Optimal distribution of the playing cards on print templates

Question

I have a mathematical Problem: I plan to have about 1200 cards printed for the open source game INCANTATA from ZeroNet.

In doing so, I now come across the problem of distributing the cards on the different DIN A3 sheets (18 cards on each sheet). I would like to print different quantities of all cards, but it is cheapest if you print each sheet e.g. exactly 6 times. Of course, a few cards that you only want to print less than 6 times are printed too many in the end, but that is cheaper than having some sheets printed less than the others.

Is there an algorithm into which you can throw the number of cards into, in order to get out the optimal number of cards on each sheet?

This would be my list (e.g. print 98 times the card with the ID 8 ):

[Number of cards] [card-ID]
98 8
42 9
42 7
36 2
34 3
29 12
26 275
24 56
24 51
22 67
22 147
21 24
20 363
20 13
18 1
17 483
17 376
17 364
17 352
17 340
17 298
17 231
17 148
15 76
14 53
14 127
13 97
13 70
13 163
13 143
12 428
11 91
11 6
11 57
11 5
11 466
11 210
11 206
11 111
11 11
10 98
10 85
10 74
10 58
10 44
10 32
10 166
10 162
9 88
9 72
9 4
9 39
9 35
9 195
8 90
8 87
8 64
8 46
8 360
8 233
8 21
8 207
8 18
8 151
7 69
7 54
7 42
7 271
7 177
7 169
7 156
6 814
6 78
6 75
6 48
6 420
6 417
6 41
6 341
6 305
6 278
6 23
6 223
6 209
6 194
6 17
6 167
6 115
6 101
5 99
5 29
5 216
5 190
5 19
5 129
5 124
5 113
4 93
4 89
4 81
4 77
4 63
4 49
4 399
4 37
4,336
4 302
4 251
4,237
4,220
4 20
4 188
4 185
4 158
4 155
4 154
4 15
4 144
4 14
4 136
4 117
4 114
4 100
4 10
3 95
3 83
3 80
3 783
3 779
3 71
3 66
3,505
3 487
3 391
3 36
3 348
3 34
3 332
3 270
3,267
3 244
3 226
3 222
3 213
3 199
3 196
3 191
3 175
3 171
3 165
3 16
3 152
3 141
3 138
3 131
3 120
3 118
3 109

Answer 1

This is a variant of the cutting stock problem. Because of the rule to print each sheet exactly 6 times, you can round each demand up to the next multiple of 6. Equivalently, replace each demand $d$ by $\lceil d/6 \rceil$ and use each of the resulting sheets 6 times. After this transformation, the total demand is $298$, which implies that you need at least $6 \lceil 298/18 \rceil = 6\cdot 17 = 102$ sheets. One optimal solution is to use the following 17 sheets 6 times each, where the number of copies of each card is in parentheses if it exceeds 1 (for example, sheet 3 contains 17 copies of card 8 and 1 copy of card 14):

 1: 1(3) 2(6) 3(6) 4(2) 10
 2: 5(2) 6(2) 7(7) 9(7)
 3: 8(17) 14
 4: 11(2) 12(5) 13(4) 15 16 17 18(2) 19 20
 5: 21(2) 23 24(4) 29 32(2) 34 35(2) 36 37 39(2) 41
 6: 42(2) 44(2) 46(2) 48 49 51(4) 53(3) 54(2) 63
 7: 56(4) 57(2) 58(2) 64(2) 66 67(4) 69(2) 71
 8: 70(3) 72(2) 74(2) 75 76(3) 77 78 80 81 83 85(2)
 9: 87(2) 88(2) 89 90(2) 91(2) 93 95 97(3) 98(2) 99 100
10: 101 109 111(2) 113 114 115 117 118 120 124 127(3) 129 131 136 138
11: 141 143(3) 144 147(4) 148(3) 151(2) 152 154 155 158
12: 156(2) 162(2) 163(3) 165 166(2) 167 169(2) 171 175 177(2) 185
13: 188 190 191 194 195(2) 196 199 206(2) 207(2) 209 210(2) 213 216 220
14: 222 223 226 231(3) 233(2) 237 244 251 267 270 271(2) 278 302 305
15: 275(5) 298(3) 332 336 340(3) 341 348 352(3)
16: 360(2) 363(4) 364(3) 376(3) 391 399 417 420 428(2)
17: 466(2) 483(3) 487 505 779 783 814

When the cost of each sheet is the same, this is called the bin packing problem, and I used SAS to solve it. Starting from this example, I modified the Assign variables to be >= 0 integer instead of binary. But this is really overkill when your cards are all the same size. Once you figure out the required number of sheets by summing the transformed demands and dividing by 18, you can arbitrarily assign the cards to the sheets, respecting the capacity of 18 per sheet.