algorithm by eplicitly writing down the updating equations. Sol.: A convenient choice is Newton s algorithm for equality-constrained problems, which is described by (k+) = (k) + t (k) (k), where the search direction (k) satisfies the linear system (along with the multipliers ν) 2 f o ( (k) ) A T (k) fo ( = (k) ), A 0 ν 0 and t (k) are appropriately chosen step sizes (e.g., obtained via backtracking line search). Q.2. You are given the following problem: minimize a T + b T y,y C ¹ d s.t. Ey ¹ f G + Hy ¹ p where, y are optimization variables in R n. In order to ease the solution, you are asked to distribute the optimization task over two subprocessors coordinated by a master processor. Propose both a primal and a dual decomposition solution to accomplish this task. Sol.: Primal decomposition: minimize z inf (a T )+ inf (b T y), C y C y with C = {: C ¹ d, G ¹ z} and C y = {y: Ey ¹ f, Hy ¹ p z}. Dual decomposition: maimize λ with C 0 = {: C ¹ d} and C 0 y = {y: Ey ¹ f}. inf (a T + λ T G)+ inf (b T y + λ T Hy) λ T p, C 0 y C 0

2 Q.3. Consider the strategic game defined by the following utility matri: 2, 2 0,, 0,. Find Pareto-optimal points and Nash equilibria. Repeat for the game defined by the utility matri: 3, 3, 4 4, 0, 0. Sol.: First game: Pareto-optimal point: (2, 2); Nash equilibria: {(2, 2), (, )}. Second game: Pareto-optimal points: {(3, 3), (, 4), (4, )}; Nash equilibria: {(, 4), (4, )} Q.4. Consider the paper-rock-scissor game defined by the utility matri P R S P 0, 0,, R, 0, 0, S,, 0, 0 Is there a Nash equilibrium in mied strategies? If so, calculate it. Sol.: For any finite- and discrete- strategy game there is at least a NE. It can be calculated via the indifference principle (denoting by p ij the probability that player i chooses the option j) E p [U () = P] =E p [U () = R] =E p [U () = S] p 22 ( p 2 p 22 )= p 2 +( p 2 p 22 )=p 2 p 22 p 2j =/3 and the same can be done for the utility of the second player obtaining p j =/3. PART II: Problems - P.. For each of the following three problems: (a) solve the primal problem (i.e., give p and X opt ); (b) solve the dual problem (i.e., give d and the optimal multipliers); (c) are Slater s conditions satisfied?; (d) does strong duality hold?; (e) is there a solution to the KKT conditions?. minimize s.t. 2 Sol.: Primal: p =, X opt = { } 2

3 Dual: L(, λ) = + λ( 2 ) g(λ) = 2λ = λ with dom g = {λ 6= 0} The solution of the dual problem maimize λ can be easily found by setting g(λ) =0, λ>0 which leads to λ =/2 and d =. Slater s conditions are satisfied (consider for instance =0for which 2 < ), so that we know that strong duality holds and the dual problem is attained. KKT conditions: since the KKT conditions are necessary and sufficient for optimality, the only solution is (,λ )=(, /2). 2. minimize s.t. 2 0 Sol.: Primal: p =0,X opt = {0} = C, where C represents the feasible set. Dual: = 2λ L(, λ) = + λ 2 g(λ) = with dom g = {λ 6= 0} The solution of the dual problem maimize is clearly λ>0 d =0which is not attained. Slater s conditions are not satisfied (C = {0} for which 2 =0). Strong duality holds since p = d. KKT conditions: since Slater s conditions are not satisfied but the problem is conve, the KKT conditions are necessary and sufficient for (, λ) to be primal-dual optimal with zero duality gap. Since here we know that there is no such (, λ) (the dual problem is not attained), then the KKT have no solution. 3. minimize s.t. 0 Sol.: Primal: p =0,X opt = {0} = C, where C represents the feasible set. Dual: L(, λ) = + λ g(λ) =0 = 0 with dom g = {λ >} 3

