al problems, which means solving P3 is equivalent to solving the dual problem. Notice that the dual function is max The dual problem becomes {tk},{fk},{Pk} (7.6) In the following, we first obtain the optimal solutions for the given auxiliary variables (2k,Bk) and Lagrange multipliers (ak, Mk, Ok, Nk, m). Then the Lagrange multipliers are updated via gradient descent method. Lastly, the auxiliary variables are updated as well. 7.3.1 Update Pkstk, and fk Equation (7.5) can be re-organized as (7.7) (2W)amt akthkkkmaT. To maximize the dual function, m) can be decomposed into K sub-problems. Specifically, the k-th problem is (7.8) {tk},{fk},{Pkl where denotes the constant value that is irrelevant to the optimizing variables. Proposition 7.1 The optimal transmit power and duration for the k-th UE should be respectively, where [x]+ (x,0). = max Proof: Taking the derivative of the Lagrange function 4k w.r.t. Pk yields (7.9) Let ow = 0, = the optimal PK can be obtained. Notice that the optimal P* is equal to   Simi- let ow = 0, = can get the optimal expression for fk. Remark: In order to maximize EE, user k with a higher channel gain gk should transmit with a higher power Pk. This can be seen from the optimal expression of P* Notice that the similar conclusion is also drawn in [209]. 7 Mobile Edge Computing in 5G Wireless Networks For tk, the partial derivative expression of 4k w.r.t. tk becomes (7.10) Clearly, the optimization problem is a linear function of tk. Therefore, the following problem can be solved efficiently by interior point methods. (7.11a) s.t. (7.2c), (7.2d), (7.2f), (7.2g). (7.11b) 7.3.2 Update Lagrange Multipliers Now, we proceed to update the Lagrange multipliers Ak, Mk, Nk, and m. From the problem definition, with known Pk,tk, and fk, the dual problem is always convex. Specifically, min m) is an affine function w.r.t. dual variables. Thus, we can apply Ak,Mk, Ok,Nk,M the simple gradient method for the variable update. Specifically, we choose initial ak(0), and m(0) as the center of the ellipsoid which