bv_cvxbook_extra_exercises

# G ? where g is the dual function for 12 this can be

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˜ g ( ˜ λ ), where ˜ g is the dual function for (12). This can be expressed as p ? log ˜ g ( ˜ λ ) . How do these bounds compare? Are they the same, or is one better than the other? 4.12 Variable bounds and dual feasibility. In many problems the constraints include variable bounds , as in minimize f 0 ( x ) subject to f i ( x ) 0 , i = 1 , . . . , m l i x i u i , i = 1 , . . . , n. (13) Let μ 2 R n + be the Lagrange multipliers associated with the constraints x i u i , and let 2 R n + be the Lagrange multipliers associated with the constraints l i x i . Thus the Lagrangian is L ( x, λ , μ, ) = f 0 ( x ) + m X i =1 λ i f i ( x ) + μ T ( x - u ) + T ( l - x ) . (a) Show that for any x 2 R n and any λ , we can choose μ 0 and 0 so that x minimizes L ( x, λ , μ, ). In particular, it is very easy to find dual feasible points. (b) Construct a dual feasible point ( λ , μ, ) by applying the method you found in part (a) with x = ( l + u ) / 2 and λ = 0. From this dual feasible point you get a lower bound on f ? . Show that this lower bound can be expressed as f ? f 0 (( l + u ) / 2) - (( u - l ) / 2) T | r f 0 (( l + u ) / 2) | where | · | means componentwise. Can you prove this bound directly? 4.13 Deducing costs from samples of optimal decision. A system (such as a firm or an organism) chooses a vector of values x as a solution of the LP minimize c T x subject to Ax b, with variable x 2 R n . You can think of x 2 R n as a vector of activity levels, b 2 R m as a vector of requirements, and c 2 R n as a vector of costs or prices for the activities. With this interpretation, the LP above finds the cheapest set of activity levels that meet all requirements. (This interpretation is not needed to solve the problem.) We suppose that A is known, along with a set of data ( b (1) , x (1) ) , . . . , ( b ( r ) , x ( r ) ) , where x ( j ) is an optimal point for the LP, with b = b ( j ) . (The solution of an LP need not be unique; all we say here is that x ( j ) is an optimal solution.) Roughly speaking, we have samples of optimal decisions, for di erent values of requirements. You do not know the cost vector c . Your job is to compute the tightest possible bounds on the costs c i from the given data. More specifically, you are to find c max i and c min i , the maximum and minimum possible values for c i , consistent with the given data. Note that if x is optimal for the LP for a given c , then it is also optimal if c is scaled by any positive factor. To normalize c , then, we will assume that c 1 = 1. Thus, we can interpret c i as the relative cost of activity i , compared to activity 1. 36 Subscribe to view the full document.

(a) Explain how to find c max i and c min i . Your method can involve the solution of a reasonable number (not exponential in n , m or r ) of convex or quasiconvex optimization problems. (b) Carry out your method using the data found in deducing_costs_data.m . You may need to determine whether individual inequality constraints are tight; to do so, use a tolerance threshold of = 10 - 3 . (In other words: if a T k x - b k 10 - 3 , you can consider this inequality as tight.) Give the values of c max i and c min i , and make a very brief comment on the results.  {[ snackBarMessage ]}

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