问一个排队论的问题

对于一个M/G/1系统
一个包A到达的时候  server正在服务另一个包B 包B有一个剩余的服务时间R
包A的等待时间记为W
根据数学推导 我们可以知道 E(R)=E(W)*(1-rou)  其中 rou=lamda/mu
1-rou 是server 空闲的概率
请问这个式子有啥物理解释
囧 完全想不到 ：（
.175

It shows the simple fact the waiting time in queue for A is exactly the
residual service time for B if A is facing an empty queue upon arriving
at the system.

A little confusing... Since the probability that A is facing an empty server is  1-\rou, in this case R=0.
A possible interpretation is \rouE[W] is the cumulative service time for the queued jobs, but it is not straightforward and needs Little's law.

Sorry, my statement was imcomplete. The event I described is meant to say that
the queue is empty but the server is busy. In any case, a direct interpretation
is not that obvious.

One thing fotgot to mention (too sleepy last night): the formula is just a
variation of Pollaczek-Khintchine formula, so any interpretation that applies
to PK would work for this one :)