The is certainly no optimal solution here and there are multiple trade-offs involved. For the simpler goal of maximizing the bonus in the case that the available BC always exceeds the 7.5% of the amount already invested, it is clear that the excess should go to the project with the maximum expected lifetime. The rationale is that the longer a project lasts, the more compound interest bonus any additional BC will accrue. So the allocation strategy for the excess is to minimize the maximum expected duration d of the available projects under the assumption that future investments will exactly match the 7.5% threshold. d is of the form d = -ln(f)/ln(1.225) + t_pop, with f being the fraction of completion such the f=1 is the lightbulb being exactly full and t_pop is the expected time it will take from completion to breakthrough.
So basically, the strategy is, to use the excess money to equalize the lightbulbs. But maximizing the research bonus is not the goal in play. Why? Because spending BC (beyond the 1st to keep the investment from decaying, which we will ignore from now on) on a project already in the percentages, even if it gets tripled, is not always the best course of action, as you can get the tech for free just by waiting, with the expected waiting time t_pop = 1/(f-1). (This assumes chance to pop p=f-1; might in fact be twice that, not sure). So here, we have a time/BC tradeoff at hand.
Assuming a fixed research budget, the goal is not to maximize the research bonus (and thus the total research investment), but to minimize the time to get the techs we need. So after having spent our 7.5% on all projects in the lightbulb phase (f<1), we should consider to invest the remaining BCs where they will buy us the most time. For a project with a base cost c, an additional BC will buy us (x is cost in BC and f'=df/dx)
-d' = f'/(f*ln(1.225)) = 1/(c*f*ln(1.225)) years for a project in the lightbulb phase (f<1, f'=1/c, no bonus)
(-1/(f-1))' = 3/(c*(f-1)^2) years for a project in the percentage phase (f>1, f'=3/c, triple bonus)
However, time on a cheap project is clearly worth less than time on an expensive project. If we assume the the value of the saved time is proportional to the base cost c, then our allocation depends only on f, the fraction of completion. Both marginal utility functions are strictly decreasing with f, so we will again fill up from below, allocating each additional BC to the project with the darkest lightbulb or lowest pop-percentage p=f-1 resp.
For the regime described above i.e. the LEDs of all lightbulb projects are off (no additional bonus), and the LEDs of all percentage projects are on (triple bonus), we get the indifference equation p = 0.78*sqrt(f). So assume, we have put our 7.5% into all lightbulb projects, and put all percentage projects on hold (1 BC) and let A and B be the lowest-completion lightbulb (f) and percentage project (p), then out next BC should go to A if p>0.78*sqrt(f) and to B if p<0.78*sqrt(f).
Let B be at 25% to pop, then we should fill up all lightbulb projects to 10% completion (even without bonus) before we start putting BC into B (even if it would get tripled). OTOH if A is already 40% complete, we are better off bonus-boosting our about to pop projects up to 50%, before investing more than the bonus amount into A.
Obviously, this analysis completely disregards opportunity costs, the relative value and urgency of techs and the dependencies of the tech ladder.
ignatius
So basically, the strategy is, to use the excess money to equalize the lightbulbs. But maximizing the research bonus is not the goal in play. Why? Because spending BC (beyond the 1st to keep the investment from decaying, which we will ignore from now on) on a project already in the percentages, even if it gets tripled, is not always the best course of action, as you can get the tech for free just by waiting, with the expected waiting time t_pop = 1/(f-1). (This assumes chance to pop p=f-1; might in fact be twice that, not sure). So here, we have a time/BC tradeoff at hand.
Assuming a fixed research budget, the goal is not to maximize the research bonus (and thus the total research investment), but to minimize the time to get the techs we need. So after having spent our 7.5% on all projects in the lightbulb phase (f<1), we should consider to invest the remaining BCs where they will buy us the most time. For a project with a base cost c, an additional BC will buy us (x is cost in BC and f'=df/dx)
-d' = f'/(f*ln(1.225)) = 1/(c*f*ln(1.225)) years for a project in the lightbulb phase (f<1, f'=1/c, no bonus)
(-1/(f-1))' = 3/(c*(f-1)^2) years for a project in the percentage phase (f>1, f'=3/c, triple bonus)
However, time on a cheap project is clearly worth less than time on an expensive project. If we assume the the value of the saved time is proportional to the base cost c, then our allocation depends only on f, the fraction of completion. Both marginal utility functions are strictly decreasing with f, so we will again fill up from below, allocating each additional BC to the project with the darkest lightbulb or lowest pop-percentage p=f-1 resp.
For the regime described above i.e. the LEDs of all lightbulb projects are off (no additional bonus), and the LEDs of all percentage projects are on (triple bonus), we get the indifference equation p = 0.78*sqrt(f). So assume, we have put our 7.5% into all lightbulb projects, and put all percentage projects on hold (1 BC) and let A and B be the lowest-completion lightbulb (f) and percentage project (p), then out next BC should go to A if p>0.78*sqrt(f) and to B if p<0.78*sqrt(f).
Let B be at 25% to pop, then we should fill up all lightbulb projects to 10% completion (even without bonus) before we start putting BC into B (even if it would get tripled). OTOH if A is already 40% complete, we are better off bonus-boosting our about to pop projects up to 50%, before investing more than the bonus amount into A.
Obviously, this analysis completely disregards opportunity costs, the relative value and urgency of techs and the dependencies of the tech ladder.
ignatius