4 We have d =0and the set of optimal values for λ is {λ >}. Slater s conditions are not satisfied (C = {0} for which =0). Strong duality holds since p = d. KKT conditions: the constraint is not differentiable so that we cannot write the KKT conditions. P.2. In the multihop network in the figure below, each ith link has capacity C i (in bits/ sec) and it is operated for a fraction 0 i of the total time ( i =). The effective rate on the ith link is then i C i. Moreover, the end-to-end rate from node to node M + is easily shown to be min i=,2,...,m { i C i }, that is, it is limited by the worst link. C C 2 C3 CM i= M+ Figure : (a) Write the problem of maimizing the end-to-end rate (min i=,2,...,m { i C i })overthefraction of times. Is this problem conve? Does it satisfy Slater s conditions? Sol.: The problem is maimize min i=,2,...,m { i C i } s.t. i, i 0, i = i= which is easily shown to be conve since the constraints are affine and the objective is concave (the pointwise minimum of affine functions is concave). Moreover, since all the constraints are affine and the problem is feasible, it satisfies Slater s conditions. (b) Write the problem as LP and the corresponding KKT conditions. Verify that the only solution of the KKT conditions is j = /C j Mi= for all j =,..., M. (Hint: can an optimal C i solution have i =0for some i?) Finally, can we have optimal solutions for the problem at hand other than j = /C j Mi=? Why? C i Sol.: The equivalent LP problem is minimize t,t t i C i 0 s.t. i = i= i 0 4

5 Noticing that the optimal solution must have i > 0 (otherwise the end-to-end rate would be zero), we can write the Lagrangian as Ã! L(t,, λ,ν) = t + λ i (t i C i )+ν i The KKT conditions are: i= i= λ i = 0 (a) i= λ i C i + ν = 0 (b) i = (c) i= t i C i 0 (d) i > 0 (e) λ i ( i C i t) = 0 (f) λ i 0 (g) From (b) and (a) we have ν = Mi= and λ i = ν C i > 0. Since λ i > 0, from (f) and (c), C i we obtain t = Mi= and i = /C i Mj=. Since the problem satisfies the Slater s conditions, C i C j KKT conditions are necessary and sufficient for optimality, so that no other optimal solutions can be found. (c) Consider now an alternative scenario where C i is not a constant but depends on i as µ C i ( i )=log + P, (2) N i where signal (P ) andnoise(n) powersaregiven andfied. Equation (2) accounts for the fact that if a link is operated for less time it can employ more power (P/ i ) without violating the long-term power constraint of P. Is the problem of maimizing the capacity min i=,2,...,m { i C i ( i )} conve? Sol.: Yes, since i C i ( i ) is concave (it can be proved by taking the second derivative or using the fact that it is the perspective function of log + P N, which is concave). P.3. Two nodes transmit packets at rates and 2 respectively on a given wired network. Node has a greater need of bandwidth so that the utilities of the two nodes are U ( )=2 and U 2 ( 2 )= 2, respectively. Moreover, the finite capacity of the links in the network poses the following constraints on 0 and 2 0: (3a) (3b) (a) Plot the region of achievable utility values θ, and identify the Pareto optimal points. 5

6 Sol.: The region θ of achievable utility values is given by the equations (U 0 and U 2 0): U + U 2 U 2 +2U 2, fromtheproblemconstraints andthedefinitions of the utility functions. It follows that region θ is a polytope with vertices (0, 0), (, 0), (2/3, /3), (0, /2). The Pareto optimal points are easily shown to be given by the equations: U 2 = 2 U 4 for 0 U 2/3 U 2 = U for 2/3 U. (b) Consider the scalarization method for finding the Pareto-optimal points. Are all the Pareto-optimal points solutions of a scalar problem? If so, give the corresponding values of the weights α that provide the Pareto-optimal points. Sol.: Since the problem is conve (utilities and constraints are linear), all the Pareto optimal points are solutions of the scalar problem maimize α U ()+α 2 U 2 () 0, 2 0 s.t for some α º 0. In particular, it is clear that by choosing α =[] T and α =[/2 2] T (or scalar multiples of these vectors), we obtain all the Pareto-optimal points. (c) Formulate this scenario as a strategic game, where the strategy set is defined by all the 0 and 2 0 that satisfy (3). Is there at least a Nash equilibrium? If so, find the Nash equilibria. Sol.: Thegameathandis< {, 2}, C, {U i ()} i=,2 >, where the set of strategy C = { R 2 +: 2 + 2, +2 2 }. Since the set of strategies is compact and the utility U i () is quasiconcave in i and continuous in, we know that the game has at least one Nash equilibrium. From the definition of Nash equilibrium, it is easy to see that all the Pareto optimal points of the problem at hand are also Nash equilibria. 6