With this assumption,the constant u is calculated (by substituting (0.11)into (0.12)as Then we have a"a-pP+空 (0.15a) Using this value of u,the power allocated to the ith subchannel is given by -(-) i=1,2,m-p+1 (0.15b) If the power allocated to the channel with the lowest gain is negative (i.e.P<0), then we discard this channel by setting P=0 and return the algorithm with the iteration count p=p+1.即迭代执行(0.15a)和(0.15b),将总功率P在剩余的(m-p叶1)个 子信道之间进行分配。迭代计算直到获得的所有P≥0或p=m为止。 >7 使用。With this assumption, the constant μ is calculated (by substituting (0.11) into (0.12)) as 1 0 1 m p i i N μ P λ − + = ⎛ ⎞ ⎜ ⎟ − = ⎝ ⎠ ∑ Then we have 1 0 1 1 1 1 m p i i P N m p μ λ − + = ⎛ ⎞ = + ⎜ ⎟ − + ⎝ ⎠ ∑ (0.15a) Using this value of μ , the power allocated to the ith subchannel is given by 0 , 1,2,., 1 i i N P i mp μ λ ⎛ ⎞ = − = −+ ⎜ ⎟ ⎝ ⎠ (0.15b) If the power allocated to the channel with the lowest gain is negative (i.e., 1 0 P m p − + < ), then we discard this channel by setting 1 0 P m p − + = and return the algorithm with the iteration count p = p+1. 即迭代执行(0.15a)和(0.15b)，将总功率 P 在剩余的(m-p+1)个 子信道之间进行分配。迭代计算直到获得的所有 0 Pi ≥ 或 p=m 为止