Skip to content

Latest commit

 

History

History
141 lines (94 loc) · 4.61 KB

spec.md

File metadata and controls

141 lines (94 loc) · 4.61 KB
Raw

Language specification

This file specifies the language used for describing systems in systems. There are three primary kinds of objects to specify:

  • stocks hold values, and
  • flows transition values from one stock to another.
  • finally, formula are used to describe initial and maximum values for stocks, and the magnitude of flows.

Specifying stocks

Stocks are specified on their own line, or implicitly in flow declarations:

MyStock

This would create a stock named MyStock with an initial value of zero and a maximum value of infinity:

OtherStock(10)

You can also specify maximum values:

ThirdStock(0, 10)

This would create ThirdStock with an initial value of zero, and a maximum value of ten.

Going back to OtherStock for a moment, you can also use the special literal inf to explicitly specify its maximum value:

OtherStock(10, inf)

This is a more explicit way to specify a stock with an infinite maximum. Generally it's a strange indicator if you're using the inf literal directly, and instead you'd use the special syntax for infinite flows:

[InfiniteFlow]

This InfiniteFlow would have initial and maximum values of infinity.

Without going too far into the details, initial and maximums can be specified using any legal formula, more on formulas below:

Managers(2)
Engineers(Managers * 4, Managers * 8)

In many cases, though, you'll end up specifying your stocks inline in your flows, as opposed to doing them on their own lines, but the syntax is the same.

Flows

For example, this would have both a and b would initialize at zero, and both would have infinite maximum values, in addition there would be a flow of one unit per round from a to b (assuming that a is above zero):

a > b @ 1

In the above example, a has an initial value of zero, so it would never do anything. Most working systems address that problem by starting with an infinite stock:

[a] >  b  @ 5
 b  > [c] @ 3

In the above, a and c would be infinite, and b would start with a value of zero. You can also solve the empty start problem by specifying non-zero initial values for your stocks:

a(10) > b(3)  @ 5
b     > c(12) @ 1
c     > a

In this example, a is initialized at 10, b at 3, and c at 12. Note that you don't have to set the value at first reference. It is legal to initialize a value at a later definition of a stock, e.g. this is fine:

a(1) > b @ 5
b(2) > c @ 3
c(3) > a @ 1

However, it is illegal to initialize the same stock multiple times.

a(1) > b(2) @ 1
b(3) > a    @ 1

This will throw an error, because you can't initialize b twice with different values!

Rates, Conversions and Leaks

Each line specifies two nodes and the link between them. Links are described following the @ character. The most common type of flow is a rate, which is a fixed transfer of values in one stock to another.

For example, moving two units per round between a and b:

# these are equivalent
a > b @ 2
a > b @ Rate(2)

Up to two units will be transfered from a to b each round.

Another common kind of flow is the conversion flow, which takes the entire contents of the source stock and multiplies that value against the conversion rate, adding the result to the next flow.

# these are equivalent
a(10) > b @ 0.5
a(10) > b @ Conversion(0.5)

The above would multiple 0.5 against 10 and move 5 units to b, with the other 5 units being lost to the conversion rate (e.g. disappearing). A common example of a conversion rate would be the offer acceptance rate in a hiring funnel.

The third kind of flow is the leak, which combines properties of the rate and conversion flows. It moves a fixed percentage of the source flow into the destination flow, while leaving the remainder intact.

a(10) > b @ Leak(0.2)

Considering the difference between the conversion and leak, if the above were a conversion, then the value of a after one round would be 0, but if it's a leak, then the value would be 8.

Formulas

Any flow value, initial value and maximum value can be a formula:

Recruiters(3)
Engineers(Managers * 4, Managers * 8)
[Candidates] > Engineers @ Recruiters * 6
[Candidates] > Managers  @ Recruiters * 3

The above system shows that Engineers has an initial value of Managers * 4, a maximum value of Managers * 8 and then shows that both Engineers and Managers grow at multiples of the value of the Recruiters stock.

This is also a good example of using the Recruiters stock as a variable, as it doesn't' actually change over time.