Doing merchant job's, giving bonus discounts

[Moodles]

I was thinking, too much experience is given when you defeat bandits annoying a town, or escorting a merchant caravan.. I think you should recieve maybe half the experience what it is now in .704 and give you a further 2% discount for every successful job in the town, discount applying to the merchant, and the weapon and armor seller in the town (because the merchant has put a good word in for you).

 

[Narcissus]

Not a bad idea at all, but there would need to be a cap on the amount the discount could be. I could see a maximum of ... maybe 10% off?

I don't think that it should also apply to the armorsmith and weaponsmith. I can understand why it might be reasonable for them to offer you a discount ... but I think it would make it easy to end up with the best weapons too early in the game. A 10% discount would be too much, in that case.

Just my thoughts,

Narcissus

 

[Ingolifs]

If you want something to give a bit of a decrease per activity, but you don't want it to go below a certain value, the exponential function is your best friend.

percent of normal price = normal price*(0.9+0.1^n) where n is the number of times you've done it.

If you don't understand, go back to high school to learn math.

 

[Manitas]

percent of normal price = normal price*(0.9+0.1^n) where n is the number of times you've done it.

Ingo, from now on consider me your fan.

Let's wage a war on linear relations in games!

 

[Ingolifs]

percent of normal price = normal price*(0.9+0.1^n) where n is the number of times you've done it.

Ingo, from now on consider me your fan.

Let's wage a war on linear relations in games!

Damn right, brotha!

I always saw it as pointless that in games you have certain hard caps on things. If a sword requires 180 strength to use, nd i have 179, why can't i at least lift it with a struggle? And why does a sword have a dexterity requirement of 180? what exactly prevents me from wrapping my hand around the sword handle, and just swing with it?

If i ever made a game, you'd never see any of these restrictions. The game would be a forest of interrelated differential equations!

 

[Dynamite_cow]

Hmm how about having factions also affect the prices? Like, if you are fighting for the Vaegirs, you get a little discount in Vaegir towns while Swadian towns are more expensive... Maybe it could even have to do with the relations, so the more you fight your enemy, the cheaper/more expensive the prices would get. Of course, a limit here too, like 10% or something.

 

[thorins]

hi all,

i thnk instead of a percent off maybe we could look a t just a few denar off. if the item costs 100 denars and the merchant usually asks at least double which is 200 denars. then you could do 100 jobs for him and earn your discount. and if the item cost more then there is a larger number of jobs you can do for your discount. the thing is the computer must know not to drop the price below its set value or cost. so you cant do 1000 jobs for a merchant and get every thing for free lol

 

[svart]

What do escorting merchant trains have to do with trading discounts ??
The escort is a mercenery/bodyguard job, why would a mercenery be interested in discounts ?

 

[leo]

If you want something to give a bit of a decrease per activity, but you don't want it to go below a certain value, the exponential function is your best friend.

percent of normal price = normal price*(0.9+0.1^n) where n is the number of times you've done it.

If you don't understand, go back to high school to learn math.

I don't like that formula too much.

If n=0, then the right hand side equals 1.9*(normal price).
If n=1, then the right hand side equals 1*(normal price)

Sounds sort of crazy that the merchant would charge you 90% more if you've never done her any favors than if you've done exactly one job.

I suggest a minor addition to the formula:

Percent of normal price = normal price * (0.9+.1^(n+1))


Meaning that if you've never done a job, you will be charged normal price * (.9 + .1^(1)), or normal price.

 

[Ingolifs]

Heh. To be honest, I wasn't thinking too hard when I was writing up that formula.

Let's have another tackle of the problem. Now that I think about it, a 10^-n style exponential is too sharp. 2^-n would be better.
The relative percentage of price reduction would be: 1-2^-n

Now we just subtract, say 20% of that from the original price S.
S(0.8 + 0.2*2^(-n))
That would be better overall.

The mistake I made in the old formula was that I forgot to multiply the exponent by 0.1. the formula should've been normal price*(0.9+0.1*0.1^n)

If you're good, you'll notice that's basically the same as your formula. Thanks for pointing it out.

 

[leo]

You're right about the steepness. That's something I hadn't considered. For that matter, 2^(-n) is probably still a bit steep.

The function price= (.8 + .2*2^(-n)) yields the following vallues:

price(0) = 1
price(1) = .9
price(2) = .85
price(3) = .825
price(4) = .8125

Each successive completed task yields half the reduction of the completed task. The fourth task will yield only 1/8th the reduction that the first task gave.

Perhaps (5/4) would be a better base than 2, since each successive task would yield 80% of the reduction of the previous task, and would therefore not be so small by comparison.



A minor note: The trading formula was updated recently by Armagan to make sure that, regardless of how skilled your merchant is, you will never be able to sell an item for more than (or as much as) the merchant would charge for it. In that same vein, it's important that any suggested formula ensures that players cannot obtain free wealth and bankrupt shops by buying items from a merchant and selling them back for a profit.

Because of the way that the trading skill formula affects the trade penalty now [presumably, from the skill information, it is an exponential equation of the form: penalty = k *.9^(Trading skill], it may be easier to simply increase a player's effective trading skill when speaking with merchants for whom he or she has done favors.