### Evaluation of resource utilization

The probability that UE transmits data in a scheduling cycle is (1 − *p*^{K}) which implies the average number of subframes in which UE successfully transmits data in a scheduling cycle is *L*(1 − *p*^{K}). Therefore, we have

$$ \rho \left(K,L\right)=\frac{L\left(1-{p}^K\right)}{L+K-1}. $$

(1)

### The optimum setting

In this sub-section, we determine the optimum selection of *K* and *L* which maximizes *ρ*(*K*, *L*). We have the following lemmas.

**Lemma 1.** Given *K*, *ρ*(*K*, *L*) is an increasing function of *L*.

Proof: We shall prove, for a given *K*, *ρ*(*K*, *L*) is an increasing function of *L* by showing that *ρ*(*K*, *L* + 1)/*ρ*(*K*, *L*) ≥ 1. We have

$$ {\displaystyle \begin{array}{c}\ \frac{\rho \left(K,L+1\right)}{\rho \left(K,L\right)}=\frac{\left(L+1\right)\left(L+K-1\right)}{L\left(L+K\right)}\\ {}=\frac{N\left(L+K+d\right)+\left(L+K\right)-\left(L+1\right)}{N\left(L+K\right)}\\ {}=\frac{L\left(L+K+d\right)+K-1}{L\left(L+K\right)}\ge 1.\end{array}} $$

(2)

This completes the proof of Lemma 1.

According to Lemma 1, to increase the resource utilization of the LAA network, *L* should be chosen as large as possible. In some countries, there is a restriction on the maximum channel occupancy time (MCOT) in each scheduling cycle. For example, in Japan, MCOT is set to be 4 ms [14]. To maximize *ρ*(*K*, *L*) without violating regulations, the value of *L* should be chosen to be MCOT.

**Lemma 2.** It holds that, given *L*, *ρ*(*K* + 1, *L*) > *ρ*(*K*, *L*) if and only if 1 + *p* + … + *p*^{K} < (*L* + *K*)*p*^{K}.

Proof: Since *ρ*(*K*, *L*) = *L*(1 − *p*^{K})/(*L* + *K* − 1), we have

$$ {\displaystyle \begin{array}{c}\ \rho \left(K+1,N\right)>\rho \left(K,N\right)\iff \frac{L\left(1-{p}^{K+1}\right)}{L+K}>\frac{L\left(1-{p}^K\right)}{L+K-1}\\ {}\iff \left(1-{p}^{K+1}\right)\left(L+K-1\right)>\left(1-{p}^K\right)\left(L+K\right)\\ {}\iff {p}^{K+1}-1>\left(L+K\right)\left(1-{p}^K-1+{p}^{K+1}\right)\\ {}\iff 1-{p}^{K+1}<\left(L+K\right){p}^K\left(1-p\right)\\ {}\iff 1+p+\dots +{p}^K<\left(L+K\right){p}^K.\end{array}} $$

(3)

Therefore, Lemma 2 is true. Note that in the above proof, the symbol ⇔ denotes if and only if.

**Lemma 3.** Given *L*, *ρ*(*K*, *L*) ≤ *ρ*(*K* − 1, *L*) implies *ρ*(*K* + 1, *L*) ≤ *ρ*(*K*, *L*).

Proof: Assume that *ρ*(*K*, *L*) ≤ *ρ*(*K* − 1, *L*).According to Lemma 2, it is true that 1 + *p* + … + *p*^{K − 1} ≥ (*L* + *K* − 1)*p*^{K − 1}. Since (*L* + *K* − 1)*p*^{K − 1} ≥ (*L* + *K* − 1)*p*^{K}, we have

$$ {\displaystyle \begin{array}{c}1+p+\dots +{p}^K=\left(1+p+\dots +{p}^{K-1}\right)+{p}^K\\ {}\ge \left(L+K-1+d\right){p}^K+{p}^K\\ {}=\left(L+K\right){p}^K,\end{array}} $$

(4)

which, according to Lemma 2 again, implies that *ρ*(*K* + 1, *L*) ≤ *ρ*(*K*, *L*). End of proof, Lemma 3.

Based on Lemma 1 to Lemma 3, we have the following theorem for finding the optimum values of *K* and *L* which maximize *ρ*(*K*, *L*).

**Theorem 1.** *The optimum values of K and L, denoted by K*^{∗}*and L*^{∗}*, which maximize ρ*(*K*, *L*) *satisfy L*^{∗} = MCOT *and K*^{∗} = min {*k*| *ρ*(*k* + 1, *L*^{∗}) ≤ *ρ*(*k*, *L*^{∗})}.