The tie-break criteria Board Count and Berlin are both used in FIDE team competitions.
The Berlin and Board Count tie-break systems both assign weights to the points scored on each board. Points obtained on the first board are considered more important, for tie-break purposes, than those obtained on the last board. Therefore, a more appropriate name for these methods would be “Weighted Board”. Normally one would expect higher weights to be assigned to the first board, but the Board Count does this in a rather indirect way.
Orion implements only the Berlin system and not the Board Count as defined in the FIDE Handbook C.07 because it is more intuitive for players. Below it is shown that, although the two methods produce different numerical values, they actually generate the same ranking under the conditions in which Board Count is applicable. Therefore, the two methods are perfectly interchangeable.
The article also provides the conversion formula between the Berlin score and the Board Count score:
Berlin = (B+1)Game_Points − Board Count
where B is the number of boards.
Equivalence between Board Count (BC) and the Berlin weighted sum
Let a team competition be played on $B$ boards.
For a given team $T$ and a board $b \in {1,\dots,B}$, denote by
$$ v_b(T) $$
the total number of game points scored by team $T$ on board $b$ over all rounds, regardless of who was playing on that board.
FIDE Board Count (BC)
According to the FIDE Handbook, C.07 Tie-Break Regulations, Article 12.1 ("Board Count (BC)"), the tie-break is computed as
$$ BC(T) = \sum_{b=1}^{B} b \, v_b(T) $$
and the lower the sum, the higher the ranking of the team.
Moreover, it can only be used when all tied teams have scored the same number of game points.
Berlin weighted sum
Define the Berlin-type weighted sum
$$ Ber(T) = \sum_{b=1}^{B} (B-b+1)\, v_b(T) $$
where higher boards receive larger weight (board 1 gets weight $B$, board $B$ gets weight 1).
Teams are ranked by descending $Ber(T)$.
Theorem (Equivalence on tied teams)
Consider two teams $A$ and $C$ that are tied on total game points:
$$ \sum_{b=1}^{B} v_b(A) = \sum_{b=1}^{B} v_b(C) $$
Then
$$ BC(A) < BC(C) \Longleftrightarrow Ber(A) > Ber(C) $$
Hence, within any tie group with equal total game points (the condition required by the FIDE rule for using BC), ranking by BC (ascending) and ranking by Berlin (descending) are identical.
Proof
Let
$$ GP(T) = \sum_{b=1}^{B} v_b(T) $$
be the total game points scored by team $T$.
Using the identity
$$ (B-b+1) = (B+1) - b $$
we obtain
$$ Ber(T) = \sum_{b=1}^{B} ((B+1)-b)\, v_b(T) $$
$$ Ber(T) = (B+1)\sum_{b=1}^{B} v_b(T) - \sum_{b=1}^{B} b\, v_b(T) $$
$$ Ber(T) = (B+1)\,GP(T) - BC(T) $$
Now suppose $A$ and $C$ are tied on total game points, so
$$ GP(A) = GP(C) $$
Then
$$ Ber(A) > Ber(C) $$
$$ \Longleftrightarrow (B+1)GP(A) - BC(A) > (B+1)GP(C) - BC(C) $$
$$ \Longleftrightarrow -BC(A) > -BC(C) $$
$$ \Longleftrightarrow BC(A) < BC(C) $$
which proves the equivalence.