from last time as the number of cells waiting at the heads of all
the input lines, destined for output j,
but not served at time i.
The following example should illustrate the intuitive meaning of
.
at time = i
for some fixed output j.
and
.
Now consider 
at time = i+1
for the same output j.
Considering the figure below,
at time i+1 we now have
and
.
End Example
We can write (note typo: the second = should be a +)
(eq. 1) 
where
which looks a lot like the state evolution equation for the queueing system analyzed before during our Output Queuing study.
Thus we take E{ } and let i go to infinity. Like the n's in output queuing analysis, here the B's drop out and we get
As before, in order to have B not drop out, we square to both sides to get
When we let i go to infinity, the B squared terms cancel out and the S term goes to the mean of A. We also use
,
,
,
and
to get (as i goes to infinity)
Rearranging, we get
Since A is Poisson the mean is equal to the variance so
can be simplified as
(eq. 2) =
But we also have
Hence
Since the Bi's all have the same distribution, we can write
(eq. 3) 
Setting (eq. 2) = (eq. 3), we get
Solving for E{A}, we get
But with the plus sign it would be greater than 1 which would make the system unstable.
Thus in steady state, we must have E{A} = 0.586 That is, the maximum throughput of input queueing under heavy load is 0.586.