Suppose in the waterfilling algorithm that instead of log(1 + a_iP_i) the utilit
ID: 3828174 • Letter: S
Question
Suppose in the waterfilling algorithm that instead of log(1 + a_iP_i) the utility function was tan^-1(a_iP_i). i.e. the waterfulling optimisation problem became max Sigma_i tan^-1(a_i P_i) Subject to P_i greaterthanorequalto I = 1, ellipsis, N P_1 + ellipsis + P_N = P i. Determine the equivalent of equation (1) on slide 13 of the notes on waterfilling. ii. What interpretation can be given to the nature of the utility function used? Explain. Solution Thus, if 1/lambda - 1/a_i > 0 then P_i = 1/lambda - 1/a_i and if 1/lambda/1/a_iExplanation / Answer
thus, if 1/lamda - 1/ a1 > 0 then p1 =1 / landa -1 / a1 and
if 1/lamda - 1/ a1 < 0 then p1 = 0
we can write this as
p1 = max ( 1/lamda - 1/ a1 ,0) = [ 1/lamda - 1/ a1] max (x,0) = x+
to solve this need to find lamda