Investigation of Bribery Costs

**QwertyMIDX** · 03-01-2005, 07:19

Has anyone done any serious research to determine where the costs come from? I was wondering if the price of the units being bribed was a factor in determining the costs, for example, if I were to increase the price of Iberian Infantry to 500/250 would the cost of bribing an army with units of Iberian Infantry increase? I will check into this on my own and post anything I find, but any input would be much appreciated.

**hung41584** · 03-01-2005, 08:35

seems you have to pay a base price for bribes and then the rest of the units are moderately more expensive than usual. I'd say about 2-3 times their value. Plus the base. Its actually better to bribe larger armies than smaller ones, which is quite the opposite in reality.

**khelvan** · 03-02-2005, 07:55

Bump - anyone have information on this?

**Pode** · 03-02-2005, 16:29

In one of the homebrew mods that was trying to force better recruiting, the guy reversed the cost ratio of units, such that peasants cost much more than cohorts. He threw in an offhand comment that bribing peasant rebels became ludicrously expensive.

**QwertyMIDX** · 03-03-2005, 00:25

Ok, I did some testing, I bribed a rebel army with 2 units of peasents, 1 iberian infantry, and 1 belric slingers. With vanilla unit costs it cost 8640. I then increased the cost of iberian infantry to 1000 (roughly 4x) and the cost of bribing the army increased to 15040.

**Simetrical** · 03-03-2005, 04:39

Interesting. Vanilla recruitment costs would put the cost for that army at 100 + 100 + 240 + 470 = 910 denarii, and bribing it cost 8640 (or about 9.5 times the cost). Changing that to 100 + 100 + 1000 + 470 = 1670 denarii, the bribe cost became 15040, or about 9 times the cost. This does indeed suggest a reduction in cost for more expensive armies.

Now, let's assume the bribe cost function is linear. That would mean that it could be expressed as y = mx + b. The two data points give us 8640 = 910m + b and 15040 = 1670m + b. Subtract the former from the latter, and we know that 6400 = 760m, or m ≈ 8.42. Then we can solve for b: 8640 ≈ 7663.16 + b, which means that b ≈ 976.84. The final function, therefore, should be y ≈ 8.42x + 976.84. Get another data point, and we can figure out if this is right. (I'd get the point myself, but I have to go to bed soon, so I don't have time.)

Edit: Or, of course, it could be based on custom battle recruitment cost. This would make more sense, since it would take into account two-turn training times, higher-level barracks, and similar cost reducers.

-Simetrical