How can I come up with a simple diminishing return equation?

Question

There are formulas out there for a diminishing return equation; however, those usually involve exponential. What other ways are there for coming up with such an equation? Take, for example, the following test case -- One farm produces 10 food, for every 10 farms produced, the production rate drops by 5%.

Answer 1

For formulating a diminishing returns equation, I'd immediately think fractions.

This is a graph of \$y=\frac{1}{F}\$

\$y\$ will get smaller as \$F\$ gets larger. This will give you a steady drop-off that never reaches 0. From this you can transform it to get the sort of curve that you want. Using numbers > 0 will always give positive output that is never 0.

Honestly, I'd recommend going to WolframAlpha and putting in some equations and looking at the graphs that it draws to see if it gives the curve that you want. Other than that, read up of linear and quadratic equations to be able to quickly figure out what it is that you want to alter in a formula. This is because modelling graphs through equations is a bit of a big topic, and if I could explain it here perfectly I'd sell that explanation to some maths teachers first.

Basically, for linear graphs, remember \$y=mx+c\$. \$m\$ is the gradient, and can be positive or negative depending on what you need, and \$c\$ is the point at which it intercepts the y axis. \$x\$ is your input variable and \$y\$ is your output.

This is a graph of \$y=mx+c\$ where \$m=1\$ and \$c=0\$

For quadratic graphs, it gets a bit more complicated, so I'll be a bit vague and you'll have to read up on the specifics yourself. Khan Academy is a really good resource for teaching this. It's of the general form \$y=ax^2+bx+c\$. \$c\$ is still the y intercept, and you can tweak it to "lift" the graph. \$a\$ and \$b\$ both affect the curve similarly, but to different degrees.

This is \$y=-x^2+2x+10\$. Note the \$-x^2\$, which makes the curve inverted.

Basically, play around with the graphs until you get what you want, although I highly recommend reading up on it more if you want to design the experience quickly and cleanly. Basic equations are important to games and really interesting.

Other things to note are exponential and logarithmic graphs, i.e. graphs of \$y=e^x\$ and \$y=ln(x)\$ to get rapidly increasing and rapidly decreasing graphs depending on the transformation. As well as this, vectors and transformations are helpful, as they describe what you're doing the the "base" graph.

Answer 2

Diminishing returns = decreasing derivative

• Since you still want some returns even at higher levels means that the derivative should be positive, otherwise building more farms would decrease the food production (which might even make sense if you take into account logistics and upkeep costs)
• It should approach zero assymptotically, if it goes towards a non-zero value you will end up with constant increase per farm at some level
• depending on how fast it goes to zero you can have an upper limit or an unbounded function

So what do you need to do? Pick a function that fits the above criteria and integrate it.
The simplest choice for this task is g(t,n) = 1/(t+1)^n where n=1 marks the boundary between ever increasing and bounded functions.
The integral of g from 0 to x is what you need: f(x,n) = ((x+1)**(1-n) - 1)/(1-n)
Here is how it looks for different n

And here its normalized to the same final value

By changing the exponent n you can easily adjust the balancing
Note: The derivative here is the production per farm, while the integral is the total production for a number of farms

Answer 3

In general, a linear equation will start with y = mx + b, where b is your starting value, and mx is how you adjust the starting value as x increases.

So the first part of your equation, the b, will be 10 because you want farms to start at 10 food.

y = mx + 10

Next, in your case, you want to adjust the food by produced by every ten farms. So you will need to divide by ten to get an equation that works for every ten farms (assuming that the x / 10 returns an integer, ie 13 / 10 = 1:

y = m * (x / 10) + 10.

So finally, we need to figure out how we want the food to change for every x / 10 farms. In your case you want it to decrease by 0.5 (5% of 10), which is linear. So we get:

y = -0.5 * ( x / 10 ) + 10.

So for farm x = 5, we get 5 / 10 = 0, 0 * -0.5 = 0, 0 + 10 = 10. For farm x = 11 we get 11 / 10 = 1, 1 * -0.5 = -0.5, -0.5 + 10 = 9.5, for farm 23, we would get 9.0.

Then you just need to compute the total food for all farms.

y = 0
for( x = 0; x < totalFarms; x++ )
{
    y += -5 * ( x / 10 ) + 10;
}

But maybe you by 5%, you wanted it to decrease by 5% of the previous value. Ie, 10, 10 * 0.95 = 9.5, 9.5 * 0.95 = 9.025 (in this case, the amount we decrease by gets less and less). So lets modify the equation. 5%is an exponential type increase, and the exponential formula is y = b*m^x.

We still have b = 10, and we need to do our divide by 10 trick. So we have y = 10*m^(x/10). m is 0.95, since we want to take 95% of the value each time. So the equation for farm x is y = 10*0.95^(x/10).

Answer 4

Would a linear diminishing return do? \$production\ per\ farm = (1 - (0.05 \frac{ f}{10}) ) * production\ rate\$.

This gives a total production (rate * # of farms) peak at f = 100.

Answer 5

You might want to consider an algorithmic solution that matches the situation.

That is, consider why there are diminishing returns in your game situation, and model those.

Multiple facilities of the same type might have diminishing returns in that there might be other resources or facilities which they depend on, or which result in bottlenecks, or other limiting situations, such as a road network, or available workers or transportation or fresh water or electricity or whatever.

One farm can produce 10 food per day in ideal circumstances, but it requires two farmer-hours per day as well. It also requires one fresh water per food per day, and its own well only provides up to 5 water per day. The rest must be taken from an adjacent stream or river or brought in by transportation. And getting the food to where it needs to be to be useful may also be an issue. Etc. Remove some or add more depending on what you want to represent, but these can be much more interesting and meaningful reasons, which add interest and value to your other game systems, as opposed to an artificial mathematical formula not based on other game elements.

Answer 6

If you want a generalized equation to mess around with, you could use a cosine graph: A*cos(Bx+C)+D

But modify it to half a period, so it would include the pseudo exponential rise at the beginning, then a short period of linear increases, to a finally diminishing point of returns. The only problem with this is that it would require to to create a impassable ceiling. So after a certain amount of farms you would see no increase.

The Image below is a graph of increases in pace for 30 min runs, doing the exact same workout in preparation. It's obviously not perfect, but you might be able to work off this to find what you're looking for.

Answer 7

Here is standard capacitor charge voltage equation which looks like diminish return: $$\large{v(t)=V_0 * (1 - e^{\frac{-t}{t_0}})}$$