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where <math>b = \Omega_{FB} T_{FB}</math> is the feedback resource consisted by multiplying the feedback frequency resource and the frequency temporal resource subsequently and <math>\rho_{FB}</math> is SNR of the feedback channel. Then, the required feedback resource to satisfy <math>\Delta R \leq \log_2 g</math> is
:<math> b_{FB} \geq \frac{B}{\log_2(1+\rho_{FB})} = \frac{(M-1) \log_2 \rho_{b,m} - (M-1) \log_2 (g-1)}{\log_2(1+\rho_{FB})} </math>.
Note that differently from the feedback bits case, the required feedback resource is a function of both downlink and uplink
:<math> b_{FB,min}^* = \lim_{\rho_{FB} \to \infty } \frac{(M-1) \log_2 \rho_{b,m} - (M-1) \log_2 (g-1)}{\log_2(1+\rho_{FB})} = M - 1</math>.
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