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Assume we have 50 variables that we want to split into 3 groups and each group should have at least 5 variables. How to formulate that all variables in the same group take the same values? I used the binary variable y to assign the variable to each group.

$$\max \sum_i Q_i C_i $$ $$Q_i \in \lbrace 0, 1, ... 20 \rbrace \quad \forall i = 1, ... 50 $$ $$\sum_i y_{i1} \ge 5$$ $$\sum_i y_{i2} \ge 5$$ $$\sum_i y_{i3} \ge 5$$ $$y_{i1}, y_{i2}, y_{i3} \in \lbrace 0, 1 \rbrace \quad \forall i=1, ..., 50$$ $$y_{i1} + y_{i2} + y_{i3} = 1 \quad \forall i=1, ..., 50$$

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    $\begingroup$ Is $C_i$ a decision variable and $Q_i$ an input parameter, or vice versa? $\endgroup$
    – RobPratt
    Commented May 19, 2023 at 1:19
  • 2
    $\begingroup$ Possible duplicate of or.stackexchange.com/questions/7552/… $\endgroup$
    – RobPratt
    Commented May 19, 2023 at 1:23
  • 2
    $\begingroup$ Please use MathJax on this site, don't upload text or equations as images. Thanks. $\endgroup$
    – Rob
    Commented May 19, 2023 at 11:39
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    $\begingroup$ What is the range of $C$? $\endgroup$
    – Reinderien
    Commented May 20, 2023 at 14:25
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    $\begingroup$ What's stopping all $Q$ from being 20? $\endgroup$
    – Reinderien
    Commented May 20, 2023 at 15:04

2 Answers 2

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You want to enforce $y_{ik}=y_{jk}\implies Q_i=Q_j$ for all $i\neq j$ and all $k=1,2,3.$ Given the domain of $Q,$ the most any two $Q$ variables can differ by is 20, so you can do this with the following two constraints for all $i,j,k$ combinations: $$Q_i - Q_j \le 20(2 - y_{ik} - y_{jk})$$ and $$Q_j - Q_i \le 20(2 - y_{ik} - y_{jk}).$$

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  • $\begingroup$ What are the tradeoffs between having 2*n*n*m == 15,000 constraints, as you've shown here; and adding 3 group variables and having 2*m*n == 300 constraints? $\endgroup$
    – Reinderien
    Commented May 22, 2023 at 14:07
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    $\begingroup$ It's actually 7,350 added constraints for my answer. As for the tradeoffs, that's an empirical question (run both and see). I suspect, though, that your approach of using three group variables would solve faster. $\endgroup$
    – prubin
    Commented May 22, 2023 at 15:37
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This problem is trivial and $Q_i = 20 \, \forall i$ if $C_i \ge 0 \, \forall i$. If $C$ may be negative, then keep your $y_{ij}$ binary assignment variables, add three group value variables $G_j$, and only enforce that the group value variables are constrained to integers - the original $Q_{i}$ are easier solved as bounded continuous. Then add constraints like the following:

Error: $$\epsilon_{ij} := Q_i - G_j $$

Tolerance: $$\tau_{ij} := 20 \left( 1 - y_{ij} \right) $$

$$\epsilon_{ij} \le \tau_{ij}$$ $$-\epsilon_{ij} \le \tau_{ij}$$

import pulp
from numpy.random import default_rng


n = 50
m = 3
group_bound = 5
Qbound = 20
rand = default_rng(seed=399)
C = rand.uniform(low=-1, high=1, size=n)

Q = pulp.LpVariable.matrix(
    name='Q', indices=range(n), cat=pulp.LpContinuous, lowBound=0, upBound=Qbound)
Qgroup = pulp.LpVariable.matrix(
    name='Qgroup', indices=range(m), cat=pulp.LpInteger, lowBound=0, upBound=Qbound)
y = pulp.LpVariable.matrix(
    name='y', indices=(range(n), range(m)), cat=pulp.LpBinary)

prob = pulp.LpProblem(name='var_grouping', sense=pulp.LpMaximize)
prob.objective = pulp.lpDot(Q, C)

for i in range(n):
    prob.addConstraint(
        name=f'groupexcl{i}',
        constraint=pulp.lpSum(y[i]) == 1)
for j in range(m):
    prob.addConstraint(
        name=f'groupmin{j}',
        constraint=pulp.lpSum([
            y[i][j] for i in range(n)
        ]) >= group_bound)

