A27 Disadvantageous farming tech upgrade time cost for Han

[stevenlau]

Problem:

As in the new release of A27, the farmstead farming tech upgrade time costs are 40s, 50s and 60s respectively.  However, for Han they become 50s, 60s and 70s respectively.  This is a disadvantage for Han.

Cause:

In early revisions of A26, the bonuses for farming tech in Han was +25% (in contrast with +20% for other civs), justifying the extra time cost.  However, the bonuses were later standardized to +20%.  Consequently, the extra time cost becomes redundant.

Relevant commit:

https://gitea.wildfiregames.com/0ad/0ad/commit/b4155900d3356bd62af7061b0890e7744332ba5e

 

[Player of 0AD]

3 hours ago, stevenlau said:

Problem:

As in the new release of A27, the farmstead farming tech upgrade time costs are 40s, 50s and 60s respectively.  However, for Han they become 50s, 60s and 70s respectively.  This is a disadvantage for Han.

Cause:

In early revisions of A26, the bonuses for farming tech in Han was +25% (in contrast with +20% for other civs), justifying the extra time cost.  However, the bonuses were later standardized to +20%.  Consequently, the extra time cost becomes redundant.

Relevant commit:

https://gitea.wildfiregames.com/0ad/0ad/commit/b4155900d3356bd62af7061b0890e7744332ba5e

 

So what? Hans farm faster because they are only 3 units on a field maximum, not 5.  So they still have an advantage.

[Feldfeld]

19 minutes ago, Player of 0AD said:

So what? Hans farm faster because they are only 3 units on a field maximum, not 5.  So they still have an advantage.

Well, while taking fewer diminishing returns is an advantage, it's dissociated from farming tech research time. I think they should be harmonized. And this is clearly an oversight as demonstrated by OP. These reports should always be encouraged regardless of the decision we take from it.

By the way, in A26 release, the Han farming technologies had no effect at all, so after all it's still an improvement

[stevenlau]

Side tracking a bit.  The two replies above mentioned "diminishing returns" and "farm faster".  These are based on the rule saying "assigning multiple farmers on a field would make them farm less efficiently".  While I do see this description in the in game Structure Tree, in theory all farmers should farm at a rate of 0.5 grain per second.  Is the "diminishing return" an actual rule in the game?  If so, to be precise how is the diminishing calculated?

[Atrik]

44 minutes ago, stevenlau said:

Side tracking a bit.  The two replies above mentioned "diminishing returns" and "farm faster".  These are based on the rule saying "assigning multiple farmers on a field would make them farm less efficiently"

Interesting question I only looked it up now. The template indicate a diminishing return of 0.9. Asking a llm to translate the code in ResourceSupply.js it helps to define the formula: (omg can't make laTex code insert, seems not working)

 

Pn = P0 * (0.9)^(n-1)

Where:

    Pn is the productivity of the nth worker
    P0 is the productivity of the first worker

So for the 5th gatherer for example, given that base gathering is 0.5 and no upgrades:

P5=0.32805

Edited February 7 by Atrik

[Feldfeld]

Han farming techs have been adjusted https://gitea.wildfiregames.com/0ad/0ad/commit/5dcef8c17a11c52817bbfb44b569cc8d29adde4a

4 hours ago, Atrik said:

Pn = P0 * (0.9)^(n-1)

Where:

    Pn is the productivity of the nth worker
    P0 is the productivity of the first worker

If P0 is the productivity of the first worker, then Pn would be the productivity of the (n+1)th worker I think. AI still has ways to go

[Atrik]

9 minutes ago, Feldfeld said:

If P0 is the productivity of the first worker, then Pn would be the productivity of the (n+1)th worker I think. AI still has ways to go

Not sure I understood what you meant, but if we take n = 1, this perfectly check out as P0 = Pn:

Pn = P0 * (0.9)^(n-1)
P1 = P0 * (0.9)^(1-1)
P1 = P0 * (0.9)^0
P1 = P0 * 1

This is how I though confident the ai didn't hallucinate .
In that case Pn is indeed the correct worker n?
 

[Feldfeld]

7 minutes ago, Atrik said:

Not sure I understood what you meant, but if we take n = 1, this perfectly check out as P0 = Pn:

Pn = P0 * (0.9)^(n-1)
P1 = P0 * (0.9)^(1-1)
P1 = P0 * (0.9)^0
P1 = P0 * 1

This is how I though confident the ai didn't hallucinate .
In that case Pn is indeed the correct worker n?
 

Uh yeah you're right. It's just not intuitive that P0 and P1 both refer to the same worker in the explanation.