lpSolve - "Must Not Be Together" Constraint?

Question

The Problem

I am using lpSolve to find the optimal lineup for a fantasy baseball team - a knapsack problem involving the price SALARY and projected points DK of each player PLAYERID within the given constraints of the contest.

The current code works great, but I have a constraint I would like to add that I can't quite figure out. The new constraint is to not have any players in the lineup facing one of the pitchers SP in the same lineup.

What I Have So Far

I created a column called MNBT (Must Not Be Together) which defines the opposing pitcher's PLAYERID that must not be found in the same lineup as each player, but I am stuck there. The first 20 rows of the data.frame slate_players are as follows (I can provide all 91 rows for this specific contest if needed):

   PLAYERID POS TEAM OPP SALARY    DK TEAM_O    MNBT
1     37584  SP  LAD OAK  10000 18.42    0SP   13170
2     11292  SP  TEX HOU   9300 18.41    0SP 1452665
3   1452665  SP  HOU TEX   7400 15.22    0SP   11292
4     11168  SP  BAL BOS   6900  9.06    0SP   13502
5     13170  SP  OAK LAD   6800  6.06    0SP   37584
6     13502  SP  BOS BAL   6700 13.52    0SP   11168
7   2038873  SP  KCR DET   6600 18.45    0SP   34649
8     34649  SP  DET KCR   6500  7.46    0SP 2038873
9     11446   C  KCR DET   5300  7.55    KCR   34649
10  1054004   C  LAD OAK   5000  8.25    LAD   13170
11    15541   C  BOS BAL   4500  7.08    BOS   11168
12  1252110   C  OAK LAD   4100  5.07    OAK   37584
13    22667   C  BAL BOS   3400  7.09    BAL   13502
14    10290   C  TEX HOU   2900  4.08    TEX 1452665
15    13171   C  DET KCR   2800  5.45    DET 2038873
16    17552   C  HOU TEX   2600  4.47    HOU   11292
17    36727  1B  LAD OAK   5800  9.09    LAD   13170
18    17648  1B  LAD OAK   5400  8.57    LAD   13170
19    17887  1B  OAK LAD   4900  7.30    OAK   37584
20    17851  1B  KCR DET   4400  7.24    KCR   34649
[...]

The Current lpSolve Code

# count the unique players and teams on the slate
unique_teams = unique(slate_players$TEAM_O)
unique_players = unique(slate_players$PLAYERID)

# define the objective for the solver
obj = slate_players$DK

# create a constraint matrix for the solver
con = rbind(t(model.matrix(~ POS + 0, slate_players)), #Positions
            t(model.matrix(~ PLAYERID + 0, slate_players)), #DupPlayers
            t(model.matrix(~ TEAM_O + 0, slate_players)), #SameTeam
            rep(1,nrow(slate_players)), #TotPlayers
            slate_players$SALARY) #MaxSalary

# set the direction for each of the constraints
dir = c("==", #1B
        "==", #2B
        "==", #3B
        "==", #C
        "==", #OF
        "==", #SP
        "==", #SS
        rep('<=',length(unique_players)), #DupPlayers
        rep('<=',length(unique_teams)), #SameTeam
        "==", #TotPlayers
        "<=") #MaxSalary

# set the limits for the right-hand side of the constraints
rhs = c(1, #1B
        1, #2B
        1, #3B
        1, #C
        3, #OF
        2, #SP
        1, #SS
        rep(1,length(unique_players)), #DupPlayers
        rep(5,length(unique_teams)), #SameTeam
        10, #TotPlayers
        50000) #MaxSalary

# find the optimal solution using the solver
result = lp("max", obj, con, dir, rhs, all.bin = TRUE)

# create a table for the players that are in optimal solution
solindex = which(result$solution==1)
optsolution = slate_players[solindex,]

The Question

How do I code this new constraint? I have been doing these kinds of adjustments manually, but I would really appreciate it if there were a solution available to automate this process. Thank you!

I'm not sure how you would code this in R, but the typical way to do an "either-or" type of constraint is to add the selection variables together as: `x + x' <= 1 for all (x, x') in set of prohibited combos` — AirSquid, Sep 25 '20 at 14:38
@AirSquid Yeah, I have that type of logic in place to ensure there are no duplicate players in the lineup (some players have multiple position eligibility), but I haven't quite figured out how to translate that into what I need here. I'm sure it's the right line of thinking - I'm just not quite to the solution yet. — Eric_Alan, Sep 26 '20 at 05:43
@Eric_Alan Does the script above run for you? I get the following error. I think this may be because you cut the list of players short. I also had to change some of the constraints before attempting the solver as there's only three player types in this data frame. Warning message: In rbind(const.mat, const.dir.num, const.rhs) : number of columns of result is not a multiple of vector length (arg 2) — windyvation, Oct 01 '20 at 16:03
When I removed the duplicate player constraint, I was able to run the code. See my proposed idea below. Since the all.bin = TRUE, the duplicate player constraint was redundant anyway. — windyvation, Oct 01 '20 at 17:52
@windyvation You are correct - the above script does not run properly without all 91 rows of the `slate_players` dataframe, which includes all seven player types. The duplicate player constraint comes into play in the full dataset as there are a handful of players that are eligible to play multiple positions but can only appear in the optimal solution one time. — Eric_Alan, Oct 03 '20 at 14:58

score 1 · Accepted Answer · answered Oct 03 '20 at 15:23

What I ended up doing, rather than a single MNBT column, was creating a helper column for each individual pitcher SP to indicate which hitters he may not appear in the optimal solution with. In these columns, I assigned a value for the pitcher of 5 and a value of 1 for each hitter he must not appear with. The constraint then became that the sum of each of these columns would be <= 5. The logic is that a maximum of 5 hitters may be in the same lineup facing any individual pitcher, but if that same pitcher appeared in the optimal solution, none of the hitters he was facing would be.

score 0 · Answer 2 · answered Oct 01 '20 at 17:58

0

While this doesn't create the set of constraints for you, this example is of one of those constraints that @AirSquid mentioned might help.

In the example above, the 6th player (13502) could not play against the 13th player (22667).

Add to the constraint:

c(0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0)

Add to the directions:

"<="

Add to the right hand side:

The next trick is how to generate all of these sets of constraints in R. Cheerio.

answered Oct 01 '20 at 17:58

windyvation

497
3
13

Yes. This is the kind of solution I had been looking for (and used a variation of in the duplicate player constraint), but hadn't quite landed on how to put into practice for the "must not be together" players. – Eric_Alan Oct 03 '20 at 15:00
@Eric_Alan how did you end up coding the constraint matrix for restricting batters vs. pitchers based on windyvation's response? – On_an_island Apr 26 '22 at 16:29

lpSolve - "Must Not Be Together" Constraint?

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