2.2.1
Convex Optimization Problems
Normally, an optimization problem has the following form [59, 60]
min f0(x) (2.28a)
subject to: fi(x)≤bi, i= 1, . . . , m (2.28b)
hi(x) = 0, i= 1, . . . , p. (2.28c)
Here, the vector x = [x1, . . . , xn] is the optimization variable of the problem, the
function f0 : Rn → R is the objective function, the functions fi : Rn → R, i =
1, . . . , m, are the inequality constraint funtions, the constantsb1, . . . , bmare the limits
for the constraints, and the functionshi :Rn→R,i= 1, . . . , pare called the equality
constraint functions [59]. If a vectorx∗ provides the minimum objective value among all feasible vectors which satisfy the constraints, then it is an optimal solution.
Then, a convex optimization problem has the following form [59]
min f0(x) (2.29a)
subject to: fi(x)≤0, i= 1, . . . , m (2.29b)
aTi x=bi, i= 1, . . . , p, (2.29c)
wheref0, . . . , fm are convex functions. Comparing (2.29) with the general form (2.28),
one can notice that the convex problem has the following requirements: 1) the objec- tive function must be convex; 2) the inequality constraint functions must be convex; 3) the equality constraint functions hi(x) = aTi x−bi must be affine [59]. Here, we
note that a function fi :Rn→R is convex if its domaindom fi is a convex set and
A functionfi is strictly convex if strict inequality holds in (2.30) whenever x6=yand
0 < α < 1. Further, a function fi is concave if −fi is convex, and strictly concave
if −fi is strictly convex. Since an affine function always holds the equality in (2.30),
therefore one can note that an affine function is both convex and concave [60]. Normally, if we can formulate a practical problem as a convex optimization prob- lem, then we can solve it efficiently. However, sometimes the formulations can be nonconvex [60]. For example, if f0 is quasiconvex instead of convex, then the prob-
lem (2.29) becomes a quasiconvex optimization problem [61]. Here, we note that a function f : Rn → R is called quasiconvex if its domain and all its sublevel sets Sα ={x ∈dom f | f(x)≤ α}, for α ∈R, are convex. The sublevel sets of convex
functions are convex, therefore convex functions are quasiconvex. But the converse is not true. For some quasiconvex problems following specific structures, e.g., convex fractional programming [62,63], they can be transformed into equivalent convex prob- lems and then get solved efficiently. The calculated optimal solutions can be proved to be optimal for the original quasiconvex problems [62, 63].
2.2.2
Lagrangian Dual and KKT Conditions
In this section, we briefly introduce the Lagrangian dual and the Karush-Kuhn-Tucker (KKT) condtions, which will be applied in Chapter 3 and Chapter 4 to solve the optimization problems and to derive the optimal power allocation strategies.
The basic idea in Lagrangian duality is to take the constraints in (2.28) into account by augmenting the objective function with a weighted sum of the constraint functions [59]. The LagrangianL:Rn×Rm×Rp →Rassociated with the problem (2.28) is defined as follows [59] L(x, λ, v) =f0(x) + m X i=1 λifi(x) + p X i=1 vihi(x). (2.31)
Here, λi is the Lagrangian multiplier associated with the ith inequality constraint
fi(x) ≤ 0, and vi is the Lagrangian multiplier associated with the ith equality con-
straint hi(x) = 0. The vectors λ and v are called the Lagrangian multiplier vectors
associated with the optimization problem [59].
Then, the Lagrangian dual functiong :Rm×Rp →Ris defined as the minimum
value of the Lagrangian over x: forλ∈Rm, v ∈Rp,
g(λ, v) = inf x L(x, λ, v) = infx f0(x) + m X i=1 λifi(x) + p X i=1 vihi(x). (2.32)
The Lagrangian dual functiong(λ, v) is concave even when the orginal problem (2.28) is not convex, since the dual function is the pointwise infimum of a family of affine functions of (λ, v) [59]. For each pair (λ, v) withλ014, the Lagrangian dual function
provides a lower bound on the optimal value p∗ of the optimization problem (2.28).
Then, in order to find the best lower bound that can be obtained from the Lagrangian function, the following optimization problem needs to be solved [59]:
max g(λ, v) (2.33a)
subject to: λ0. (2.33b)
This problem is called the Lagrangian dual problem associated with the problem (2.28). Correspondingly, the orginal problem (2.28) can be called as the primal prob- lem. Apparently, the Lagrangian dual problem (2.33) is a convex optimization prob- lem, since the objective function is concave and the constraint function is convex [59]. This does not depend on the convexity of the primal problem (2.28) [59].
Then, we assume that the functionsf0, . . . , fm,h1, . . . , hp are differentiable. From
[59], we note that if a convex optimization problem with differentiable objective and constraint functions satisfies Slater’s condition, then the KKT conditions provide necesseary and sufficient conditions for optimality. Hence, by assuming thatfi func-
tions are convex andhi functions are affine, and x∗,λ∗, v∗ are any points that satisfy
the KKT conditions fi(x∗)≤0, i= 1, . . . , m (2.34a) hi(x∗) = 0, i= 1, . . . , p (2.34b) λ∗i ≥0, i= 1, . . . , m (2.34c) λ∗ifi(x∗) = 0, i= 1, . . . , m (2.34d) ∇f0(x∗) + m X i=1 λ∗i∇fi(x∗) + p X i=1 vi∗∇hi(x∗) = 0, (2.34e)
then x∗, λ∗,v∗ are primal and dual optimal, with zero duality gap [59].