A server system with a finite queue is the following Markovian system: the state is defined as the length of the queue. At each time unit with probability $p$ regardless of the system’s state and other jobs, a new job is added. Unless the queue is full (with $N$ jobs) then the new job is disregarded and forgotten. The time it takes to accomplish a job is a geometric random variable with probability $q$, which has the following pmf:
$$f(k) = \begin{cases} q(1-q)^k & k \geq 0, \\ 0 & k<0 \end{cases}$$
I need to draw the state diagram for $N =4$, but my computations don’t match the answers of $p (1-q)$.
My attempt:
$$ P_{i,i+1} = P(\{\text{job not complete}\} \cup \{\text{new job}\}) =\\ p \cdot P(k \neq 1)= \\ p(1 - (1-q)q)=\\p (1-q-q^2) \neq p (1-q)$$
I don’t see what I’m doing wrong here. My other calculations are similarly not matching.
Other results according to the answers: $$ \begin{align} P_{N,N-1} &= 1 - (1-p)q\\ P_{i,i-1} &= q(1-p)\\ P_{00} &= 1 - p(1-q)\\ P_{ii} &= 1 - p(1-q)-(1-p)q \end{align} $$
I believe the question is using the other definition of Geometric Random Variable, that is, $f(k)=(1-q)^{k-1}q$ for $k\in\{1,2,3,\dots\}$.
Now you should get your answer of $P(i$ people in queue at time $t$, $i + 1$ people in queue at time $t+1) = p(1-q)$.