for i in range(n):
    for j in range(m):
        # if y == 0, Q == 0
        # if y == 1, Q == Qgroup

        error = Q[i] - Qgroup[j]
        tol = Qbound*(1 - y[i][j])
        prob.addConstraint(name=f'assign_{i}_{j}_hi', constraint=error <= tol)
        # Only needed if C may be negative
        prob.addConstraint(name=f'assign_{i}_{j}_lo', constraint=-error <= tol)

print(prob)
prob.solve()
assert prob.status == pulp.LpStatusOptimal

print('Qgroup =')
print([int(Qgroup[j].value()) for j in range(m)])

print('y =')
for j in range(m):
    print([int(y[i][j].value()) for i in range(n)])
print()

print('Q =')
print([int(Q[i].value()) for i in range(n)])
print()
var_grouping:
MAXIMIZE
0.3626220607951214*Q_0 + 0.42196536951115493*Q_1 + ...
SUBJECT TO
groupexcl0: y_0_0 + y_0_1 + y_0_2 = 1

groupexcl1: y_1_0 + y_1_1 + y_1_2 = 1

groupexcl2: y_2_0 + y_2_1 + y_2_2 = 1

groupexcl3: y_3_0 + y_3_1 + y_3_2 = 1
...

groupmin0: y_0_0 + y_10_0 + y_11_0 + y_12_0 + y_13_0 + y_14_0 + y_15_0
 + y_16_0 + y_17_0 + y_18_0 + y_19_0 + y_1_0 + y_20_0 + y_21_0 + y_22_0
 + y_23_0 + y_24_0 + y_25_0 + y_26_0 + y_27_0 + y_28_0 + y_29_0 + y_2_0
 + y_30_0 + y_31_0 + y_32_0 + y_33_0 + y_34_0 + y_35_0 + y_36_0 + y_37_0
 + y_38_0 + y_39_0 + y_3_0 + y_40_0 + y_41_0 + y_42_0 + y_43_0 + y_44_0
 + y_45_0 + y_46_0 + y_47_0 + y_48_0 + y_49_0 + y_4_0 + y_5_0 + y_6_0 + y_7_0
 + y_8_0 + y_9_0 >= 5

groupmin1: y_0_1 + y_10_1 + y_11_1 + y_12_1 + y_13_1 + y_14_1 + y_15_1
 + y_16_1 + y_17_1 + y_18_1 + y_19_1 + y_1_1 + y_20_1 + y_21_1 + y_22_1
 + y_23_1 + y_24_1 + y_25_1 + y_26_1 + y_27_1 + y_28_1 + y_29_1 + y_2_1
 + y_30_1 + y_31_1 + y_32_1 + y_33_1 + y_34_1 + y_35_1 + y_36_1 + y_37_1
 + y_38_1 + y_39_1 + y_3_1 + y_40_1 + y_41_1 + y_42_1 + y_43_1 + y_44_1
 + y_45_1 + y_46_1 + y_47_1 + y_48_1 + y_49_1 + y_4_1 + y_5_1 + y_6_1 + y_7_1
 + y_8_1 + y_9_1 >= 5

groupmin2: y_0_2 + y_10_2 + y_11_2 + y_12_2 + y_13_2 + y_14_2 + y_15_2
 + y_16_2 + y_17_2 + y_18_2 + y_19_2 + y_1_2 + y_20_2 + y_21_2 + y_22_2
 + y_23_2 + y_24_2 + y_25_2 + y_26_2 + y_27_2 + y_28_2 + y_29_2 + y_2_2
 + y_30_2 + y_31_2 + y_32_2 + y_33_2 + y_34_2 + y_35_2 + y_36_2 + y_37_2
 + y_38_2 + y_39_2 + y_3_2 + y_40_2 + y_41_2 + y_42_2 + y_43_2 + y_44_2
 + y_45_2 + y_46_2 + y_47_2 + y_48_2 + y_49_2 + y_4_2 + y_5_2 + y_6_2 + y_7_2
 + y_8_2 + y_9_2 >= 5

assign_0_0_hi: Q_0 - Qgroup_0 + 20 y_0_0 <= 20

assign_0_0_lo: - Q_0 + Qgroup_0 + 20 y_0_0 <= 20

assign_0_1_hi: Q_0 - Qgroup_1 + 20 y_0_1 <= 20

assign_0_1_lo: - Q_0 + Qgroup_1 + 20 y_0_1 <= 20

assign_0_2_hi: Q_0 - Qgroup_2 + 20 y_0_2 <= 20

assign_0_2_lo: - Q_0 + Qgroup_2 + 20 y_0_2 <= 20

...

