 problem that can be easily solved. Given the fixed Lagrange multipliers A, v, and , the partial derivative of rn and dn are constant, which means that the objective function is monotone increasing or decreasing with respect to r and dn. Therefore, the optimal solution is (9.9) 9 Access Control and Handoff Policy Design for RAN Slicing (9.10) We now study the solution to P3(2). We give the Lagrange dual problem of P3(2) with respect to constraints (9.2-6) and (9.2-7) as follows: P4 v, , N, 0) Ill inf L3 (A, v, , n, 0,x) (9.11) s.t. Constraints (9.2-1), (9.2-6) and (9.2-7) where (9.12) jes KEB and vectors n and O are Lagrange multipliers containing the elements Nj 0 and 0j,k 0 respectively. Here, P3(2) is a linear programming, and thus strong duality holds again, i.e. inf = sup h(A, v, , n, 0). Based on (9.12), we rewrite L3 (A, v, , N, 0,x) as (9.13) where (,,,,,)= (9.14) jES kEB As there is no cross-term of x in L3 (A, v, , n, 0,x), we can change the computation order as h(A, v, , = , + (9.15) jes keB Hence, solving P4 is equivalent to solving the following subproblem P4(n) for each UE n separately. P4(n) : (9.16) s.t. Constraints (9.2-6) and (9.2-7) Rewriting function we have Dn(2,u (Amtun) (9.17) 9.1 A Framework of User Access Control for RAN Slicing Since we want to find a binary solution of xisk for fixed Lagrange multipliers A, v, , N, 0, solving P4(n) can be described as: for UE n, we choose an NS j* and BS k* from the admissible set of NSs and BSs respectively to maximize the value of (An - + Then we design a QoS-AC policy based on the solution to P2 to find the QoS-AUS in Def- inition 9.2. Let Y and dn be the optimal solution of UE N. If n = 0 and dn = 0, it means that there is no QoS degradation of UE n. In other words, this UE can be admitted for the net- work. Hence, based on this observation, we design the QoS-AC policy, where the UEs with = 0 and dn = 0 can be accepted by the network, and others are rejected due to limited resources. Hence, the admissible set of UEs can be expressed by AQ-A = {n : =