Author Topic: A formal look at various efficiencies  (Read 7057 times)

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Iranon

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Re: A formal look at various efficiencies
« Reply #15 on: May 27, 2017, 05:15:54 AM »
For the upcoming C# version, we can incorporate fuel efficiency by engine size since that now scales consistently. The first graph (and performance-optimal fuel-to-engine ratio of 2:5) still stands if size of an individual engine remains fixed; if the number of engines remains fixed we get

(((x(1-x)^2.5)/(10/(1-x)^0.5))^0.4)/(72804525*(1.5)^0.2/487671196)

with a performance-optimal ratio of 1:3; this also applies to missiles.
 

Iranon

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Re: A formal look at various efficiencies
« Reply #16 on: June 20, 2019, 04:11:33 PM »
Something always seemed off when applying the concepts in this thread to my commercial designs. Smaller propulsion plants consistently looked better than the supposed optimum. Apparently I'm an idiot, the problem was the underlying assumption that speed is relevant in itself.
That assumption is fine for how I design warships. Too slow and we can't control an engagement, doubling our firepower may not be enough to compensate. Too weak and our speed is only useful for running away, leaving behind shattered fleets and glassed worlds.
However, When the expected service is hauling goods or colonists year after year, speed and capacity are interchangeable.

Assume we have a given tonnage available for engines or cargo capacity. Defining x as proportion of engine tonnage, our throughput is proportional to  x(1-x); 4x(1-x) if standardising for the maximum, achieved at x=0.5. Using a given type of engine, our relative fuel efficiency is (1-x): A colony ship with 3 size-50 engines and 1 cryogenic storage has the same throughput as the reverse, but needs to make 3 trips instead of one, consuming 3 times as much fuel.

If we can freely adjust our power multiplier, throughput scales linearly with it while fuel use scales with a power of 2.5. Therefore, we can achieve the best fuel efficiency by maximising (x(1-x))^2.5*(1-x), achieved at x=5/12 or 41.67%; deviating from this is less fuel-efficient than adjusting power multiplier. Standardised for 1 as the optimum: (x*(1-x))^2.5*(1-x)/0.01699

Of course, we don't have to stick to this slavishly. I may choose less tonnage in engines if I don't have ultra-low-power engines researched yet but still want low fuel use, or when even our lowest-power engines are expensive compared to the payload - quite likely for freighters. If we can't adjust power multiplier downwards, we may simply look at throughput*fuel-efficiency: (x(1-x)*(1-x)/0.148148148 .

Similarly, I may choose more tonnage in engines if I can spare a little fuel and am stuck at the 0.5 limit for commercial engines. Or when the ship is held in reserve for sudden priority use, rather than continuous service. Or when the payload is much more expensive per ton than engines of the optimal fuel efficiency (likely for colony ships).

All graphs in one: 1) throughput*fuel efficiency with fixed-power engines, 2) unadjusted throughput, 3) throughput at constant fuel efficiency with freely scaleable engines:
x(1-x)*(1-x)/0.148148148, 4(x)(1-x), ((x)(1-x))^(5/2)*(1-x)/0.01699
 

Offline misanthropope

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Re: A formal look at various efficiencies
« Reply #17 on: June 20, 2019, 07:46:01 PM »
mission tonnage times speed / (weighted cost)^2 is the way to go imo

msrp beeps don't really cut it as a measure of cost; corbomite is a whole lot less valuable than gallicite.  if you're such a horrible penny-pincher that you are generally zero slack in your build capacity some overhead needs to be applied too- which can *really* change your optimum!  of course, if you're NOT constrained by capacity forget about optimizing your ships, you've got bigger inefficiencies to worry about.
 

Iranon

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Re: A formal look at various efficiencies
« Reply #18 on: June 22, 2019, 05:39:34 PM »
Could you elaborate on that? If I understand you correctly, that looks very odd to me:
Using ICF engines (equivalent to something lower if you weigh Gallicite cost higher than other minerals) and multipliers of 0.2 to 0.5 I'd have

x*(1-x)*0.5/(500*x*0.5^2+5*(1-x))^2, x*(1-x)*0.4/(500*x*0.4^2+5*(1-x))^2, x*(1-x)*0.3/(500*x*0.3^2+5*(1-x))^2, x*(1-x)*0.2/(500*x*0.2^2+5*(1-x))^2
for freighters and

x*(1-x)*0.5/(500*x*0.5^2+100*(1-x))^2, x*(1-x)*0.4/(500*x*0.4^2+100*(1-x))^2, x*(1-x)*0.3/(500*x*0.3^2+100*(1-x))^2, x*(1-x)*0.2/(500*x*0.2^2+100*(1-x))^2
for colony ships.

I suppose I don't get why the weighted cost in the denominator should be squared.
 

Offline misanthropope

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Re: A formal look at various efficiencies
« Reply #19 on: June 22, 2019, 09:23:22 PM »
i spend too much non-recreational time staring at a screen these days to try to parse through the formulae you posted, even though that is the kind of thing that generally appeals to me.  i suspect the optimization criterion i mooted doesn't admit a closed analytical form.

you're absolutely right about cost not being squared; i merged a condition from another type of game into the idea i was trying to express (where speed is essentially fixed, as in aurora's forerunner starfire, you want to optimize offense*defense/cost^2 so as to not want to duct tape two small ships together in order to improve them).

as well, using "mission tonnage" was not just vague but misleading.  it's value of the mission package, not the tonnage.  generally to simplify analysis i will design a ship, and look at only the hull, engineering, quarters, and propulsion in terms of "cost per (free tonnage*kkps)" and then hash out the mission package in a subsequent step.

obviously im putting forward a plan that is highly sensitive to subjective inputs.
 what can i say?  the entire field of finance operates the same way.  the alternative is "a solution that is neat, simple, and wrong."