You have used j as both a constraint index and a summation index in the xMaxP constraint:
con xMaxP{j in Ts}:sum{i in Ps} inflow[i,j]<=InspectionCapacity[j]*sum{j in Ts}InspectDecision[j];
Corrected version:
con xMaxP{j in Ts}:sum{i in Ps} inflow[i,j]<=InspectionCapacity[j]*InspectDecision[j];
To see that it does what you want, check the two cases InspectDecision[j] = 0 and InspectDecision[j] = 1.
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I did see that tech support track. The issue has been fixed in the latest release (SAS/OR 14.2), and I updated the track with a couple of workarounds for 14.1.
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Yes, the capacity constraint destroys the block structure, and the corresponding optimality cuts are weaker. You would need a single Eta variable instead of Eta[p] for each p. And you would also need master Benders feasibility cuts because the subproblem would not necessarily be feasible without them. What values of max_ORDERs_per_LOCATION are you interested in for the three scenarios?
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Those additional errors are because you switched back to the old declarations for FixedCost and Y. Also, you still need the new UseMinimum and UseMaximum constraints.
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Your parentheses on the summand are incorrect. You can use either one of the following:
minimize TotalCost = sum {i in Plant} ( sum {j in Product} ( X[i,j]*VariaCost[i,j] + Y[i,j]*FixedCost[i,j] ));
minimize TotalCost = sum {i in Plant, j in Product} ( X[i,j]*VariaCost[i,j] + Y[i,j]*FixedCost[i,j] );
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Those error messages are telling you that you can't use Y[i] and FixedCost[i] because those arrays now require two subscripts each. You need Y[i,j] and FixedCost[i,j] instead.
The Producing and UseMinimum constraints that I commented out are obsolete, and M is now also obsolete because it appears only in Producing.
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From the data in the spreadsheet, it looks like you want:
number FixedCost {Plant,Product} = [
And:
var Y {Plant,Product} binary;
With corresponding changes in the objective declaration. And then:
/*con Producing {i in Plant} : sum {j in Product} X[i,j] <= M * Y[i];*/
/*con UseMinimum {i in Plant} : sum {j in Product} X[i,j] >= MinUse[i] * Y[i];*/
con UseMinimum {i in Plant, j in Product}: X[i,j] >= MinUse[i,j] * Y[i,j];
con UseMaximum {i in Plant, j in Product}: X[i,j] <= MaxUse[i,j] * Y[i,j];
You don't need M.
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The following PROC OPTMODEL statements will generate such an error:
proc optmodel;
num mynum {1..2} init [10 20 30];
print mynum;
quit;
The numeric parameter mynum has two elements in its index set but has been initialized to three values.
Please check your code for a similar issue.
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In other words, you want inflow[i,j] > 0 to force InspectDecision[j] = 1. Hint: You can do that by introducing InspectDecision[j] into the xMaxP constraint while keeping that constraint linear.
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Your Trigger constraint contains a product of decision variables and hence is nonlinear. Also, it is redundant because the left hand side is a sum of nonnegative summands. If you remove this constraint, the MILP solver will return an optimal solution with InspectDecision[j] = 0 for all j.
What logical condition on the other decision variables do you want to force InspectDecision[j] = 1?
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Here is a modification that maximizes the maximum percentage. The same optimization model works if you change the number of bins or the minimum number of items per bin.
proc optmodel;
set ITEMS;
num count {ITEMS};
read data freqdata into ITEMS=[fee] count;
num total_count = sum {i in ITEMS} count[i];
set BINS = 1..3;
/* Assign[i,b] = 1 if item i is assigned to bin b, 0 otherwise */
var Assign {ITEMS, BINS} binary;
/* assign each item to exactly one bin */
con AssignOnce {i in ITEMS}:
sum {b in BINS} Assign[i,b] = 1;
/* AssignedBin[i] = b if and only if Assign[i,b] = 1 */
impvar AssignedBin {i in ITEMS} = sum {b in BINS} b*Assign[i,b];
/* smaller fee implies same or smaller bin */
con Monotonic {i in ITEMS, j in ITEMS: i < j}:
AssignedBin[i] <= AssignedBin[j];
/* maximize the maximum percentage */
var MaxBinCountPercentage >= 0;
max Objective = MaxBinCountPercentage;
var BinCount {BINS} >= 3;
con BinCountCon {b in BINS}:
BinCount[b] = sum {i in ITEMS} count[i]*Assign[i,b];
impvar BinCountPercentage {b in BINS} =
BinCount[b] / total_count;
con MaxCon1 {b in BINS}:
MaxBinCountPercentage >= BinCountPercentage[b];
var IsMaxBin {BINS} binary;
con OneMax:
sum {b in BINS} IsMaxBin[b] = 1;
/* if IsMaxBin[b] = 1 then MaxBinCountPercentage <= BinCountPercentage[b] */
con MaxCon2 {b in BINS}:
MaxBinCountPercentage - BinCountPercentage[b] <= 1 - IsMaxBin[b];
solve;
print BinCount BinCountPercentage;
print count AssignedBin;
quit;
[1]
BinCount
BinCountPercentage
1
14
0.70
2
3
0.15
3
3
0.15
[1]
count
AssignedBin
330
4
1
350
2
1
370
1
1
390
3
1
430
1
1
450
1
1
480
1
1
500
1
1
530
3
2
540
1
3
550
2
3
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In an off-line private message, I understood the objective to be to minimize the maximum percentage. Do you instead want to maximize the maximum percentage?
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I think the following code does what you want:
data indata;
input ID fee;
datalines;
2 330
4 330
15 330
17 330
1 350
5 350
18 370
8 390
12 390
19 390
20 430
3 450
13 480
7 500
6 530
10 530
11 530
9 540
14 550
16 550
;
proc freq data=indata;
tables fee / out=freqdata;
run;
proc optmodel;
set ITEMS;
num count {ITEMS};
read data freqdata into ITEMS=[fee] count;
num total_count = sum {i in ITEMS} count[i];
set BINS = 1..3;
/* Assign[i,b] = 1 if item i is assigned to bin b, 0 otherwise */
var Assign {ITEMS, BINS} binary;
/* assign each item to exactly one bin */
con AssignOnce {i in ITEMS}:
sum {b in BINS} Assign[i,b] = 1;
/* AssignedBin[i] = b if and only if Assign[i,b] = 1 */
impvar AssignedBin {i in ITEMS} = sum {b in BINS} b*Assign[i,b];
/* smaller fee implies same or smaller bin */
con Monotonic {i in ITEMS, j in ITEMS: i < j}:
AssignedBin[i] <= AssignedBin[j];
/* minimize the maximum percentage */
var MinMax >= 0;
min Objective = MinMax;
var BinCount {BINS} >= 3;
con BinCountCon {b in BINS}:
BinCount[b] = sum {i in ITEMS} count[i]*Assign[i,b];
impvar BinCountPercentage {b in BINS} =
BinCount[b] / total_count;
con MinMaxCon {b in BINS}:
MinMax >= BinCountPercentage[b];
solve;
print BinCount BinCountPercentage;
print count AssignedBin;
quit;
[1]
BinCount
BinCountPercentage
1
6
0.30
2
7
0.35
3
7
0.35
[1]
count
AssignedBin
330
4
1
350
2
1
370
1
2
390
3
2
430
1
2
450
1
2
480
1
2
500
1
3
530
3
3
540
1
3
550
2
3
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