Network throughput with random delay selected from uniform distribution
Let $f(t)$ be the probability distribution function for the probability that the current cycle (from the moment when node 1 broadcast the last message to the moment when node 1 will start broadcasting the next message) has duration $A+t$. 

For $d\leq t\leq D$, $f(t)$ must be proportional to $A+t$. We can use the fact that $\int_d^D f(t)\mathrm{d}t=1$ to deduce that $f(t)=\frac{A+t}{\lambda}$ for $d\leq t\leq D$, where $\lambda=D(\frac{D}2+A)-d(\frac{d}2+A)$. $f(t)=0$ if $t&gt;D$ or if $t&lt;d$.

Denote by $P_s$ the probability that node 1 is not broadcasting at any time during a given time period of duration $A$. It can be shown, by considering the conditional probability that node 1 does not broadcast during the given time period given that node 1 is in a cycle of length $t$ at the start of the time period, that $P_s=\int_A^D f(t) \frac{t-A}{t+A} \mathrm{d}t=\frac{1}{\lambda}\left(\frac{\max(A,D)^2}2-A\max(A,D)-\frac{\max(A,d)^2}2+A\max(A,d)\right)$. 

The probability that a given message is transmitted successfully is $P_s^{N-1}$. The throughput is then $\frac{NP_s^{N-1}}{A+\frac{D+d}2}$, since the average length of a cycle is $A+\frac{D+d}2$.