Let c1 = K1+ββ
1 K
PK k=1ˆgk
2
and c2 = 1+β1
1 K
PK k=1gˆ2k
, then the Lagrangian of the problem (3.33) is
L(Pt, µ1, µ2) =− ζ2c1(P − Pt)Pt
ζ2c2(P − Pt)Pt+ ζPt+ ζK(P − Pt) + K + µ1(Pt− P ) − µ2Pt
and the associated KKT conditions are
− ζ2c1[ζ(K− 1)Pt2− 2K(ζP + 1)Pt+ KP (ζP + 1)]
(ζ2c2(P − Pt)Pt+ ζPt+ ζK(P − Pt) + K)2 + µ1− µ2 = 0 (G.1)
µ1(Pt− P ) = 0, µ1 ≥ 0 (G.2)
µ2Pt = 0, µ2 ≥ 0 (G.3)
Since the training power have to be greater than 0, we have µ2 = 0. If µ1 > 0, then Pt = P , but then (G.1) leads to µ1 < 0 a contradiction. Therefore, we have µ1 = µ2 = 0 and P > Pt > 0. From (G.1), we have
ζ(K− 1)Pt2− 2K(ζP + 1)Pt+ KP (ζP + 1) = 0
⇒ Pt= K(ζP + 1)±p
K(ζP + 1)(ζP + K) ζ(K− 1)
where we take negative term since positive term can not satisfy the constraint P ≥ Pt. Let a = K(ζP + 1) and b = ζP + K, then we have Ptopt = (a−√
ab)/[ζ(K − 1)] and
P − Ptopt = (b−√
ab)/[ζ(K− 1)], and from (3.32), (3.35) follows:
J(P, K) = σ2θ 1 + c1[(a + b)√
ab− 2ab]
c2[(a + b)√
ab− 2ab] + (K − 1)2√ ab
!−1
= σ2θ
1 + c1
c2+ (√(K−1)2 a−√
b)2
−1
where the first equality uses that
(K− 1)(a −√
ab) + K(K − 1)(√
ab− b) + K(K − 1)2
=(K − 1)[(K − 1)√
ab + a− Kb + K2− K
| {z }
=0
] = (K − 1)2√ ab.
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