| 1 |
Well, don't be fooled by "only if you have 2 mexes".
|
1 |
Well, don't be fooled by "only if you have 2 mexes".
|
| 2 |
\n
|
2 |
\n
|
| 3 |
Look at this example with 1 big mex and 20 smaller ones:
|
3 |
Look at this example with 1 big mex and 20 smaller ones:
|
| 4 |
http://pastebin.com/6HWf5VFN
|
4 |
http://pastebin.com/6HWf5VFN
|
| 5 |
(You can convince yourself that the W|A inputs are correct by writing the equations down for all the mexes individually and then seeing how you can merge the terms).
|
5 |
(You can convince yourself that the W|A inputs are correct by writing the equations down for all the mexes individually and then seeing how you can merge the terms).
|
| 6 |
\n
|
6 |
\n
|
| 7 |
The optimal solution here would still be to give all e to the big mex, but the current algorithm allocates a significant amount to the small mexes. This results in about 20% less OD.
|
7 |
The optimal solution here would still be to give all e to the big mex, but the current algorithm allocates a significant amount to the small mexes. This results in about 20% less OD.
|
| 8 |
\n
|
8 |
\n
|
| 9 |
Now, realistically, you won't have all your small mexes linked to the grid in most cases, but you would actually gain quite a bit more metal from keeping your e linked only to the big mex, not the smaller ones.
|
9 |
Now, realistically, you won't have all your small mexes linked to the grid in most cases, but you would actually gain quite a bit more metal from keeping your e linked only to the big mex, not the smaller ones.
|
| 10 |
\n
|
10 |
\n
|
| 11 |
The saving grace for the current gadget here is that OD figures will always be pretty small in comparison to the base mex output unless you invest quite a bit of energy, in which case the gadget's error becomes smaller.
|
11 |
The saving grace for the current gadget here is that OD figures will always be pretty small in comparison to the base mex output unless you invest quite a bit of energy, in which case the gadget's error becomes smaller.
|
| 12 |
\n
|
12 |
\n
|
| 13 |
\n
|
13 |
\n
|
| 14 |
To give a more roundabout answer to the OP question: If you have a set of n mexes that, in total, have the same base income as one big mex, then you need n times as much energy to get the same amount of OD metal out of the smaller ones as the big one.
|
14 |
To give a more roundabout answer to the OP question: If you have a set of n mexes that, in total, have the same base income as one big mex, then you need n times as much energy to get the same amount of OD metal out of the smaller ones as the big one.
|
| 15 |
\n
|
15 |
\n
|
| 16 |
PS: @Kyubey, get your minus signs straight:
|
16 |
PS: @Kyubey, get your minus signs straight:
|
| 17 |
1.077-1.194=0.117m <- wrong, that's -0.117 (similar for many parts of your calculation). You're getting the correct figure but do some wonky stuff with the signs.
|
17 |
1.077-1.194=0.117m <- wrong, that's -0.117 (similar for many parts of your calculation). You're getting the correct figure but do some wonky stuff with the signs.
|
What's the difference between a +10 mex and 5 +2 mexes?