VARIABLES
Q_0 <= 20 Continuous
Q_1 <= 20 Continuous
Q_10 <= 20 Continuous
Q_11 <= 20 Continuous
...
0 <= Qgroup_0 <= 20 Integer
0 <= Qgroup_1 <= 20 Integer
0 <= Qgroup_2 <= 20 Integer
0 <= y_0_0 <= 1 Integer
0 <= y_0_1 <= 1 Integer
0 <= y_0_2 <= 1 Integer
0 <= y_10_0 <= 1 Integer
0 <= y_10_1 <= 1 Integer
0 <= y_10_2 <= 1 Integer
...
At line 2 NAME          MODEL
At line 3 ROWS
At line 358 COLUMNS
At line 1915 RHS
At line 2269 BOUNDS
At line 2473 ENDATA
Problem MODEL has 353 rows, 203 columns and 1200 elements
Coin0008I MODEL read with 0 errors
Option for timeMode changed from cpu to elapsed
Continuous objective value is 207.128 - 0.01 seconds
Cgl0004I processed model has 353 rows, 203 columns (153 integer (150 of which binary)) and 1200 elements
Cbc0038I Initial state - 100 integers unsatisfied sum - 4.63636
Cbc0038I Pass   1: suminf.    0.00000 (0) obj. -207.128 iterations 129
Cbc0038I Solution found of -207.128
Cbc0038I Relaxing continuous gives -207.128
Cbc0038I Cleaned solution of -207.128
Cbc0038I Before mini branch and bound, 52 integers at bound fixed and 50 continuous
Cbc0038I Mini branch and bound did not improve solution (0.04 seconds)
Cbc0038I After 0.04 seconds - Feasibility pump exiting with objective of -207.128 - took 0.01 seconds
Cbc0012I Integer solution of -207.12836 found by feasibility pump after 0 iterations and 0 nodes (0.04 seconds)
Cbc0001I Search completed - best objective -207.1283647272854, took 0 iterations and 0 nodes (0.04 seconds)
Cbc0035I Maximum depth 0, 0 variables fixed on reduced cost
Cuts at root node changed objective from -207.128 to -207.128
Probing was tried 0 times and created 0 cuts of which 0 were active after adding rounds of cuts (0.000 seconds)
Gomory was tried 0 times and created 0 cuts of which 0 were active after adding rounds of cuts (0.000 seconds)
Knapsack was tried 0 times and created 0 cuts of which 0 were active after adding rounds of cuts (0.000 seconds)
Clique was tried 0 times and created 0 cuts of which 0 were active after adding rounds of cuts (0.000 seconds)
MixedIntegerRounding2 was tried 0 times and created 0 cuts of which 0 were active after adding rounds of cuts (0.000 seconds)
FlowCover was tried 0 times and created 0 cuts of which 0 were active after adding rounds of cuts (0.000 seconds)
TwoMirCuts was tried 0 times and created 0 cuts of which 0 were active after adding rounds of cuts (0.000 seconds)
ZeroHalf was tried 0 times and created 0 cuts of which 0 were active after adding rounds of cuts (0.000 seconds)

Result - Optimal solution found

Objective value:                207.12836473
Enumerated nodes:               0
Total iterations:               0
Time (CPU seconds):             0.05
Time (Wallclock seconds):       0.04

Option for printingOptions changed from normal to all
Total time (CPU seconds):       0.06   (Wallclock seconds):       0.06

Qgroup =
[0, 20, 20]
y =
[0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1]
[1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0]
[0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

Q =
[20, 20, 20, 0, 20, 20, 0, 20, 20, 0, 0, 0, 20, 20, 0, 20, 20, 0, 0, 20, 0, 0, 20, 0, 20, 20, 20, 0, 20, 0, 0, 0, 0, 20, 20, 20, 0, 20, 0, 0, 0, 0, 0, 0, 20, 20, 20, 0, 0, 0]
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