Consider your line
N_UE_rows=ceil(sqrt(randi([2,numel(unused_rows)])));
Together with your latest sample code that does [3 6] and [12 15].
With the randi starting at 2, sqrt(2) is 1.4-ish, ceil() of that is 2. So you never allocate fewer than 2 at a time.
With 3 as the maximum, ceil(sqrt(2)) -> 2 and ceil(sqrt(3)) -> 2 are the only possibilities, and the line after that one that tests the random number plus 1 against the limit will kick in pushing the value to 3. So with 3, you always allocate all 3 the first time through. Only the q=1 slot of EP{t,r,q}=E_part will be written into for this case.
With 6 as the maximum, the random choices 2, 3, and 4 map into 2, and the random choices 5 and 6 map into 3 (their square roots are between 2 and 3 and the ceil drives them to 3). So about 3/5 = 60 percent of the time you allocate 2 rows initially. When you do so that reduces the available rows to 4 and the next round sqrt(2), sqrt(3), sqrt(4) all map to 2, so the next round would be 2 as well, and that would leave 2 for the third round -- so 3/5 of the time you use up to slot q=3 of EP{t,r,q}=E_part, packing them as 2, 2, 2.
The other 2/5 = 40 percent of the time, you allocate 3 rows initially. When you do so, that reduces the available rows to 3, and the next round sqrt(2), sqrt(3) map to 2 but the special +1 case will cause it to grab the full 3 remaining. So if 3 are allocated the first time, then you use up to slot q=2 of EP{t,r,q}=E_part, packing them as 3, 3.
But when you pack as 3, 3, you do not erase any existing q=3 cell that had the 2 put into it from a 2, 2, 2 packing. So if you previously had a 2, 2, 2 packing, then after the 3, 3 packing, the cells have 3, 3 (just written) and then a left-over which is 3, 3, 2 = 8 -- two more than what you are expecting.
Now, you do effectively erase throughput1 each time through (you allocate zeros but you allocate less spaces than will be written to.) So this throughput1 that is 2 rows too large will not affect the next iteration. But... every once in a while, the 3, 3, 2 packing happens on the last iteration you do, the one for the maximum it, t and r. And when it does, you see throughput1 as being 2 rows too large after the end of the routine.
With larger array sizes, the number of extra rows could end up larger -- but you are biasing the random integer generation towards lower values, so the extra will tend to be 2 when it is there at all. Could be more though.
