Water allocation problem via MILP

main.py

from pulp import *

problem=LpProblem(name='WaterAllocationProblem',sense=const.LpMaximize)


# for f(x) i.e. user1's reward function
user1_rewards_dict={
    0:0,
    1:5,
    2:8,
    3:9,
    4:8,
    5:5,
    6:0
}

b1is=[]
user1_allocation=LpAffineExpression()
user1_reward=LpAffineExpression()
user1_constraint=LpAffineExpression()
for key in user1_rewards_dict:
    b1i=LpVariable(name=f'b1{key}',cat=const.LpBinary)
    user1_allocation+=key*b1i
    user1_reward+=user1_rewards_dict[key]*b1i
    user1_constraint+=b1i
    b1is.append(b1i)

problem+=user1_constraint==1

# for g(x) i.e. user2's reward function
user2_rewards_dict={
    0:0,
    1:5,
    2:6,
    3:3,
    4:-4,
    5:-15,
    6:-30
}

b2is=[]
user2_allocation=LpAffineExpression()
user2_reward=LpAffineExpression()
user2_constraint=LpAffineExpression()
for key in user2_rewards_dict:
    b2i=LpVariable(name=f'b2{key}',cat=const.LpBinary)
    user2_allocation+=key*b2i
    user2_reward+=user2_rewards_dict[key]*b2i
    user2_constraint+=b2i
    b2is.append(b2i)

problem+=user2_constraint==1

# for h(x) i.e. user3's reward function
user3_rewards_dict={
    0:0,
    1:7,
    2:12,
    3:15,
    4:16,
    5:15,
    6:12
}

b3is=[]
user3_allocation=LpAffineExpression()
user3_reward=LpAffineExpression()
user3_constraint=LpAffineExpression()
for key in user3_rewards_dict:
    b3i=LpVariable(name=f'b3{key}',cat=const.LpBinary)
    user3_allocation+=key*b3i
    user3_reward+=user3_rewards_dict[key]*b3i
    user3_constraint+=b3i
    b3is.append(b3i)

problem+=user3_constraint==1

problem+=user1_allocation+user2_allocation+user3_allocation==6
problem+=user1_reward+user2_reward+user3_reward

problem.solve()

print([value(b1i) for b1i in b1is])
print([value(b2i) for b2i in b2is])
print([value(b3i) for b3i in b3is])
print(value(problem.objective))