\mathbb{E}[g_i] = w_i P.\end{aligned}$$

The number of servers serving exactly $i$ users, $g_i$, can be written as $$\begin{aligned}
g_i=\sum_{p=1}^{P} \mathbbm{1}_{\{q_p=i\}}.\end{aligned}$$ Taking expectation on both sides, we have $$\begin{aligned}
\mathbb{E}[g_i] &= \sum_{p=1}^{P} \mathrm{Pr}\{ q_p=i \}\\
&=w_i P.\end{aligned}$$

The following theorem presents the average normalized worst-case delivery latency of the proposed scheme under successive transmissions, which follows by taking the expectation of both sides of Eq. and Lemma \[lemma\_prob\].

The average worst-case normalized delivery latency of the proposed scheme over all topologies under random user-server association is given by $$\begin{aligned}
\mathbb{E}[T_{sd}]=\frac{1}{\alpha} \frac{(K-t)}{(t+1)} - \frac{1}{\alpha {K\choose t}}\sum_{i=0}^{K} w_i {K-i\choose t+1} .\label{eq:avgrate}\end{aligned}$$

Since we have assumed uniform random connectivity, we have $w_i=\sfrac{{K\choose i}{P-1 \choose \rho -1}^{i} {P-1 \choose \rho}^{K-i}}{{P\choose \rho}^K}={K\choose i} \alpha^{i} (1-\alpha)^{K-i}$. The average worst-case latency is given in the following corollary.

The average worst-case normalized delivery latency with successive transmissions under uniformly random user-server association is given by $$\begin{aligned}
\mathbb{E}[T_{sd}]= \frac{K-t}{t+1} \left[ \frac{1-(1-\alpha)^{t+1}}{\alpha} \right].\end{aligned}$$

By plugging in $w_i={K\choose i} \alpha^{i} (1-\alpha)^{K-i}$ in Eq. , we obtain the above simplified expression.

Redundancy in Server Storage Capacity {#sec:incservercache}
-------------------------------------

In the analysis above, we have set the server storage capacity to $M_S=\frac{N}{\rho}$. On the other hand, for a given user cache capacity $M_U$, the 