Intuition of the Mean wait time in queuing system In queuing theory, (with a single queue and a single server) writing $A$ for the service rate (of customers) and $B$ for the arrival rate (of customers) we know that the average time a customer waits in the system is given by,

$$W=\frac{1}{A-B}$$

What is the intuitive interpretation of this equation? 

What I understand from basic intuition is that $A$ customers gets service in 1 sec (as $A$ is the service rate) so, one customer should get service in $1/A$ seconds. 

Now, $B$ means, $B$ customers comes in 1 second (the arrival might be bursty or not depending on the probability distribution). So, $(A-B)$ is , how many more customers can the server serve per second, right? 

So, why is the waiting time $W=1/(A-B)$?

How can I understand this relationship intuitively?
I am canoeing up a river at a speed (relative to the water) of $A$ miles per hour, and the current in the other direction is $B$ miles per hour. If $A\gt B$, my speed relative to the shore line is $A-B$, so it will take me $\frac{1}{A-B}$ hours to travel $1$ mile. 

Or if you prefer I am running up a down escalator.
The intuition behind the delay formula for a M/M/1 queue is based on the stability assumption requiring that during my waiting delay same numbers of customers got service and arrived to the queue. Customers arrive with a rate of $B$ while being serviced at a rate of $A$. I joined the queue and noticed some customers in front of me. Let my aggregate queueing and service delay is $W$, then during my waiting time, $A\times W$ customers were served (including myself). Provided the system is stable the same number of customers shall arrive during time W. However, this number equals to $1 + W \times B$, where "1" counts for myself. Equating both numbers (what flows in shall flow out for stability) we get:

$1 + WB = WA$

and

$W = 1/(A - B)$