C.29.BB:5.3 - One shared transfer preserves the computational total

Two computational cells store amounts 10 and 20. One step adds 5 from outside to cell A, removes 2 from cell B, and transfers 3 from A to B. Using one transfer amount gives

A' = 10 + 5 - 3 = 12
B' = 20 + 3 - 2 = 21
A' + B' = 33 = 30 + 5 - 2.

Suppose A subtracts 3 but B adds 2.9. The resulting total is 32.9; the inconsistent interface updates lose 0.1 in that step. Repairing the shared amount restores the aggregate balance. Whether the transfer approximates the intended transport well remains a separate question.

Now let both sides use 2.9. The result is A’=12.1, B’=20.9 and total 33. The aggregate balance holds, but the supplied transfer of 3 requires the local values 12 and 21. Agreement between the two interface amounts preserves the total; their accuracy must still be established for the requested local result.

If the cells store average densities, first multiply by their volumes to obtain the amounts. Summing unequal-volume cells’ densities would test another quantity.