When should consume(+) not be played?

Solution 1:

Definition: effective orb count.

Define your effective orb count n to be the number of currently active orbs plus the number of loops. If you have gold plated cables, then the loops are counted twice. Define some variables:

n    effective orb count
nt   effective orb count (this turn)
g    gold plated cables 
l    number of loops 
o    number of orbs 
f    current focus
u    is the card upgraded?

By noting how the cables and loops interact with the first orb, there can be a notion of 'effective orb count' (or the equivalent number of regular orbs).

Because loop only triggers at the start of your turn, there can sometimes be cases where the benefit is a net positive on the next turn, but not yet on the current turn. This can be important in life-or-death situations. First, let's consider the 'over the long run' effect. Let's consider block only for now. (the reasoning for damage will be the same) The benefit is gaining focus, giving block for each effective orb. Of course, we have to reduce the number of orbs by one, so we actually gain block via for n-1 orbs (so multiply n-1 by our focus gained, 2 for the base card, 3 for the upgraded version). The loss is the loss of a single orb, providing block equal to focus plus 2 (for block).

Solution:

Simplifying the versions of the equation described above results in:

Select the proper criterion to use from equations (4, ..., 7). Iff it is true, then playing a consume will raise your block by one or more. Want to know the exact value gained? Subtract the rhs from the lhs in these same equations.

These same equations for lightning orbs:

Subtract 3 more from the lhs to get the solution for dark orbs:

As for combinations of multiple orbs: there's only a single orb that actually is 'lost': your leftmost orb. Thus, look at your left-most orb and apply the solution for that. This is of course assuming that, were frost and damage orbs to be mixed, that you value each thing (damage, block) equally. Which might not be exactly the case, in which case for the generic, unequally valued mixed case the solution would be probably more along the lines of an individual analysis of the specific situation, and not in the scope of this generic answer to solve.