We had this discussion a while back. The problem you posed
was a single port decision, not the sum decision of a set
of ports. I know for 100% certainty that I'm correct here,
I can prove it mathematically. I've actually studied this
class of problems in depth. I'd be happy to go thru it if
you want, but this isn't a good forum for that depth level.
Your assumption is that the actual net cash of each port
matters in the decision. When you have to trade the same
ports regardless, this no longer matters. The actual value
of the port does not matter when you have 10 ports to trade,
and must trade all 10 of them.
Think about it. Lets say have enough product to sell off
to 10 ports. They produce the following cash values:
5.0m
3.4m
4.0m
3.6m
5.2m
4.1m
3.7m
3.8m
4.2m
4.7m
How much money will all of those ports produce? 41.7m
It doesn't matter what order you trade them, whether 1 thru
10, or 1 thru 10 odds, then evens, or whatever. No
combination will produce more than the sum. That is a
mathematical fact.
Now you can argue for removing ports from the list, but that
isn't the same thing. We're talking about having X ports that
you must trade, removing a port from the list reduces the
ore cost, but also reduces X. You still must trade the new
list as efficiently as possible, regardless of list size.
Cash/ore is not the the appropriate measure here. Cash/unit
fuel only matters when you have discretion over whether to
trade a port or not. In that, then, it becomes only a filter
for the list and not a criteria for order. Ie: If you have
5 ports to trade, and only 1 or 2 must be hit, which ones
do you hit?
This all goes to the "greedy" assumption. The idea that the
sum of all decisions yields the optimal result. Neither mine
nor yours will do so, that I can prove, but my approach will
get closer to the optimal sum than yours will.
Quote:
If you don't consider these things then a bfs ptrader will
go to the EXACT SAME PORTS every time if you start in same place.
Assuming the ports are still on the list, yes. And while there
are cases that might give B an advantage over A in terms of
efficiency if you only must choose one, if you must trade both
B and A, the cash result does not matter.
Quote:
Also Increased distance can result in increased cash from the
run by filtering the mcic's and this gives you a cash/turn or cash/gas
used equation which is closer to the mark for a ptrader than product
sold/gas used.
No, it won't. If you must trade all of the ports on the list,
then it does not matter what the next choice pays. Since it is
compulsory to trade it, it must be traded either soon or later.
Total profit cannot exceed the sum profit of the ports. Yes you
can filter the list as you go, but this only reduces the number
of ports on the list, it does not reduce the need to trade the
remaining ports efficienctly.
Quote:
this produces gas for less cash than buying down from a
full port at the beginning or end of run. Also the need to move
the planet to or from such a gas port is using gas.
This is tangential to the problem at hand, but relevant to the net
cash question. In that sense, what matters is that you have
enough ore to reach one of the ore ports on the list. For my mega
rob script, I only have it buy ore from megarob ports prior to
the mega. Still, that does not affect the net ore cost for the
run, it only ensures that there's enough ore to meet that cost.
As such, this particular issue does not have anything to do with
the method of searching for the next target, but rather only has
to do with list selection. List selection is a completely different
issue.